ChemistryCBSE Class 12 · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 12 Chemistry sample paper 2026-27: question 32(A)

From the official CBSE sample question paper for 2026-27, section E, question 32(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(A) · 5 marks · Long answerCoordination Compounds

Answer the following questions: (5x1)
(I) Two test tubes, X and Y, contain solutions of the complexes $[C r (H_{2} O)_{3} C l_{3}]$ and $[C r (e n)_{3}]^{3 +}$ respectively. If plane-polarized light is passed through these solutions, which solution will rotate the plane of polarization and why?
(II) $[N i (C O)_{4}]$ and $[N i (H_{2} O)_{6}]^{2 +}$ are two complexes of Ni. Which complex will have higher crystal field splitting energy, and which factor determines this difference?
(III) A solution of the coordination compound $P d C l_{2} .4 N H_{3}$ is treated with $A g N O_{3}$ , and 2 mole of AgCl precipitate form. Determine the primary valency and secondary valency of Pd in the compound.
(IV) The complex $[F e (C N)_{6}]^{4 -}$ exhibits $d^{2} s p^{3}$ hybridisation. Based on this, comment on the strength of the ligand $C N^{-}$ and its effect on the geometry of the complex.
(V) Why is anhydrous $C u S O_{4}$ white, but $C u S O_{4} .5 H_{2} O$ blue in colour?

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

Optical activity needs a molecule with no plane of symmetry. Which complex has chelating ligands?

Hint 2

Use the spectrochemical series for (II) and (IV).

Hint 3

For (III), the number of AgCl moles tells you how many Cl⁻ ions lie outside the coordination sphere.

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.

(I) Solution Y, $[C r (e n)_{3}]^{3 +}$ , rotates the plane of polarised light. Its three bidentate en ligands form a chiral octahedral arrangement with no plane of symmetry, so it exists as non-superimposable d and l forms. The fac and mer isomers of $[C r (H_{2} O)_{3} C l_{3}]$ each have a plane of symmetry, so they are optically inactive.
1 mark
(II) $[N i (C O)_{4}]$ has the higher crystal field splitting.
½ mark
The determining factor is the nature (field strength) of the ligand. CO is a strong field ligand (it also π-accepts), while $H_{2} O$ is a weak field ligand in the spectrochemical series.
½ mark
(III) 2 mol AgCl means 2 ionisable $C l^{-}$ , so the compound is $[P d (N H_{3})_{4}] C l_{2}$ . Primary valency = oxidation state of Pd = 2.
½ mark
Secondary valency = coordination number = 4 (four $N H_{3}$ ligands).
½ mark
(IV) $d^{2} s p^{3}$ (inner-orbital) hybridisation shows that $C N^{-}$ is a strong field ligand. It forces the six 3d electrons of $F e^{2 +}$ to pair up in $t_{2 g}$ ( $t_{2 g}^{6} e_{g}^{0}$ ), leaving two vacant 3d orbitals.
½ mark
These two 3d orbitals, together with 4s and 4p orbitals, give six $d^{2} s p^{3}$ hybrids. The complex is therefore octahedral, low-spin and diamagnetic.
½ mark
(V) In $C u S O_{4} \cdot 5 H_{2} O$ , water molecules coordinate to $C u^{2 +}$ ( $[C u (H_{2} O)_{4}]^{2 +}$ ) and split the d orbitals. A $d$ – $d$ transition of the $d^{9}$ ion absorbs red/orange light, so the compound appears blue. In anhydrous $C u S O_{4}$ there are no water ligands to cause this splitting, so no d–d absorption occurs in the visible region and it appears white.
1 mark

Answer: (I) Y – [Cr(en)₃]³⁺ (chiral); (II) [Ni(CO)₄], decided by ligand field strength; (III) primary valency 2, secondary valency 4; (IV) CN⁻ is a strong field ligand → low-spin octahedral; (V) H₂O ligands cause d-orbital splitting and d–d transitions (blue); none without water (white).

Where the marks are

(I) Y with chirality/no plane of symmetry argument (1 mark)
(II) [Ni(CO)₄] (½ mark) and nature/strength of ligand (½ mark)
(III) Primary valency 2 (½ mark), secondary valency 4 (½ mark)
(IV) CN⁻ strong field causing pairing (½ mark), octahedral low-spin geometry (½ mark)
(V) Water as ligand causes splitting/d–d transition; absent in the anhydrous salt (1 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 paperSample paper 2026-27: every question, by section.

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CBSE Class 12 Chemistry sample paper 2026-27: question 32(A)

Hints

Step-by-step solution

Revise the chapter

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