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

Options

A. `K_(2)Cr_(2)O_(7)`
B. `(NH_(4))_(2)(TiCl_(6))`
C. `CoSO_(4)`
D. `K_(3)[Cu(CN)_(4)]`

Answer

C

Explanation

In `(NH_(4))_(2)(TiCl_(6))`, the O.S. of Ti is +4, i.e., `Ti^(4+)`. The configuration of `Ti^(4+)`, is `3d^(0), 4s^(0)`, i.e., is has no unpaired electron, hence it is diamagnetic and colourless (because of absence of d-electrons).
In `K_(2)Cr_(2)O_(7)`, O.S. of Cr is +6 i.e. `Cr^(6+)`, the electronic configuration of `Cr^(6+)` is `(3d^(0) 4s^(0))` i.e. it has no unpaired electron. Thus it is diamagnetic and colourless (absence of d-electrons)
In `CoSO_(4)`, the O.S of Co is +2 i.e., `Co^(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) [Cu(CN)_(4)]`, the O.S. of Cu is +1, i.e., `Cu^(2+)`. Its configuration is `3d^(0) 4s^(0)`. It has no unpaired electron so it is diamagnetic.