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Coordination Compounds · Sections 5.5.1–5.5.3

Valence Bond
& Hybridisation

Thirty NEET-pattern questions on hybridisation types, inner versus outer orbital complexes, spin-only magnetic moments and the limitations of VBT. Each is worked through Given, Asked, Concept, Formula, Baby Steps and a Shortcut. Attempt with solutions closed, then open to check the reasoning — not just the answer.

30Questions
0801
0702
0803
0704
30Shortcuts

Suggested timing. Nearly every question here runs the same four-step chain: oxidation state → d-electron count → ligand strength → unpaired electrons. Once that chain is automatic, target 40–45 seconds per question, or about 22 minutes for the set. If a question is taking longer, the delay is almost always at step one or step three, not at the arithmetic.

Log your attempt before opening the solutions. Record each wrong answer against one of three causes — this is more useful than the raw score.
SectionQCorrectDidn't knowMisreadToo slow
Hybridisation01–08
Inner / outer09–15
Magnetic moment16–23
Prediction & limits24–30
01

Hybridisation Types & Geometry

08 Q
[ 01 ]H NEET styled2sp3

The hybridisation and geometry of [Co(NH3)6]3+ are: (Z of Co = 27)

  • (a)sp3d2, octahedral
  • (b)d2sp3, octahedral
  • (c)dsp2, square planar
  • (d)sp3, tetrahedral

Answer — (b) d2sp3, octahedral

Given
[Co(NH3)6]3+, ZCo = 27, coordination number 6.
Asked
Hybridisation and geometry.
Concept
Coordination number 6 always gives an octahedron, but the hybridisation depends on which d orbitals are used. NH3 is strong enough for Co(III) to pair the 3d electrons, freeing two inner 3d orbitals.
Formula
Inner (n−1)d used → d2sp3; outer nd used → sp3d2
Baby steps
  1. Oxidation state of Co = +3 ⇒ Co3+ = [Ar]3d6.
  2. NH3 with a +3 metal acts as a strong field ligand → all six electrons pair into three 3d orbitals.
  3. Two 3d orbitals are now empty and available for hybridisation.
  4. Two 3d + one 4s + three 4p = d2sp3, octahedral, diamagnetic.
ShortcutWrite the d orbital used before sp3 for inner (3d), after for outer (4d). The position of the letter d tells you the whole story.
[ 02 ]H NEET stylesp3d

The hybridisation of iron in [Fe(CO)5] and the distribution of hybrid orbitals in space are:

  • (a)sp3d, trigonal bipyramidal
  • (b)d2sp3, octahedral
  • (c)sp3, tetrahedral
  • (d)dsp2, square planar

Answer — (a) sp3d, trigonal bipyramidal

Given
[Fe(CO)5], a neutral homoleptic carbonyl with coordination number 5.
Asked
Hybridisation and shape.
Concept
Each coordination number maps to a hybridisation and a shape. Coordination number 5 requires five hybrid orbitals, obtained by mixing one s, three p and one d orbital.
Formula
CN 4 → sp3 or dsp2; CN 5 → sp3d; CN 6 → d2sp3 or sp3d2
Baby steps
  1. CO is neutral and the complex is neutral, so Fe is in the zero oxidation state.
  2. Fe(0) = 3d64s2; the strong field CO pushes the 4s electrons into 3d → 3d84s0, then pairs them to 3d10... in NCERT terms, all electrons pair and five orbitals are freed.
  3. Five hybrid orbitals are needed for five CO ligands.
  4. One s + three p + one d = sp3d → trigonal bipyramidal.
ShortcutMemorise the NCERT Table 5.2 row by row: 4/sp3/tetrahedral, 4/dsp2/square planar, 5/sp3d/trigonal bipyramidal, 6/sp3d2 and 6/d2sp3/octahedral.
[ 03 ]H NEET styledsp2

Which hybridisation gives rise to a square planar geometry?

  • (a)sp3
  • (b)dsp2
  • (c)sp3d
  • (d)d2sp3

Answer — (b) dsp2

Given
Four hybridisation schemes.
Asked
The one producing square planar geometry.
Concept
Square planar geometry needs four hybrid orbitals directed to the corners of a square, which requires the dx²−y² orbital to participate along with s and two p orbitals.
Formula
dsp2 → square planar (4 orbitals in one plane)
Baby steps
  1. Coordination number 4 can be either sp3 or dsp2.
  2. sp3 uses no d orbital and directs the four bonds tetrahedrally at 109.5°.
  3. dsp2 includes one d orbital, which forces all four bonds into one plane at 90°.
  4. Hence dsp2 → square planar.
Shortcutdsp2 requires an empty (n−1)d orbital, so every dsp2 complex must have all its d electrons paired — making d8 square planar complexes diamagnetic.
[ 04 ]H NEET styleOrbital count

The number of atomic orbitals mixed in d2sp3 hybridisation and the number of hybrid orbitals produced are respectively:

  • (a)5 and 5
  • (b)6 and 6
  • (c)6 and 4
  • (d)4 and 6

Answer — (b) 6 and 6

Given
d2sp3 hybridisation.
Asked
Number of atomic orbitals used and hybrid orbitals formed.
Concept
Hybridisation conserves the number of orbitals: n atomic orbitals always give exactly n hybrid orbitals of equivalent energy and definite geometry.
Formula
d2sp3 = 2d + 1s + 3p = 6 orbitals
Baby steps
  1. Read the superscripts: d2 means two d orbitals.
  2. s with no superscript means one s orbital.
  3. p3 means three p orbitals.
  4. 2 + 1 + 3 = 6 atomic orbitals, giving 6 hybrid orbitals for six ligands.
ShortcutThe number of hybrid orbitals must always equal the coordination number. If they disagree, the hybridisation assigned is wrong.
[ 05 ]H NEET styled10

The hybridisation and magnetic nature of [Zn(NH3)4]2+ are: (Z of Zn = 30)

  • (a)dsp2, diamagnetic
  • (b)sp3, diamagnetic
  • (c)sp3, paramagnetic
  • (d)d2sp3, diamagnetic

Answer — (b) sp3, diamagnetic

Given
[Zn(NH3)4]2+, ZZn = 30, coordination number 4.
Asked
Hybridisation and magnetic behaviour.
Concept
Zn2+ has a completely filled 3d subshell. With no vacant 3d orbital available, dsp2 is impossible however strong the ligand, so only sp3 remains.
Formula
μ = √n(n+2); d10 ⇒ n = 0
Baby steps
  1. Zn2+ = [Ar]3d10 — all five 3d orbitals are full.
  2. No 3d orbital can be emptied, so dsp2 cannot occur.
  3. Hybridise 4s and three 4p → sp3, tetrahedral.
  4. All electrons are paired → n = 0 → diamagnetic, and also colourless.
Shortcutd10 ions (Zn2+, Cu+, Ag+, Cd2+) are always sp3 at CN 4, always diamagnetic and always colourless.
[ 06 ]H NEET styled9

[Cu(NH3)4]2+ is square planar and paramagnetic. Its hybridisation and number of unpaired electrons are: (Z of Cu = 29)

  • (a)sp3, 1
  • (b)dsp2, 1
  • (c)dsp2, 0
  • (d)d2sp3, 2

Answer — (b) dsp2, 1

Given
[Cu(NH3)4]2+, square planar, paramagnetic, ZCu = 29.
Asked
Hybridisation and unpaired electron count.
Concept
Square planar geometry fixes the hybridisation as dsp2. A d9 ion can free one d orbital by promoting a single electron, but it is left with one unpaired electron — so unlike d8, it stays paramagnetic.
Formula
μ = √n(n+2) = √3 = 1.73 BM
Baby steps
  1. Cu2+ = [Ar]3d9.
  2. Square planar is stated ⇒ hybridisation must be dsp2.
  3. Nine electrons cannot all pair in four orbitals, so one remains unpaired.
  4. μ = √1(1+2) = 1.73 BM, paramagnetic.
ShortcutSquare planar d8 → diamagnetic. Square planar d9 (Cu2+) → exactly one unpaired electron. This pair is a favourite contrast.
[ 07 ]H NEET styleComparison

Which of the following pairs of complexes have the same hybridisation?

  • (a)[NiCl4]2− and [Ni(CN)4]2−
  • (b)[NiCl4]2− and [Ni(CO)4]
  • (c)[Ni(CN)4]2− and [Ni(NH3)6]2+
  • (d)[Ni(CO)4] and [Ni(CN)4]2−

Answer — (b) [NiCl4]2− and [Ni(CO)4]

Given
Four nickel complexes with different ligands and coordination numbers.
Asked
The pair sharing a hybridisation.
Concept
Same coordination number does not guarantee same hybridisation — the ligand field strength decides whether a d orbital is available. Compare geometry, not just CN.
Formula
CN 4: sp3 (tetrahedral) or dsp2 (square planar)
Baby steps
  1. [NiCl4]2−: Ni2+ d8, Cl− weak → no pairing → sp3.
  2. [Ni(CO)4]: Ni(0) becomes 3d10 under CO → 4s and 4p free → sp3.
  3. [Ni(CN)4]2−: strong field pairs the d8 electrons → dsp2.
  4. [Ni(NH3)6]2+: CN 6 → sp3d2. Matching pair = option B.
ShortcutBoth sp3 nickel complexes reach it for opposite reasons: one because the ligand is too weak to pair, the other because the metal is already d10.
[ 08 ]H NEET styled3

[Cr(NH3)6]3+ and [Cr(H2O)6]3+ both have d2sp3 hybridisation because:

  • (a)Both ligands are strong field ligands
  • (b)Cr3+ is d3, so two 3d orbitals are vacant without any pairing being needed
  • (c)Chromium always uses 4d orbitals
  • (d)Both complexes are diamagnetic

Answer — (b) Cr3+ is d3, so two 3d orbitals are vacant without any pairing being needed

Given
Two Cr(III) octahedral complexes, one with a strong and one with a weak field ligand.
Asked
Reason both are inner orbital complexes.
Concept
For metal ions with up to three d electrons (d1, d2, d3), the electrons occupy three separate orbitals under Hund's rule, leaving two 3d orbitals already vacant. Inner orbital hybridisation is then possible regardless of ligand strength.
Formula
d1–d3 ⇒ two vacant d orbitals available ⇒ d2sp3 always
Baby steps
  1. Cr3+ = [Ar]3d3.
  2. Three electrons occupy three of the five 3d orbitals singly.
  3. Two 3d orbitals remain empty without any pairing.
  4. Both complexes use d2sp3 and both have 3 unpaired electrons, μ = 3.87 BM.
ShortcutFor d1, d2 and d3 ions the magnetic behaviour of the free ion and its complexes is the same — ligand strength makes no difference.
02

Inner vs Outer Orbital Complexes

07 Q
[ 09 ]I NEET styleOuter orbital

Which of the following is an outer orbital complex?

  • (a)[Co(NH3)6]3+
  • (b)[Fe(CN)6]3−
  • (c)[CoF6]3−
  • (d)[Mn(CN)6]3−

Answer — (c) [CoF6]3−

Given
Four octahedral complexes of first-series metals.
Asked
The outer orbital complex.
Concept
An outer orbital complex uses the outer nd orbitals (sp3d2) because a weak field ligand fails to pair the (n−1)d electrons. It is also called high spin or spin-free.
Formula
Weak field ligand → sp3d2 → outer orbital
Baby steps
  1. NH3 (with M3+) and CN− are strong field → pairing occurs → inner orbital.
  2. F− sits near the weak end of the spectrochemical series.
  3. Co3+ = d6 stays high spin as t2g4eg2 → no vacant 3d orbital.
  4. It must use 4d → sp3d2, outer orbital, 4 unpaired electrons.
ShortcutHalide ligand in the formula → assume outer orbital / high spin. CN− or CO → inner orbital / low spin.
[ 10 ]I NEET styleUnpaired count

The number of unpaired electrons in [Mn(CN)6]3− is: (Z of Mn = 25)

  • (a)4
  • (b)3
  • (c)2
  • (d)0

Answer — (c) 2

Given
[Mn(CN)6]3−, ZMn = 25, CN− strong field.
Asked
Number of unpaired electrons.
Concept
A strong field ligand forces maximum pairing within the three lower d orbitals before any electron occupies the higher pair.
Formula
Strong field dn: fill three orbitals first, pairing as needed
Baby steps
  1. Oxidation state: x − 6 = −3 ⇒ Mn3+.
  2. Mn3+ = [Ar]3d4.
  3. CN− is strong field → all four electrons crowd into three 3d orbitals as t2g4.
  4. Two orbitals hold pairs and one holds a single electron... giving 2 unpaired electrons, μ = 2.83 BM, d2sp3.
ShortcutContrast with [MnCl6]3−: weak field leaves all 4 electrons unpaired (μ = 4.90 BM). Same metal, same charge, different ligand.
[ 11 ]I NEET styled8 octahedral

[Ni(NH3)6]2+ is an outer orbital complex because:

  • (a)NH3 is a weak field ligand for every metal ion
  • (b)Ni2+ is d8, and even with maximum pairing two 3d orbitals cannot be made vacant
  • (c)Nickel prefers 4d orbitals to 3d orbitals
  • (d)The complex is diamagnetic

Answer — (b) Ni2+ is d8, and even with maximum pairing two 3d orbitals cannot be made vacant

Given
[Ni(NH3)6]2+, Ni2+ = 3d8, coordination number 6.
Asked
Reason for outer orbital character.
Concept
d2sp3 hybridisation needs two empty (n−1)d orbitals. Eight d electrons can at best be squeezed into four orbitals, leaving only one vacant — never two.
Formula
d8 octahedral ⇒ sp3d2 always
Baby steps
  1. Ni2+ = 3d8 with two unpaired electrons.
  2. Maximum pairing would fill four 3d orbitals and leave one empty.
  3. Two vacant inner d orbitals are required but unavailable.
  4. So 4d orbitals must be used → sp3d2, outer orbital, paramagnetic with 2 unpaired electrons.
Shortcutd8 octahedral is always outer orbital, no matter how strong the ligand. Only at CN 4 can d8 go inner, as dsp2.
[ 12 ]I NEET styleTerminology

Inner orbital complexes are also known as:

  • (a)High spin or spin-free complexes
  • (b)Low spin or spin-paired complexes
  • (c)Outer orbital complexes
  • (d)Neutral complexes

Answer — (b) Low spin or spin-paired complexes

Given
Terminology of valence bond theory.
Asked
Alternative names for inner orbital complexes.
Concept
The three names describe the same situation from different angles: inner orbital (which d orbitals are used), low spin (few unpaired electrons), spin-paired (electrons have been forced to pair).
Formula
Inner orbital = low spin = spin-paired = d2sp3
Baby steps
  1. A strong field ligand forces the (n−1)d electrons to pair.
  2. Pairing empties inner d orbitals, which then hybridise → inner orbital.
  3. Pairing also reduces the number of unpaired electrons → low spin, spin-paired.
  4. So all three terms describe the same complex.
ShortcutThe mirror set is equally examinable: outer orbital = high spin = spin-free = sp3d2.
[ 13 ]I NEET styleDiamagnetic

Which of the following is a diamagnetic inner orbital complex?

  • (a)[Fe(CN)6]4−
  • (b)[Fe(CN)6]3−
  • (c)[FeF6]3−
  • (d)[Fe(H2O)6]2+

Answer — (a) [Fe(CN)6]4−

Given
Four iron complexes, ZFe = 26.
Asked
The diamagnetic inner orbital complex.
Concept
Diamagnetic means zero unpaired electrons. Only a low spin d6 configuration achieves this among these options, since all six electrons pair into three orbitals.
Formula
μ = 0 ⇒ n = 0
Baby steps
  1. [Fe(CN)6]4−: Fe2+ = d6, strong field → t2g6 → 0 unpaired, d2sp3, diamagnetic.
  2. [Fe(CN)6]3−: Fe3+ = d5, strong field → 1 unpaired, paramagnetic.
  3. [FeF6]3−: weak field d5 → 5 unpaired.
  4. [Fe(H2O)6]2+: weak field d6 → 4 unpaired.
ShortcutOnly low spin d6 is diamagnetic in an octahedral field: [Fe(CN)6]4−, [Co(NH3)6]3+, [Co(C2O4)3]3−.
[ 14 ]I NEET styleOrbitals used

In an inner orbital octahedral complex of a first transition series metal, the d orbitals involved in hybridisation belong to which shell?

  • (a)4d
  • (b)3d
  • (c)5d
  • (d)3p

Answer — (b) 3d

Given
Inner orbital complex of a 3d series metal.
Asked
Which d orbitals hybridise.
Concept
‘Inner’ refers to the (n−1)d orbitals, which for the first transition series is 3d. ‘Outer’ refers to the nd orbitals, which is 4d.
Formula
Inner = (n−1)d = 3d; outer = nd = 4d
Baby steps
  1. The valence shell of a first-series metal is n = 4.
  2. Inner orbital hybridisation uses (n−1)d = 3d, with 4s and 4p.
  3. Written d2sp3, with the d written first because it comes from a lower shell.
  4. Outer orbital hybridisation uses 4d, written sp3d2.
ShortcutThe written order encodes the shell: d2sp3 means 3d comes before 4s; sp3d2 means 4d comes after 4p.
[ 15 ]I NEET styleComparison

[MnCl6]3− has four unpaired electrons while [Mn(CN)6]3− has two. Valence bond theory explains this as:

  • (a)The first is an outer orbital complex and the second an inner orbital complex
  • (b)Manganese has different oxidation states in the two
  • (c)The first is tetrahedral and the second octahedral
  • (d)Chloride is a stronger ligand than cyanide

Answer — (a) The first is an outer orbital complex and the second an inner orbital complex

Given
Two Mn(III) octahedral complexes with different magnetic moments.
Asked
The VBT explanation.
Concept
Both contain Mn3+ (d4). The ligand decides whether the 3d electrons pair, and therefore whether inner (3d) or outer (4d) orbitals are used.
Formula
μ = √n(n+2)
Baby steps
  1. Cl− is weak field → no pairing → 4 unpaired → no vacant 3d → sp3d2, outer orbital.
  2. CN− is strong field → pairing occurs → 2 unpaired.
  3. Two 3d orbitals are freed → d2sp3, inner orbital.
  4. μ values: 4.90 BM and 2.83 BM respectively.
ShortcutWhenever a question gives the same metal in the same oxidation state but different μ, the answer is always inner versus outer orbital.
03

Spin-Only Magnetic Moment

08 Q
[ 16 ]M NEET styleSpin-only

The spin-only magnetic moment of a complex with three unpaired electrons is:

  • (a)1.73 BM
  • (b)2.83 BM
  • (c)3.87 BM
  • (d)4.90 BM

Answer — (c) 3.87 BM

Given
n = 3 unpaired electrons.
Asked
Spin-only magnetic moment.
Concept
Only unpaired electron spins contribute to the magnetic moment; orbital contribution is ignored in the spin-only approximation.
Formula
μ = √n(n+2) Bohr magneton
Baby steps
  1. Substitute n = 3 into the formula.
  2. μ = √3(3+2) = √15.
  3. √15 = 3.873.
  4. μ = 3.87 BM.
ShortcutLearn the ladder and skip the arithmetic every time: n = 1 → 1.73, 2 → 2.83, 3 → 3.87, 4 → 4.90, 5 → 5.92 BM.
[ 17 ]M NEET styleReverse

A complex ion has a spin-only magnetic moment of 2.83 BM. The number of unpaired electrons present is:

  • (a)1
  • (b)2
  • (c)3
  • (d)4

Answer — (b) 2

Given
μ = 2.83 BM.
Asked
Number of unpaired electrons.
Concept
The spin-only formula can be inverted: square the moment and solve the resulting quadratic for n.
Formula
μ2 = n(n+2) ⇒ n2 + 2n − μ2 = 0
Baby steps
  1. μ2 = (2.83)2 ≈ 8.
  2. n(n+2) = 8 ⇒ n2 + 2n − 8 = 0.
  3. (n + 4)(n − 2) = 0.
  4. n = 2 (the negative root is rejected).
ShortcutNever solve the quadratic in the exam. Recognise 2.83 BM as n = 2 straight from the memorised ladder.
[ 18 ]M NEET styled5 high spin

The spin-only magnetic moment of [Mn(H2O)6]2+ is: (Z of Mn = 25)

  • (a)1.73 BM
  • (b)3.87 BM
  • (c)4.90 BM
  • (d)5.92 BM

Answer — (d) 5.92 BM

Given
[Mn(H2O)6]2+, ZMn = 25, H2O weak field.
Asked
Spin-only magnetic moment.
Concept
A weak field ligand leaves the free-ion configuration intact, so Hund's rule governs and every d orbital is singly occupied first.
Formula
μ = √n(n+2)
Baby steps
  1. Mn2+ = [Ar]3d5.
  2. H2O is weak field → no pairing → all five orbitals singly occupied → n = 5.
  3. μ = √5(5+2) = √35.
  4. μ = 5.92 BM — the maximum possible for the 3d series.
ShortcutHigh spin d5 (Mn2+, Fe3+) gives the largest magnetic moment of any 3d ion. If an option says 5.92 BM, look for d5 with a weak ligand.
[ 19 ]M NEET styleDiamagnetic

The spin-only magnetic moment of [Ni(CN)4]2− is:

  • (a)0 BM
  • (b)1.73 BM
  • (c)2.83 BM
  • (d)3.87 BM

Answer — (a) 0 BM

Given
[Ni(CN)4]2−, square planar, Ni2+ = 3d8.
Asked
Spin-only magnetic moment.
Concept
Square planar geometry requires dsp2 hybridisation, which needs one empty 3d orbital. For a d8 ion this can only be achieved by pairing every electron.
Formula
μ = √n(n+2); n = 0 ⇒ μ = 0
Baby steps
  1. Ni2+ = 3d8, normally with two unpaired electrons.
  2. CN− is strong field → the two unpaired electrons pair up.
  3. This empties one 3d orbital for dsp2 hybridisation.
  4. n = 0 → μ = 0 BM, diamagnetic.
ShortcutDiamagnetic complexes to recognise on sight: [Ni(CN)4]2−, [Ni(CO)4], [Co(NH3)6]3+, [Fe(CN)6]4−.
[ 20 ]M NEET styled6 high spin

The spin-only magnetic moment of [CoF6]3− is: (Z of Co = 27)

  • (a)0 BM
  • (b)1.73 BM
  • (c)4.90 BM
  • (d)5.92 BM

Answer — (c) 4.90 BM

Given
[CoF6]3−, ZCo = 27, F− weak field.
Asked
Spin-only magnetic moment.
Concept
Co3+ is d6. With a weak field ligand it stays high spin: five orbitals filled singly, then the sixth electron pairs.
Formula
μ = √n(n+2)
Baby steps
  1. Co3+ = [Ar]3d6.
  2. F− is weak field → high spin → t2g4eg2.
  3. Unpaired electrons: 4 (one orbital holds a pair).
  4. μ = √4(4+2) = √24 = 4.90 BM, paramagnetic, sp3d2.
ShortcutSame d6 ion, opposite answers: [CoF6]3− gives 4.90 BM but [Co(NH3)6]3+ gives 0 BM. The ligand is the whole question.
[ 21 ]M NEET styleRanking

Which of the following has the lowest spin-only magnetic moment?

  • (a)[Fe(H2O)6]3+
  • (b)[Cr(H2O)6]3+
  • (c)[Fe(CN)6]3−
  • (d)[Mn(H2O)6]2+

Answer — (c) [Fe(CN)6]3−

Given
Four octahedral complexes with the given ligands.
Asked
The lowest magnetic moment.
Concept
Count the unpaired electrons in each after applying the ligand field strength, then compare.
Formula
μ = √n(n+2)
Baby steps
  1. [Fe(H2O)6]3+: d5 weak field → n = 5 → 5.92 BM.
  2. [Cr(H2O)6]3+: d3 → n = 3 → 3.87 BM.
  3. [Fe(CN)6]3−: d5 strong field → pairing → n = 1 → 1.73 BM.
  4. [Mn(H2O)6]2+: d5 weak field → n = 5 → 5.92 BM.
ShortcutTwo of the options are d5 with weak ligands and give the identical answer — a clue that the intended answer is the strong field one.
[ 22 ]M NEET styleTetrahedral

The spin-only magnetic moment of the tetrahedral complex [CoCl4]2− is: (Z of Co = 27)

  • (a)1.73 BM
  • (b)2.83 BM
  • (c)3.87 BM
  • (d)5.92 BM

Answer — (c) 3.87 BM

Given
[CoCl4]2−, tetrahedral, ZCo = 27.
Asked
Spin-only magnetic moment.
Concept
Tetrahedral complexes are essentially always high spin, because the splitting is too small to force pairing. So use the free-ion configuration directly.
Formula
μ = √n(n+2); tetrahedral ⇒ high spin
Baby steps
  1. Oxidation state: x − 4 = −2 ⇒ Co2+.
  2. Co2+ = [Ar]3d7.
  3. High spin d7: five orbitals filled singly, then two pair → 3 unpaired.
  4. μ = √3(3+2) = 3.87 BM, sp3 hybridisation.
ShortcutFor any tetrahedral complex, skip the ligand strength question entirely and just count unpaired electrons in the free ion.
[ 23 ]M NEET styleIdentification

A complex shows a spin-only magnetic moment of 1.73 BM. Which of the following could it be?

  • (a)[Ti(H2O)6]3+
  • (b)[Cr(NH3)6]3+
  • (c)[Ni(NH3)6]2+
  • (d)[Mn(H2O)6]2+

Answer — (a) [Ti(H2O)6]3+

Given
μ = 1.73 BM; four candidate complexes.
Asked
The complex matching this moment.
Concept
Convert the moment to an unpaired electron count first, then find which metal ion can produce that count under the stated ligand field.
Formula
μ = 1.73 BM ⇒ n = 1
Baby steps
  1. 1.73 BM corresponds to n = 1.
  2. [Ti(H2O)6]3+: Ti3+ = 3d1 → exactly 1 unpaired. Match.
  3. [Cr(NH3)6]3+: d3 → n = 3 → 3.87 BM.
  4. [Ni(NH3)6]2+: d8 outer orbital → n = 2 → 2.83 BM. [Mn(H2O)6]2+: n = 5.
ShortcutConvert μ to n first, then hunt for the matching ion. Going the other way round wastes time on every option.
04

Geometry from Magnetic Data & Limitations of VBT

07 Q
[ 24 ]L NEET stylePrediction

A four-coordinate complex of Ni(II) is found to be diamagnetic. Its geometry and hybridisation are:

  • (a)Tetrahedral, sp3
  • (b)Square planar, dsp2
  • (c)Octahedral, d2sp3
  • (d)Trigonal bipyramidal, sp3d

Answer — (b) Square planar, dsp2

Given
Ni(II) complex, coordination number 4, magnetic moment zero.
Asked
Geometry and hybridisation.
Concept
Magnetic data is the experimental route to geometry in VBT. For d8 Ni(II) at CN 4, tetrahedral leaves 2 unpaired electrons while square planar requires complete pairing.
Formula
μ = 0 ⇒ n = 0
Baby steps
  1. Ni2+ = 3d8, with two unpaired electrons in the free ion.
  2. Tetrahedral sp3 would retain those two → μ = 2.83 BM, paramagnetic.
  3. Diamagnetic means they have paired, emptying a 3d orbital.
  4. That empty 3d orbital enables dsp2 → square planar.
ShortcutAt CN 4: paramagnetic (μ ≈ 2.83) → tetrahedral; diamagnetic (μ = 0) → square planar. The magnetic moment decides the shape.
[ 25 ]L NEET stylePrediction

The spin-only magnetic moment of a four-coordinate Mn(II) complex is 5.9 BM. Its geometry is:

  • (a)Square planar
  • (b)Tetrahedral
  • (c)Octahedral
  • (d)Linear

Answer — (b) Tetrahedral

Given
Mn(II) complex, coordination number 4, μ = 5.9 BM.
Asked
Geometry.
Concept
If all five d electrons remain unpaired, no d orbital can be vacated, so dsp2 hybridisation is impossible and the complex must be sp3.
Formula
μ = 5.9 BM ⇒ n = 5
Baby steps
  1. Mn2+ = 3d5.
  2. μ = 5.9 BM ⇒ n(n+2) = 35 ⇒ n = 5 unpaired electrons.
  3. All five 3d orbitals are singly occupied, so none is available for dsp2.
  4. Only 4s and 4p can hybridise → sp3, tetrahedral.
ShortcutHigh magnetic moment at CN 4 rules out square planar automatically, because dsp2 demands a vacated d orbital.
[ 26 ]L NEET styleLimitation

Which of the following is a limitation of valence bond theory?

  • (a)It cannot explain the colour of coordination compounds
  • (b)It cannot predict the geometry of any complex
  • (c)It cannot explain the formation of coordinate bonds
  • (d)It cannot account for the existence of complex ions

Answer — (a) It cannot explain the colour of coordination compounds

Given
Shortcomings of VBT.
Asked
A genuine limitation.
Concept
VBT successfully explains formation, geometry and (qualitatively) magnetic behaviour. It fails on colour, on quantitative magnetic and stability data, and on distinguishing weak from strong ligands.
Formula
—
Baby steps
  1. VBT does explain bond formation, structure and shape → options B, C and D are false.
  2. Colour arises from d–d electronic transitions between split energy levels.
  3. VBT has no concept of d orbital splitting at all, so it cannot address colour.
  4. Option A is a real limitation, later fixed by crystal field theory.
ShortcutThe six VBT limitations to recall: many assumptions, no quantitative magnetic data, no colour, no stability data, no exact prediction for CN 4 shapes, and no weak/strong ligand distinction.
[ 27 ]L NEET styleLimitation

Valence bond theory does not:

  • (a)Explain why some ligands cause pairing of electrons while others do not
  • (b)Explain the geometry of [Ni(CO)4]
  • (c)Describe the nature of the metal–ligand bond
  • (d)Account for the diamagnetism of [Fe(CN)6]4−

Answer — (a) Explain why some ligands cause pairing of electrons while others do not

Given
Statements about what VBT can and cannot do.
Asked
The failure of the theory.
Concept
VBT can describe the outcome of pairing once you tell it which ligands are strong, but it offers no reason why CN− pairs electrons and F− does not. That ranking comes from experiment, and is explained by CFT via Δo versus P.
Formula
—
Baby steps
  1. VBT explains geometry from hybridisation → option B is within its scope.
  2. It describes the coordinate bond as ligand lone pair donation → option C is fine.
  3. It accounts for diamagnetism once pairing is assumed → option D is fine.
  4. It cannot predict or explain which ligands are strong field → option A.
ShortcutVBT takes the spectrochemical series as an input it cannot justify. CFT explains it by comparing Δo with the pairing energy P.
[ 28 ]L NEET styleNature

Regarding hybrid orbitals in coordination compounds, the correct statement is:

  • (a)Hybrid orbitals are real orbitals that can be isolated experimentally
  • (b)Hybridisation is a mathematical manipulation of the wave equations of atomic orbitals
  • (c)Hybrid orbitals always have lower energy than the original atomic orbitals
  • (d)Hybridisation occurs only in the absence of ligands

Answer — (b) Hybridisation is a mathematical manipulation of the wave equations of atomic orbitals

Given
Nature of hybridisation as described in the NCERT text.
Asked
The correct statement.
Concept
NCERT states explicitly that hybrid orbitals do not actually exist. Hybridisation is a mathematical combination of the wave functions of the atomic orbitals involved, used to rationalise observed geometry.
Formula
—
Baby steps
  1. Hybrid orbitals cannot be isolated or observed → option A is false.
  2. They are constructed by combining atomic orbital wave functions → option B is correct.
  3. Hybrid orbitals have an energy intermediate between the mixing orbitals → option C is false.
  4. Hybridisation is invoked precisely because ligands are approaching → option D is false.
ShortcutThis is a direct one-line lift from the NCERT text and appears as a statement-based question. Learn the wording: ‘a mathematical manipulation of the wave equation’.
[ 29 ]L NEET styleCarbonyl

[Ni(CO)4] is diamagnetic although Ni2+ normally has two unpaired electrons. The reason is:

  • (a)Nickel is in the zero oxidation state and CO forces the 4s electrons into 3d, giving 3d10
  • (b)CO is a weak field ligand
  • (c)The complex is square planar
  • (d)Nickel is in the +2 state here as well

Answer — (a) Nickel is in the zero oxidation state and CO forces the 4s electrons into 3d, giving 3d10

Given
[Ni(CO)4], tetrahedral and diamagnetic.
Asked
Reason for diamagnetism.
Concept
In a neutral homoleptic carbonyl the metal is in the zero oxidation state. Free Ni(0) is 3d84s2, and the very strong field CO drives the 4s electrons into the 3d subshell, completely filling it.
Formula
μ = √n(n+2); n = 0 ⇒ diamagnetic
Baby steps
  1. CO is neutral and the complex is neutral ⇒ Ni is in the 0 oxidation state.
  2. Ni(0) = 3d84s2.
  3. The strong CO field shifts the 4s pair into 3d → 3d104s0, all paired.
  4. Empty 4s and three 4p hybridise → sp3, tetrahedral, diamagnetic.
ShortcutNever assume the metal is +2 just because it usually is. In a neutral carbonyl the oxidation state is always zero, which changes the d count entirely.
[ 30 ]L NEET stylePrediction

Which of the following statements about the prediction of geometry using magnetic data is correct?

  • (a)Magnetic moment gives the number of unpaired electrons, from which the hybridisation and hence the geometry can be deduced
  • (b)Magnetic moment directly gives the coordination number
  • (c)Diamagnetic complexes are always tetrahedral
  • (d)Magnetic moment depends on the number of paired electrons

Answer — (a) Magnetic moment gives the number of unpaired electrons, from which the hybridisation and hence the geometry can be deduced

Given
The role of magnetic measurements in valence bond theory.
Asked
The correct statement.
Concept
Magnetic susceptibility experiments give the magnetic moment, which yields the number of unpaired electrons. That number reveals whether inner d orbitals were vacated, and hence which hybridisation and geometry apply.
Formula
μ → n → orbital availability → hybridisation → geometry
Baby steps
  1. Measure μ and invert μ = √n(n+2) to obtain n.
  2. Compare n with the free-ion value to see whether pairing occurred.
  3. Pairing frees inner d orbitals → dsp2 or d2sp3; no pairing → sp3 or sp3d2.
  4. The hybridisation then fixes the geometry → option A.
ShortcutOnly unpaired electrons contribute to μ, since paired spins cancel. That is why the formula is called the spin-only magnetic moment.