BITSAT 2016 Paper

(a) $\mathrm{V}=3d^{3}4s^{2};\mathrm{V}^{2+}=3d^{3}=3$ unpaired electrons$\mathrm{Cr}=3d^{5}4s^{1};\mathrm{Cr}^{2+}=3d^{4}=4$ unpaired electrons$\mathrm{Mn}=3d^{5}4s^{2}$ ;$\mathrm{Mn}^{2+}=3d^{5}=5$ unpaired electrons$\mathrm{Fe}=3d^{6}4s^{2};\mathrm{Fe}^{2+}=3d^{6}=4$ unpaired electronsHence the correct order of paramagnetic behaviour$\mathrm{V}^{2+} < \mathrm{Cr}^{2+}=\mathrm{Fe}^{2+} < \mathrm{Mn}^{2+}$(b) For the same oxidation state, the ionic radii generally decreases as the atomic number increases in a particular transition series. hence the order is$\mathrm{Mn}^{++}>\mathrm{Fe}^{++}>\mathrm{Co}^{++}>\mathrm{Ni}^{++}$(c) In solution, the stability of the compound depends upon electrode potentials, SEP of the transitions metal ions are given as $\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+}=-0.41,\mathrm{Sc}^{3+}$is highly stable as it does not show $+2$ O. S.(d) $\mathrm{Sc}-(+2),(+3)$$\mathrm{Ti}-(+2),(+3)(+4)$i.e. $\mathrm{Sc} < \mathrm{Ti} < \mathrm{Cr}=\mathrm{Mn}$