The correct order of the complexes $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}$ (A), $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ (B), $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}$ (C) and $\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}$ (D) in terms wavelength of light absorbed is:

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Coordination Compounds

The correct order of the complexes

[Co (NH_{3})_{5} (H_{2} O)]^{3 +}

(A),

[Co (NH_{3})_{6}]^{3 +}

(B),

[Co (CN)_{6}]^{3 -}

[CoCl (NH_{3})_{5}]^{2 +}

(D) in terms wavelength of light absorbed is:

(A)D $>$ A $>$ B $>$ C✓ Correct
(B)C $>$ B $>$ D $>$ A
(C)D $>$ C $>$ B $>$ A
(D)C $>$ B $>$ A $>$ D

Answer: (A)

Solution

Absorbed energy varies inversely with wavelength, so the complex with the largest crystal-field splitting absorbs the shortest wavelength. With cobalt(III) in every case, the splitting follows the ligand-field order

CN^{-} > NH_{3} > H_{2} O > Cl^{-}

[Co(CN)_{6}]^{3 -}

splits most, then

[Co(NH_{3})_{6}]^{3 +}

, then

[Co(NH_{3})_{5} (H_{2} O)]^{3 +}

, and

[CoCl(NH_{3})_{5}]^{2 +}

least. The wavelength absorbed therefore rises in the reverse order:

[CoCl(NH_{3})_{5}]^{2 +} > [Co(NH_{3})_{5} (H_{2} O)]^{3 +} > [Co(NH_{3})_{6}]^{3 +} > [Co(CN)_{6}]^{3 -}

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Coordination Compounds

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