Question

In aqueous solution, $C{r}^{2+}$ is stronger reducing agent than $F{e}^{2+}$. This is because

• A. $C{r}^{2+}$ ion is more stable than $F{e}^{2+}$
• B. $C{r}^{3+}$ ion with $d^3$ configuration has favourable crystal field stabilization energy
• C. $C{r}^{3+}$ has half-filled configuration and hence more stable
• D. $F{e}^{3+}$ in aqueous solution is more stable than $C{r}^{3+}$
• E. $F{e}^{2+}$ ion with $d^6$ configuration has favourable crystal field stabilization energy

Answer 1

Answer: (B)

$\because C{r}^{3+}$ ion has $d^3$ configuration $\left(t_{2g}^3{e}_g^0\right)$, which makes $C{r}^{3+}$ stable by providing crystal field stabilisation energy. Thus, $C{r}^{2+}$ ion can easily changes to $C{r}^{3+}$ (i.e. get oxidised) and acts as a strong reducing agent.

$C{r}^{2+}⟶C{r}^{3+}+e^{-}$