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. $K_3\left[Cu\right(CN{)}_4]$

Answer 1

Answer: (B)

$VOS{O}_4$

(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 $\left[Ar\right]3d^{0}4s^0$. 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 $\left[Ar\right]3d^{1}4s^0$, the salt is paramagnetic.
(c) $(N{H}_4{)}_2\cdot [TiC{l}_6]\cdot 2H_{2}O$ (ammonium chlorotitantate) is a yellow coloured crystal and $Ti$ has $+4$ oxidation state. As electronic configuration of $T{i}^{4+}$ is $\left[Ar\right]3d^{0}4s^0$, the salt is diamagnetic.
(d) $K_3\left[Cu\right(CN{)}_4]$ (Potassium tetracyanocuprate(I)) is a white soluble complex, In it, $Cu$ has $+1$ oxidation state and $\left[Ar\right]3d^0$ configuration. Hence, it is paramagnetic.