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

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

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

Q32(B) · 5 marks · Long answerCoordination Compounds

Answer the following questions: (5x1)
(I) Three coordination compounds of cobalt(III) are given:
Compound A: $[C o (N H_{3})_{6}] C l_{3}$
Compound B: $[C o (N H_{3})_{5} C l] C l_{2}$
Compound C: $[C o (N H_{3})_{4} C l_{2}] C l$
Which of the three compounds will show maximum electrolytic conductivity and why?
(II) On the basis of crystal field theory, write the electronic configuration for $d^{5}$ ion if $Δ_{0} > P$ .
(III) Using IUPAC norms, write the formulae of Diaquadiammine(oxalato) chromium(III) chloride.
(IV) Compare the optical activity of $[C o (N H_{3})_{2} C l_{2} (e n)] C l$ and $[C o (N H_{3})_{4} C l_{2}] C l$ . Which one is optically active and why?
(V) Explain why $[C u (N H_{3})_{4}]^{2 +}$ shows a blue colour while $[C u (H_{2} O)_{4}]^{2 +}$ shows a greenish colour. Which factor influences the colour difference?

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

For conductivity, count how many ions each compound gives when only the ionisable Cl outside the bracket separates.

Hint 2

When $Δ_{0} > P$ , decide whether electrons prefer to pair in the lower set or move up to $e_{g}$ .

Hint 3

For optical activity, check whether a chelating ligand removes every plane of symmetry from the octahedron.

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) Compound A, $[C o (N H_{3})_{6}] C l_{3}$ , shows the maximum conductivity. In water:
- A: $[C o (N H_{3})_{6}] C l_{3} \to [C o (N H_{3})_{6}]^{3 +} + 3 C l^{-}$ gives 4 ions
- B: $[C o (N H_{3})_{5} C l] C l_{2} \to [C o (N H_{3})_{5} C l]^{2 +} + 2 C l^{-}$ gives 3 ions
- C: $[C o (N H_{3})_{4} C l_{2}] C l \to [C o (N H_{3})_{4} C l_{2}]^{+} + C l^{-}$ gives 2 ions
The more ions there are in solution, the higher the electrolytic conductivity. So A > B > C.
1 mark
(II) When $Δ_{0} > P$ , the ligand is strong field. Pairing in the $t_{2 g}$ set costs less energy than promoting an electron to $e_{g}$ . All five electrons therefore enter $t_{2 g}$ , giving the low-spin configuration $t_{2 g}^{5} e_{g}^{0}$ , with one unpaired electron.
1 mark
(III) Chromium is in the +3 state. The ligands are 2 NH $_{3}$ (neutral), 2 H $_{2}$ O (neutral) and C $_{2}$ O $_{4}^{2 -}$ (−2), so the complex ion is $[C r (N H_{3})_{2} (H_{2} O)_{2} (C_{2} O_{4})]^{+}$ . One Cl $^{-}$ balances the charge.
Formula: $[C r (N H_{3})_{2} (H_{2} O)_{2} (C_{2} O_{4})] C l$
1 mark
(IV) $[C o (N H_{3})_{2} C l_{2} (e n)] C l$ is optically active and $[C o (N H_{3})_{4} C l_{2}] C l$ is not.
- The bidentate en ligand can give an octahedral arrangement with no plane of symmetry. This happens in the isomer with the two Cl cis and the two NH $_{3}$ cis. Its mirror image cannot be superimposed on it, so it exists as a pair of enantiomers.
- $[C o (N H_{3})_{4} C l_{2}]^{+}$ is of the type $M A_{4} B_{2}$ with only monodentate ligands. Both its cis and trans forms have a plane of symmetry, so they are achiral.
1 mark
(V) NH $_{3}$ is a stronger field ligand than H $_{2}$ O.
- It therefore causes a larger crystal field splitting ( $Δ$ ) in $[C u (N H_{3})_{4}]^{2 +}$ .
- The two complexes absorb light of different wavelengths in their d–d transitions.
- The complementary colours seen are different: deep blue for the ammine complex and greenish for the aqua complex.
The factor that decides the colour difference is the nature (field strength) of the ligand, which controls the magnitude of $Δ$ .
1 mark

Answer: (I) A, $[C o (N H_{3})_{6}] C l_{3}$ (4 ions); (II) $t_{2 g}^{5} e_{g}^{0}$ ; (III) $[C r (N H_{3})_{2} (H_{2} O)_{2} (C_{2} O_{4})] C l$ ; (IV) $[C o (N H_{3})_{2} C l_{2} (e n)] C l$ is optically active (chiral, no plane of symmetry); (V) NH $_{3}$ is a stronger ligand, gives larger $Δ$ and a different absorbed wavelength; ligand strength decides the colour.

Where the marks are

Identifies A as most conducting because it gives 4 ions, compared with 3 and 2 ions for B and C (1 mark)
Writes the low-spin configuration $t_{2 g}^{5} e_{g}^{0}$ (1 mark)
Writes the correct formula $[C r (N H_{3})_{2} (H_{2} O)_{2} (C_{2} O_{4})] C l$ (1 mark)
States that the en complex is optically active because it is chiral (no mirror plane), while the $M A_{4} B_{2}$ complex is achiral (1 mark)
Explains that different ligand field strength gives a different $Δ$ , so different wavelengths are absorbed; names ligand strength as the factor (1 mark)

Revise the chapter

Chapter: Coordination Compounds (notes, formulas, where marks are lost and every past question)
Topic: IUPAC Nomenclature of Coordination Compounds

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(B)

Hints

Step-by-step solution

Revise the chapter

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