Question

$C{r}^{2+}$ is more easily oxidised than $N{i}^{2+}$.

If true enter 1, else enter 0.

Answer 1

$C{r}^{2+}$ is more easily oxidised, than $N{i}^{2+}$.

$C{r}^{2+}$ is strongly reducing agent in nature.

The electronic configuration of $C{r}^{2+}\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^4$.

It has a $d^4$ configuration. While acting as a reducing agent, it gets oxidized to $C{r}^{3+}$ (electronic configuration, $d^3$). This $d^3$ configuration can be written as $t_{2g}^3$ configuration, which is a more stable configuration.

$N{i}^{2+}$ is an oxidizing agent.

$N{i}^{2+}\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^8$.

It tends to accept two electrons and get reduced to achieve $d^{10}$ configuration which is a more stable configuration.