Question

Among the given ions in aqueous solution, $Z{n}^{2+},N{i}^{2+}$ and $C{r}^{3+}$ (atomic number $Zn=30,Ni=28,Cr=24$):

• A. Only $Z{n}^{2+}$ is colourless and $N{i}^{2+}$ and $C{r}^{3+}$ are coloured
• B. All three are colourless
• C. All three are coloured
• D. Only $N{i}^{2+}$ is coloured and $Z{n}^{2+}$ and $C{r}^{3+}$ are colourless

Answer 1

The correct option is A Only $Z{n}^{2+}$ is colourless and $N{i}^{2+}$ and $C{r}^{3+}$ are coloured

The electronic configuration of ${}_{30}Z{n}^{2+}$ is $\left[Ar\right]3d^{10}4s^0$. As there are no unpaired electrons in $Z{n}^{2+}$, thus there is no scope for $d-d$ transition in $Z{n}^{2+}$ ion. Hence, it is colourless in aqueous solution.

The electronic configuration of ${}_{28}N{i}^{2+}$ is $\left[Ar\right]3d^{8}4s^0$. As there are $2$ unpaired electrons in $N{i}^{2+}$ ion, $d-d$ transition of electron is possible here. So, $N{i}^{2+}$ is coloured in aqueous solution.

The electronic configuration of ${}_{24}C{r}^{3+}$ is $\left[Ar\right]3d^{3}4s^0$. As there are $3$ unpaired electrons, the $d-d$ transition of electrons is possible and thus $C{r}^{3+}$ shows colour in aqueous solution.