The Valence Bond Theory (VBT) explains the formation, magnetic behaviour and geometry of coordination compounds. The Crystal Field Theory (CFT) of coordination compounds is based on the effect of different crystal fields (provided by the ligands taken as point charges), on the degeneracy of d-orbital energies of the central metal atom/ion. The splitting of the d-orbitals provides different electronic arrangements in strong and weak crystal fields. Answer the following questions : (a) In octahedral crystal field, energies of which d-orbitals will be raised when ligands approach the central metal atom/ion ? Give reason in support of your answer. (b) Using crystal field theory, write the electronic configuration of central metal atom/ion of the following : (i) [CoF_6]³- (ii) [Co(NH_3)_6]³+ [At. No. : Co = 27] (c) [NiCl_4]²- is paramagnetic while [Ni(CO)_4] is diamagnetic though both are tetrahedral. Why ? [Atomic No. : Ni = 28] (c) Write hybridization and magnetic behaviour of the complex [Fe(CN)_6]³-. [Atomic No. : Fe = 26]

ChemistryCBSE Class 12 · Board exam 2026, set 56/1/1 · 4 marks · Case-based

CBSE Class 12 Chemistry 2026: question 30

Set in the CBSE Class 12 board exam of 2026, paper 56/1/1, section D, question 30, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q30 · 4 marks · Case-basedCompetency-basedCoordination Compounds

The Valence Bond Theory (VBT) explains the formation, magnetic behaviour and geometry of coordination compounds. The Crystal Field Theory (CFT) of coordination compounds is based on the effect of different crystal fields (provided by the ligands taken as point charges), on the degeneracy of d-orbital energies of the central metal atom/ion. The splitting of the d-orbitals provides different electronic arrangements in strong and weak crystal fields.
Answer the following questions :

(a)
In octahedral crystal field, energies of which d-orbitals will be raised when ligands approach the central metal atom/ion ? Give reason in support of your answer.
2 marks
(b)
Using crystal field theory, write the electronic configuration of central metal atom/ion of the following :
(i) $[Co F_{6}]^{3 -}$
(ii) $[Co (N H_{3})_{6}]^{3 +}$
[At. No. : Co = 27]
1 mark
(c)
$[NiC l_{4}]^{2 -}$ is paramagnetic while $[Ni (CO)_{4}]$ is diamagnetic though both are tetrahedral. Why ? [Atomic No. : Ni = 28]
1 mark
OR
(c)
Write hybridization and magnetic behaviour of the complex $[Fe (CN)_{6}]^{3 -}$ . [Atomic No. : Fe = 26]
1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 56/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Which d-orbitals have lobes lying along the axes?

Hint 2

Find the oxidation state of Co (+3, d⁶), then decide whether each ligand is weak or strong field.

Hint 3

Strong-field ligands pair up electrons; count the unpaired electrons left.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(a) The energies of the $e_{g}$ orbitals, ** $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ **, are raised.
1 mark
Reason: in an octahedral field the six ligands approach along the x, y and z axes. The lobes of $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ point along these axes, so they are repelled more strongly than $d_{x y}$ , $d_{y z}$ and $d_{x z}$ , whose lobes lie between the axes.
1 mark
(b)(i) In $[Co F_{6}]^{3 -}$ , $C o^{3 +}$ is $3 d^{6}$ . $F^{-}$ is a weak-field ligand ( $Δ_{o} < P$ ), giving the high-spin configuration ** $t_{2 g}^{4} e_{g}^{2}$ **.
½ mark
(b)(ii) In $[Co (N H_{3})_{6}]^{3 +}$ , $C o^{3 +}$ is $3 d^{6}$ . $N H_{3}$ is a strong-field ligand ( $Δ_{o} > P$ ), giving the low-spin configuration ** $t_{2 g}^{6} e_{g}^{0}$ **.
½ mark
(c) [first alternative] In $[NiC l_{4}]^{2 -}$ , $N i^{2 +}$ is $3 d^{8}$ . $C l^{-}$ is a weak-field ligand and does not pair the electrons, so two unpaired electrons remain ( $s p^{3}$ ), making it paramagnetic. In $[Ni (CO)_{4}]$ , Ni(0) is $3 d^{8} 4 s^{2}$ . CO is a strong-field ligand and pushes the 4s electrons into 3d to give $3 d^{10}$ with all electrons paired ( $s p^{3}$ ), making it diamagnetic.
½ mark
(c) [second alternative] In $[Fe (CN)_{6}]^{3 -}$ , $F e^{3 +}$ is $3 d^{5}$ . Strong-field $C N^{-}$ pairs the electrons into three 3d orbitals, leaving two 3d orbitals empty. Hybridisation is ** $d^{2} s p^{3}$ (inner-orbital octahedral) and, with one unpaired electron, the complex is paramagnetic**.
½ mark

Answer: (a) $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ , whose lobes lie along the axes of ligand approach; (b)(i) $t_{2 g}^{4} e_{g}^{2}$ , (ii) $t_{2 g}^{6} e_{g}^{0}$ ; (c) Cl⁻ weak field leaves 2 unpaired electrons (paramagnetic), CO strong field pairs them all (diamagnetic) / [Fe(CN)₆]³⁻: d²sp³, paramagnetic.

Where the marks are

Identifies d_{x²−y²} and d_{z²} (1 mark)
Reason: ligands approach along the axes, greater repulsion (1 mark)
t_{2g}⁴e_g² for [CoF₆]³⁻ (½ mark) and t_{2g}⁶e_g⁰ for [Co(NH₃)₆]³⁺ (½ mark)
Weak-field Cl⁻ vs strong-field CO explanation (1 mark) / d²sp³ (½ mark) and paramagnetic (½ mark)

Revise the chapter

Chapter: Coordination Compounds (notes, formulas, where marks are lost and every past question)
Topic: Crystal Field Theory

Practise this chapterBoard-style questions from Coordination Compounds, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 56/1/1: every question, by section.

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CBSE Class 12 Chemistry 2026: question 30

Hints

Step-by-step solution

Revise the chapter

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