CBSE Class 12 Chemistry 2025: question 32(A)

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1. (a) In tetrahedral complexes the crystal field splitting is small (${\Delta }_t\approx \frac{4}{9}{\Delta }_o$). It is never large enough to exceed the pairing energy and force electrons to pair, so low spin tetrahedral complexes are not formed.

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2. (b) $\mathrm{C}{\mathrm{o}}^{2+}$ is $3d^7$. With a strong field ligand, ${\Delta }_o$ is large and one electron goes into a high-energy $e_g$ orbital, which it easily loses. This gives $\mathrm{C}{\mathrm{o}}^{3+}$ ($3d^6$, $t_{2g}^6{e}_g^0$), which has a very high crystal field stabilisation energy and is more stable. So $\mathrm{C}{\mathrm{o}}^{2+}$ is readily oxidised.

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3. (c) $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6][\mathrm{Cr}(\mathrm{CN}{)}_6]$ shows coordination isomerism, because the ligands can be interchanged between the cationic and anionic complex entities, e.g. $[\mathrm{Cr}(\mathrm{N}{\mathrm{H}}_3{)}_6][\mathrm{Co}(\mathrm{CN}{)}_6]$.

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4. (d) Ni²⁺ is $3d^8$. ${\mathrm{H}}_2\mathrm{O}$ is a weak field ligand, so $[\mathrm{Ni}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$ has 2 unpaired electrons, and d–d transitions absorb red light, so the solution looks green. $\mathrm{C}{\mathrm{N}}^{-}$ is a strong field ligand that pairs all the electrons (dsp², square planar). $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ has no unpaired electrons and no visible d–d absorption, so it is colourless.

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5. (e) $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{C}{\mathrm{O}}_3)]\mathrm{Cl}$: pentaamminecarbonatocobalt(III) chloride

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Answer: (a) Δt < pairing energy; (b) strong field gives stable d⁶ Co³⁺; (c) coordination isomerism; (d) unpaired electrons in [Ni(H₂O)₆]²⁺, none in [Ni(CN)₄]²⁻; (e) Pentaamminecarbonatocobalt(III) chloride

Revise the chapter

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