• # question_answer Consider the hydrates ions of $T{{i}^{2+}},{{V}^{2+}},T{{i}^{3+}}$and $S{{c}^{3+}}.$The correct order of their spin-only magnetic moments is : [JEE Main 10-4-2019 Morning] A) $S{{c}^{3+}}<T{{i}^{3+}}<T{{i}^{2+}}<{{V}^{2+}}$B) $T{{i}^{3+}}<T{{i}^{2+}}<S{{c}^{3+}}<{{V}^{2+}}$C) $S{{c}^{3+}}<T{{i}^{3+}}<{{V}^{2+}}<T{{i}^{2+}}$D) ${{V}^{2+}}<T{{i}^{2+}}<T{{i}^{3+}}<S{{c}^{3+}}$

$T{{i}^{+2}}=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{2}}$ unpaired electrons = 2. spin only magnetic moment$(\mu )=\sqrt{2(2+2)}$                                                 $=\sqrt{8}B.M$ $T{{i}^{+3}}=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}$ unpaired electrons = 1 $\mu =\sqrt{1(1+2)}=\sqrt{3}B.M$ ${{V}^{+2}}=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{3}}$ $\mu =\sqrt{3(3+2)}=\sqrt{15}B.M$ $S{{c}^{+3}}=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$ $\mu =0$