Question

Simplified absorption spectra of three complexes ((i), (ii) and (iii)) of \(\mathrm{Mn^{+}}\) ion are provided below; their \(\lambda_{max}\) values are marked as A, B and C respectively. The correct match between the complexes and their \(\lambda_{max}\) values is

(i) \(\mathrm{[M(NCS)_6]^{(-6+n)}}\)
(ii) \(\mathrm{[MF_6]^{(-6+n)}}\)
(iii) \(\mathrm{[M(NH_3)_6]^{n+}}\)

• A. A-(i), B-(ii), C-(iii)
• B. A-(ii), B-(iii), C-(i)
• C. A-(ii), B-(i), C-(iii)
• D. A-(iii), B-(i), C-(ii)

Hint:

Use the spectrochemical series: stronger field ligand gives larger splitting, higher energy absorbed and shorter \(\lambda_{max}\). On the wavelength axis A is shortest and C is longest.

Answer 1

The Correct Option is D

Spectrochemical series: \(\mathrm{F^-} \lt \mathrm{H_2O} \lt \mathrm{NCS^-} \lt \mathrm{NH_3}\), so the crystal field splitting \(\Delta_o\) order is \(\mathrm{MF_6} \lt \mathrm{M(NCS)_6} \lt \mathrm{M(NH_3)_6}\), i.e. (ii) \(\lt\) (i) \(\lt\) (iii).
\(\lambda_{max} \propto 1/\Delta_o\) (since \(\Delta_o = hc/\lambda\)), so the order of \(\lambda_{max}\) is (ii) \(\gt\) (i) \(\gt\) (iii).
In the spectrum the wavelength increases left to right, so A (leftmost, shortest \(\lambda\)) is (iii), B is (i) and C (longest \(\lambda\)) is (ii).
Answer: (4) A-(iii), B-(i), C-(ii)