Question

Simplified absorption spectra of three complexes ((i), (ii) and (iii)) of $M^{n+}$ 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) $\left[M\right(NCS{)}_6{]}^{(-6+n)}$
(ii) $[M{F}_6{]}^{(-6+n)}$
(iii) $\left[M\right(N{H}_3{)}_6{]}^{n+}$

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

Answer 1

The correct option is B A - (iii) , B - (i), C - (ii)

$△=\frac{hc}{{\lambda }_{absorbedf\left(max\right)}}$
$A\to N{H}_3$ compound (iii)
$B\to NCS$ compound (i)
$C\to F^{-}$ compound (ii)
Using spectrochemical series of ligand,
$F^{-}<NC{S}^{-}<N{H}_3$ order of $△$ positive
crystal field splitting energy.
So, $N{H}_3$ complex $\to A$
$F^{-}$ complex $-C$
$NC{S}^{-}$ complex $\to B$