Question

For which of the following pairs, magnetic moment is same?

• A. $MnCl _{2}, CuSO _{4}$
• B. $CuCl _{2}, TiCl _{3}$
• C. $TiO _{2}, CuSO _{4}$
• D. $TiCl _{3}, NiCl _{2}$

Answer 1

The Correct Option is B

Species having the same number of unpaired electrons, have same magnetic moment. (a) $MnCl _{2} \Rightarrow Mn ^{2+}=[ Ar ] 3 d^{5}, 4 s^{0}$ (five unpaired electrons) $CuSO _{4} \Rightarrow Cu ^{2+}=[ Ar ] 3 d^{\frac{1}{9}}, 4 s^{0}$ (one unpaired electron) (b) $CuCl _{2} \Rightarrow Cu ^{2+}=[ Ar ] 3 d^{9}, 4 s^{0}$ (one unpaired electron) $TiCl _{3} \Rightarrow Ti ^{3+}=[ Ar ] 3 d^{1}, 4 s$ (one unpaired electron) $\therefore CuCl _{2}$ and $TiCl _{3}$ have same magnetic moment. $\mu=\sqrt{1(1+2)}=\sqrt{3}=1.73 BM$ (c) $TiO _{2} \Rightarrow Ti ^{4+}=[ Ar ] 3 d^{0}, 4 s^{0}$ (no unpaired electron) $CuSO _{4} \Rightarrow Cu ^{2+}=[ Ar ] 3 d^{9}, 4 s^{0}$ (one unpaired electron) (d) $TiCl _{3} \Rightarrow Ti ^{3+}=[ Ar ] 3 d^{1}, 4 s^{0}$ (one unpaired electron) $NiCl _{2} \Rightarrow Ni ^{2+}=[ Ar ] 3 d^{8}, 4 s^{0}$ (two unpaired electrons)