c
\(\text { Element } \quad \quad \;\;\;\quad \quad \quad \mathrm{Ti}<\mathrm{V}<\mathrm{Cr}<\mathrm{Mn}\)

\(\text {No. of oxidation states} \; 3 \quad\quad 4 \quad \quad 5 \quad \quad 6\) Given order is correct.

Magnetic moment \((\mu)=\sqrt{n(n+2)}\)  \(B.M\).

For \(\mathrm{Ti}^{3+} n=1, \mu=\sqrt{1(1+2)}=\sqrt{3} \,\mathrm\,\,{B} . \mathrm{M}\)

For \(V^{3+} n=2, \mu=\sqrt{2(2+2)}=\sqrt{8}\,\, \mathrm{B} . \mathrm{M}\)

For \(\mathrm{Cr}^{3+} n=3, \mu=\sqrt{3(3+2)}=\sqrt{15} \,\,\mathrm{B} \cdot \mathrm{M}\)

For \(\mathrm{Mn}^{3+} n=4, \mu=\sqrt{4(4+2)}=\sqrt{24} \,\,\mathrm{B} \cdot \mathrm{M}\)

Thus magnetic moment:

\(\mathrm{Ti}^{3+} < \mathrm{V}^{3+} < \mathrm{Cr}^{3+} < \mathrm{Mn}^{3+}\)

Melting point order

\(\mathrm{Mn}\quad < \quad \mathrm{Ti}\quad < \quad \mathrm{Cr}\quad <\quad \mathrm{V}\)

\(1245\,^{o} \mathrm{C} \;\;1668\,^{o} \mathrm{C} \;\;\;\;1875\,^{o} \mathrm{C} \;\;\;\;1900\,^{o} \mathrm{C}\)

\(2^{\text {nd }}\) ionisation enthalpy order

\(\quad \quad \quad \quad \quad \quad \mathrm{Ti}\; < \;\mathrm{V} < \;\mathrm{Mn}\; < \;\mathrm{Cr}\)

\((\text { in }\, \mathrm{kJ} / \mathrm{mol}): 1320\;\; 1376\;\;1513 \;\; 1635\)