(a) $K _{2} Cr _{2} O _{7}$ (potassium dichromate) forms orange- red triclinic crystals and dichromate ion $\left( Cr _{2} O _{7}^{2-}\right)$ in $K _{2} Cr _{2} O _{7}$ has
two tetrahedral units shared through an oxygen atom. The central atom $Cr$ is in $+6$ oxidation state which has electronic configuration $[ Ar ] 3 d^{0} 4 s^{0} .$ As it does not have any unpaired electron, it is diamagnetic.
(b) $NOSO _{4}$ (vanadyl sulphate) is a hygroscopic blue solid. It features $VO ^{2+}$ ion, which has vanadium in its $+4$ oxidation state. Since, $V ^{4+}$ has electronic configuration $[ Ar ] \,3 d ^{1} \,4 s ^{0}$, the salt is paramagnetic.
(c) $\left( NH _{4}\right)_{2} \cdot\left[ TiCl _{6}\right] \cdot 2 H _{2} O \quad$ (Ammonium
chlorotitantate) is a yellow coloured crystal and Ti has $+4$ oxidation state. As electronic configuration of $Ti ^{4+}$ is $[ Ar ] 3 d ^{0}\, 4 s ^{0}$, 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 $[ Ar ] 3 d ^{10}$ configuration. Hence, it is paramagnetic.