Question

Among the following compounds both coloured and paramagnetic one is

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

Answer 1

Answer: (B)

(a) $K_{2}C{r}_2{O}_7$ (potassium dichromate) forms orange- red triclinic crystals and dichromate ion $\left(C{r}_2{O}_7^{2-}\right)$ in $K_{2}C{r}_2{O}_7$ has

two tetrahedral units shared through an oxygen atom. The central atom $Cr$ is in $+6$ oxidation state which has electronic configuration $[Ar]3d^{0}4s^0.$ As it does not have any unpaired electron, it is diamagnetic.

(b) $NOS{O}_4$ (vanadyl sulphate) is a hygroscopic blue solid. It features $V{O}^{2+}$ ion, which has vanadium in its $+4$ oxidation state. Since, $V^{4+}$ has electronic configuration $[Ar]3d^{1}4s^0$, the salt is paramagnetic.

(c) ${\left(N{H}_4\right)}_2\cdot \left[TiC{l}_6\right]\cdot 2H_{2}O$ (Ammonium

chlorotitantate) is a yellow coloured crystal and Ti has $+4$ oxidation state. As electronic configuration of $T{i}^{4+}$ is $[Ar]3d^{0}4s^0$, the salt is diamagnetic.

(d) $K_3\left[Cu(CN{)}_4\right]$ (Potassium tetracyanocuprate (I)) is a white soluble complex. In it, Cu has $+1$ oxidation state and $[Ar]3d^{10}$ configuration. Hence, it is paramagnetic.