Homoleptic octahedral complexes of a metal ion '$\mathrm{M}^{3+}$' with three monodentate ligands $\mathrm{L}_{1}$, $\mathrm{L}_{2}$ and $\mathrm{L}_{3}$ absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is

(A) $\mathrm{L}_{3}<\mathrm{L}_{1}<\mathrm{L}_{2}$

(B) $\mathrm{L}_{3}<\mathrm{L}_{2}<\mathrm{L}_{1}$

(C) $\mathrm{L}_{1}<\mathrm{L}_{2}<\mathrm{L}_{3}$

(D) $\mathrm{L}_{2}<\mathrm{L}_{1}<\mathrm{L}_{3}$

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Coordination Compounds

Homoleptic octahedral complexes of a metal ion '

M^{3 +}

' with three monodentate ligands

L_{1}

L_{2}

and

L_{3}

absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is

(A) $L_{3} < L_{1} < L_{2}$ ✓ Correct
(B) $L_{3} < L_{2} < L_{1}$
(C) $L_{1} < L_{2} < L_{3}$
(D) $L_{2} < L_{1} < L_{3}$

Answer: (A)

Solution

A larger splitting

Δ

means the complex absorbs shorter-wavelength light, since

Δ \propto \frac{1}{λ}

. Here

L_{2}

absorbs blue (shortest

λ

, largest

Δ

L_{1}

absorbs green (middle), and

L_{3}

absorbs red (longest

λ

, smallest

Δ

). Because stronger ligands give larger

Δ

, the increasing order of ligand strength is

L_{3} < L_{1} < L_{2}

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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