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Q.
Among the following, the compound that is both paramagnetic and coloured is
Coordination Compounds
Solution:
In $\left( NH _{4}\right)_{2} TiCl _{6},$ the oxidation state of $Ti$ is $+4$ i.e., $Ti ^{4+}$. The configuration of $Ti ^{4+}$ is $3 d^{0} 4 s^{0},$ i.e., it has no unpaired electrons, hence it is diamagnetic and colourless (because of absence of $d$ -electrons).
In $K _{2} Cr _{2} O _{7},$ oxidation state of $Cr$ is $+6$ i.e. $Cr ^{6+},$ the electronic configuration of $Cr ^{6+}$ is $\left(3 d^{0} 4 s^{0}\right)$ i.e. it has no unpaired electron. Thus, it is diamagnetic and colourless (due to absence of $d$ -electrons).
In $\operatorname{CoSO}_{4},$ the oxidation state of $C o$ is $+2$ i.e. $Co ^{2+}$. Its configuration is $3 d^{7}$ i.e. it has unpaired electrons in $3 d$ -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., $Cu ^{+}$. Its configuration is $3 d^{0} 4 s^{0}$. It has no unpaired electron so, it is diamagnetic.