For electronic spectra of $\mathrm{K}_{2} \mathrm{CrO}_{4}(\mathrm{A})$ and $\mathrm{K}_{2} \mathrm{MoO}_{4}(\mathrm{B})$ the correct combination is
(a)Transition is d-d and $\lambda_{\max }$ has for $\mathrm{A}<\mathrm{B}$
(b)Transition is LMCT and $\lambda_{\max }$ for $\mathrm{A}<\mathrm{B}$
(c)Transition is LMCT and $\lambda_{\max }$ for $\mathrm{A}>\mathrm{B}$
(d)Transition is MLCT and $\lambda_{\max }$ for $\mathrm{A}>\mathrm{B}$
Answer
Answer: C ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Both are d⁰, colour from O→M LMCT. Cr(VI) 3d lies lower than Mo(VI) 4d, so chromate absorbs at longer wavelength (yellow) than molybdate (colourless): λmax(A) > λmax(B).