a. Using, Valence Bond Theory identify A, B, C, D, E and F in the following table (3) S.No Complex central metal ion configuration of metal ion Hybridization of Metal ion Geometry of the Complex Number Of Unpaired Electron Magnetic Behaviour --- --- --- --- --- --- --- --- i [CoF_4]²- A 3d^7 sp³ tetrahedral B Paramagnetic ii [Cr(H_2O)_2C_2O_4)_2]^- Cr³+ 3d³ C octahedral 3 D iii [Ni(CO)_4] Ni 3d^84s² E F 0 Diamagnetic b. Write the ionic equations for the reaction of acidified K_2Cr_2O_7 with (i) H_2S and (ii) FeSO_4 (2)

ChemistryCBSE Class 12 · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 12 Chemistry sample paper 2024-25: question 31(b)

From the official CBSE sample question paper for 2024-25, section E, question 31(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(b) · 5 marks · Long answerCoordination Compounds

a. Using, Valence Bond Theory identify A, B, C, D, E and F in the following table (3)

S.No	Complex	central metal ion	configuration of metal ion	Hybridization of Metal ion	Geometry of the Complex	Number Of Unpaired Electron	Magnetic Behaviour
i	$[C o F_{4}]^{2 -}$	A	$3 d^{7}$	$s p^{3}$	tetrahedral	B	Paramagnetic
ii	$[C r (H_{2} O)_{2} C_{2} O_{4})_{2}]^{-}$	$C r^{3 +}$	$3 d^{3}$	C	octahedral	3	D
iii	$[N i (C O)_{4}]$	Ni	$3 d^{8} 4 s^{2}$	E	F	0	Diamagnetic

b. Write the ionic equations for the reaction of acidified $K_{2} C r_{2} O_{7}$ with (i) $H_{2} S$ and (ii) $F e S O_{4}$ (2)

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

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

Hint 1

Find each metal's oxidation state from the charge on the complex and its ligands.

Hint 2

For d³ octahedral, count the vacant inner d orbitals. For Ni(CO)₄, remember CO forces pairing.

Hint 3

Dichromate is reduced to Cr³⁺ (gaining 6e⁻); balance with H⁺ and water.

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. In $[C o F_{4}]^{2 -}$ : $x + 4 (- 1) = - 2$ , so $x = + 2$ . **A = $C o^{2 +}$ ** ( $3 d^{7}$ ).
½ mark
F⁻ is a weak-field ligand, so there is no pairing and $3 d^{7}$ has B = 3 unpaired electrons.
½ mark
$C r^{3 +}$ ( $3 d^{3}$ ) leaves two empty inner 3d orbitals for the six ligands, so the hybridisation is **C = $d^{2} s p^{3}$ ** (inner orbital, octahedral).
½ mark
It has 3 unpaired electrons, so it is D = paramagnetic.
½ mark
In $[N i (C O)_{4}]$ , strong-field CO makes the $4 s^{2}$ electrons pair into 3d, giving $3 d^{10}$ . The empty 4s and three 4p orbitals hybridise: **E = $s p^{3}$ **.
½ mark
$s p^{3}$ hybridisation gives a F = tetrahedral geometry.
½ mark
b(i) $C r_{2} O_{7}^{2 -} + 8 H^{+} + 3 H_{2} S \to 2 C r^{3 +} + 3 S ↓ + 7 H_{2} O$
1 mark
b(ii) $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 F e^{2 +} \to 2 C r^{3 +} + 6 F e^{3 +} + 7 H_{2} O$
1 mark

Answer: A = Co²⁺, B = 3, C = d²sp³, D = paramagnetic, E = sp³, F = tetrahedral; (i) Cr₂O₇²⁻ + 8H⁺ + 3H₂S → 2Cr³⁺ + 3S + 7H₂O; (ii) Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

Where the marks are

Each of A–F correct (½ mark each, 3 marks)
Balanced ionic equation with H₂S (1 mark)
Balanced ionic equation with Fe²⁺ (1 mark)

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

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 2024-25: every question, by section.

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CBSE Class 12 Chemistry sample paper 2024-25: question 31(b)

Hints

Step-by-step solution

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