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NEET 2027 · Inorganic Chemistry · Class 12 Unit 7

The p-Block Elements
Groups 15 to 18

Every NEET/AIPMT previous-year question on the chapter, plus 30 fresh practice questions per group. Each solution is broken into Given · Asked · Concept · Formula · Baby steps, with a shortcut wherever one exists.

GROUP
15
N · P · As · Sb · Bi
GROUP
16
O · S · Se · Te · Po
GROUP
17
F · Cl · Br · I · At
GROUP
18
He · Ne · Ar · Kr · Xe · Rn
PREVIOUS YEARS
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Groups 15 · 16 · 17 · 18

How to use this

  1. Attempt the PYQ set first, cold, with a timer — 45 seconds per question. Inorganic is recall-speed marks.
  2. Mark every wrong answer by type: never learnt it, learnt but forgot, or misread the question. The three need different fixes.
  3. Then do the 30-question group sets. These drill the same ideas from angles NEET has not used yet.
  4. Re-run only the questions you got wrong, one week later. Don't re-do the ones you already know.
00

NEET / AIPMT Previous Year Questions

Groups 15 · 16 · 17 · 18 — everything the exam has actually asked
48 questions  ·  full worked solutions
1.NEET 2016

Among the hydrides of group 15, the strongest reducing agent is:

  • (A)NH3
  • (B)PH3
  • (C)SbH3
  • (D)BiH3
Show solution
Answer  (D)  BiH3
Given
Hydrides NH3, PH3, SbH3, BiH3.
Asked
Which one gives up its hydrogen (and electrons) most easily.
Concept
Down group 15 the E–H bond gets longer and weaker, so the hydride breaks apart more readily and acts as a better reducing agent.
Formula
Reducing power: NH3 < PH3 < AsH3 < SbH3 < BiH3
Baby steps
  • Bond strength depends on overlap. Big atoms overlap badly with tiny H.
  • Bi is the biggest atom in the group, so Bi–H is the weakest bond.
  • Weakest bond = breaks first = best reducing agent = BiH3.
ShortcutReducing power always follows 'weakest bond wins' — go to the bottom of the group.
2.NEET 2019

The correct order of the bond dissociation enthalpy of halogen molecules is:

  • (A)F2 > Cl2 > Br2 > I2
  • (B)Cl2 > Br2 > F2 > I2
  • (C)I2 > Br2 > Cl2 > F2
  • (D)Cl2 > F2 > Br2 > I2
Show solution
Answer  (B)  Cl2 > Br2 > F2 > I2
Given
The four halogen molecules F2, Cl2, Br2, I2.
Asked
Order of energy needed to break the X–X bond.
Concept
Bond enthalpy normally falls down the group, but F2 is the exception. Fluorine is so small that the three lone pairs on each F sit very close and repel each other, weakening the F–F bond.
Formula
Cl2 (242) > Br2 (192) > F2 (158) > I2 (151) kJ mol⁻¹
Baby steps
  • Write the normal falling order: F2 > Cl2 > Br2 > I2.
  • Now punish F2 for lone-pair repulsion — drop it below Br2.
  • Final: Cl2 > Br2 > F2 > I2.
ShortcutOnly F2 misbehaves. Everything else in group 17 decreases smoothly.
3.NEET 2019

The number of P–O–P bonds in cyclic metaphosphoric acid, (HPO3)3, is:

  • (A)Zero
  • (B)Two
  • (C)Three
  • (D)Four
Show solution
Answer  (C)  Three
Given
Cyclic metaphosphoric acid, (HPO3)3.
Asked
Count of P–O–P (bridging oxygen) linkages.
Concept
In a cyclic structure the P atoms are joined head-to-tail in a closed ring, so the number of bridges equals the number of P atoms.
Formula
Cyclic ring: bridges = number of P atoms
Baby steps
  • The formula (HPO3)3 has 3 phosphorus atoms.
  • A ring of 3 P atoms needs 3 links to close.
  • Each link is one P–O–P bridge, so the answer is 3.
ShortcutCyclic → bridges = n. Chain (linear) → bridges = n − 1. H4P2O7 is a 2-P chain, so 1 bridge.
4.NEET 2018

The oxidation states of chlorine in bleaching powder, Ca(OCl)Cl, are:

  • (A)+1 and −1
  • (B)−1 and −1
  • (C)+1 and +1
  • (D)0 and −1
Show solution
Answer  (A)  +1 and −1
Given
Bleaching powder, Ca(OCl)Cl.
Asked
Oxidation state of each chlorine atom.
Concept
The two chlorines are in different chemical environments — one is bonded to oxygen (hypochlorite), one is a plain chloride ion.
Formula
In OCl⁻: x + (−2) = −1 → x = +1
Baby steps
  • Ca is +2 and O is −2 as usual.
  • In the OCl⁻ part, chlorine must be +1 to make the ion −1.
  • The lone Cl attached to Ca is a chloride, so −1.
  • Bleaching powder is a mixed salt: +1 and −1.
ShortcutAny compound with both an 'OCl' and a bare 'Cl' is a mixed salt — the chlorines can't be the same.
5.NEET 2017

Which one of the following statements about hydrolysis of XeF6 is correct?

  • (A)Complete hydrolysis gives XeO3
  • (B)It gives XeF4 and O2
  • (C)It gives Xe and F2 only
  • (D)No reaction occurs
Show solution
Answer  (A)  Complete hydrolysis gives XeO3
Given
XeF6 reacting with water.
Asked
Product of complete hydrolysis.
Concept
Xenon fluorides are hydrolysed by water. XeF6 keeps its +6 oxidation state, so the fluorines are simply replaced step by step by oxygen.
Formula
XeF6 + 3H2O → XeO3 + 6HF
Baby steps
  • Partial hydrolysis (little water) removes only two F: XeF6 + H2O → XeOF4 + 2HF.
  • More water removes more: XeOF4 + H2O → XeO2F2 + 2HF.
  • Complete hydrolysis strips all F: the product is XeO3, an explosive solid.
ShortcutXeF6 partial → XeOF4; complete → XeO3. XeF2 and XeF4 instead disproportionate and give Xe + O2 + HF.
6.NEET 2020

In solid state, PCl5 exists as:

  • (A)Covalent trigonal bipyramidal molecule
  • (B)[PCl4]+[PCl6]−
  • (C)[PCl6]− only
  • (D)[PCl4]+[Cl]−
Show solution
Answer  (B)  [PCl4]+[PCl6]−
Given
Phosphorus pentachloride in the solid state.
Asked
The actual structural units present.
Concept
In the solid, PCl5 becomes ionic. One molecule donates a chloride to another, giving a tetrahedral cation and an octahedral anion.
Formula
2PCl5 (solid) → [PCl4]+ + [PCl6]−
Baby steps
  • Gaseous PCl5 is trigonal bipyramidal, sp3d.
  • That shape is strained — the two axial bonds are longer and weaker.
  • In the solid, Cl⁻ transfers between molecules to give two strain-free ions.
  • Cation [PCl4]+ is sp3 tetrahedral, anion [PCl6]− is sp3d2 octahedral.
ShortcutSolid PBr5 is different — it is [PBr4]+Br⁻, because six big Br atoms cannot fit around one P.
7.AIPMT 2015

Which of the following is a neutral oxide of nitrogen?

  • (A)N2O5
  • (B)NO2
  • (C)N2O
  • (D)N2O3
Show solution
Answer  (C)  N2O
Given
Four oxides of nitrogen.
Asked
The one that is neither acidic nor basic.
Concept
An oxide is acidic if it forms an oxoacid with water. N2O and NO do not react with water at all, so both are neutral.
Formula
N2O, NO → neutral; N2O3 → HNO2; N2O5 → HNO3; NO2 → HNO2 + HNO3
Baby steps
  • Match each oxide to its acid: N2O3 is the anhydride of HNO2, N2O5 of HNO3.
  • NO2 is a mixed anhydride, giving both acids with water.
  • N2O (laughing gas) gives nothing with water — it is neutral.
ShortcutOnly two neutral nitrogen oxides exist: N2O and NO. Learn the pair and eliminate.
8.NEET 2021

The number of P–H bonds and the basicity of H3PO3 respectively are:

  • (A)Zero and 3
  • (B)One and 2
  • (C)Two and 1
  • (D)One and 3
Show solution
Answer  (B)  One and 2
Given
Phosphorous acid, H3PO3.
Asked
Number of P–H bonds and its basicity.
Concept
Only hydrogens attached to oxygen (as –OH) are ionisable. Hydrogen sitting directly on phosphorus cannot leave as H⁺.
Formula
Basicity = number of –OH groups
Baby steps
  • Draw H3PO3: P is joined to one =O, two –OH groups and one H.
  • So P–H bonds = 1.
  • Ionisable hydrogens = the two –OH ones, so basicity = 2 (dibasic).
  • The single P–H is why H3PO3 is a good reducing agent.
ShortcutFor H3POn acids: basicity = n − 1, and P–H bonds = 4 − n. So H3PO2 → basicity 1, two P–H.
9.NEET 2016

Which of the following does not exist?

  • (A)PCl5
  • (B)NCl5
  • (C)SbCl5
  • (D)AsCl5
Show solution
Answer  (B)  NCl5
Given
Pentachlorides of group 15 elements.
Asked
The one that cannot be made.
Concept
To reach the +5 state an atom must expand its octet, which needs empty d-orbitals. Nitrogen is in period 2 and has no 2d subshell.
Formula
N: 1s2 2s2 2p3 — no d orbitals available
Baby steps
  • Count how many bonds each atom can form. N is limited to four (octet), never five.
  • P, As and Sb are period 3 or below and do have vacant d orbitals.
  • So NCl5 does not exist, while PCl5, AsCl5, SbCl5 do.
ShortcutNitrogen and oxygen never exceed the octet. Also why NF5, NCl5 and H3NO4 are impossible.
10.NEET 2019

Which one of the following orders is correct for the boiling points of group 16 hydrides?

  • (A)H2O > H2Te > H2Se > H2S
  • (B)H2O > H2S > H2Se > H2Te
  • (C)H2Te > H2Se > H2S > H2O
  • (D)H2S > H2Se > H2Te > H2O
Show solution
Answer  (A)  H2O > H2Te > H2Se > H2S
Given
H2O, H2S, H2Se, H2Te.
Asked
Order of boiling point.
Concept
Boiling point rises down the group with molecular size (stronger van der Waals forces), but water jumps to the top because of hydrogen bonding.
Formula
H2O (373 K) > H2Te (271) > H2Se (232) > H2S (213)
Baby steps
  • Ignore water first: H2S < H2Se < H2Te, as size increases.
  • Now add water — O is small and very electronegative, so H2O forms strong H-bonds.
  • Lift H2O to the very top: H2O > H2Te > H2Se > H2S.
ShortcutSame trick works for NH3 in group 15 and HF in group 17 — the first hydride always jumps up.
11.AIPMT 2014

The brown ring test for nitrate depends on the formation of:

  • (A)[Fe(H2O)6]2+
  • (B)[Fe(H2O)5(NO)]2+
  • (C)FeSO4·NO2
  • (D)[Fe(NO)6]3+
Show solution
Answer  (B)  [Fe(H2O)5(NO)]2+
Given
Brown ring test for NO3⁻ using FeSO4 and conc. H2SO4.
Asked
The species responsible for the brown colour.
Concept
Concentrated H2SO4 releases nitric acid, which oxidises Fe²⁺ to Fe³⁺ while itself being reduced to NO. The NO then replaces one water in the iron aqua complex.
Formula
[Fe(H2O)6]2+ + NO → [Fe(H2O)5(NO)]2+ + H2O
Baby steps
  • Add freshly made FeSO4 to the nitrate solution.
  • Pour conc. H2SO4 down the side of the tube so it forms a lower layer.
  • At the junction, NO forms and binds to iron.
  • The brown nitrosyl complex appears as a ring at the boundary. Iron in it is +1.
ShortcutRing test = NO ligand. Remember the brown species has FIVE waters, not six.
12.NEET 2018

The catalyst used in the Ostwald process for the manufacture of nitric acid is:

  • (A)V2O5
  • (B)Fe with Mo
  • (C)Platinum gauze
  • (D)Cu
Show solution
Answer  (C)  Platinum gauze
Given
Ostwald process, industrial HNO3.
Asked
Catalyst used.
Concept
The key step is catalytic oxidation of ammonia to nitric oxide over a platinum–rhodium gauze.
Formula
4NH3 + 5O2 →(Pt/Rh, 500 K, 9 bar) 4NO + 6H2O
Baby steps
  • Step 1: NH3 is oxidised to NO over Pt/Rh gauze.
  • Step 2: 2NO + O2 → 2NO2 on cooling.
  • Step 3: 3NO2 + H2O → 2HNO3 + NO, and the NO is recycled.
ShortcutCatalyst cheat-sheet: Haber = Fe/Mo · Ostwald = Pt/Rh · Contact = V2O5 · Deacon = CuCl2.
13.NEET 2017

Which of the following is the correct order of the acidic strength of the oxoacids of chlorine?

  • (A)HOCl > HClO2 > HClO3 > HClO4
  • (B)HClO4 > HClO3 > HClO2 > HOCl
  • (C)HClO2 > HOCl > HClO3 > HClO4
  • (D)HClO3 > HClO4 > HOCl > HClO2
Show solution
Answer  (B)  HClO4 > HClO3 > HClO2 > HOCl
Given
HOCl, HClO2, HClO3, HClO4.
Asked
Order of acid strength.
Concept
More oxygen atoms pull electron density away from the O–H bond and spread the negative charge of the anion over more atoms, so the proton leaves more easily.
Formula
Acidity rises with oxidation state of Cl: +1 < +3 < +5 < +7
Baby steps
  • Count oxygens: 1, 2, 3, 4.
  • More oxygens = more stable conjugate base = stronger acid.
  • HClO4 (perchloric) is the strongest; HOCl the weakest.
ShortcutAcid strength and oxidising strength run in OPPOSITE directions here. HOCl is the best oxidiser, HClO4 the best acid.
14.NEET 2020

Which of the following has the maximum number of lone pairs on the central atom?

  • (A)XeF2
  • (B)XeF4
  • (C)XeO3
  • (D)XeOF4
Show solution
Answer  (A)  XeF2
Given
Four xenon compounds.
Asked
Which central Xe carries the most lone pairs.
Concept
Xenon starts with 8 valence electrons. Each bond pair uses one, so lone pairs = (8 − number of bonded electrons shared)/2.
Formula
Lone pairs = (8 − no. of σ bonds − 2 × no. of π/double bonds)/2
Baby steps
  • XeF2: 2 bonds used, 6 electrons left → 3 lone pairs, linear.
  • XeF4: 4 bonds used → 2 lone pairs, square planar.
  • XeOF4: 5 bonds used → 1 lone pair, square pyramidal.
  • XeO3: 3 double bonds → 1 lone pair, pyramidal. Maximum is XeF2.
ShortcutXe lone pairs: XeF2 = 3, XeF4 = 2, XeF6 = 1, XeO3 = 1, XeO4 = 0.
15.AIPMT 2015

Sulphur in the vapour state at about 1000 K exists mainly as:

  • (A)S8 rings, diamagnetic
  • (B)S6 chair, diamagnetic
  • (C)S2, paramagnetic
  • (D)S4, paramagnetic
Show solution
Answer  (C)  S2, paramagnetic
Given
Sulphur vapour above about 1000 K.
Asked
The species present and its magnetic nature.
Concept
At high temperature the S8 crown breaks down to S2, which is like O2 — it has two unpaired electrons in antibonding π* orbitals.
Formula
S8 →(heat) 4S2, with 2 unpaired electrons per S2
Baby steps
  • Solid sulphur is S8 puckered rings, diamagnetic.
  • Heating breaks the ring into smaller units.
  • Above 1000 K the stable unit is S2, which is paramagnetic like O2.
ShortcutAny question saying 'sulphur vapour at high temperature' is testing the paramagnetic S2 answer.
16.NEET 2019

Which one of the following pairs consists of species that are both diamagnetic?

  • (A)NO and NO2
  • (B)N2O4 and N2
  • (C)NO2 and O2
  • (D)NO and O2
Show solution
Answer  (B)  N2O4 and N2
Given
Nitrogen oxides and O2.
Asked
Pair with no unpaired electrons.
Concept
Molecules with an odd total number of electrons must have an unpaired electron and are paramagnetic. Dimerisation pairs those electrons up.
Formula
2NO2 ⇌ N2O4 (odd-electron → even-electron)
Baby steps
  • NO has 15 electrons (odd) → paramagnetic. NO2 has 23 (odd) → paramagnetic.
  • N2O4 is the dimer of NO2; the two odd electrons pair into an N–N bond → diamagnetic.
  • N2 has a full triple bond with all electrons paired → diamagnetic.
  • So N2O4 and N2 is the diamagnetic pair. (O2 is paramagnetic by MO theory.)
ShortcutOdd electron count = paramagnetic, no exceptions. O2 is the trap — it is even but still paramagnetic.
17.NEET 2016

The shape of XeF4 and the hybridisation of Xe in it are:

  • (A)Tetrahedral, sp3
  • (B)Square planar, sp3d2
  • (C)See-saw, sp3d
  • (D)Square pyramidal, sp3d2
Show solution
Answer  (B)  Square planar, sp3d2
Given
XeF4.
Asked
Shape and hybridisation.
Concept
Count the steric number (bond pairs + lone pairs) to get hybridisation, then let lone pairs decide the visible shape.
Formula
Steric number = 4 bp + 2 lp = 6 → sp3d2
Baby steps
  • Xe has 8 valence electrons; 4 go into bonds with F.
  • Remaining 4 electrons = 2 lone pairs.
  • Total 6 electron domains → octahedral arrangement, sp3d2.
  • The 2 lone pairs sit opposite each other (axial), leaving the 4 F atoms in a square plane.
ShortcutOctahedral base with 2 lone pairs is ALWAYS square planar — they go trans to minimise repulsion.
18.NEET 2021

Which of the following elements does not show allotropy?

  • (A)Nitrogen
  • (B)Phosphorus
  • (C)Sulphur
  • (D)Oxygen
Show solution
Answer  (A)  Nitrogen
Given
N, P, S, O.
Asked
The element with no allotropes.
Concept
Allotropy needs different ways of linking atoms. Nitrogen forms only the N≡N triple-bonded molecule and nothing else.
Formula
N2 only; P: white, red, black; S: rhombic, monoclinic; O: O2, O3
Baby steps
  • Nitrogen's pπ–pπ overlap is so effective that N2 is the only stable form.
  • Phosphorus cannot do that, so it catenates into P4 and polymer chains — hence allotropes.
  • Sulphur gives S8 rings in two crystal forms; oxygen gives O2 and O3.
ShortcutWithin group 15 and 16, only nitrogen lacks allotropes.
19.AIPMT 2014

Which is the correct order of the thermal stability of the hydrides of group 16?

  • (A)H2O < H2S < H2Se < H2Te
  • (B)H2O > H2S > H2Se > H2Te
  • (C)H2S > H2O > H2Se > H2Te
  • (D)H2Te > H2Se > H2S > H2O
Show solution
Answer  (B)  H2O > H2S > H2Se > H2Te
Given
H2O, H2S, H2Se, H2Te.
Asked
Order of thermal stability.
Concept
Stability depends on E–H bond strength, which falls as the central atom gets bigger and orbital overlap with H worsens.
Formula
Bond enthalpy O–H > S–H > Se–H > Te–H
Baby steps
  • Small atom = short strong bond = hard to decompose.
  • O is smallest, so H2O is the most stable (it does not decompose on heating).
  • H2Te is so unstable it decomposes on standing.
ShortcutThermal stability and reducing power are always opposites. Stability falls down; reducing power rises down.
20.NEET 2018

Ammonia is not manufactured by the Haber process at very high temperature because:

  • (A)The reaction is endothermic
  • (B)The catalyst decomposes
  • (C)The forward reaction is exothermic, so high T lowers the yield
  • (D)NH3 is insoluble
Show solution
Answer  (C)  The forward reaction is exothermic, so high T lowers the yield
Given
N2 + 3H2 ⇌ 2NH3, ΔH = −92 kJ mol⁻¹.
Asked
Why very high temperature is avoided.
Concept
Le Chatelier's principle — heating an exothermic equilibrium pushes it backwards, reducing the amount of product.
Formula
N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol⁻¹
Baby steps
  • Forward reaction releases heat, so extra heat favours the reverse.
  • But at low temperature the rate is uselessly slow.
  • Compromise: about 700 K, 200 atm, iron catalyst with Mo as promoter.
ShortcutHaber conditions to memorise: 200 atm · 700 K · Fe catalyst · Mo promoter. High pressure helps because moles decrease (4 → 2).
21.NEET 2017

Which noble gas was used by Neil Bartlett to prepare the first noble gas compound?

  • (A)Helium
  • (B)Argon
  • (C)Xenon
  • (D)Krypton
Show solution
Answer  (C)  Xenon
Given
Bartlett's 1962 discovery.
Asked
The noble gas that reacted.
Concept
Bartlett noticed that PtF6 oxidised O2 to O2⁺. Since the ionisation enthalpy of Xe is almost the same as that of O2, he predicted Xe would react too.
Formula
Xe + PtF6 → Xe+[PtF6]−
Baby steps
  • IE of O2 = 1175 kJ mol⁻¹; IE of Xe = 1170 kJ mol⁻¹ — nearly identical.
  • So whatever oxidises O2 should oxidise Xe.
  • The red solid Xe⁺[PtF6]⁻ was the first true noble gas compound.
ShortcutOnly Xe (and one Kr compound, KrF2) form real compounds. He, Ne and Ar form none.
22.NEET 2020

The correct order of the acidic strength of the hydrides of group 17 is:

  • (A)HF > HCl > HBr > HI
  • (B)HI > HBr > HCl > HF
  • (C)HCl > HF > HBr > HI
  • (D)HBr > HI > HCl > HF
Show solution
Answer  (B)  HI > HBr > HCl > HF
Given
HF, HCl, HBr, HI in water.
Asked
Order of acid strength.
Concept
Acidity of a binary hydride is governed by bond strength, not electronegativity. The weaker the H–X bond, the easier H⁺ leaves.
Formula
Acid strength ∝ 1/(H–X bond enthalpy)
Baby steps
  • H–F is the shortest and strongest bond, so HF is a weak acid.
  • Going down, bonds lengthen and weaken.
  • H–I is weakest, so HI is the strongest acid.
ShortcutBond strength beats electronegativity for binary acids. HF being 'weak' despite F being most electronegative is the classic trap.
23.AIPMT 2015

Which of the following statements about ozone is not correct?

  • (A)It has a bent structure with a bond angle of about 117°
  • (B)Both O–O bond lengths are equal due to resonance
  • (C)It is diamagnetic
  • (D)The O–O bond length in O3 is shorter than in O2
Show solution
Answer  (D)  The O–O bond length in O3 is shorter than in O2
Given
Structure and properties of O3.
Asked
The incorrect statement.
Concept
Ozone is a resonance hybrid of two structures, so both bonds are identical and intermediate between a single and a double bond — hence longer than the double bond in O2.
Formula
O–O in O3 = 128 pm; O=O in O2 = 121 pm
Baby steps
  • Central O is sp2 with one lone pair → bent, 117°.
  • Resonance makes both bonds equal at 128 pm.
  • O2 has a true double bond at 121 pm, which is shorter.
  • So the statement that O3's bond is shorter than O2's is wrong.
ShortcutBond order in O3 is 1.5, in O2 it is 2. Higher bond order = shorter bond, always.
24.NEET 2019

Aqua regia, which dissolves gold and platinum, is a mixture of concentrated HCl and HNO3 in the ratio:

  • (A)1 : 3 by volume
  • (B)3 : 1 by volume
  • (C)1 : 1 by volume
  • (D)2 : 1 by volume
Show solution
Answer  (B)  3 : 1 by volume
Given
Aqua regia.
Asked
Volume ratio of conc. HCl to conc. HNO3.
Concept
HNO3 supplies the oxidising power to convert Au to Au³⁺, while HCl supplies Cl⁻ to lock it up as a stable complex, driving the reaction forward.
Formula
Au + 4H⁺ + NO3⁻ + 4Cl⁻ → [AuCl4]− + NO + 2H2O
Baby steps
  • Neither acid alone dissolves gold.
  • HNO3 oxidises, HCl complexes — together they work.
  • The mixture is 3 parts conc. HCl to 1 part conc. HNO3.
ShortcutRemember it as 'three parts of the common acid, one part of the strong oxidiser'.
25.NEET 2016

The bleaching action of chlorine is due to:

  • (A)Reduction
  • (B)Oxidation
  • (C)Hydrolysis
  • (D)Its acidic nature
Show solution
Answer  (B)  Oxidation
Given
Bleaching by Cl2 in the presence of moisture.
Asked
The chemical process responsible.
Concept
Chlorine reacts with water to give hypochlorous acid, which releases nascent oxygen. That oxygen oxidises the coloured substance, destroying the chromophore permanently.
Formula
Cl2 + H2O → HCl + HOCl; HOCl → HCl + [O]
Baby steps
  • Dry chlorine cannot bleach — moisture is essential.
  • HOCl decomposes to nascent oxygen [O].
  • [O] oxidises the dye irreversibly, so the bleaching is permanent.
ShortcutCl2 bleaches by oxidation (permanent). SO2 bleaches by reduction (temporary — colour returns in air).
26.NEET 2021

The number of peroxide (–O–O–) linkages in peroxodisulphuric acid, H2S2O8, is:

  • (A)Zero
  • (B)One
  • (C)Two
  • (D)Three
Show solution
Answer  (B)  One
Given
H2S2O8 (Marshall's acid).
Asked
Number of peroxide linkages.
Concept
The prefix 'peroxo' tells you one –O–O– bridge has replaced a normal oxygen bridge. Both sulphurs stay at +6.
Formula
HO–SO2–O–O–SO2–OH
Baby steps
  • Draw two SO2(OH) units.
  • Join them not by a single O but by an O–O bridge.
  • That gives exactly one peroxide linkage.
ShortcutH2SO5 (Caro's acid) also has 1 peroxide linkage. H2S2O7 (oleum) has none — it has an S–O–S bridge instead.
27.AIPMT 2014

Which of the following elements does not form a stable compound with a +5 oxidation state readily, because of the inert pair effect?

  • (A)Phosphorus
  • (B)Arsenic
  • (C)Antimony
  • (D)Bismuth
Show solution
Answer  (D)  Bismuth
Given
Group 15 elements in the +5 state.
Asked
The element for which +5 is least stable.
Concept
Going down the group, the ns² electrons become reluctant to take part in bonding because the intervening d and f electrons shield the nucleus poorly. This is the inert pair effect.
Formula
Stability of +5 falls, of +3 rises, down the group
Baby steps
  • N and P readily show +5.
  • By Sb the +5 state is weaker.
  • In Bi the 6s² pair is effectively locked, so Bi³⁺ is the stable state and BiCl5 is unknown.
  • Bi(V) compounds are powerful oxidising agents because they revert to Bi(III).
ShortcutInert pair effect shows up in groups 13–16 for the heaviest member: Tl⁺, Pb²⁺, Bi³⁺ are the stable ions.
28.NEET 2018

Which one of the following is used as an oxidising agent in the contact process and what is the catalyst?

  • (A)O2 with Fe
  • (B)O2 with V2O5
  • (C)H2O2 with Pt
  • (D)O3 with CuCl2
Show solution
Answer  (B)  O2 with V2O5
Given
Manufacture of H2SO4 by the contact process.
Asked
Oxidant and catalyst for SO2 → SO3.
Concept
Sulphur dioxide is oxidised by atmospheric oxygen over vanadium(V) oxide, a heterogeneous catalyst that resists poisoning better than platinum.
Formula
2SO2 + O2 ⇌(V2O5, 720 K, 2 bar) 2SO3, ΔH = −196 kJ mol⁻¹
Baby steps
  • Burn S or a sulphide ore to get SO2.
  • Oxidise SO2 to SO3 over V2O5.
  • Absorb SO3 in conc. H2SO4 to get oleum, H2S2O7.
  • Dilute the oleum with water to get H2SO4.
ShortcutSO3 is never absorbed directly in water — it forms a fog. Always oleum first.
29.NEET 2017

The shape of ClF3 is:

  • (A)Trigonal planar
  • (B)Pyramidal
  • (C)T-shaped
  • (D)Trigonal bipyramidal
Show solution
Answer  (C)  T-shaped
Given
Interhalogen ClF3.
Asked
Molecular shape.
Concept
Steric number 5 gives a trigonal bipyramidal arrangement, and the lone pairs always occupy equatorial positions where repulsion is least.
Formula
Steric number = 3 bp + 2 lp = 5 → sp3d
Baby steps
  • Cl has 7 valence electrons; 3 are used in bonds to F.
  • Remaining 4 electrons = 2 lone pairs.
  • 5 domains → trigonal bipyramidal skeleton, sp3d.
  • Both lone pairs sit equatorial, leaving 3 F in a T shape (slightly bent, about 87°).
ShortcutAB3 interhalogens (ClF3, BrF3, ICl3) are all T-shaped. AB5 (BrF5, IF5) are square pyramidal. AB7 (IF7) is pentagonal bipyramidal.
30.NEET 2020

Which of the following has the highest first ionisation enthalpy?

  • (A)He
  • (B)Ne
  • (C)Ar
  • (D)Xe
Show solution
Answer  (A)  He
Given
Noble gases He, Ne, Ar, Xe.
Asked
Highest first ionisation enthalpy.
Concept
Ionisation enthalpy falls down a group as the outermost electron gets further from the nucleus and better shielded. Helium's electron is in the tiny 1s shell with no shielding at all.
Formula
IE: He (2372) > Ne (2080) > Ar (1520) > Kr (1351) > Xe (1170) kJ mol⁻¹
Baby steps
  • All noble gases have full shells, so all are hard to ionise.
  • Helium is the smallest atom in the entire periodic table apart from H.
  • Its electron is closest to the nucleus, so He has the highest IE of any element.
ShortcutHelium holds the record for the highest ionisation enthalpy of all elements.
31.AIPMT 2015

White phosphorus is stored under water because:

  • (A)It reacts violently with air and ignites at about 303 K
  • (B)It dissolves in water
  • (C)It is a liquid
  • (D)It reacts with nitrogen
Show solution
Answer  (A)  It reacts violently with air and ignites at about 303 K
Given
Storage of white phosphorus.
Asked
Reason for keeping it under water.
Concept
White phosphorus is P4 — a strained tetrahedron with 60° bond angles. That angle strain makes it extremely reactive, and it catches fire in air at only about 303 K.
Formula
P4 + 5O2 → P4O10
Baby steps
  • The P4 tetrahedron has 6 P–P bonds and 60° angles, far from the ideal 109.5°.
  • The strain makes it burst into flame in air.
  • It is insoluble in water but soluble in CS2, so water is a safe storage medium.
ShortcutWhite P: reactive, poisonous, glows in dark, 60° angle. Red P: polymeric, safe, used in matchboxes. Black P: most stable.
32.NEET 2019

Which of the following gives PH3 on reaction with water and is used in Holme's signals?

  • (A)Ca3P2
  • (B)P4O10
  • (C)PCl3
  • (D)H3PO3
Show solution
Answer  (A)  Ca3P2
Given
A compound used in Holme's signals.
Asked
Identify the compound.
Concept
Calcium phosphide hydrolyses to phosphine. Traces of P2H4 in the product ignite spontaneously in air, lighting up the escaping gas.
Formula
Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3
Baby steps
  • Containers with CaC2 and Ca3P2 are thrown into the sea.
  • Water hydrolyses Ca3P2 to PH3 and CaC2 to C2H2.
  • Impure PH3 ignites on its own, and the burning gas marks the spot.
ShortcutCa3P2 → phosphine → Holme's signal. Also used in smoke screens.
33.NEET 2016

Which of the following statements about fluorine is not correct?

  • (A)It shows only the −1 oxidation state
  • (B)It has the highest electronegativity
  • (C)It has the most negative electron gain enthalpy of the halogens
  • (D)It forms only one oxoacid, HOF
Show solution
Answer  (C)  It has the most negative electron gain enthalpy of the halogens
Given
Properties of fluorine.
Asked
The incorrect statement.
Concept
Fluorine is so small that an incoming electron feels strong repulsion from the electrons already crowded in the 2p subshell. Chlorine, being larger, accommodates it better.
Formula
Electron gain enthalpy: Cl (−349) > F (−328) kJ mol⁻¹ in magnitude
Baby steps
  • F is indeed the most electronegative element and shows only −1 (no d orbitals).
  • HOF is its only oxoacid.
  • But its electron gain enthalpy is LESS negative than that of chlorine because of small-size repulsion.
  • So statement 3 is the wrong one.
ShortcutSame anomaly in group 16: S has a more negative electron gain enthalpy than O. Period 2 is always overcrowded.
34.NEET 2021

The hybridisation of iodine in IF7 is:

  • (A)sp3d
  • (B)sp3d2
  • (C)sp3d3
  • (D)sp3
Show solution
Answer  (C)  sp3d3
Given
IF7.
Asked
Hybridisation of the central iodine.
Concept
Steric number is the count of sigma bonds plus lone pairs; with seven bonds and no lone pairs the count is 7.
Formula
Steric number 7 → sp3d3, pentagonal bipyramidal
Baby steps
  • I has 7 valence electrons, all used in bonds to seven F atoms.
  • So lone pairs = 0 and bond pairs = 7.
  • Steric number 7 needs one s, three p and three d orbitals → sp3d3.
ShortcutSteric number to hybridisation: 2 = sp, 3 = sp², 4 = sp³, 5 = sp³d, 6 = sp³d², 7 = sp³d³.
35.AIPMT 2014

Which of the following oxoacids of phosphorus contains a P–P bond?

  • (A)H3PO4
  • (B)H4P2O7
  • (C)H4P2O6
  • (D)H3PO3
Show solution
Answer  (C)  H4P2O6
Given
Oxoacids of phosphorus.
Asked
The one with a direct P–P linkage.
Concept
Hypophosphoric acid, H4P2O6, is the only common oxoacid in which two phosphorus atoms are bonded straight to each other rather than through an oxygen bridge.
Formula
(HO)2(O)P–P(O)(OH)2, with P in +4
Baby steps
  • H4P2O7 (pyrophosphoric) has a P–O–P bridge, not P–P.
  • H3PO4 and H3PO3 have only one P atom.
  • H4P2O6 has the two P atoms joined directly, giving oxidation state +4.
Shortcut'Pyro' = oxygen bridge (P–O–P). 'Hypophosphoric' = direct P–P bond.
36.NEET 2018

Which of the following noble gases forms clathrate compounds?

  • (A)He
  • (B)Ne
  • (C)Ar
  • (D)Both He and Ne
Show solution
Answer  (C)  Ar
Given
Clathrate formation with quinol or water.
Asked
Which noble gas can be trapped.
Concept
A clathrate is a cage compound — the gas atom is physically trapped in the lattice, not chemically bonded. The atom must be big enough not to escape through the cage.
Formula
Ar, Kr, Xe form clathrates; He and Ne do not
Baby steps
  • Quinol crystallises with cavities of a fixed size.
  • He and Ne are too small — they slip straight out.
  • Ar, Kr and Xe are large enough to be held, so they form clathrates.
ShortcutClathrates are physical trapping, not bonding. Size is the only criterion — the small ones fail.
37.NEET 2020

Sulphur hexafluoride, SF6, is inert towards hydrolysis whereas SF4 is readily hydrolysed. This is because:

  • (A)SF6 has no d orbitals
  • (B)S in SF6 is sterically protected and coordinatively saturated
  • (C)SF6 is ionic
  • (D)SF4 is a gas
Show solution
Answer  (B)  S in SF6 is sterically protected and coordinatively saturated
Given
SF6 vs SF4 towards water.
Asked
Reason for the inertness of SF6.
Concept
Six fluorine atoms pack tightly around the sulphur, blocking any approach by a water molecule. There is also no room for extra coordination and no lone pair to attack.
Formula
SF6: 6 bp, 0 lp, octahedral, sp3d2
Baby steps
  • SF4 is see-saw with one lone pair, so sulphur is exposed.
  • Water can attack that exposed site: SF4 + 2H2O → SO2 + 4HF.
  • In SF6 the sulphur is completely shielded, so hydrolysis is only thermodynamically feasible, never kinetically possible.
ShortcutNote the wording — SF6's inertness is KINETIC, not thermodynamic. NEET has tested this exact distinction.
38.NEET 2017

Which one of the following reactions of xenon compounds is not feasible?

  • (A)XeF6 + 3H2O → XeO3 + 6HF
  • (B)3XeF4 + 6H2O → 2Xe + XeO3 + 12HF + 1.5O2
  • (C)2XeF2 + 2H2O → 2Xe + 4HF + O2
  • (D)XeF6 + RbF → Rb[XeF7]
Show solution
Answer  (D)  XeF6 + RbF → Rb[XeF7]
Given
Reactions of xenon fluorides.
Asked
Statement that is chemically wrong (the NEET version marks the non-feasible one).
Concept
Xenon fluorides act as fluoride acceptors with strong Lewis acids, forming cations such as [XeF5]⁺. With fluoride donors like RbF, XeF6 does accept F⁻ — so watch the direction of transfer carefully in each option.
Formula
XeF6 + PF5 → [XeF5]+[PF6]−
Baby steps
  • Check hydrolysis products first — options 1 to 3 are all standard and correct.
  • XeF6 reacting with a strong Lewis acid gives [XeF5]⁺.
  • Whenever an option reverses the acid/base role of the fluoride, it is the odd one out.
ShortcutLearn the two directions: XeF6 + Lewis acid (PF5, SbF5) → [XeF5]⁺ salt. XeF6 + strong base (NaOH) → sodium perxenate Na4XeO6.
39.AIPMT 2015

Which one of the following anions is present in the chain structure of silicates? (p-block cross-check)

  • (A)SiO44−
  • (B)Si2O76−
  • (C)(SiO32−)n
  • (D)Si2O52−
Show solution
Answer  (C)  (SiO32−)n
Given
Silicate structural types.
Asked
Anion of a chain silicate.
Concept
Silicates are classified by how many corners of the SiO4 tetrahedron are shared. Sharing two corners in a row builds an infinite chain.
Formula
Chain silicate: (SiO32−)n, two corners shared
Baby steps
  • Zero corners shared → orthosilicate SiO4⁴⁻.
  • One corner shared → pyrosilicate Si2O7⁶⁻.
  • Two corners shared → chain (pyroxene), (SiO3²⁻)n.
  • Three corners shared → sheet, (Si2O5²⁻)n.
ShortcutCorners shared: 0, 1, 2, 3, 4 → ortho, pyro, chain/ring, sheet, 3-D. This is group 14, but NEET mixes it into p-block sets.
40.NEET 2019

The correct increasing order of the bond angles of NH3, PH3, AsH3 and SbH3 is:

  • (A)NH3 < PH3 < AsH3 < SbH3
  • (B)SbH3 < AsH3 < PH3 < NH3
  • (C)PH3 < NH3 < AsH3 < SbH3
  • (D)AsH3 < NH3 < SbH3 < PH3
Show solution
Answer  (B)  SbH3 < AsH3 < PH3 < NH3
Given
Hydrides of group 15.
Asked
Increasing order of bond angle.
Concept
As the central atom becomes larger and less electronegative, bond pairs sit further away and repel each other less, so the angle shrinks towards 90° (nearly pure p-orbital bonding).
Formula
NH3 107.8° > PH3 93.6° > AsH3 91.8° > SbH3 91.3°
Baby steps
  • Nitrogen is small and very electronegative, so bond pairs are pulled close and repel strongly → widest angle.
  • Down the group electronegativity drops and size rises, so the angle contracts.
  • Increasing order is therefore SbH3 < AsH3 < PH3 < NH3.
ShortcutSame pattern in group 16: H2O 104.5° > H2S 92° > H2Se 91° > H2Te 90°. Only the first member is far from 90°.
41.NEET 2021

Which of the following can be used to prepare very pure nitrogen?

  • (A)Heating NH4Cl with NaNO2
  • (B)Thermal decomposition of Ba(N3)2
  • (C)Liquefaction of air
  • (D)Passing NH3 over CuO
Show solution
Answer  (B)  Thermal decomposition of Ba(N3)2
Given
Methods of preparing N2.
Asked
Method giving the purest product.
Concept
Sodium or barium azide decomposes cleanly to the metal and nitrogen gas, with no other gaseous by-product to contaminate the sample.
Formula
Ba(N3)2 →(heat) Ba + 3N2
Baby steps
  • NH4Cl + NaNO2 gives N2 but is contaminated with NO and HNO3.
  • Liquefaction of air always leaves traces of noble gases.
  • Azide decomposition gives only a solid metal plus N2, so the gas is very pure.
ShortcutLab prep = NH4Cl + NaNO2. Very pure prep = azide. Industrial = liquefaction of air.
42.NEET 2016

The oxidation state of sulphur in the thiosulphate ion, S2O32−, in terms of its two distinct sulphur atoms, is:

  • (A)+2 and +2
  • (B)+6 and −2
  • (C)+4 and 0
  • (D)+5 and −1
Show solution
Answer  (B)  +6 and −2
Given
S2O3²⁻.
Asked
Oxidation state of each sulphur.
Concept
The two sulphurs are not equivalent — one is the central atom surrounded by oxygens, the other is a terminal sulphur that has replaced an oxygen of sulphate.
Formula
Average = +2; actual = +6 (central) and −2 (terminal)
Baby steps
  • Thiosulphate is sulphate with one O swapped for an S.
  • The central S keeps its sulphate value of +6.
  • The substituting S takes the place of an oxygen, so it carries −2.
  • Their average, (+6 − 2)/2 = +2, matches the simple formula calculation.
Shortcut'Thio' means an S has replaced an O. The replacing S always takes the oxidation state that O had, i.e. −2.
43.AIPMT 2014

Which of the following is the strongest oxidising agent?

  • (A)F2
  • (B)Cl2
  • (C)Br2
  • (D)I2
Show solution
Answer  (A)  F2
Given
The halogens.
Asked
Strongest oxidising agent.
Concept
Oxidising power in solution depends on three steps: breaking the X–X bond, adding an electron, and hydrating the ion. Fluorine wins because its bond is weak and its hydration enthalpy is enormous.
Formula
F2 > Cl2 > Br2 > I2; E° for F2/F⁻ = +2.87 V
Baby steps
  • F2 has a low bond dissociation enthalpy (only 158 kJ mol⁻¹).
  • F⁻ is tiny, so its hydration enthalpy is very high (−515 kJ mol⁻¹).
  • Both factors make the overall electrode potential the largest of any element.
  • Note this happens despite F having a less negative electron gain enthalpy than Cl.
ShortcutF2 is the strongest oxidising agent among ALL elements. Its power comes from bond enthalpy + hydration, not electron gain enthalpy.
44.NEET 2020

Which of the following gives sulphuryl chloride when reacted with chlorine in the presence of activated charcoal?

  • (A)SO3
  • (B)SO2
  • (C)H2SO4
  • (D)H2S
Show solution
Answer  (B)  SO2
Given
Reaction with Cl2 over activated charcoal.
Asked
The reactant that gives SO2Cl2.
Concept
Sulphur dioxide acts as a reducing agent and adds chlorine directly, with charcoal serving as the catalyst.
Formula
SO2 + Cl2 →(charcoal) SO2Cl2
Baby steps
  • Charcoal activates the chlorine.
  • SO2 is oxidised from +4 to +6 as it takes on two chlorines.
  • The product SO2Cl2 (sulphuryl chloride) is a fuming liquid that hydrolyses to H2SO4 + HCl.
ShortcutDo not confuse SO2Cl2 (sulphuryl chloride, S is +6) with SOCl2 (thionyl chloride, S is +4).
45.NEET 2018

Which of the following elements has the least negative electron gain enthalpy in group 16?

  • (A)O
  • (B)S
  • (C)Se
  • (D)Te
Show solution
Answer  (A)  O
Given
Group 16 elements.
Asked
Element with the least negative electron gain enthalpy.
Concept
Oxygen's 2p subshell is very compact, so an incoming electron faces strong inter-electronic repulsion. Sulphur, with a larger 3p subshell, accepts the electron more comfortably.
Formula
ΔegH: O (−141) is less negative than S (−200) kJ mol⁻¹
Baby steps
  • Normally electron gain enthalpy becomes less negative down a group.
  • But the first member of the group breaks the rule because of its small size.
  • So S is the most negative and O the least negative among O, S, Se, Te.
ShortcutPeriod 2 exceptions to remember: O < S and F < Cl in magnitude of electron gain enthalpy.
46.NEET 2017

The number of S–O–S bonds present in pyrosulphuric acid (oleum), H2S2O7, is:

  • (A)Zero
  • (B)One
  • (C)Two
  • (D)Three
Show solution
Answer  (B)  One
Given
H2S2O7.
Asked
Number of S–O–S bridges.
Concept
A 'pyro' acid is formed when two molecules of the parent acid lose one water molecule between them, leaving a single oxygen bridge.
Formula
2H2SO4 → H2S2O7 + H2O
Baby steps
  • Take two H2SO4 molecules side by side.
  • Remove one H2O from between them.
  • The remaining oxygen forms a single S–O–S bridge. So the answer is 1.
ShortcutChain of n identical units → (n − 1) bridges. H2S2O7 has 2 S atoms → 1 bridge; H4P2O7 has 2 P → 1 bridge.
47.AIPMT 2015

Which of the following oxides is amphoteric?

  • (A)N2O5
  • (B)P4O10
  • (C)Sb2O3
  • (D)Bi2O3
Show solution
Answer  (C)  Sb2O3
Given
Oxides of group 15 elements.
Asked
The amphoteric one.
Concept
Oxide character shifts from acidic to basic down the group as the element becomes more metallic. The middle members are amphoteric.
Formula
N, P oxides acidic → As, Sb oxides amphoteric → Bi oxide basic
Baby steps
  • N2O5 and P4O10 are non-metal oxides → strongly acidic.
  • Bi2O3 is a metal oxide → basic.
  • Sb sits at the metalloid boundary, so Sb2O3 reacts with both acids and bases → amphoteric.
ShortcutAmphoteric = the metalloid in the middle. In group 15 that is As2O3 and Sb2O3.
48.NEET 2019

Which of the following is used in the preparation of UF6 for uranium enrichment in the nuclear industry?

  • (A)ClF3
  • (B)ICl
  • (C)IF5
  • (D)BrCl
Show solution
Answer  (A)  ClF3
Given
Interhalogen compounds and their uses.
Asked
The one used to make UF6.
Concept
Chlorine trifluoride is a powerful fluorinating agent because the Cl–F bonds are weak and easily donate fluorine atoms.
Formula
U + 3ClF3 → UF6 + 3ClF
Baby steps
  • Interhalogens are more reactive than the parent halogens because A–B bonds are weaker than A–A bonds.
  • ClF3 releases fluorine readily.
  • It converts uranium metal to volatile UF6, which is then separated by gaseous diffusion.
ShortcutInterhalogen uses: ClF3 and BrF3 → fluorinating agents for nuclear fuel; ICl → Wijs reagent for iodine number of fats.

Answer key — NEET / AIPMT Previous Year Questions

1 D2 B3 C4 A5 A6 B7 C8 B9 B10 A11 B12 C13 B14 A15 C16 B17 B18 A19 B20 C21 C22 B23 D24 B25 B26 B27 D28 B29 C30 A31 A32 A33 C34 C35 C36 C37 B38 D39 C40 B41 B42 B43 A44 B45 A46 B47 C48 A
15

Nitrogen Family

N · P · As · Sb · Bi
30 questions  ·  full worked solutions
1.Periodic trends

The general outer electronic configuration of group 15 elements is:

  • (A)ns2np2
  • (B)ns2np3
  • (C)ns2np4
  • (D)ns2np5
Show solution
Answer  (B)  ns2np3
Given
Group 15 of the periodic table.
Asked
General valence shell configuration.
Concept
Group number 15 means 5 valence electrons. Two fill the s subshell and the remaining three go one each into the three p orbitals (Hund's rule), giving a half-filled, extra-stable p subshell.
Formula
Group 15 → ns2np3
Baby steps
  • Valence electrons = 15 − 10 = 5.
  • Fill s first: ns2.
  • Put the last three in p, singly: np3.
  • The half-filled p3 explains the unusually high ionisation enthalpy of this group.
ShortcutFor p-block: valence electrons = group number − 10.
2.Trends

Which element of group 15 shows the maximum tendency for catenation?

  • (A)Nitrogen
  • (B)Phosphorus
  • (C)Arsenic
  • (D)Bismuth
Show solution
Answer  (B)  Phosphorus
Given
N, P, As, Bi.
Asked
Element with greatest catenation.
Concept
Catenation needs a strong element–element single bond. The N–N single bond is very weak because lone pairs on the two small nitrogen atoms repel each other strongly.
Formula
Bond enthalpy: P–P (201) > N–N (159) > As–As (146) kJ mol⁻¹
Baby steps
  • N is small, so its lone pairs crowd each other → weak N–N single bond.
  • P is bigger, lone pairs are further apart → strongest single bond in the group.
  • Below P the bonds weaken again with poor overlap.
  • So catenation is maximum in phosphorus.
ShortcutNitrogen is great at multiple bonding (N≡N) but poor at single-bond catenation. The two are opposite skills.
3.Bonding

Nitrogen exists as a diatomic gas while phosphorus exists as P4 because:

  • (A)N is more electronegative
  • (B)N forms strong pπ–pπ multiple bonds, P does not
  • (C)P is a metal
  • (D)N has d orbitals
Show solution
Answer  (B)  N forms strong pπ–pπ multiple bonds, P does not
Given
N2 gas vs P4 solid.
Asked
Reason for the difference.
Concept
Only period-2 elements have small enough p orbitals for effective sideways (pπ–pπ) overlap. Phosphorus's 3p orbitals are too large and diffuse, so P prefers three single bonds instead.
Formula
N≡N bond enthalpy = 941 kJ mol⁻¹
Baby steps
  • Nitrogen achieves its octet with one triple bond → a small, stable N2 molecule.
  • Phosphorus cannot form pπ–pπ bonds, so each P forms three single bonds.
  • Three single bonds per atom builds the P4 tetrahedron.
ShortcutPeriod 2 (C, N, O) does pπ–pπ. Period 3 and below (Si, P, S) prefers single bonds and catenation.
4.Hydrides

The correct order of basic strength of group 15 hydrides is:

  • (A)NH3 > PH3 > AsH3 > SbH3 > BiH3
  • (B)BiH3 > SbH3 > AsH3 > PH3 > NH3
  • (C)PH3 > NH3 > AsH3 > BiH3
  • (D)NH3 > AsH3 > PH3 > SbH3
Show solution
Answer  (A)  NH3 > PH3 > AsH3 > SbH3 > BiH3
Given
NH3, PH3, AsH3, SbH3, BiH3.
Asked
Order of basicity.
Concept
Basicity means willingness to donate the lone pair. In a small atom the lone pair is concentrated in a compact orbital and is easy to donate; in a large atom it is spread out and diffuse.
Formula
Basicity ∝ 1/size of central atom
Baby steps
  • N is the smallest, so its lone pair has the highest charge density.
  • Down the group the lone pair spreads over a bigger orbital and becomes less available.
  • BiH3 is barely basic at all.
ShortcutBasicity, bond angle, bond strength and thermal stability all DECREASE down group 15. Only reducing power increases.
5.Hydrides

Which of the following hydrides has the lowest boiling point?

  • (A)NH3
  • (B)PH3
  • (C)AsH3
  • (D)SbH3
Show solution
Answer  (B)  PH3
Given
Group 15 hydrides.
Asked
Lowest boiling point.
Concept
Boiling point rises with molecular mass, but NH3 is lifted out of order by hydrogen bonding. PH3 has neither the H-bonding of NH3 nor the large mass of AsH3.
Formula
Boiling point: PH3 < AsH3 < NH3 < SbH3
Baby steps
  • Rank by mass first: PH3 < AsH3 < SbH3.
  • NH3 would be lowest by mass, but H-bonding pushes it up above AsH3.
  • That leaves PH3 at the bottom.
ShortcutWhenever the first hydride of a group is pulled upward by H-bonding, the SECOND member becomes the minimum.
6.Nitrogen

Nitrogen is chemically inert at room temperature mainly because:

  • (A)It is a gas
  • (B)Of the very high N≡N bond dissociation enthalpy
  • (C)It has no lone pair
  • (D)It has empty d orbitals
Show solution
Answer  (B)  Of the very high N≡N bond dissociation enthalpy
Given
Molecular nitrogen at room temperature.
Asked
Reason for its inertness.
Concept
Any reaction of N2 must start by breaking the triple bond, which costs an enormous amount of energy. The activation barrier is therefore very high.
Formula
N≡N → 2N, ΔH = +941 kJ mol⁻¹
Baby steps
  • The triple bond has one sigma and two pi bonds.
  • 941 kJ mol⁻¹ is one of the highest bond enthalpies known.
  • At room temperature no ordinary reaction supplies that energy, so N2 is inert.
  • At high temperature it does react — e.g. with Mg to give Mg3N2.
ShortcutN2 reacts with only three things easily: Li (Li3N), Mg (Mg3N2), and H2 under Haber conditions.
7.Preparation

In the laboratory, dinitrogen is prepared by heating:

  • (A)NH4NO3
  • (B)NH4Cl with NaNO2
  • (C)NH4NO2 alone at 800 K
  • (D)(NH4)2Cr2O7
Show solution
Answer  (B)  NH4Cl with NaNO2
Given
Laboratory preparation of N2.
Asked
The standard reagent pair.
Concept
Ammonium nitrite is unstable, so it is generated in situ from ammonium chloride and sodium nitrite and decomposes immediately to nitrogen and water.
Formula
NH4Cl + NaNO2 → N2 + 2H2O + NaCl
Baby steps
  • Mix aqueous NH4Cl and NaNO2.
  • NH4NO2 forms and decomposes on warming.
  • N2 is collected over water. (NH4NO3 would give N2O instead, and (NH4)2Cr2O7 gives N2 too but is the volcano demonstration.)
ShortcutNH4NO2 → N2 (dinitrogen). NH4NO3 → N2O (laughing gas). One letter, completely different product.
8.Ammonia

The shape of the ammonia molecule and the hybridisation of nitrogen are:

  • (A)Trigonal planar, sp2
  • (B)Pyramidal, sp3
  • (C)Tetrahedral, sp3
  • (D)Bent, sp3
Show solution
Answer  (B)  Pyramidal, sp3
Given
NH3 molecule.
Asked
Shape and hybridisation.
Concept
Steric number counts bond pairs plus lone pairs to fix the hybridisation; the visible shape then ignores the lone pair positions.
Formula
Steric number = 3 bp + 1 lp = 4 → sp3, bond angle 107.8°
Baby steps
  • N has 5 valence electrons; 3 form bonds with H, leaving one lone pair.
  • 4 electron domains → sp3 hybridisation, tetrahedral electron geometry.
  • The lone pair occupies one corner, so the atoms form a trigonal pyramid.
  • Lone pair–bond pair repulsion squeezes the angle from 109.5° to 107.8°.
ShortcutElectron geometry uses all domains; molecular shape counts only atoms. sp³ with 1 lp = pyramidal, with 2 lp = bent.
9.Ammonia

Ammonia forms a deep blue colour with which of the following?

  • (A)FeSO4 solution
  • (B)Cu2+ solution
  • (C)AgNO3 only
  • (D)BaCl2 solution
Show solution
Answer  (B)  Cu2+ solution
Given
Test for ammonia with metal salt solutions.
Asked
The ion giving a deep blue colour.
Concept
Excess ammonia acts as a ligand and replaces water in the copper aqua ion, forming an intensely coloured tetraammine complex.
Formula
Cu2+ + 4NH3 → [Cu(NH3)4]2+ (deep blue)
Baby steps
  • A little NH3 first precipitates pale blue Cu(OH)2.
  • Excess NH3 dissolves the precipitate.
  • The deep blue [Cu(NH3)4]²⁺ complex forms — a standard NH3 test.
ShortcutNH3 also dissolves AgCl as [Ag(NH3)2]⁺ (colourless) — that's the other classic ammonia test.
10.Nitric acid

Dilute nitric acid reacting with copper gives mainly:

  • (A)NO2
  • (B)NO
  • (C)N2O
  • (D)N2
Show solution
Answer  (B)  NO
Given
Cu + dilute HNO3.
Asked
Main nitrogen-containing gas produced.
Concept
The extent of reduction of nitrogen depends on acid concentration. Concentrated acid is reduced only to +4 (NO2); dilute acid, with fewer H⁺ ions per site, is reduced further to +2 (NO).
Formula
3Cu + 8HNO3 (dilute) → 3Cu(NO3)2 + 2NO + 4H2O
Baby steps
  • Nitrogen in HNO3 starts at +5.
  • Concentrated acid: reduced by 1 → NO2, brown fumes.
  • Dilute acid: reduced by 3 → NO, colourless (turns brown in air as it forms NO2).
  • Very dilute acid with active metals like Zn can go all the way to NH4⁺.
ShortcutMore dilute = deeper reduction. Conc. → NO2, dilute → NO, very dilute with active metal → N2O or NH4NO3.
11.Nitric acid

Which metal becomes passive in concentrated nitric acid?

  • (A)Cu
  • (B)Zn
  • (C)Cr
  • (D)Mg
Show solution
Answer  (C)  Cr
Given
Metals treated with conc. HNO3.
Asked
The metal rendered passive.
Concept
Concentrated nitric acid oxidises the metal surface to a thin, continuous, non-porous oxide layer that blocks further attack.
Formula
Cr, Al, Fe, Co, Ni → passive in conc. HNO3
Baby steps
  • Conc. HNO3 is a strong oxidiser.
  • On Cr, Al and Fe it forms an adherent oxide film.
  • The film seals the metal, so the reaction stops — this is passivity.
  • Cu, Zn and Mg have no such protective film and dissolve.
ShortcutPassive metals in conc. HNO3: Cr, Al, Fe, Co, Ni. This is why conc. HNO3 is transported in aluminium tankers.
12.Oxides

The anhydride of nitric acid is:

  • (A)N2O3
  • (B)NO2
  • (C)N2O5
  • (D)N2O
Show solution
Answer  (C)  N2O5
Given
HNO3 and its oxides.
Asked
Corresponding acid anhydride.
Concept
An anhydride is what remains after removing water from the acid, keeping the oxidation state unchanged.
Formula
2HNO3 − H2O → N2O5
Baby steps
  • Take two HNO3: H2N2O6.
  • Remove one H2O: N2O5.
  • Check oxidation state — N is +5 in both, so this is correct.
ShortcutAnhydride pairs: HNO2 → N2O3, HNO3 → N2O5, H2SO3 → SO2, H2SO4 → SO3, H3PO4 → P4O10.
13.Oxides

Which oxide of nitrogen is known as laughing gas?

  • (A)NO
  • (B)N2O
  • (C)NO2
  • (D)N2O4
Show solution
Answer  (B)  N2O
Given
Nitrogen oxides.
Asked
The one called laughing gas.
Concept
Dinitrogen oxide is a neutral, colourless gas with mild anaesthetic properties, made by carefully heating ammonium nitrate.
Formula
NH4NO3 →(heat, 523 K) N2O + 2H2O
Baby steps
  • N in N2O has oxidation state +1.
  • It is neutral to litmus and does not react with water.
  • It is used as a mild anaesthetic and as a propellant in whipped cream.
ShortcutN2O and NO are the only neutral oxides of nitrogen; both are also the ones with the lowest oxidation states (+1, +2).
14.Allotropes

The bond angle in the P4 molecule of white phosphorus is:

  • (A)109.5°
  • (B)120°
  • (C)60°
  • (D)90°
Show solution
Answer  (C)  60°
Given
P4 tetrahedron.
Asked
P–P–P bond angle.
Concept
In the tetrahedral P4 unit the four phosphorus atoms sit at the corners of a regular tetrahedron and are bonded along the edges, forming equilateral triangular faces.
Formula
Each face is an equilateral triangle → 60°
Baby steps
  • Draw the P4 tetrahedron: 4 atoms, 6 P–P bonds.
  • Each face is an equilateral triangle, so every angle is 60°.
  • The ideal angle for sp3 would be 109.5°, so P4 carries huge angular strain.
  • That strain is exactly why white phosphorus is so reactive.
ShortcutDon't confuse the 60° in the P4 tetrahedron with the 109.5° of a tetrahedral molecule like CH4. Here the atoms bond along edges, not to a centre.
15.Allotropes

Which allotrope of phosphorus is the most thermodynamically stable?

  • (A)White
  • (B)Red
  • (C)Black
  • (D)Yellow
Show solution
Answer  (C)  Black
Given
Allotropes of phosphorus.
Asked
Most stable form.
Concept
Stability rises as the structure becomes more polymeric and layered, since more bonds are formed per atom and angular strain is relieved.
Formula
Stability: white < red < black
Baby steps
  • White P is discrete strained P4 units — least stable, most reactive.
  • Red P is a polymeric chain of linked P4 units — more stable, non-poisonous.
  • Black P has a layered, graphite-like structure and even conducts electricity — the most stable form.
ShortcutReactivity order is exactly reversed: white > red > black.
16.Phosphine

Phosphine is prepared in the laboratory by heating white phosphorus with:

  • (A)Dilute HCl
  • (B)Concentrated NaOH in an inert atmosphere
  • (C)Water
  • (D)Conc. H2SO4
Show solution
Answer  (B)  Concentrated NaOH in an inert atmosphere
Given
Laboratory preparation of PH3.
Asked
Reagent used.
Concept
White phosphorus disproportionates in hot concentrated alkali — some phosphorus is reduced to PH3 (−3) and some is oxidised to hypophosphite (+1).
Formula
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
Baby steps
  • Heat white P with conc. NaOH in a CO2 atmosphere.
  • P goes from 0 to −3 in PH3 and from 0 to +1 in NaH2PO2 — a disproportionation.
  • An inert atmosphere is essential because traces of P2H4 make the gas catch fire.
ShortcutAny time an element ends up in two different oxidation states from one starting material, call it disproportionation.
17.Phosphine

PH3 is a weaker base than NH3 because:

  • (A)P is more electronegative
  • (B)The lone pair on P is in a larger, more diffuse orbital
  • (C)PH3 has no lone pair
  • (D)PH3 is a solid
Show solution
Answer  (B)  The lone pair on P is in a larger, more diffuse orbital
Given
Basicity comparison of NH3 and PH3.
Asked
Reason PH3 is the weaker base.
Concept
Donating a lone pair is easy when that pair is compact and has high electron density. In P the lone pair sits in a large 3sp3 orbital and is spread thin.
Formula
Basicity: NH3 ≫ PH3 > AsH3 > SbH3 > BiH3
Baby steps
  • Both have one lone pair on the central atom.
  • N is small (65 pm), so its lone pair is concentrated.
  • P is much bigger (110 pm), so the same pair is diluted over more space.
  • Less concentrated lone pair = poorer electron donor = weaker base.
ShortcutPH3 is so weakly basic it does not turn moist red litmus blue the way NH3 does.
18.Halides

PCl3 on hydrolysis gives:

  • (A)H3PO4 and HCl
  • (B)H3PO3 and HCl
  • (C)H3PO2 and Cl2
  • (D)POCl3 and H2
Show solution
Answer  (B)  H3PO3 and HCl
Given
PCl3 + water.
Asked
Hydrolysis products.
Concept
Hydrolysis is a substitution, not a redox reaction, so phosphorus keeps its oxidation state of +3. Each Cl is replaced by an OH group.
Formula
PCl3 + 3H2O → H3PO3 + 3HCl
Baby steps
  • P is +3 in PCl3.
  • Replace each of the three Cl atoms with OH.
  • That gives P(OH)3, which rearranges to H3PO3 (phosphorous acid, +3).
  • Similarly PCl5 + 4H2O → H3PO4 + 5HCl, keeping P at +5.
ShortcutHydrolysis never changes oxidation state. Match +3 chloride → +3 acid, +5 chloride → +5 acid.
19.Halides

In gaseous PCl5, the axial P–Cl bonds are longer than the equatorial ones because:

  • (A)Axial bonds are ionic
  • (B)Axial bonds face repulsion from three equatorial bond pairs each
  • (C)Equatorial bonds use d orbitals
  • (D)Axial Cl atoms are larger
Show solution
Answer  (B)  Axial bonds face repulsion from three equatorial bond pairs each
Given
Trigonal bipyramidal PCl5.
Asked
Reason axial bonds are longer.
Concept
In a trigonal bipyramid, an axial bond pair is at 90° to three equatorial bond pairs, while an equatorial pair is at 90° to only two axial pairs. More close-range repulsion means a longer, weaker bond.
Formula
Axial P–Cl = 240 pm; equatorial P–Cl = 202 pm
Baby steps
  • Count 90° neighbours: axial has 3, equatorial has 2.
  • More repulsion pushes the axial chlorines further out.
  • So the axial bonds are longer and weaker, and PCl5 loses them first to become PCl3 + Cl2.
ShortcutSame reasoning explains why lone pairs always take equatorial positions in sp³d molecules like ClF3 and SF4.
20.Oxoacids

The basicity of hypophosphorous acid, H3PO2, is:

  • (A)1
  • (B)2
  • (C)3
  • (D)4
Show solution
Answer  (A)  1
Given
H3PO2.
Asked
Basicity (number of replaceable H).
Concept
Only hydrogen atoms bonded to oxygen can ionise. In H3PO2 the phosphorus carries two hydrogens directly, and only one OH group is present.
Formula
Structure: H–P(=O)(H)–OH → 1 ionisable H
Baby steps
  • Draw H3PO2: P has one =O, one –OH and two P–H bonds.
  • P–H hydrogens cannot leave as H⁺.
  • So basicity = 1 (monobasic).
  • The two P–H bonds make H3PO2 a strong reducing agent — it reduces AgNO3 to silver.
ShortcutFor H3POn: basicity = n − 1, P–H bonds = 4 − n. Check with H3PO4 (basicity 3, no P–H).
21.Oxoacids

On heating, H3PO3 undergoes disproportionation to give:

  • (A)H3PO4 and PH3
  • (B)H4P2O7 and H2O
  • (C)P4O10 and H2
  • (D)H3PO2 and P4
Show solution
Answer  (A)  H3PO4 and PH3
Given
Thermal decomposition of phosphorous acid.
Asked
Products formed.
Concept
Phosphorus is at +3 in H3PO3, an intermediate oxidation state. On heating, part of it is oxidised to +5 and part is reduced to −3.
Formula
4H3PO3 →(heat) 3H3PO4 + PH3
Baby steps
  • Starting oxidation state of P is +3.
  • Some P rises to +5 → H3PO4.
  • Some P falls to −3 → PH3.
  • One species going both up and down means disproportionation.
ShortcutElements in intermediate oxidation states are disproportionation candidates. Check: +3 sits between −3 and +5.
22.Trends

The correct order of the first ionisation enthalpy of group 15 elements is:

  • (A)N < P < As < Sb < Bi
  • (B)N > P > As > Sb > Bi
  • (C)P > N > As > Bi > Sb
  • (D)Bi > Sb > As > P > N
Show solution
Answer  (B)  N > P > As > Sb > Bi
Given
N, P, As, Sb, Bi.
Asked
Order of first ionisation enthalpy.
Concept
Ionisation enthalpy falls down a group because atomic size increases and the outer electrons are better shielded, so they are held less tightly.
Formula
N (1402) > P (1012) > As (947) > Sb (834) > Bi (703) kJ mol⁻¹
Baby steps
  • Size increases smoothly down the group.
  • Bigger atom = outer electron further from nucleus = easier to remove.
  • So the enthalpy decreases steadily from N to Bi.
  • Group 15 values are unusually high for their period because of the stable half-filled p3 configuration.
ShortcutCompare across period 2: N has a higher IE than O, because of the half-filled p³ stability.
23.Trends

Which of the following is the most metallic element in group 15?

  • (A)N
  • (B)P
  • (C)As
  • (D)Bi
Show solution
Answer  (D)  Bi
Given
Group 15 elements.
Asked
Most metallic element.
Concept
Metallic character increases down a group as ionisation enthalpy falls and the atom loses electrons more readily.
Formula
N, P non-metals → As, Sb metalloids → Bi metal
Baby steps
  • N and P are typical non-metals.
  • As and Sb sit on the boundary as metalloids.
  • Bi is a genuine metal, shiny and a conductor.
  • So metallic character is maximum in Bi.
ShortcutGroup 15 is the only group containing non-metals, metalloids AND a metal — good for 'which group shows all three' questions.
24.Oxides

Which pair of oxides is neutral?

  • (A)N2O and NO
  • (B)NO2 and N2O3
  • (C)N2O5 and NO
  • (D)N2O3 and N2O5
Show solution
Answer  (A)  N2O and NO
Given
Oxides of nitrogen.
Asked
The neutral pair.
Concept
An oxide is acidic only if it reacts with water to give an acid. N2O and NO do not react with water at all.
Formula
N2O (+1) and NO (+2) → neutral; N2O3, NO2, N2O5 → acidic
Baby steps
  • Test each with water. N2O3 → HNO2, N2O5 → HNO3, NO2 → mixture of both.
  • N2O and NO give nothing.
  • So the neutral pair is N2O and NO.
ShortcutLow oxidation states (+1, +2) → neutral; high oxidation states (+3, +4, +5) → acidic. This rule generalises across the p-block.
25.Structures

The number of P–P bonds in the P4 molecule is:

  • (A)4
  • (B)5
  • (C)6
  • (D)3
Show solution
Answer  (C)  6
Given
White phosphorus, P4.
Asked
Number of P–P bonds.
Concept
In a tetrahedron each vertex is joined to every other vertex, so the number of bonds equals the number of edges.
Formula
Edges of a tetrahedron = 6
Baby steps
  • Each P atom forms 3 bonds (it needs 3 to complete its octet).
  • Total bond-ends = 4 × 3 = 12.
  • Each bond has 2 ends, so bonds = 12/2 = 6.
ShortcutCount bonds as (atoms × bonds per atom)/2. Works for P4O6 (12 P–O bonds) and P4O10 (16 P–O bonds) too.
26.Structures

In P4O10, the number of P=O (terminal double bond) linkages is:

  • (A)Zero
  • (B)Two
  • (C)Four
  • (D)Six
Show solution
Answer  (C)  Four
Given
Phosphorus pentoxide, P4O10.
Asked
Number of terminal P=O bonds.
Concept
P4O10 is built from the P4 tetrahedron: one oxygen inserts into each of the 6 P–P bonds as a bridge, and each of the 4 phosphorus atoms also carries one terminal oxygen.
Formula
10 O = 6 bridging + 4 terminal
Baby steps
  • Start with the P4 tetrahedron (6 edges).
  • Insert one O into each edge → 6 bridging O, giving P4O6.
  • Add one terminal O to each of the 4 P atoms → 4 P=O bonds.
  • Total oxygens 6 + 4 = 10, matching P4O10.
ShortcutP4O6 = 6 bridging O only, P is +3. P4O10 = 6 bridging + 4 terminal, P is +5.
27.Uses

Nitric acid is stored in dark coloured bottles because:

  • (A)It reacts with glass
  • (B)It decomposes in light to NO2, O2 and water
  • (C)It evaporates in light
  • (D)It absorbs CO2
Show solution
Answer  (B)  It decomposes in light to NO2, O2 and water
Given
Storage of HNO3.
Asked
Reason for using dark bottles.
Concept
Nitric acid is photochemically unstable — light supplies enough energy to break the N–O bonds, and the brown NO2 formed dissolves in the acid and colours it yellow.
Formula
4HNO3 →(light) 4NO2 + O2 + 2H2O
Baby steps
  • Light triggers decomposition.
  • NO2 formed is brown and dissolves in the acid, turning it yellow.
  • Dark bottles cut out light and keep the acid colourless.
ShortcutSame reason AgBr and AgCl are kept in dark bottles — anything that decomposes photochemically.
28.Fertilisers

Which of the following is the correct set of conditions for the Haber process?

  • (A)500 K, 1 atm, Pt catalyst
  • (B)700 K, 200 atm, Fe catalyst with Mo promoter
  • (C)1000 K, 10 atm, V2O5
  • (D)300 K, 500 atm, Ni
Show solution
Answer  (B)  700 K, 200 atm, Fe catalyst with Mo promoter
Given
Industrial synthesis of ammonia.
Asked
Optimum conditions.
Concept
The forward reaction is exothermic and reduces the number of gas moles, so high pressure helps the yield while temperature is a compromise between yield and rate.
Formula
N2 + 3H2 ⇌ 2NH3, ΔH = −92 kJ mol⁻¹
Baby steps
  • 4 moles of gas become 2, so raising pressure shifts the equilibrium right — use about 200 atm.
  • The reaction is exothermic, so low T favours yield but is too slow.
  • Compromise at about 700 K with iron oxide catalyst; Mo acts as a promoter.
  • NH3 is removed by liquefaction and unreacted gases are recycled.
ShortcutA promoter is not a catalyst — it makes the catalyst more effective. Fe is the catalyst, Mo the promoter.
29.Trends

The correct order of the −ve electron gain enthalpy (magnitude) in group 15 is:

  • (A)N > P > As > Sb
  • (B)P > As > Sb > N
  • (C)Sb > As > P > N
  • (D)As > P > N > Sb
Show solution
Answer  (B)  P > As > Sb > N
Given
Group 15 elements.
Asked
Order of magnitude of electron gain enthalpy.
Concept
Nitrogen breaks the trend: its 2p subshell is so compact that adding an extra electron causes severe repulsion, so nitrogen actually has a slightly positive electron gain enthalpy.
Formula
ΔegH: N (+ve, ≈ 0) < Sb < As < P (most negative)
Baby steps
  • Normally the magnitude falls down a group, so expect P > As > Sb.
  • But N is anomalous because of its very small size and half-filled p3 stability.
  • N drops to the bottom of the list, giving P > As > Sb > N.
ShortcutTwo reasons N resists an extra electron: tiny size and a stable half-filled p³. This is why N3⁻ is rare compared to P3⁻.
30.Reactions

Ammonia acts as a Lewis base because it:

  • (A)Accepts a proton only
  • (B)Donates its lone pair of electrons
  • (C)Accepts electrons
  • (D)Releases OH⁻ ions
Show solution
Answer  (B)  Donates its lone pair of electrons
Given
NH3 as a base.
Asked
The Lewis definition of its basicity.
Concept
A Lewis base is an electron pair donor. Nitrogen in NH3 has one lone pair that it can hand to any electron-deficient species.
Formula
NH3 + BF3 → H3N→BF3 (adduct)
Baby steps
  • Lewis base = electron pair donor; Lewis acid = electron pair acceptor.
  • NH3 has one lone pair on N.
  • It donates that pair to BF3 (electron deficient) or to metal ions to form complexes.
  • Its Brønsted basicity (accepting H⁺) is really the same lone pair at work.
ShortcutAnything with a lone pair can be a Lewis base: NH3, H2O, PH3, F⁻. Anything with an empty orbital is a Lewis acid: BF3, AlCl3, metal ions.

Answer key — Nitrogen Family

1 B2 B3 B4 A5 B6 B7 B8 B9 B10 B11 C12 C13 B14 C15 C16 B17 B18 B19 B20 A21 A22 B23 D24 A25 C26 C27 B28 B29 B30 B
16

Oxygen Family

O · S · Se · Te · Po
30 questions  ·  full worked solutions
1.Periodic trends

The general outer electronic configuration of the chalcogens is:

  • (A)ns2np3
  • (B)ns2np4
  • (C)ns2np5
  • (D)ns2np6
Show solution
Answer  (B)  ns2np4
Given
Group 16 (chalcogens).
Asked
General valence configuration.
Concept
Group 16 elements have six valence electrons: two in the s subshell and four in the p subshell, leaving two unpaired p electrons.
Formula
Group 16 → ns2np4
Baby steps
  • Valence electrons = 16 − 10 = 6.
  • Fill s: ns2. Remaining 4 go to p: np4.
  • Two unpaired electrons explain the common valency of 2 and the −2 oxidation state.
ShortcutChalcogen means 'ore forming' — most metal ores are oxides or sulphides.
2.Trends

Oxygen shows an oxidation state of +2 only in:

  • (A)H2O
  • (B)OF2
  • (C)O3
  • (D)Na2O
Show solution
Answer  (B)  OF2
Given
Compounds of oxygen.
Asked
Compound with O in +2.
Concept
Oxygen is the second most electronegative element, so it is negative in almost every compound. Only fluorine, being more electronegative, can force oxygen positive.
Formula
In OF2: 2(−1) + x = 0 → x = +2
Baby steps
  • F is always −1 in any compound.
  • In OF2 there are two F, contributing −2 in total.
  • For the neutral molecule, O must be +2.
  • In H2O, O3 and Na2O oxygen is −2 or 0.
ShortcutOnly fluorine makes oxygen positive. In O2F2 oxygen is +1; in OF2 it is +2.
3.Trends

The correct order of atomic radii in group 16 is:

  • (A)O > S > Se > Te
  • (B)Te > Se > S > O
  • (C)S > O > Se > Te
  • (D)Se > Te > S > O
Show solution
Answer  (B)  Te > Se > S > O
Given
O, S, Se, Te.
Asked
Order of atomic radius.
Concept
Down a group a new principal shell is added each time, so the radius increases despite the rising nuclear charge.
Formula
Radius (pm): O 66 < S 104 < Se 117 < Te 137
Baby steps
  • Each step down adds one shell.
  • The added shell shields the outer electrons from the nucleus.
  • So size increases: O < S < Se < Te, i.e. Te > Se > S > O.
  • The jump from O to S is the largest of the group.
ShortcutThe first-to-second member jump is always the biggest in any p-block group.
4.Hydrides

Which of the following hydrides is the strongest acid in aqueous solution?

  • (A)H2O
  • (B)H2S
  • (C)H2Se
  • (D)H2Te
Show solution
Answer  (D)  H2Te
Given
Group 16 hydrides.
Asked
Strongest acid.
Concept
For binary hydrides, acidity is decided by how easily the E–H bond breaks. Bigger central atom means longer, weaker bond and easier release of H⁺.
Formula
Acidity: H2O < H2S < H2Se < H2Te
Baby steps
  • Te is the largest atom in the list.
  • The Te–H bond is therefore the longest and weakest.
  • The proton leaves most easily, making H2Te the strongest acid.
  • Water is so weak an acid it is effectively neutral.
ShortcutDown a group, acidity of hydrides ALWAYS increases. Across a period, electronegativity takes over instead.
5.Hydrides

The decreasing order of the bond angle in group 16 hydrides is:

  • (A)H2O > H2S > H2Se > H2Te
  • (B)H2Te > H2Se > H2S > H2O
  • (C)H2S > H2O > H2Se > H2Te
  • (D)H2Se > H2S > H2O > H2Te
Show solution
Answer  (A)  H2O > H2S > H2Se > H2Te
Given
H2O, H2S, H2Se, H2Te.
Asked
Decreasing order of bond angle.
Concept
In heavier members, bonding uses almost pure p orbitals (which are at 90°) instead of sp3 hybrids, because s–p mixing becomes poor with increasing size.
Formula
H2O 104.5° > H2S 92.1° > H2Se 91° > H2Te 90°
Baby steps
  • Water uses sp3 hybrids, giving an angle near the tetrahedral value.
  • In H2S and below, the s and p orbitals differ too much in energy to mix well.
  • Bonding uses nearly pure p orbitals, so the angle collapses to about 90°.
ShortcutAny hydride angle near 90° means unhybridised p-orbital bonding. Only the period-2 hydride keeps a big angle.
6.Oxygen

Dioxygen is paramagnetic. According to molecular orbital theory, this is because:

  • (A)It has a double bond
  • (B)It has two unpaired electrons in π* orbitals
  • (C)It has a lone pair
  • (D)It is a gas
Show solution
Answer  (B)  It has two unpaired electrons in π* orbitals
Given
O2 molecule.
Asked
Reason for paramagnetism.
Concept
Filling the molecular orbitals of O2 places the last two electrons singly into the two degenerate π* antibonding orbitals, following Hund's rule.
Formula
O2: σ2s² σ*2s² σ2pz² π2px² π2py² π*2px¹ π*2py¹
Baby steps
  • O2 has 16 electrons in total.
  • After filling the bonding orbitals, 2 electrons remain for the two π* orbitals.
  • Hund's rule puts them singly with parallel spins.
  • Two unpaired electrons = paramagnetic. Bond order = (10 − 6)/2 = 2.
ShortcutMO theory's biggest win — Lewis structures wrongly predict O2 is diamagnetic.
7.Ozone

The bond angle and hybridisation of the central oxygen atom in ozone are:

  • (A)120°, sp2
  • (B)117°, sp2
  • (C)109.5°, sp3
  • (D)180°, sp
Show solution
Answer  (B)  117°, sp2
Given
O3 molecule.
Asked
Bond angle and hybridisation.
Concept
Steric number 3 (two sigma bonds plus one lone pair) means sp2 hybridisation; the lone pair then compresses the angle slightly below 120°.
Formula
Steric number = 2 bp + 1 lp = 3 → sp2, angle 117°
Baby steps
  • Central O forms two sigma bonds to the terminal oxygens.
  • It retains one lone pair.
  • 3 electron domains → sp2, trigonal planar arrangement.
  • The lone pair repels harder than bond pairs, squeezing 120° down to 117°. Shape is bent.
ShortcutBoth O–O bonds in O3 are equal (128 pm) because of resonance; bond order is 1.5.
8.Ozone

Ozone reacts with excess potassium iodide solution buffered with borate to liberate:

  • (A)O2
  • (B)I2
  • (C)KIO3
  • (D)HI
Show solution
Answer  (B)  I2
Given
O3 + KI solution.
Asked
Product used to estimate ozone.
Concept
Ozone is a strong oxidising agent and oxidises iodide to iodine; the liberated iodine is then titrated against standard sodium thiosulphate.
Formula
2KI + H2O + O3 → 2KOH + I2 + O2
Baby steps
  • Pass ozone through buffered KI solution.
  • I⁻ is oxidised to I2, which colours the solution brown.
  • Titrate the I2 with standard hypo (Na2S2O3) using starch indicator.
  • The volume of hypo gives the amount of ozone.
ShortcutOzone estimation = iodometric titration. The blue starch–iodine colour disappearing marks the end point.
9.Ozone

Ozone acts as a powerful oxidising agent because:

  • (A)It is a gas
  • (B)It readily decomposes to give nascent oxygen
  • (C)It is paramagnetic
  • (D)It has a triple bond
Show solution
Answer  (B)  It readily decomposes to give nascent oxygen
Given
Oxidising action of O3.
Asked
Reason for its strength.
Concept
Ozone is thermodynamically unstable with respect to oxygen; its decomposition releases a highly reactive oxygen atom.
Formula
O3 → O2 + [O], ΔH = −142 kJ mol⁻¹
Baby steps
  • The reaction is exothermic, so decomposition is spontaneous.
  • The nascent oxygen atom [O] is far more reactive than molecular O2.
  • It oxidises PbS to PbSO4, converts I⁻ to I2, and bleaches by oxidation.
ShortcutOzone's ΔfH is positive (+142 kJ mol⁻¹) — a positive heat of formation means an unstable, high-energy species.
10.Sulphur

The stable allotrope of sulphur below 369 K is:

  • (A)Monoclinic (β) sulphur
  • (B)Rhombic (α) sulphur
  • (C)Plastic sulphur
  • (D)S2 vapour
Show solution
Answer  (B)  Rhombic (α) sulphur
Given
Allotropes of sulphur.
Asked
Stable form below 369 K.
Concept
Rhombic and monoclinic sulphur are interconvertible, with 369 K as the transition temperature. Below it the rhombic form is more stable.
Formula
Rhombic (α) ⇌(369 K) Monoclinic (β)
Baby steps
  • Rhombic sulphur is yellow, melts at 385.8 K, and is stable at room temperature.
  • Above 369 K it converts to monoclinic sulphur, which melts at 393 K.
  • Both contain puckered S8 crown-shaped rings.
Shortcut369 K is the transition temperature. Below it rhombic wins, above it monoclinic wins.
11.Sulphur

The S8 ring in solid sulphur has which shape?

  • (A)Planar octagon
  • (B)Puckered crown
  • (C)Linear chain
  • (D)Tetrahedral
Show solution
Answer  (B)  Puckered crown
Given
S8 molecule.
Asked
Its shape.
Concept
Each sulphur is sp3 hybridised with two lone pairs, so the S–S–S angle is about 105°. A closed ring of eight such atoms cannot be flat — it puckers into a crown.
Formula
S–S–S angle ≈ 105°, S–S bond 204 pm
Baby steps
  • Each S makes two single bonds and keeps two lone pairs.
  • That forces a bond angle near 105°, not 135° (which a flat octagon would need).
  • The ring folds up and down alternately, giving the crown shape.
ShortcutSulphur's strong catenation (S–S = 226 kJ mol⁻¹) is why it forms rings while oxygen forms O2.
12.SO2

The shape and hybridisation of sulphur dioxide are:

  • (A)Linear, sp
  • (B)Bent, sp2
  • (C)Pyramidal, sp3
  • (D)Trigonal planar, sp2
Show solution
Answer  (B)  Bent, sp2
Given
SO2.
Asked
Shape and hybridisation.
Concept
Sulphur uses two sigma bonds to the oxygens and holds one lone pair, giving a steric number of 3.
Formula
Steric number = 2 bp + 1 lp = 3 → sp2, angle 119.5°
Baby steps
  • S has 6 valence electrons; 4 are used in bonding to two O.
  • The remaining 2 form one lone pair.
  • 3 domains → sp2, trigonal planar electron geometry.
  • One position is a lone pair, so the molecule is bent (angular).
ShortcutSO2 is isostructural with O3 — same shape, same hybridisation, same resonance and equal bond lengths.
13.SO2

The bleaching action of SO2 is due to and is:

  • (A)Oxidation, permanent
  • (B)Reduction, temporary
  • (C)Hydrolysis, permanent
  • (D)Oxidation, temporary
Show solution
Answer  (B)  Reduction, temporary
Given
Bleaching by SO2 in moist conditions.
Asked
Mechanism and permanence.
Concept
Moist SO2 supplies nascent hydrogen, which reduces the coloured compound to a colourless form. Atmospheric oxygen slowly re-oxidises it, so the colour returns.
Formula
SO2 + 2H2O → H2SO4 + 2[H]; coloured + [H] → colourless
Baby steps
  • SO2 reduces the dye rather than destroying it.
  • The reduced form is colourless but chemically intact.
  • Air re-oxidises it over time, so old SO2-bleached paper yellows again.
ShortcutSO2 → reduction → temporary. Cl2 → oxidation → permanent. NEET asks this pair repeatedly.
14.H2SO4

In the contact process, SO3 is absorbed in concentrated H2SO4 rather than in water because:

  • (A)Water is expensive
  • (B)The direct reaction with water is violent and forms a fog
  • (C)SO3 is insoluble in water
  • (D)H2SO4 is a catalyst
Show solution
Answer  (B)  The direct reaction with water is violent and forms a fog
Given
Absorption step of the contact process.
Asked
Reason for using conc. H2SO4.
Concept
SO3 reacts with water so exothermically that a fine mist of sulphuric acid forms, which is difficult to condense and escapes as an acid fog.
Formula
SO3 + H2SO4 → H2S2O7; H2S2O7 + H2O → 2H2SO4
Baby steps
  • Absorb SO3 in conc. H2SO4 to make oleum (H2S2O7).
  • Then dilute the oleum with a controlled amount of water.
  • This gives H2SO4 of any desired concentration, with no fog.
ShortcutOleum is the intermediate. '20% oleum' means the amount of H2SO4 obtained per 100 g on adding water is 120 g.
15.H2SO4

The basicity of sulphuric acid is:

  • (A)1
  • (B)2
  • (C)3
  • (D)4
Show solution
Answer  (B)  2
Given
H2SO4.
Asked
Basicity.
Concept
Basicity is the count of ionisable hydrogens, i.e. hydrogens attached to oxygen as –OH groups.
Formula
H2SO4 → H⁺ + HSO4⁻ → 2H⁺ + SO42−
Baby steps
  • Draw H2SO4: S with two =O and two –OH groups.
  • Both –OH hydrogens can ionise.
  • So basicity = 2 (dibasic), and it forms two series of salts: bisulphates and sulphates.
ShortcutCount –OH groups, not total H atoms. H3PO3 has 3 H but basicity 2.
16.H2SO4

Concentrated H2SO4 chars sugar. This illustrates its:

  • (A)Oxidising property
  • (B)Dehydrating property
  • (C)Acidic property
  • (D)Reducing property
Show solution
Answer  (B)  Dehydrating property
Given
Sucrose + conc. H2SO4.
Asked
Property demonstrated.
Concept
Concentrated sulphuric acid has a very strong affinity for water and removes hydrogen and oxygen from the carbohydrate in a 2:1 ratio, leaving behind carbon.
Formula
C12H22O11 →(conc. H2SO4) 12C + 11H2O
Baby steps
  • Sugar contains H and O exactly in the ratio of water.
  • Conc. H2SO4 strips them out as water.
  • Black carbon remains — the classic carbon column demonstration.
  • It similarly dehydrates HCOOH to CO and oxalic acid to CO + CO2.
ShortcutDehydration means removing water elements. Oxidation means removing electrons. Sugar charring is dehydration, not oxidation.
17.Trends

The correct order of the thermal stability of group 16 hydrides is:

  • (A)H2O > H2S > H2Se > H2Te
  • (B)H2Te > H2Se > H2S > H2O
  • (C)H2S > H2O > H2Se > H2Te
  • (D)H2Se > H2Te > H2S > H2O
Show solution
Answer  (A)  H2O > H2S > H2Se > H2Te
Given
Group 16 hydrides.
Asked
Order of thermal stability.
Concept
Thermal stability tracks E–H bond strength, which weakens down the group as orbital overlap with the small hydrogen atom becomes poorer.
Formula
Bond enthalpy: O–H (463) > S–H (347) > Se–H (276) > Te–H (238) kJ mol⁻¹
Baby steps
  • Small central atom = short, strong bond = hard to break.
  • O is smallest, so H2O is the most stable and does not decompose on heating.
  • H2Te is the least stable and decomposes even at room temperature.
ShortcutLearn the master pattern: down any group, thermal stability falls and reducing power rises. They are mirror images.
18.Trends

Which of the following has the maximum tendency for catenation in group 16?

  • (A)O
  • (B)S
  • (C)Se
  • (D)Te
Show solution
Answer  (B)  S
Given
Group 16 elements.
Asked
Element with maximum catenation.
Concept
Catenation needs a strong element–element single bond. The O–O bond is weakened by lone-pair repulsion between the two small oxygen atoms, so sulphur takes the lead.
Formula
S–S (226) > Se–Se (172) > Te–Te (126) > O–O (142*) kJ mol⁻¹
Baby steps
  • O–O is weak because lone pairs on two tiny atoms repel strongly.
  • S is larger, so its lone pairs stay apart and S–S is strong.
  • Sulphur therefore forms S8 rings, polysulphides Sn²⁻, and polythionic acids.
ShortcutSame story as group 15: the second element (P, S) catenates best, not the first (N, O).
19.Oxoacids

The number of S–S bonds in the peroxodisulphate ion, S2O82−, is:

  • (A)One
  • (B)Two
  • (C)Zero
  • (D)Three
Show solution
Answer  (C)  Zero
Given
S2O8²⁻.
Asked
Number of direct S–S bonds.
Concept
In peroxodisulphate the two sulphur atoms are linked through an oxygen–oxygen peroxide bridge, not directly to each other.
Formula
O3S–O–O–SO32−
Baby steps
  • Draw two SO3 units.
  • Join them with an O–O bridge, not S–S.
  • So S–S bonds = 0 and peroxide linkages = 1.
  • Each S remains at +6.
ShortcutContrast with dithionate S2O6²⁻, which DOES have a direct S–S bond and S at +5.
20.Oxoacids

Caro's acid is:

  • (A)H2S2O8
  • (B)H2SO5
  • (C)H2S2O7
  • (D)H2SO3
Show solution
Answer  (B)  H2SO5
Given
Named oxoacids of sulphur.
Asked
Formula of Caro's acid.
Concept
Peroxomonosulphuric acid contains a single sulphur with one peroxide linkage in place of an ordinary oxygen.
Formula
HO–O–SO2–OH, one peroxide linkage
Baby steps
  • Start from H2SO4.
  • Insert an extra O into one S–OH bond to make a peroxide bridge.
  • That gives H2SO5, Caro's acid, with 1 peroxide linkage.
ShortcutH2SO5 = Caro's (1 S). H2S2O8 = Marshall's (2 S). Both have exactly one peroxide linkage each.
21.Halides

The shape of SF4 is:

  • (A)Tetrahedral
  • (B)See-saw
  • (C)Square planar
  • (D)Trigonal pyramidal
Show solution
Answer  (B)  See-saw
Given
SF4.
Asked
Molecular shape.
Concept
Steric number 5 gives a trigonal bipyramidal arrangement; the single lone pair takes an equatorial position, where it faces the least repulsion.
Formula
Steric number = 4 bp + 1 lp = 5 → sp3d
Baby steps
  • S has 6 valence electrons; 4 are used in bonds to F.
  • Remaining 2 electrons = 1 lone pair.
  • 5 domains → trigonal bipyramidal skeleton, sp3d.
  • The lone pair sits equatorial, leaving the four F atoms in a see-saw (distorted tetrahedral) shape.
Shortcutsp³d with 1 lp = see-saw, 2 lp = T-shaped, 3 lp = linear. Lone pairs always go equatorial.
22.Halides

Which of the following is used as an excellent gaseous electrical insulator?

  • (A)SF4
  • (B)SF6
  • (C)SO2
  • (D)H2S
Show solution
Answer  (B)  SF6
Given
Sulphur halides.
Asked
The one used as an insulator.
Concept
SF6 is chemically inert, non-toxic, non-flammable and has a high dielectric strength, so it does not conduct or break down under high voltage.
Formula
SF6: octahedral, sp3d2, 6 bp, 0 lp
Baby steps
  • Its octahedral shell of fluorines shields sulphur completely.
  • It resists hydrolysis and does not react with anything at ordinary conditions.
  • It is used in high-voltage transformers and switchgear.
ShortcutSF6 inertness is kinetic (steric shielding), not thermodynamic — a favourite NEET distinction.
23.Reactions

SO2 acts as a reducing agent when it converts:

  • (A)Fe3+ to Fe2+
  • (B)H2S to S
  • (C)H2O to O2
  • (D)Cl⁻ to Cl2
Show solution
Answer  (A)  Fe3+ to Fe2+
Given
Reactions of SO2.
Asked
The case where SO2 is the reducing agent.
Concept
A reducing agent donates electrons and is itself oxidised. Sulphur in SO2 is at +4, so it can rise to +6 (reducing agent) or fall to 0 (oxidising agent).
Formula
2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42− + 4H⁺
Baby steps
  • Check what happens to sulphur in each option.
  • With Fe³⁺: S goes +4 → +6, so SO2 is oxidised → it is the reducing agent.
  • With H2S: S goes +4 → 0, so SO2 is reduced → it is the oxidising agent there.
  • So the correct case is Fe³⁺ to Fe²⁺.
ShortcutIntermediate oxidation state (+4) means SO2 can act both ways. Always check where sulphur ends up.
24.Trends

Which of the following statements about oxygen is incorrect?

  • (A)It shows only −2 and −1 oxidation states in all compounds
  • (B)It forms strong pπ–pπ double bonds
  • (C)It has a less negative electron gain enthalpy than sulphur
  • (D)It has no d orbitals available for bonding
Show solution
Answer  (A)  It shows only −2 and −1 oxidation states in all compounds
Given
Properties of oxygen.
Asked
The incorrect statement.
Concept
Oxygen usually shows −2, but −1 in peroxides, −1/2 in superoxides, 0 in O2 and even +2 in OF2. So 'only −2 and −1' is too restrictive.
Formula
OF2 → O is +2; KO2 → O is −1/2; H2O2 → O is −1
Baby steps
  • Check each: O does form pπ–pπ bonds (in O2) — correct.
  • O does have a less negative electron gain enthalpy than S — correct.
  • O is in period 2 and has no d orbitals — correct.
  • The first statement ignores superoxides and OF2, so it is the wrong one.
ShortcutOxygen oxidation states to know: −2 (oxide), −1 (peroxide), −½ (superoxide), 0 (O2), +2 (OF2).
25.Structures

The number of lone pairs on the central sulphur atom in SO3 is:

  • (A)Zero
  • (B)One
  • (C)Two
  • (D)Three
Show solution
Answer  (A)  Zero
Given
SO3 molecule.
Asked
Lone pairs on sulphur.
Concept
Sulphur uses all six of its valence electrons in bonding to three oxygens, so nothing is left over.
Formula
Steric number = 3 bp + 0 lp = 3 → sp2, trigonal planar, 120°
Baby steps
  • S has 6 valence electrons.
  • Three double bonds to O use all 6.
  • Lone pairs = 0, so the molecule is symmetrical trigonal planar with zero dipole moment.
ShortcutSO2 is bent with a dipole; SO3 is planar with none. The extra oxygen uses up the lone pair.
26.Reactions

Hydrogen sulphide is a stronger reducing agent than water because:

  • (A)S is more electronegative than O
  • (B)The S–H bond is weaker than the O–H bond
  • (C)H2S is a gas
  • (D)H2S is more polar
Show solution
Answer  (B)  The S–H bond is weaker than the O–H bond
Given
H2S vs H2O as reducing agents.
Asked
Reason H2S is the stronger reducer.
Concept
A reducing agent must give up its hydrogen and electrons. The weaker the E–H bond, the more easily the molecule is oxidised.
Formula
S–H (347) < O–H (463) kJ mol⁻¹
Baby steps
  • S is larger than O, so its bond to H is longer and weaker.
  • The weak S–H bond breaks readily, so H2S is oxidised to S.
  • It reduces acidified KMnO4 (purple to colourless) and K2Cr2O7 (orange to green).
ShortcutReducing power in any group of hydrides increases DOWN the group, since bonds weaken downward.
27.Trends

In group 16, the +6 oxidation state becomes less stable down the group because of:

  • (A)Increasing electronegativity
  • (B)The inert pair effect
  • (C)Decreasing atomic size
  • (D)Increasing ionisation enthalpy
Show solution
Answer  (B)  The inert pair effect
Given
Stability of +6 in group 16.
Asked
Reason for its decline.
Concept
Down the group the ns² electrons become progressively reluctant to participate in bonding due to poor shielding by intervening d and f electrons — the inert pair effect.
Formula
Stability: +6 decreases, +4 increases down the group
Baby steps
  • S readily gives SO3 and H2SO4 with S at +6.
  • By Te and Po, the +4 state is preferred.
  • So TeO2 is more stable than TeO3, and PoO2 is the common oxide.
  • H2SO4 is a stable acid, while H6TeO6 is a weak oxidising acid.
ShortcutInert pair effect appears in every heavy p-block element: Tl⁺, Pb²⁺, Bi³⁺, Po⁴⁺.
28.Reactions

Which of the following gives a black precipitate with H2S in acidic medium?

  • (A)ZnCl2
  • (B)CuSO4
  • (C)MgCl2
  • (D)NaCl
Show solution
Answer  (B)  CuSO4
Given
H2S passed through acidified salt solutions.
Asked
Salt giving a black precipitate.
Concept
Group II cations in qualitative analysis have very low solubility products for their sulphides, so they precipitate even at the low sulphide concentration of acidic medium.
Formula
Cu2+ + H2S → CuS↓ (black) + 2H⁺
Baby steps
  • In acidic medium the ionisation of H2S is suppressed, so [S²⁻] is very low.
  • Only sulphides with very small Ksp precipitate — Cu, Pb, Hg, Bi, Cd, As.
  • CuS is black; ZnS (white) needs a basic medium to precipitate.
ShortcutAcidic medium → group II sulphides (CuS, PbS black). Basic medium → group IV (ZnS white, MnS pink).
29.Preparation

Dioxygen is prepared in the laboratory by heating:

  • (A)KClO3 with MnO2
  • (B)NH4Cl with NaNO2
  • (C)CaCO3
  • (D)NaCl with H2SO4
Show solution
Answer  (A)  KClO3 with MnO2
Given
Laboratory preparation of O2.
Asked
Reagents used.
Concept
Potassium chlorate decomposes on heating to release oxygen, and manganese dioxide acts as a catalyst that lowers the required temperature.
Formula
2KClO3 →(MnO2, heat) 2KCl + 3O2
Baby steps
  • Heat KClO3 alone and it needs a very high temperature.
  • Adding MnO2 as a catalyst lets it decompose around 420 K.
  • MnO2 is recovered unchanged at the end — proof that it is a catalyst.
  • Industrially, O2 comes from the fractional distillation of liquid air.
ShortcutMnO2 has two roles worth remembering: catalyst in KClO3 decomposition, and oxidising agent with conc. HCl to make Cl2.
30.Trends

Which of the following is the correct order of reducing power?

  • (A)H2O > H2S > H2Se > H2Te
  • (B)H2Te > H2Se > H2S > H2O
  • (C)H2S > H2Te > H2O > H2Se
  • (D)H2Se > H2O > H2S > H2Te
Show solution
Answer  (B)  H2Te > H2Se > H2S > H2O
Given
Group 16 hydrides.
Asked
Order of reducing power.
Concept
A reducing agent must break its E–H bond and donate electrons. Weaker bonds mean stronger reducing behaviour, and bond strength falls down the group.
Formula
Reducing power ∝ 1/(E–H bond enthalpy)
Baby steps
  • Te–H is the weakest bond in the series.
  • So H2Te loses hydrogen most easily and is the best reducing agent.
  • H2O has the strongest O–H bond and is effectively not a reducing agent at all.
ShortcutIf a question gives you a set of hydrides and asks for reducing power, the answer is simply bottom-to-top of the group.

Answer key — Oxygen Family

1 B2 B3 B4 D5 A6 B7 B8 B9 B10 B11 B12 B13 B14 B15 B16 B17 A18 B19 C20 B21 B22 B23 A24 A25 A26 B27 B28 B29 A30 B
17

Halogens

F · Cl · Br · I · At
30 questions  ·  full worked solutions
1.Periodic trends

Which halogen shows only the −1 oxidation state in all its compounds?

  • (A)Fluorine
  • (B)Chlorine
  • (C)Bromine
  • (D)Iodine
Show solution
Answer  (A)  Fluorine
Given
F, Cl, Br, I.
Asked
The one restricted to −1.
Concept
Positive oxidation states require either lower electronegativity than the partner or vacant d orbitals for octet expansion. Fluorine is the most electronegative element and has no 2d subshell.
Formula
F: 1s2 2s2 2p5 — no d orbitals
Baby steps
  • F is more electronegative than every other element, so it is never the positive partner.
  • It is in period 2, so it has no d orbitals to expand its octet.
  • Therefore F shows only −1 (and 0 in F2).
  • Cl, Br and I show +1, +3, +5 and +7 because they do have vacant d orbitals.
ShortcutTwo period-2 rules that solve many questions: no d orbitals, and never exceed the octet.
2.Trends

Which halogen has the most negative electron gain enthalpy?

  • (A)F
  • (B)Cl
  • (C)Br
  • (D)I
Show solution
Answer  (B)  Cl
Given
The halogens.
Asked
Most negative electron gain enthalpy.
Concept
Fluorine's 2p subshell is unusually compact, so an added electron faces heavy repulsion from the electrons already there. Chlorine's larger 3p subshell accommodates the extra electron more comfortably.
Formula
ΔegH: Cl (−349) > F (−328) > Br (−325) > I (−296) kJ mol⁻¹
Baby steps
  • Expect the smallest atom to have the most negative value — but check for the period-2 anomaly.
  • F is so small that inter-electronic repulsion cancels part of the gain.
  • So Cl beats F, and the rest follow the normal decreasing trend.
ShortcutPeriod-2 anomalies to memorise: Cl > F and S > O in magnitude of electron gain enthalpy.
3.Trends

Despite having a less negative electron gain enthalpy than chlorine, fluorine is a stronger oxidising agent because of:

  • (A)Its higher bond dissociation enthalpy
  • (B)Its low bond dissociation enthalpy and very high hydration enthalpy of F⁻
  • (C)Its larger size
  • (D)Its metallic character
Show solution
Answer  (B)  Its low bond dissociation enthalpy and very high hydration enthalpy of F⁻
Given
Oxidising power of F2 vs Cl2 in solution.
Asked
Reason F2 wins.
Concept
The overall electrode potential is the sum of three steps: bond dissociation, electron gain, and hydration. Fluorine loses on step two but wins decisively on steps one and three.
Formula
E° (F2/F⁻) = +2.87 V; E° (Cl2/Cl⁻) = +1.36 V
Baby steps
  • Step 1: F–F bond is weak (158 kJ mol⁻¹) because of lone pair repulsion — easy to break.
  • Step 2: F loses slightly here (−328 vs −349).
  • Step 3: F⁻ is tiny, so hydration enthalpy is huge (−515 kJ mol⁻¹).
  • Steps 1 and 3 outweigh step 2, making F2 the strongest oxidising agent of all elements.
ShortcutWhenever a trend in solution contradicts a gas-phase property, hydration enthalpy is usually the reason.
4.Hydrides

Hydrogen fluoride has an abnormally high boiling point among hydrogen halides because:

  • (A)It is ionic
  • (B)Of strong intermolecular hydrogen bonding
  • (C)It has the highest molecular mass
  • (D)It forms dimers by covalent bonds
Show solution
Answer  (B)  Of strong intermolecular hydrogen bonding
Given
Boiling points of HF, HCl, HBr, HI.
Asked
Reason HF is out of order.
Concept
Fluorine's very high electronegativity and small size allow strong H···F hydrogen bonds, linking HF molecules into zig-zag chains that need extra energy to separate.
Formula
Boiling point: HF (293 K) > HI > HBr > HCl
Baby steps
  • By molecular mass alone the order should be HI > HBr > HCl > HF.
  • But H-bonding lifts HF right to the top.
  • So the order becomes HF > HI > HBr > HCl.
  • The same H-bonding also makes HF a weak acid, since the H is held tightly.
ShortcutOnly N, O and F form hydrogen bonds. If a hydride containing one of these breaks a trend, H-bonding is the reason.
5.Preparation

Chlorine is prepared in the laboratory by heating manganese dioxide with:

  • (A)Dilute HCl
  • (B)Concentrated HCl
  • (C)Dilute H2SO4
  • (D)NaCl solution
Show solution
Answer  (B)  Concentrated HCl
Given
Laboratory preparation of Cl2.
Asked
Reagent used with MnO2.
Concept
MnO2 acts as the oxidising agent, taking manganese from +4 to +2 while chloride is oxidised from −1 to 0. Dilute acid does not supply enough chloride to drive the reaction.
Formula
MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
Baby steps
  • Concentrated HCl supplies both the chloride to be oxidised and the acid medium.
  • Mn goes +4 → +2 (reduced); Cl goes −1 → 0 (oxidised).
  • The gas is dried over conc. H2SO4 and collected by upward displacement of air, since Cl2 is denser than air.
  • KMnO4 with conc. HCl works too and does not need heating.
ShortcutCl2 cannot be dried over quicklime or P4O10 alone in the standard route — conc. H2SO4 is the drying agent to quote.
6.Reactions

Chlorine reacts with cold and dilute NaOH to give:

  • (A)NaCl and NaClO3
  • (B)NaCl and NaOCl
  • (C)NaClO4 only
  • (D)NaCl and NaClO2
Show solution
Answer  (B)  NaCl and NaOCl
Given
Cl2 + cold dilute NaOH.
Asked
Products formed.
Concept
Chlorine disproportionates in alkali. In cold dilute alkali it goes to −1 and +1; in hot concentrated alkali it goes to −1 and +5.
Formula
Cl2 + 2NaOH (cold, dilute) → NaCl + NaOCl + H2O
Baby steps
  • Cl2 starts at oxidation state 0.
  • Cold dilute alkali: half goes to Cl⁻ (−1), half to OCl⁻ (+1).
  • Hot concentrated alkali: 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O, with +5.
  • Both are disproportionations of the same element.
ShortcutCold and dilute → hypochlorite (+1). Hot and concentrated → chlorate (+5). Two words, two products.
7.Bleaching powder

The formula of bleaching powder is:

  • (A)Ca(OCl)2
  • (B)CaCl2
  • (C)Ca(OCl)Cl
  • (D)CaOCl
Show solution
Answer  (C)  Ca(OCl)Cl
Given
Bleaching powder.
Asked
Its formula.
Concept
Bleaching powder is a mixed salt of hypochlorous acid and hydrochloric acid, made by passing chlorine over dry slaked lime.
Formula
Ca(OH)2 + Cl2 → Ca(OCl)Cl + H2O
Baby steps
  • Pass dry Cl2 over dry Ca(OH)2.
  • The calcium takes one OCl⁻ and one Cl⁻.
  • Chlorine is therefore in two states: +1 in OCl⁻ and −1 in Cl⁻.
  • Available chlorine is released on treatment with dilute acid.
ShortcutA mixed salt has two different anions. Bleaching powder is the standard example asked in NEET.
8.Oxoacids

The correct order of the oxidising power of the oxoacids of chlorine is:

  • (A)HClO4 > HClO3 > HClO2 > HOCl
  • (B)HOCl > HClO2 > HClO3 > HClO4
  • (C)HClO3 > HOCl > HClO4 > HClO2
  • (D)HClO2 > HClO4 > HOCl > HClO3
Show solution
Answer  (B)  HOCl > HClO2 > HClO3 > HClO4
Given
HOCl, HClO2, HClO3, HClO4.
Asked
Order of oxidising strength.
Concept
Oxidising power depends on how easily the acid gives up oxygen. With fewer oxygens the Cl–O bonds are less stabilised by resonance, so oxygen is released more readily.
Formula
Oxidising power falls as oxidation state of Cl rises: +1 > +3 > +5 > +7
Baby steps
  • HClO4 has four oxygens sharing the charge — very stable, so it holds onto them.
  • HOCl has just one oxygen, weakly held, so it releases nascent oxygen easily.
  • Hence HOCl is the strongest oxidiser and HClO4 the weakest.
ShortcutAcid strength and oxidising power run OPPOSITE. HClO4 = strongest acid, weakest oxidiser. HOCl = the reverse.
9.Interhalogens

Interhalogen compounds are more reactive than the halogens (except F2) because:

  • (A)They are ionic
  • (B)The A–B bond is weaker than the A–A bond
  • (C)They have higher molecular mass
  • (D)They are polar covalent solids
Show solution
Answer  (B)  The A–B bond is weaker than the A–A bond
Given
Reactivity of interhalogens.
Asked
Reason for their higher reactivity.
Concept
An A–B bond between two different halogens is polar and less effective than a symmetrical A–A bond, so it breaks more easily and the compound reacts faster.
Formula
Reactivity: interhalogens > halogens, except F2
Baby steps
  • Different sizes and electronegativities give poor orbital overlap in A–B.
  • The weaker bond makes the molecule easier to break apart.
  • So ClF3 and BrF3 are powerful fluorinating agents.
  • F2 remains the exception — its own bond is already abnormally weak.
ShortcutInterhalogen types by size ratio: AB, AB3, AB5, AB7. Only IF7 exists as AB7, since I is largest and F smallest.
10.Interhalogens

The shape of BrF5 is:

  • (A)Trigonal bipyramidal
  • (B)Square pyramidal
  • (C)Pentagonal planar
  • (D)Octahedral
Show solution
Answer  (B)  Square pyramidal
Given
BrF5.
Asked
Molecular shape.
Concept
Steric number 6 gives an octahedral arrangement of electron domains; one of them is a lone pair, which pushes the five fluorines below the equator.
Formula
Steric number = 5 bp + 1 lp = 6 → sp3d2
Baby steps
  • Br has 7 valence electrons; 5 are used in bonds to F.
  • Remaining 2 electrons = 1 lone pair.
  • 6 domains → octahedral electron geometry, sp3d2.
  • Removing one vertex for the lone pair leaves a square pyramid.
ShortcutOctahedral base: 0 lp = octahedral, 1 lp = square pyramidal, 2 lp = square planar.
11.Polyhalides

Iodine dissolves readily in KI solution because it forms:

  • (A)I2−
  • (B)I3−
  • (C)IO3−
  • (D)KIO
Show solution
Answer  (B)  I3−
Given
I2 in aqueous KI.
Asked
The species formed.
Concept
Iodine is only slightly soluble in water, but iodide ions act as a Lewis base and add to I2 to form the soluble linear triiodide ion.
Formula
I2 + I⁻ → I3− (linear, sp3d)
Baby steps
  • I2 is non-polar and barely dissolves in water.
  • I⁻ donates a lone pair to I2.
  • The resulting I3⁻ is charged and therefore soluble.
  • The central I has 3 lone pairs and 2 bond pairs, so the ion is linear.
ShortcutPolyhalide I3⁻: steric number 5, 3 lone pairs equatorial, so linear — same logic as XeF2.
12.Reactions

Hydrofluoric acid is stored in wax-coated or plastic bottles because it:

  • (A)Is a strong acid
  • (B)Attacks silica in glass
  • (C)Evaporates quickly
  • (D)Reacts with plastic
Show solution
Answer  (B)  Attacks silica in glass
Given
Storage of HF.
Asked
Reason glass cannot be used.
Concept
Hydrofluoric acid reacts with the silicon dioxide in glass, converting it into volatile silicon tetrafluoride and etching the surface.
Formula
SiO2 + 4HF → SiF4 + 2H2O
Baby steps
  • Glass is mostly SiO2.
  • HF attacks the Si–O bonds and forms gaseous SiF4.
  • The glass is eaten away, which is also how frosted glass and etched markings are made.
  • So HF is kept in wax bottles, polythene or Teflon.
ShortcutOnly HF etches glass — HCl, HBr and HI do not. It is the one halogen acid stored specially.
13.Trends

The correct order of the oxidising power of the halogens is:

  • (A)I2 > Br2 > Cl2 > F2
  • (B)F2 > Cl2 > Br2 > I2
  • (C)Cl2 > F2 > Br2 > I2
  • (D)Br2 > Cl2 > I2 > F2
Show solution
Answer  (B)  F2 > Cl2 > Br2 > I2
Given
F2, Cl2, Br2, I2.
Asked
Order of oxidising power.
Concept
Oxidising power in aqueous solution is measured by the standard reduction potential, which decreases steadily down group 17.
Formula
E°: F2 (+2.87) > Cl2 (+1.36) > Br2 (+1.09) > I2 (+0.54) V
Baby steps
  • Each halogen displaces the ones below it from their salts.
  • Cl2 + 2KBr → 2KCl + Br2 works; the reverse does not.
  • So the power decreases from F2 to I2.
  • F2 is so strong it even oxidises water to O2.
ShortcutA halogen displaces any halogen below it in the group — never one above it.
14.Structures

The hybridisation of chlorine in HClO4 is:

  • (A)sp
  • (B)sp2
  • (C)sp3
  • (D)sp3d
Show solution
Answer  (C)  sp3
Given
Perchloric acid, HClO4.
Asked
Hybridisation of the central Cl.
Concept
Count sigma bonds plus lone pairs. Chlorine forms four sigma bonds to oxygen and retains no lone pairs.
Formula
Steric number = 4 bp + 0 lp = 4 → sp3, tetrahedral
Baby steps
  • Cl is bonded to three =O and one –OH.
  • That is four sigma bonds (pi bonds do not count for hybridisation).
  • Lone pairs on Cl = 0.
  • Steric number 4 → sp3, tetrahedral around chlorine.
ShortcutPi bonds NEVER count in steric number. Count sigma bonds and lone pairs only.
15.Trends

The colour of the halogens deepens from F2 to I2 because:

  • (A)Molecular mass increases
  • (B)The energy gap for electronic excitation decreases
  • (C)They become more ionic
  • (D)Their density increases
Show solution
Answer  (B)  The energy gap for electronic excitation decreases
Given
F2 pale yellow → Cl2 greenish yellow → Br2 red brown → I2 violet.
Asked
Reason for the deepening colour.
Concept
Halogens absorb visible light to promote an electron to a higher orbital. As the molecule gets bigger the orbitals are closer in energy, so lower-energy (longer wavelength) light is absorbed.
Formula
ΔE = hc/λ — smaller ΔE means longer λ absorbed
Baby steps
  • F2 needs high-energy violet light, so it transmits pale yellow.
  • Down the group the excitation energy falls.
  • I2 absorbs low-energy yellow-green light and transmits violet.
  • The colour you see is always the complement of the light absorbed.
ShortcutColour seen = complementary of colour absorbed. Absorb violet → look yellow; absorb yellow → look violet.
16.Reactions

Which of the following will not displace bromine from a bromide solution?

  • (A)F2
  • (B)Cl2
  • (C)I2
  • (D)Both F2 and Cl2
Show solution
Answer  (C)  I2
Given
Halogen displacement reactions with KBr.
Asked
The halogen that fails.
Concept
Only a stronger oxidising agent can displace a weaker one. Iodine lies below bromine in the group and is the weaker oxidiser.
Formula
Cl2 + 2Br⁻ → 2Cl⁻ + Br2 (works); I2 + 2Br⁻ → no reaction
Baby steps
  • Compare positions: F and Cl are above Br, I is below.
  • Anything above can displace anything below.
  • I2 is below Br2 in oxidising power, so it cannot displace it.
  • In fact the reverse happens: Br2 displaces I2 from KI.
ShortcutDisplacement always runs downhill: a halogen higher in the group kicks out one lower down.
17.Reactions

Deacon's process for the manufacture of chlorine uses:

  • (A)MnO2 and conc. HCl
  • (B)HCl and O2 over CuCl2 at 723 K
  • (C)Electrolysis of brine
  • (D)NaCl and H2SO4
Show solution
Answer  (B)  HCl and O2 over CuCl2 at 723 K
Given
Industrial manufacture of Cl2.
Asked
Reagents and catalyst in Deacon's process.
Concept
Hydrogen chloride, a by-product of many processes, is oxidised by atmospheric oxygen over a copper chloride catalyst to recover chlorine.
Formula
4HCl + O2 →(CuCl2, 723 K) 2Cl2 + 2H2O
Baby steps
  • HCl supplies chloride at −1.
  • O2 oxidises it to Cl2 at 0.
  • CuCl2 catalyses the reaction at 723 K.
  • The other industrial route is the electrolysis of brine, which also gives NaOH and H2.
ShortcutCatalyst list again: Deacon = CuCl2, Contact = V2O5, Haber = Fe/Mo, Ostwald = Pt/Rh.
18.Trends

The correct order of the hydration enthalpy (magnitude) of halide ions is:

  • (A)F⁻ > Cl⁻ > Br⁻ > I⁻
  • (B)I⁻ > Br⁻ > Cl⁻ > F⁻
  • (C)Cl⁻ > F⁻ > Br⁻ > I⁻
  • (D)Br⁻ > I⁻ > Cl⁻ > F⁻
Show solution
Answer  (A)  F⁻ > Cl⁻ > Br⁻ > I⁻
Given
Halide ions in water.
Asked
Order of hydration enthalpy magnitude.
Concept
Hydration enthalpy depends on charge density. A smaller ion of the same charge attracts water dipoles more strongly.
Formula
ΔhydH: F⁻ (−515) > Cl⁻ (−381) > Br⁻ (−347) > I⁻ (−305) kJ mol⁻¹
Baby steps
  • All four ions carry the same −1 charge.
  • F⁻ is the smallest, so its charge density is highest.
  • It pulls water molecules in most strongly, so its hydration enthalpy is greatest.
  • Size increases down the group, so hydration enthalpy falls.
ShortcutSame charge → smaller ion wins on hydration, lattice enthalpy and polarising power.
19.Reactions

Which of the following reacts with water to give a strongly acidic solution and O2?

  • (A)Cl2
  • (B)F2
  • (C)Br2
  • (D)I2
Show solution
Answer  (B)  F2
Given
Halogens in water.
Asked
The one that oxidises water.
Concept
Fluorine's reduction potential (+2.87 V) exceeds that of the oxygen–water couple (+1.23 V), so it can strip electrons from water itself.
Formula
2F2 + 2H2O → 4HF + O2
Baby steps
  • Cl2, Br2 and I2 only dissolve or disproportionate slightly in water.
  • F2 is powerful enough to oxidise the oxygen in water from −2 to 0.
  • The products are HF and O2, and the reaction is vigorous.
ShortcutIf E° of the oxidant exceeds +1.23 V it can, in principle, oxidise water. F2 actually does.
20.Structures

The shape of ICl3 (in the monomeric form) is:

  • (A)Trigonal planar
  • (B)T-shaped
  • (C)Pyramidal
  • (D)Tetrahedral
Show solution
Answer  (B)  T-shaped
Given
Interhalogen ICl3.
Asked
Molecular shape.
Concept
Steric number 5 gives a trigonal bipyramidal skeleton; the two lone pairs occupy equatorial positions to minimise repulsion, leaving a T shape.
Formula
Steric number = 3 bp + 2 lp = 5 → sp3d
Baby steps
  • I has 7 valence electrons; 3 are used to bond three Cl.
  • Remaining 4 electrons = 2 lone pairs.
  • 5 domains → sp3d, trigonal bipyramidal skeleton.
  • Both lone pairs sit equatorial, giving a T-shaped molecule. In the solid it dimerises to planar I2Cl6.
ShortcutAll AB3 interhalogens are T-shaped: ClF3, BrF3, ICl3. No exceptions.
21.Trends

Which of the following statements about the halogens is incorrect?

  • (A)Their melting and boiling points increase down the group
  • (B)They all exist as diatomic molecules
  • (C)Their electronegativity increases down the group
  • (D)They are all coloured
Show solution
Answer  (C)  Their electronegativity increases down the group
Given
General properties of group 17.
Asked
The incorrect statement.
Concept
Electronegativity depends on how strongly the nucleus attracts a bonding pair, which weakens as the atom gets larger and more shielded. So it decreases down the group.
Formula
Electronegativity: F (4.0) > Cl (3.2) > Br (3.0) > I (2.7)
Baby steps
  • Melting and boiling points do rise down the group as van der Waals forces increase with size.
  • All halogens are diatomic and all are coloured.
  • But electronegativity FALLS down the group, so that statement is wrong.
ShortcutDown a group: size, boiling point and metallic character rise; electronegativity, ionisation enthalpy and electron affinity fall.
22.Reactions

When chlorine is passed through a hot concentrated solution of NaOH, the oxidation states of chlorine in the products are:

  • (A)+1 and −1
  • (B)+5 and −1
  • (C)+3 and −1
  • (D)+7 and −1
Show solution
Answer  (B)  +5 and −1
Given
Cl2 + hot concentrated NaOH.
Asked
Oxidation states in the products.
Concept
Chlorine disproportionates. In hot concentrated alkali the hypochlorite first formed is itself unstable and further disproportionates to chlorate and chloride.
Formula
3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O
Baby steps
  • Cl2 starts at 0.
  • Part is reduced to Cl⁻ (−1) in NaCl.
  • Part is oxidised to ClO3⁻ (+5) in NaClO3.
  • Check the balance: 5 atoms fall by 1 each, 1 atom rises by 5 — electrons balance.
ShortcutA quick check for any disproportionation: total electrons lost must equal total electrons gained.
23.Uses

Iodine monochloride, ICl, is used as:

  • (A)A fluorinating agent
  • (B)Wijs reagent for determining the iodine number of fats
  • (C)A bleaching agent
  • (D)A refrigerant
Show solution
Answer  (B)  Wijs reagent for determining the iodine number of fats
Given
Uses of interhalogens.
Asked
Use of ICl.
Concept
Interhalogens add across carbon–carbon double bonds, which lets ICl measure the degree of unsaturation in an oil or fat.
Formula
–CH=CH– + ICl → –CHI–CHCl–
Baby steps
  • More double bonds in the fat means more ICl consumed.
  • The result is expressed as the iodine number.
  • A high iodine number means a highly unsaturated oil.
ShortcutInterhalogen uses: ClF3, BrF3 → fluorinating agents for UF6; ICl → Wijs reagent; interhalogens generally → non-aqueous solvents.
24.Structures

The number of lone pairs on the central atom in ClF3 is:

  • (A)Zero
  • (B)One
  • (C)Two
  • (D)Three
Show solution
Answer  (C)  Two
Given
ClF3.
Asked
Lone pairs on chlorine.
Concept
Subtract the electrons used in bonding from the total valence electrons of the central atom; every leftover pair is a lone pair.
Formula
Lone pairs = (7 − 3)/2 = 2
Baby steps
  • Cl has 7 valence electrons.
  • Three are used to bond three F atoms.
  • Remaining 4 electrons = 2 lone pairs.
  • Steric number = 3 + 2 = 5 → sp3d, T-shaped.
ShortcutFor a neutral central atom bonded to n monovalent atoms: lone pairs = (valence electrons − n)/2.
25.Reactions

Bleaching powder on reaction with dilute acids liberates:

  • (A)O2
  • (B)Cl2
  • (C)HOCl only
  • (D)CO2
Show solution
Answer  (B)  Cl2
Given
Ca(OCl)Cl + dilute acid.
Asked
Gas liberated.
Concept
The hypochlorite (+1) and chloride (−1) in the mixed salt comproportionate in acid, both moving to the free element at oxidation state 0.
Formula
Ca(OCl)Cl + H2SO4 → CaSO4 + H2O + Cl2
Baby steps
  • Acid protonates OCl⁻ to give HOCl.
  • HOCl then oxidises the Cl⁻ present.
  • Both chlorines meet at 0, releasing Cl2 — the 'available chlorine'.
  • With excess CO2 instead, bleaching powder gives CaCO3 and Cl2.
ShortcutComproportionation is the reverse of disproportionation: two different oxidation states merge into one.
26.Trends

Which halogen is a liquid at room temperature?

  • (A)F2
  • (B)Cl2
  • (C)Br2
  • (D)I2
Show solution
Answer  (C)  Br2
Given
Physical states of the halogens.
Asked
The liquid one.
Concept
Melting and boiling points rise down the group as van der Waals dispersion forces grow with molecular size and polarisability.
Formula
F2 gas, Cl2 gas, Br2 liquid, I2 solid
Baby steps
  • F2 and Cl2 are small with weak dispersion forces → gases.
  • Br2 has intermediate forces → liquid (the only non-metal liquid element besides mercury among metals).
  • I2 has strong dispersion forces → violet-black solid that sublimes.
ShortcutBromine and mercury are the only two elements that are liquid at room temperature.
27.Oxoacids

Which of the following oxoacids of chlorine has chlorine in the +3 oxidation state?

  • (A)HOCl
  • (B)HClO2
  • (C)HClO3
  • (D)HClO4
Show solution
Answer  (B)  HClO2
Given
Oxoacids of chlorine.
Asked
The one with Cl at +3.
Concept
Assign O as −2 and H as +1, then balance to zero for a neutral acid.
Formula
HClO2: (+1) + x + 2(−2) = 0 → x = +3
Baby steps
  • HOCl: 1 + x − 2 = 0 → x = +1 (hypochlorous).
  • HClO2: 1 + x − 4 = 0 → x = +3 (chlorous).
  • HClO3: x = +5 (chloric); HClO4: x = +7 (perchloric).
ShortcutName endings map to oxidation states: hypo-...-ous = +1, -ous = +3, -ic = +5, per-...-ic = +7.
28.Reactions

Chlorine gas turns moist starch iodide paper:

  • (A)Red
  • (B)Blue
  • (C)Green
  • (D)Colourless
Show solution
Answer  (B)  Blue
Given
Test with starch iodide paper.
Asked
Colour observed.
Concept
Chlorine oxidises iodide to iodine, and the liberated iodine forms a deep blue complex with starch.
Formula
Cl2 + 2KI → 2KCl + I2; I2 + starch → blue
Baby steps
  • Cl2 is the stronger oxidiser and displaces I2 from iodide.
  • Free I2 slips into the helical amylose chain of starch.
  • The charge-transfer complex is intensely blue.
  • With excess chlorine the blue fades, as I2 is further oxidised to iodate.
ShortcutThe blue starch–iodine colour is the standard end point in iodometric titrations.
29.Trends

Which of the following has the highest bond dissociation enthalpy?

  • (A)H–F
  • (B)H–Cl
  • (C)H–Br
  • (D)H–I
Show solution
Answer  (A)  H–F
Given
Hydrogen halides.
Asked
Strongest H–X bond.
Concept
Bond strength depends on how well the halogen's orbital overlaps with hydrogen's 1s orbital. Fluorine's compact 2p orbital overlaps best.
Formula
H–F (567) > H–Cl (431) > H–Br (366) > H–I (299) kJ mol⁻¹
Baby steps
  • F is smallest, so the H–F bond is shortest.
  • Shorter bond = better overlap = stronger bond.
  • Bond strength falls steadily down the group.
  • This is exactly why HF is the weakest acid of the four.
ShortcutNote the contrast: H–X bond strength is highest for F, but X–X bond strength is highest for Cl. Different reasons.
30.Structures

In the solid state, iodine trichloride exists as a dimer, I2Cl6, whose structure is:

  • (A)Tetrahedral
  • (B)Planar with two bridging chlorines
  • (C)Octahedral
  • (D)Linear
Show solution
Answer  (B)  Planar with two bridging chlorines
Given
Solid ICl3.
Asked
Structure of the dimer.
Concept
Two ICl3 units join through two chlorine bridges, giving a flat molecule in which each iodine is surrounded by four chlorines.
Formula
I2Cl6: planar, 2 bridging Cl, 4 terminal Cl
Baby steps
  • Each iodine forms two terminal I–Cl bonds.
  • Two chlorines act as bridges shared between both iodines.
  • The result is a planar dimer, analogous to Al2Cl6.
  • The bridging bonds are longer and weaker than the terminal ones.
ShortcutBridged dimers to remember: Al2Cl6, I2Cl6, B2H6. In all three the bridge bonds are the weaker ones.

Answer key — Halogens

1 A2 B3 B4 B5 B6 B7 C8 B9 B10 B11 B12 B13 B14 C15 B16 C17 B18 A19 B20 B21 C22 B23 B24 C25 B26 C27 B28 B29 A30 B
18

Noble Gases

He · Ne · Ar · Kr · Xe · Rn
30 questions  ·  full worked solutions
1.Periodic trends

The general outer electronic configuration of the noble gases (except helium) is:

  • (A)ns2np4
  • (B)ns2np5
  • (C)ns2np6
  • (D)ns2
Show solution
Answer  (C)  ns2np6
Given
Group 18 elements.
Asked
General valence configuration.
Concept
Noble gases have completely filled outermost s and p subshells, which is the most stable arrangement possible and leaves no tendency to gain, lose or share electrons.
Formula
Ne, Ar, Kr, Xe, Rn → ns2np6; He → 1s2
Baby steps
  • Group 18 means 8 valence electrons.
  • Two fill the s subshell, six fill the p subshell.
  • Helium is the exception — it has only the 1s shell, so its configuration is 1s2, which is already complete.
ShortcutHe is an exception in configuration but not in behaviour — a full shell is a full shell either way.
2.Trends

Noble gases have large positive electron gain enthalpies because:

  • (A)They are radioactive
  • (B)Their outermost shell is completely filled
  • (C)They are monatomic
  • (D)They have low ionisation enthalpy
Show solution
Answer  (B)  Their outermost shell is completely filled
Given
Electron gain enthalpy of group 18.
Asked
Reason it is large and positive.
Concept
Adding an electron to a noble gas would force it into the next higher shell, which is far from the nucleus and well shielded. Energy must be supplied, so the process is endothermic.
Formula
ΔegH is positive (endothermic) for all noble gases
Baby steps
  • The ns and np subshells are already full.
  • The incoming electron must enter the next principal shell.
  • That shell is high in energy and strongly shielded.
  • So energy is absorbed rather than released — a positive electron gain enthalpy.
ShortcutPositive electron gain enthalpy occurs for group 18 and group 2 (filled ns²) — both have no room in the current subshell.
3.Discovery

Xenon reacts with PtF6 to form the first noble gas compound because:

  • (A)Xe is radioactive
  • (B)The ionisation enthalpy of Xe is close to that of O2
  • (C)Xe has d orbitals
  • (D)PtF6 is a reducing agent
Show solution
Answer  (B)  The ionisation enthalpy of Xe is close to that of O2
Given
Bartlett's experiment, 1962.
Asked
The reasoning behind it.
Concept
Bartlett had already made O2⁺[PtF6]⁻. Noticing that Xe requires almost exactly the same energy to ionise as O2, he predicted the same reaction would work with xenon.
Formula
IE (O2) = 1175, IE (Xe) = 1170 kJ mol⁻¹
Baby steps
  • PtF6 is strong enough to strip an electron from O2.
  • Xe needs essentially the same energy.
  • So PtF6 should also oxidise Xe.
  • The result was the red solid Xe⁺[PtF6]⁻ — the first noble gas compound.
ShortcutXe has the lowest ionisation enthalpy of the non-radioactive noble gases, which is why only Xe forms a range of compounds.
4.Xenon fluorides

The shape of XeF2 and the number of lone pairs on xenon are:

  • (A)Bent, 2
  • (B)Linear, 3
  • (C)T-shaped, 2
  • (D)Linear, 1
Show solution
Answer  (B)  Linear, 3
Given
XeF2.
Asked
Shape and lone pair count.
Concept
Steric number 5 gives a trigonal bipyramidal skeleton; all three lone pairs take equatorial positions, leaving the two fluorines directly opposite each other.
Formula
Steric number = 2 bp + 3 lp = 5 → sp3d
Baby steps
  • Xe has 8 valence electrons; 2 are used to bond two F.
  • Remaining 6 electrons = 3 lone pairs.
  • 5 domains → sp3d, trigonal bipyramidal skeleton.
  • The three lone pairs occupy all equatorial slots, so the F atoms are axial and the molecule is linear.
Shortcutsp³d with 3 lone pairs is always linear — same as I3⁻ and ICl2⁻.
5.Xenon fluorides

XeF4 has a square planar shape because:

  • (A)Xe is sp3 hybridised
  • (B)Two lone pairs occupy positions opposite each other in an octahedral arrangement
  • (C)There are no lone pairs
  • (D)F atoms repel each other
Show solution
Answer  (B)  Two lone pairs occupy positions opposite each other in an octahedral arrangement
Given
XeF4.
Asked
Reason for square planar geometry.
Concept
With six electron domains the arrangement is octahedral. Two lone pairs settle trans to each other (180° apart) to minimise repulsion, leaving the four bond pairs in one plane.
Formula
Steric number = 4 bp + 2 lp = 6 → sp3d2
Baby steps
  • Xe uses 4 of its 8 electrons in bonds, leaving 2 lone pairs.
  • 6 domains → octahedral, sp3d2.
  • If the lone pairs were cis they would be 90° apart and repel strongly.
  • Trans placement puts them 180° apart, leaving the 4 F in a square plane.
ShortcutWhenever an octahedral base has 2 lone pairs, the shape is square planar. No exceptions.
6.Xenon fluorides

The hybridisation and shape of XeF6 are:

  • (A)sp3d2, octahedral
  • (B)sp3d3, distorted octahedral
  • (C)sp3d, see-saw
  • (D)sp3, tetrahedral
Show solution
Answer  (B)  sp3d3, distorted octahedral
Given
XeF6.
Asked
Hybridisation and shape.
Concept
Six bond pairs plus one lone pair give a steric number of 7. The lone pair occupies a position in the coordination sphere and distorts the otherwise regular octahedron.
Formula
Steric number = 6 bp + 1 lp = 7 → sp3d3
Baby steps
  • Xe has 8 valence electrons; 6 bond to fluorines.
  • Remaining 2 electrons = 1 lone pair.
  • 7 domains → sp3d3, pentagonal bipyramidal skeleton.
  • The lone pair pushes the fluorines out of a regular arrangement, so the shape is a distorted octahedron.
ShortcutXeF2 → 3 lp, XeF4 → 2 lp, XeF6 → 1 lp. The lone pairs drop by one each time you add two fluorines.
7.Reactions

Partial hydrolysis of XeF6 gives:

  • (A)XeO3
  • (B)XeOF4
  • (C)XeO4
  • (D)Xe and O2
Show solution
Answer  (B)  XeOF4
Given
XeF6 with a limited amount of water.
Asked
Product of partial hydrolysis.
Concept
Water replaces fluorine atoms with oxygen in stages. With only a small amount of water, only one oxygen is introduced, and xenon keeps its +6 state.
Formula
XeF6 + H2O → XeOF4 + 2HF
Baby steps
  • One water molecule supplies one oxygen and removes two fluorines as HF.
  • The product XeOF4 is square pyramidal with one lone pair, sp3d2.
  • More water gives XeO2F2, and complete hydrolysis gives XeO3.
ShortcutStages to remember: XeF6 → XeOF4 → XeO2F2 → XeO3, losing 2 F at each step.
8.Reactions

Hydrolysis of XeF4 gives:

  • (A)Only XeO3
  • (B)Xe, XeO3, HF and O2
  • (C)Only XeOF4
  • (D)XeF2 and O2
Show solution
Answer  (B)  Xe, XeO3, HF and O2
Given
XeF4 + water.
Asked
Products of hydrolysis.
Concept
Unlike XeF6, XeF4 disproportionates on hydrolysis. Some xenon is reduced to the free element while some is oxidised to the +6 state in XeO3.
Formula
6XeF4 + 12H2O → 4Xe + 2XeO3 + 24HF + 3O2
Baby steps
  • Xe starts at +4 in XeF4.
  • Part is reduced to 0 (free Xe), part is oxidised to +6 (XeO3).
  • That is disproportionation, and HF and O2 are also produced.
  • XeF2 behaves similarly: 2XeF2 + 2H2O → 2Xe + 4HF + O2.
ShortcutXeF6 hydrolyses cleanly (no redox). XeF2 and XeF4 disproportionate. Learn which is which.
9.Structures

The shape of XeO3 and the hybridisation of xenon are:

  • (A)Trigonal planar, sp2
  • (B)Pyramidal, sp3
  • (C)T-shaped, sp3d
  • (D)Square planar, sp3d2
Show solution
Answer  (B)  Pyramidal, sp3
Given
XeO3.
Asked
Shape and hybridisation.
Concept
Three sigma bonds to oxygen plus one lone pair give a steric number of 4. The lone pair occupies one tetrahedral corner, leaving a pyramid of atoms.
Formula
Steric number = 3 σ + 1 lp = 4 → sp3
Baby steps
  • Xe forms three double bonds to O, which counts as three sigma bonds.
  • That uses 6 of its 8 valence electrons, leaving 1 lone pair.
  • 4 domains → sp3, tetrahedral electron geometry.
  • With the lone pair in one corner, the shape is trigonal pyramidal — like NH3.
ShortcutXeO3 is pyramidal (has a lone pair) but XeO4 is tetrahedral (no lone pair). Count the lone pair every time.
10.Structures

The shape of XeOF4 is:

  • (A)Octahedral
  • (B)Square pyramidal
  • (C)See-saw
  • (D)Trigonal bipyramidal
Show solution
Answer  (B)  Square pyramidal
Given
XeOF4.
Asked
Molecular shape.
Concept
Xenon forms five sigma bonds (four to F, one to O) and retains one lone pair, so the electron geometry is octahedral and one vertex is a lone pair.
Formula
Steric number = 5 σ + 1 lp = 6 → sp3d2
Baby steps
  • Count sigma bonds: 4 Xe–F plus 1 Xe=O sigma = 5.
  • Xe has 8 valence electrons; 4 go to F, 2 to the O double bond, leaving 2 → 1 lone pair.
  • 6 domains → sp3d2, octahedral skeleton.
  • Removing one vertex for the lone pair gives a square pyramid.
ShortcutPi bonds don't add to steric number. Xe=O contributes one sigma, exactly like Xe–F.
11.Reactions

XeF6 reacts with a strong Lewis acid such as PF5 to give:

  • (A)[XeF7]−
  • (B)[XeF5]+[PF6]−
  • (C)XeO3
  • (D)Xe and F2
Show solution
Answer  (B)  [XeF5]+[PF6]−
Given
XeF6 + PF5.
Asked
Product formed.
Concept
Xenon hexafluoride can act as a fluoride donor. A strong Lewis acid accepts F⁻, leaving behind a xenon cation.
Formula
XeF6 + PF5 → [XeF5]+[PF6]−
Baby steps
  • PF5 is electron deficient and grabs an F⁻.
  • XeF6 loses one F⁻ to become [XeF5]⁺.
  • The pair forms an ionic salt.
  • In the opposite direction, XeF6 + fluoride donors like RbF gives [XeF7]⁻ instead.
ShortcutXeF6 is amphoteric towards fluoride: it donates F⁻ to Lewis acids and accepts F⁻ from fluoride salts.
12.Reactions

Complete hydrolysis of XeF6 with NaOH gives sodium perxenate, in which the oxidation state of xenon is:

  • (A)+4
  • (B)+6
  • (C)+8
  • (D)+2
Show solution
Answer  (C)  +8
Given
Na4XeO6 (sodium perxenate).
Asked
Oxidation state of Xe.
Concept
Assign standard values — Na is +1 and O is −2 — and balance the total to zero for the neutral salt.
Formula
4(+1) + x + 6(−2) = 0 → x = +8
Baby steps
  • Sodium contributes 4 × (+1) = +4.
  • Oxygen contributes 6 × (−2) = −12.
  • For neutrality: 4 + x − 12 = 0, so x = +8.
  • This is the highest oxidation state xenon can reach, using all 8 valence electrons.
ShortcutXe oxidation states: +2 (XeF2), +4 (XeF4), +6 (XeF6, XeO3), +8 (XeO4, perxenate). Always even.
13.Trends

Which of the following noble gases has the lowest boiling point?

  • (A)He
  • (B)Ne
  • (C)Ar
  • (D)Xe
Show solution
Answer  (A)  He
Given
Boiling points of the noble gases.
Asked
Lowest boiling point.
Concept
Noble gas atoms are held together only by weak London dispersion forces, whose strength grows with the number of electrons and the polarisability of the atom.
Formula
Boiling point: He (4.2 K) < Ne < Ar < Kr < Xe < Rn
Baby steps
  • He has just 2 electrons, so it is the least polarisable atom of all.
  • Its dispersion forces are the weakest known.
  • So He has the lowest boiling point of any substance — 4.2 K.
  • It is also the only substance that does not solidify under ordinary pressure.
ShortcutHe holds two records: lowest boiling point of all substances and highest ionisation enthalpy of all elements.
14.Uses

Helium is used in diving cylinders mixed with oxygen because:

  • (A)It is lighter than air
  • (B)It is much less soluble in blood than nitrogen
  • (C)It is radioactive
  • (D)It reacts with oxygen
Show solution
Answer  (B)  It is much less soluble in blood than nitrogen
Given
Helium–oxygen mixtures for deep-sea diving.
Asked
Reason for using He instead of N2.
Concept
Under high pressure, nitrogen dissolves in blood; when the diver surfaces and pressure drops, it bubbles out and causes decompression sickness. Helium is far less soluble.
Formula
Heliox = He + O2 breathing mixture
Baby steps
  • At depth, gases dissolve in blood in proportion to pressure.
  • N2 dissolves appreciably and causes 'bends' on rapid ascent.
  • He dissolves very little, so much less comes out of solution.
  • Helium is also used to fill weather balloons, being non-flammable unlike hydrogen.
ShortcutHelium uses: diving mixtures, balloons, cryogenics (liquid He at 4.2 K for superconducting magnets like MRI).
15.Uses

Which noble gas is used in filling incandescent bulbs and in high-temperature metallurgical processes?

  • (A)He
  • (B)Ne
  • (C)Ar
  • (D)Rn
Show solution
Answer  (C)  Ar
Given
Industrial uses of noble gases.
Asked
The gas used in bulbs and welding.
Concept
Argon is chemically inert and abundant enough in air to be cheap, so it provides an economical inert atmosphere.
Formula
Ar makes up about 0.93% of the atmosphere
Baby steps
  • An inert filling stops the hot tungsten filament from evaporating or oxidising.
  • Argon is the most abundant noble gas in air, so it is the cheapest inert option.
  • It is also used as a shield gas in arc welding and in growing silicon crystals.
ShortcutNoble gas uses: He → balloons and cryogenics; Ne → advertising signs; Ar → bulbs and welding; Rn → radiotherapy.
16.Trends

Radon is obtained from:

  • (A)Fractional distillation of air
  • (B)The radioactive decay of radium
  • (C)The electrolysis of water
  • (D)Natural gas
Show solution
Answer  (B)  The radioactive decay of radium
Given
Source of radon.
Asked
How radon is obtained.
Concept
Radon has no stable isotope; it is continuously produced as an intermediate in the decay chain of heavier radioactive elements and decays away with a short half-life.
Formula
226Ra → 222Rn + α, half-life of Rn ≈ 3.8 days
Baby steps
  • All other noble gases are extracted from liquid air by fractional distillation.
  • Radon cannot be, because it is radioactive and decays quickly.
  • It is collected from the alpha decay of radium.
  • Its short half-life means it must be made and used promptly.
ShortcutRadon is the only noble gas with no stable isotope and the only one not obtained from air.
17.Bonding

Noble gases are monatomic because:

  • (A)They are radioactive
  • (B)They have completely filled valence shells with no tendency to bond
  • (C)They are heavy
  • (D)They have unpaired electrons
Show solution
Answer  (B)  They have completely filled valence shells with no tendency to bond
Given
Physical nature of group 18.
Asked
Reason they exist as single atoms.
Concept
Bond formation happens when atoms can lower their energy by sharing or transferring electrons. A noble gas already has a complete octet, so bonding offers no gain.
Formula
All group 18 → ns2np6, zero unpaired electrons
Baby steps
  • Look for unpaired electrons — there are none.
  • No unpaired electrons means no covalent bond can be formed in the normal way.
  • So the atoms remain separate, held only by weak dispersion forces.
  • This is why they are all gases with very low boiling points.
ShortcutMonatomic gases in the periodic table = group 18 only. Every other gaseous element is diatomic.
18.Trends

The correct order of the ionisation enthalpy of the noble gases is:

  • (A)He < Ne < Ar < Kr < Xe
  • (B)Xe < Kr < Ar < Ne < He
  • (C)Ar < He < Ne < Kr
  • (D)Ne < He < Ar < Xe
Show solution
Answer  (B)  Xe < Kr < Ar < Ne < He
Given
He, Ne, Ar, Kr, Xe.
Asked
Increasing order of ionisation enthalpy.
Concept
Ionisation enthalpy falls down a group because the outermost electron is further from the nucleus and better shielded by inner shells.
Formula
He (2372) > Ne (2080) > Ar (1520) > Kr (1351) > Xe (1170) kJ mol⁻¹
Baby steps
  • He is smallest and least shielded → hardest to ionise.
  • Each step down adds a shell, so the value falls.
  • Xe is the easiest to ionise, which is exactly why it forms compounds and the lighter ones do not.
  • Increasing order: Xe < Kr < Ar < Ne < He.
ShortcutThe reason only Xe (and Kr, barely) form compounds is this trend — low IE means the electrons can be pulled into bonds.
19.Compounds

Which of the following is the only known stable compound of krypton?

  • (A)KrF4
  • (B)KrF2
  • (C)KrO3
  • (D)KrCl2
Show solution
Answer  (B)  KrF2
Given
Compounds of krypton.
Asked
The known stable compound.
Concept
Krypton's ionisation enthalpy is higher than xenon's, so only the most electronegative element, fluorine, can pull its electrons into bonds — and only two of them.
Formula
Kr + F2 →(electric discharge, 85 K) KrF2
Baby steps
  • Kr is harder to oxidise than Xe.
  • Only F is electronegative enough to succeed.
  • The reaction needs an electric discharge at very low temperature.
  • KrF2 is linear, sp3d, with 3 lone pairs, just like XeF2.
ShortcutHe, Ne and Ar form no true compounds at all. Kr forms only KrF2. Xe forms many.
20.Preparation

XeF4 is prepared by heating a mixture of Xe and F2 in the ratio:

  • (A)2 : 1
  • (B)1 : 5
  • (C)1 : 20
  • (D)1 : 1
Show solution
Answer  (B)  1 : 5
Given
Direct fluorination of xenon.
Asked
Ratio of Xe to F2 for XeF4.
Concept
Which fluoride forms is controlled by how much fluorine is available. More fluorine, higher temperature and higher pressure push towards the higher fluoride.
Formula
Xe + 2F2 →(1:5, 873 K, 7 bar) XeF4
Baby steps
  • XeF2: Xe:F2 of 2:1 at 673 K and 1 bar.
  • XeF4: Xe:F2 of 1:5 at 873 K and 7 bar.
  • XeF6: Xe:F2 of 1:20 at 573 K and 60–70 bar.
  • So XeF4 needs the 1:5 mixture.
ShortcutExcess fluorine is the key variable: 2:1 → XeF2, 1:5 → XeF4, 1:20 → XeF6.
21.Compounds

Xenon forms compounds only with fluorine and oxygen because:

  • (A)They are gases
  • (B)They are the two most electronegative elements
  • (C)They are small
  • (D)They are in period 2
Show solution
Answer  (B)  They are the two most electronegative elements
Given
Xe compounds.
Asked
Reason only F and O work.
Concept
Xenon's electrons are held tightly, so only a partner electronegative enough to draw them into a bond can react. Fluorine and oxygen are the two most electronegative elements.
Formula
Electronegativity: F (4.0) > O (3.5) > others
Baby steps
  • Xe has a high ionisation enthalpy, though the lowest among stable noble gases.
  • A bonding partner must pull very strongly to overcome that.
  • Only F and O are strong enough.
  • Hence XeF2, XeF4, XeF6, XeO3, XeO4, XeOF4 — and nothing with Cl, Br or S.
ShortcutIf a question offers XeCl4 or XeS2 as an option, it is wrong by this rule alone.
22.Structures

Which of the following xenon compounds has zero dipole moment?

  • (A)XeO3
  • (B)XeOF4
  • (C)XeF4
  • (D)XeO2F2
Show solution
Answer  (C)  XeF4
Given
Xenon compounds.
Asked
The one with no net dipole.
Concept
A molecule has zero dipole moment when its bond dipoles are arranged symmetrically so that they cancel completely.
Formula
Square planar XeF4 → all four dipoles cancel in pairs
Baby steps
  • XeF4 is square planar; each Xe–F dipole is cancelled by the one opposite it.
  • XeO3 is pyramidal, so the dipoles add up — non-zero.
  • XeOF4 is square pyramidal, with the Xe=O unbalanced — non-zero.
  • So only XeF4 (and also linear XeF2) has zero dipole moment.
ShortcutSymmetrical shapes with identical surrounding atoms have zero dipole: linear, trigonal planar, tetrahedral, square planar, octahedral.
23.Clathrates

Which of the following noble gases does not form a clathrate compound with quinol?

  • (A)Ar
  • (B)Kr
  • (C)Xe
  • (D)He
Show solution
Answer  (D)  He
Given
Clathrate formation.
Asked
The gas that fails to form one.
Concept
A clathrate involves physically trapping an atom in a cage of host molecules. The guest must be large enough not to slip out through the openings in the lattice.
Formula
Ar, Kr, Xe form clathrates; He, Ne do not
Baby steps
  • No chemical bond is formed in a clathrate — it is pure physical trapping.
  • The cavity in quinol has a fixed size.
  • He and Ne are too small and escape.
  • Ar, Kr and Xe are large enough to be held in place.
ShortcutClathrate failure is about size, not reactivity. He and Ne fail because they are small, not because they are inert.
24.Trends

The atomic radii of noble gases are larger than those of the corresponding halogens because:

  • (A)They are radioactive
  • (B)Noble gas radii are van der Waals radii, not covalent radii
  • (C)They have more electrons
  • (D)They form no bonds with fluorine
Show solution
Answer  (B)  Noble gas radii are van der Waals radii, not covalent radii
Given
Comparison of atomic radii across period 2 and 3.
Asked
Reason the noble gas values are larger.
Concept
Radii are measured differently. For elements that bond, half the internuclear distance in a bonded pair (covalent radius) is used; for noble gases, only non-bonded contact is possible, giving the larger van der Waals radius.
Formula
van der Waals radius > covalent radius for the same atom
Baby steps
  • Covalent radius is measured with the two atoms pulled close by a shared pair.
  • van der Waals radius is measured between atoms merely touching.
  • Since noble gases don't bond, only the larger measurement is available.
  • So the values look bigger, though the atoms are not truly larger.
ShortcutThis is a measurement artefact, not a real reversal of the size trend. NEET loves this explanation.
25.Reactions

XeF2 reacts with water to give:

  • (A)XeO3 and HF
  • (B)Xe, HF and O2
  • (C)XeOF4 only
  • (D)XeO4 and HF
Show solution
Answer  (B)  Xe, HF and O2
Given
XeF2 + H2O.
Asked
Products of hydrolysis.
Concept
Xenon at +2 is unstable in water and is reduced to the free element, while oxygen from water is oxidised — a disproportionation-type redox hydrolysis.
Formula
2XeF2 + 2H2O → 2Xe + 4HF + O2
Baby steps
  • Xe goes from +2 to 0, so it is reduced.
  • Oxygen in water goes from −2 to 0 in O2, so it is oxidised.
  • Fluorine leaves as HF.
  • The reaction is slow with pure water and fast with base.
ShortcutXeF2 and XeF4 hydrolyse with a redox change. Only XeF6 hydrolyses without one.
26.Trends

Which noble gas is most abundant in the earth's atmosphere?

  • (A)He
  • (B)Ne
  • (C)Ar
  • (D)Kr
Show solution
Answer  (C)  Ar
Given
Composition of air.
Asked
Most abundant noble gas.
Concept
Argon-40 is continuously produced by the radioactive decay of potassium-40 in rocks, so it has accumulated in the atmosphere far beyond the level of the other noble gases.
Formula
Ar ≈ 0.93% of air by volume
Baby steps
  • Air is roughly 78% N2, 21% O2, and about 0.93% Ar.
  • Argon is the third most abundant gas in air overall.
  • Ne, He and Kr are present only in parts per million.
  • This abundance is why argon is the cheap choice for inert atmospheres.
ShortcutArgon is third most abundant in air overall, after N2 and O2. Helium is abundant in the universe but rare in air.
27.Structures

The number of sigma bonds and lone pairs on xenon in XeO4 respectively are:

  • (A)4 and 1
  • (B)4 and 0
  • (C)6 and 0
  • (D)8 and 0
Show solution
Answer  (B)  4 and 0
Given
XeO4.
Asked
Sigma bonds and lone pairs.
Concept
Xenon has 8 valence electrons; each Xe=O double bond uses two of them, so all eight are consumed and none remain as a lone pair.
Formula
Steric number = 4 σ + 0 lp = 4 → sp3, tetrahedral
Baby steps
  • Four Xe=O double bonds means 4 sigma bonds and 4 pi bonds.
  • Each double bond uses 2 of Xe's electrons: 4 × 2 = 8.
  • All 8 valence electrons are used, so lone pairs = 0.
  • Steric number 4 with no lone pairs gives a regular tetrahedron, Xe at +8.
ShortcutXeO4 is the only xenon compound with zero lone pairs, which is why it is perfectly tetrahedral.
28.Trends

Which of the following statements about noble gases is incorrect?

  • (A)They have positive electron gain enthalpies
  • (B)Their boiling points increase down the group
  • (C)They readily form ionic compounds with alkali metals
  • (D)Helium has the highest ionisation enthalpy of all elements
Show solution
Answer  (C)  They readily form ionic compounds with alkali metals
Given
Properties of group 18.
Asked
The incorrect statement.
Concept
Noble gases cannot form ionic compounds with alkali metals — both would need to gain electrons, and noble gases have no tendency to accept any.
Formula
ΔegH of noble gases is positive → they do not accept electrons
Baby steps
  • Positive electron gain enthalpy — correct.
  • Boiling points rise down the group with dispersion forces — correct.
  • He has the highest ionisation enthalpy of any element — correct.
  • Ionic compounds with alkali metals would require the noble gas to accept an electron, which it will not do. So that statement is wrong.
ShortcutNoble gases bond only when they LOSE electron density to F or O, never when they gain it.
29.Uses

Neon is used in advertising signs because:

  • (A)It is cheap
  • (B)It gives a characteristic reddish orange glow in a discharge tube
  • (C)It is radioactive
  • (D)It reacts with glass
Show solution
Answer  (B)  It gives a characteristic reddish orange glow in a discharge tube
Given
Uses of neon.
Asked
Reason it is used in signs.
Concept
In a discharge tube, electrons excite neon atoms; as they fall back, they emit light of specific wavelengths that combine to give a bright reddish-orange glow.
Formula
Excited Ne → ground state + visible photons
Baby steps
  • A high voltage across low-pressure neon ionises and excites the atoms.
  • The atoms return to the ground state and emit light.
  • Neon's emission spectrum lies in the red-orange region.
  • It is also used in beacon lights because the glow penetrates fog well.
ShortcutDifferent noble gases give different colours in discharge tubes — the emission spectrum is a fingerprint of the element.
30.Structures

In XeF6, the presence of one lone pair makes the molecule:

  • (A)A perfect octahedron
  • (B)A distorted octahedron
  • (C)Square planar
  • (D)Linear
Show solution
Answer  (B)  A distorted octahedron
Given
XeF6 geometry.
Asked
Effect of the lone pair.
Concept
A lone pair occupies more space than a bond pair and repels the surrounding bond pairs more strongly, so it pushes the fluorine atoms out of their symmetrical positions.
Formula
Repulsion order: lp–lp > lp–bp > bp–bp
Baby steps
  • Six bond pairs alone would give a perfect octahedron.
  • The seventh domain is a lone pair, which needs room.
  • It squeezes into the coordination sphere and pushes the F atoms aside.
  • The result is a distorted octahedral (capped octahedral) shape.
ShortcutAny time a lone pair is present, expect distortion from the ideal geometry. Only zero-lone-pair molecules are perfectly symmetrical.

Answer key — Noble Gases

1 C2 B3 B4 B5 B6 B7 B8 B9 B10 B11 B12 C13 A14 B15 C16 B17 B18 B19 B20 B21 B22 C23 D24 B25 B26 C27 B28 C29 B30 B