Question 341 markSection D

(c) [NiCl4]2−[NiCl_4]^{2-} is paramagnetic while [Ni(CO)4][Ni(CO)_4] is diamagnetic though both are tetrahedral. Why? [Atomic No.: Ni=28Ni = 28]

OR

(c) Write hybridization and magnetic behaviour of the complex [Fe(CN)6]3−[Fe(CN)_6]^{3-}. [Atomic No.: Fe=26Fe = 26]

Coordination Compounds
VBT: magnetic behaviour and hybridisation of tetrahedral/octahedral complexes

Official answer

Primary answer — [NiCl4]2−[NiCl_4]^{2-} vs [Ni(CO)4][Ni(CO)_4]


The difference is the oxidation state of Ni and the field strength of the ligand.


  • [NiCl4]2−[NiCl_4]^{2-}: Ni is +2 → 3d83d^8. Cl−Cl^- is a weak-field ligand and does not pair the d-electrons. Ni2+Ni^{2+} uses sp3sp^3 hybridisation, leaving 2 unpaired electrons → paramagnetic (tetrahedral).
  • [Ni(CO)4][Ni(CO)_4]: Ni is in the 0 oxidation state → 3d84s23d^8 4s^2. CO is a strong-field ligand; it forces the 4s electrons into the 3d set giving 3d103d^{10} (all paired). Ni(0) then uses sp3sp^3 hybridisation with 0 unpaired electrons → diamagnetic (tetrahedral).

So both are tetrahedral, but weak-field Cl−Cl^- leaves unpaired electrons while strong-field CO removes them.


OR answer — [Fe(CN)6]3−[Fe(CN)_6]^{3-}


  • Fe is +3 → 3d53d^5. CN−CN^- is a strong-field ligand → pairs the electrons (t2g5eg0t_{2g}^5 e_g^0).
  • Hybridisation: d2sp3d^2sp^3 (inner-orbital, octahedral).
  • 1 unpaired electron → paramagnetic (low-spin).
Answer keywords:NiCl4 2- paramagneticNi(CO)4 diamagneticsp3 hybridisationweak field CO strong fieldoxidation state Nid²sp³ inner orbitalFe(CN)6 3-magnetic behaviour

Marking scheme

  1. 1.Primary: ½ mark — [NiCl4]2−[NiCl_4]^{2-} is Ni2+Ni^{2+} (d8d^8) with weak-field Cl−Cl^-, sp3sp^3, 2 unpaired → paramagnetic; ½ mark — [Ni(CO)4][Ni(CO)_4] is Ni(0) with strong-field CO giving d10d^{10} (all paired), sp3sp^3 → diamagnetic.
  2. 2.OR: ½ mark — [Fe(CN)6]3−[Fe(CN)_6]^{3-} has d2sp3d^2sp^3 hybridisation (strong-field CN−CN^-, inner orbital); ½ mark — paramagnetic with 1 unpaired electron.

Hint

Compare the oxidation state of Ni and the field strength of the ligand — Cl−Cl^- is weak, CO is strong.

Quick oral answer

In [NiCl4]2−[NiCl_4]^{2-} nickel is +2 (d8d^8) and weak-field Cl−Cl^- leaves two unpaired electrons, so it is sp3sp^3 and paramagnetic; in [Ni(CO)4][Ni(CO)_4] nickel is zero-valent and strong-field CO pairs everything to d10d^{10}, so it is sp3sp^3 and diamagnetic — both tetrahedral.

Common mistakes

  1. 1.Assuming Ni has the same oxidation state in both complexes — it is +2 in [NiCl4]2−[NiCl_4]^{2-} but 0 in [Ni(CO)4][Ni(CO)_4].
  2. 2.Writing sp3d2sp^3d^2 for [Fe(CN)6]3−[Fe(CN)_6]^{3-}; the strong-field CN−CN^- gives inner-orbital d2sp3d^2sp^3 hybridisation.
  3. 3.Concluding that geometry decides magnetism — both Ni complexes are tetrahedral yet differ in magnetic behaviour because of ligand field strength.

Analysis and explanation

Concept


Valence Bond Theory explains geometry and magnetism through hybridisation and electron pairing driven by the ligand.


  • In [NiCl4]2−[NiCl_4]^{2-}, Ni2+Ni^{2+} (d8d^8) with the weak-field Cl−Cl^- keeps two unpaired electrons; sp3sp^3 hybridisation gives a tetrahedral, paramagnetic complex.
  • In [Ni(CO)4][Ni(CO)_4], zero-valent Ni has a 3d84s23d^8 4s^2 arrangement; the strong π-acceptor CO promotes pairing so the 4s electrons collapse into 3d103d^{10}, leaving no unpaired electrons — sp3sp^3, tetrahedral, diamagnetic.

Exam trap


The examiner is testing whether you notice the different oxidation states (Ni2+Ni^{2+} vs Ni0Ni^0) AND the ligand strengths (Cl−Cl^- weak, CO strong). Quoting only one factor loses the mark. For the OR part, remember CN−CN^- is strong-field, so [Fe(CN)6]3−[Fe(CN)_6]^{3-} is inner-orbital d2sp3d^2sp^3, not sp3d2sp^3d^2.


Real-world


[Ni(CO)4][Ni(CO)_4] is central to the Mond process for purifying nickel; its volatility and easy thermal decomposition let nickel be separated from impurities.

Interesting facts

[Ni(CO)4][Ni(CO)_4] is a colourless, highly toxic, volatile liquid (b.p. ~43 °C) used industrially in the Mond process to refine nickel.

The strong π-back-bonding from filled metal d-orbitals into CO's empty π∗\pi^* orbitals is what makes CO such a strong-field ligand.

Question details

Q34 of 37 questions

Board
CBSE
Class
Class 12
Subject
Chemistry
Year
2026
Section
Section D
Marks
1
Chapter
Coordination Compounds
Topic
VBT: magnetic behaviour and hybridisation of tetrahedral/octahedral complexes
Bloom's level
analyze
Syllabus topic
Unit: Coordination Compounds — Valence Bond Theory: Hybridisation and Magnetic Behaviour
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why can two tetrahedral complexes differ in magnetic behaviour?

Geometry alone does not fix magnetism. [NiCl4]2−[NiCl_4]^{2-} (Ni2+Ni^{2+}, weak-field Cl−Cl^-) keeps 2 unpaired electrons and is paramagnetic, whereas [Ni(CO)4][Ni(CO)_4] (Ni0Ni^0, strong-field CO) pairs all electrons to d10d^{10} and is diamagnetic — the ligand field strength and oxidation state decide.

What is the hybridisation of [Fe(CN)6]3−[Fe(CN)_6]^{3-}?

Fe3+Fe^{3+} is d5d^5; strong-field CN−CN^- pairs the electrons to t2g5t_{2g}^5, giving inner-orbital d2sp3d^2sp^3 hybridisation, octahedral geometry, and 1 unpaired electron (paramagnetic).