Question

The heat of hydration of $C{r}^{2+}$ ion is $460Kcal/mole$. For $\left[Cr\right(H_{2}O{)}_6{]}^{2+}{\Delta }_0=13900c{m}^{-1}$. What heat of hydration would be, if there were no crystal field stabilization energy?

• A. $-436Kcal/mole$
• B. $-245Kcal/mole$
• C. $-420Kcal/mole$
• D. $+436Kcal/mole$

Answer 1

The correct option is D $+436Kcal/mole$

$\text{CFSE for}\left[Cr\right(H_{2}O{)}_6{]}^{2+}=-0.6{\Delta }_0$
$\text{Experimental value - Calculated value of CFSE}$ (i.e $-24Kcal/mole$)
$\therefore \text{Heat of hydration}=460-24=436Kcalmol{e}^{-1}$