Question

Which one of the following complexes has largest value of ${\Delta }_0?$

• A. $\left[Cr\right(H_{2}O{)}_6{]}^{+2}$
• B. $\left[Cr\right(H_{2}O{)}_6{]}^{+3}$
• C. $\left[Mn\right(H_{2}O{)}_6{]}^{+2}$
• D. $\left[Mn\right(H_{2}O{)}_6{]}^{+3}$

Answer 1

The correct option is B $\left[Cr\right(H_{2}O{)}_6{]}^{+3}$

Magnitude of CFSE depends upon the following factors.
(1) Nature of central metal cation: the the value of CFSE depends other following factors of central metal cation as given as
(a) For the Complex having same geometry and same ligands but having different numbers of d-electrons then CFSE decrease on increasing number of d-electrons in the central metal cation.
(b) When numbers of d-electrons are same then CFSE increasing on increasing Oxidation number.
(c) For same Ligand, Oxidation state, same d-electrons, CFSE increasing on increasing principle quantum Number of d- orbitals like 3d < 4d < 5d etc.
$C{r}^{2+}$ has more d electron than $C{r}^{3+}$ so it has lower CFSE.
Similarly, $M{n}^{2+}$ has more d electron than $M{n}^{3+}$ so it has lower CFSE.
$C{r}^{3+}\to t_{2g}^3{e}_g^0\to CFSE=|-6/5|{\Delta }_0$

$M{n}^{3+}\to d^3\to t_{2g}^{3}e{g}^1$
$\to CFSE=-6/5{\Delta }_0+3/5{\Delta }_0=|-3/5|{\Delta }_0$
$\therefore {\Delta }_0=C{r}^{3+}>M{n}^{3+}$