JEE PYQ: Coordination Compounds Question 16
Question 16 - 2020 (02 Sep Shift 2)
Simplified absorption spectra of three complexes ((i), (ii) and (iii)) of $\text{M}^{n+}$ 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) $[\text{M(NCS)}_6]^{(-6+n)}$ (ii) $[\text{MF}_6]^{(-6+n)}$ (iii) $[\text{M(NH}_3)_6]^{n+}$
(a) A-(iii), B-(i), C-(ii) (b) A-(ii), B-(i), C-(iii) (c) A-(ii), B-(iii), C-(i) (d) A-(i), B-(ii), C-(iii)
Show Answer
Answer: (a)
Solution
Spectrochemical series: NH$3$ > NCS$^-$ > F$^-$. Stronger ligand = greater splitting = smaller $\lambda$. So $\lambda{\text{NH}3} < \lambda{\text{NCS}} < \lambda_{\text{F}^-}$.
BETA
