Among Co ^ 3 + , Ti ^ 2 + , V ^ 2 + and Cr ^ 2 + ions, one if used...

Among the given species ${Co}^{3 +}$ cannot undergo oxidation, hence, ${Co}^{3 +}$ can't liberate ${H}_{2}$.

It has ${d}^{6}$ configuration,

Number of unpaired electrons $= 4$

$\mu = \sqrt{n\left(n + 2\right)}B . M\mu = \sqrt{4 \times 6} = 4 . 92B . M .$