Question

Among the following, the compound that is both paramagnetic and coloured is

• A. ${\left(N{H}_4\right)}_2\left(TiC{l}_6\right)$
• B. $K_{2}C{r}_2{O}_7$
• C. $Co\left(S{O}_4\right)$
• D. $K_3\left[Cu(CN{)}_4\right]$

Answer 1

Answer: (C)

In ${\left(N{H}_4\right)}_{2}TiC{l}_6,$ the oxidation state of $Ti$ is $+4$ i.e., $T{i}^{4+}$. The configuration of $T{i}^{4+}$ is $3d^{0}4s^0,$ i.e., it has no unpaired electrons, hence it is diamagnetic and colourless (because of absence of $d$ -electrons).

In $K_{2}C{r}_2{O}_7,$ oxidation state of $Cr$ is $+6$ i.e. $C{r}^{6+},$ the electronic configuration of $C{r}^{6+}$ is $\left(3d^{0}4s^0\right)$ i.e. it has no unpaired electron. Thus, it is diamagnetic and colourless (due to absence of $d$ -electrons).

In ${\mathrm{CoSO}}_4,$ the oxidation state of $Co$ is $+2$ i.e. $C{o}^{2+}$. Its configuration is $3d^7$ i.e. it has unpaired electrons in $3d$ -orbitals, so it is paramagnetic. Because of incompletely filled $d$ -orbitals, it is coloured, i.e. $(c)$ is correct answer. In $K_3\left[Cu(CN{)}_4\right],$ the oxidation state of $Cu$ is $+1,$ i.e., $C{u}^{+}$. Its configuration is $3d^{0}4s^0$. It has no unpaired electron so, it is diamagnetic.