Question

Identify the correct trend given below:
(Atomic No. = Ti : 22, Cr : 24 and Mo : 42)

• A. $\Delta \, of \, \left(\left[C r \left(H_{2} O\right)_{6}\right]\right)^{2 +}>\left(\left[M o \left(H_{2} O\right)_{6}\right]\right)^{2 +}$ and $\Delta \, of \, \left(\left[T i \left(H_{2} O\right)_{6}\right]\right)^{3 +}>\left(\left[T i \left(H_{2} O\right)_{6}\right]\right)^{2 +}$
• B. $\Delta \, of \, \left(\left[C r \left(H_{2} O\right)_{6}\right]\right)^{2 +}>\left(\left[M o \left(H_{2} O\right)_{6}\right]\right)^{2 +}$ and $\Delta \, of \, \left(\left[T i \left(H_{2} O\right)_{6}\right]\right)^{3 +} < \left(\left[T i \left(H_{2} O\right)_{6}\right]\right)^{2 +}$
• C. $\Delta \, of \, \left(\left[C r \left(H_{2} O\right)_{6}\right]\right)^{2 +} < \left(\left[M o \left(H_{2} O\right)_{6}\right]\right)^{2 +}$ and $\Delta \, of \, \left(\left[T i \left(H_{2} O\right)_{6}\right]\right)^{3 +}>\left(\left[T i \left(H_{2} O\right)_{6}\right]\right)^{2 +}$
• D. $\Delta \, of \, \left(\left[C r \left(H_{2} O\right)_{6}\right]\right)^{2 +} < \left(\left[M o \left(H_{2} O\right)_{6}\right]\right)^{2 +}$ and $\Delta \, of \, \left(\left[T i \left(H_{2} O\right)_{6}\right]\right)^{3 +} < \left(\left[T i \left(H_{2} O\right)_{6}\right]\right)^{2 +}$

Answer 1

Answer: (C)

As the oxidation state of metal increases the CFSE value increases hence $\Delta of\left(\left[\right. Ti \left(\right. H_{2} O \left.\right)_{6} \left]\right.\right)^{3 +}>\left(\left[\right. Ti \left(\right. H_{2} O \left.\right)_{6} \left]\right.\right)^{2 +}$
$\text{Δ}$ increase from 3d to 4d Series. So, $\text{Δ} \propto $ CFSE (Crystal field stabilisation energy)
$\left(\Delta of \left[Cr \left(H_{2} O\right)_{6}\right]\right)^{2 +} < \Delta of\left(\left[Mo \left(H_{2} O\right)_{6}\right]\right)^{2 +}$
Because here $\text{Δ}$ depends on Z_eff & Z_eff of 4d series is more than 3d series