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Question

Chemistry Question on coordination compounds

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

A

MnCl2,CuSO4MnCl _{2}, CuSO _{4}

B

CuCl2,TiCl3CuCl _{2}, TiCl _{3}

C

TiO2,CuSO4TiO _{2}, CuSO _{4}

D

TiCl3,NiCl2TiCl _{3}, NiCl _{2}

Answer

CuCl2,TiCl3CuCl _{2}, TiCl _{3}

Explanation

Solution

Species having the same number of unpaired electrons, have same magnetic moment. (a) MnCl2Mn2+=[Ar]3d5,4s0MnCl _{2} \Rightarrow Mn ^{2+}=[ Ar ] 3 d^{5}, 4 s^{0} (five unpaired electrons) CuSO4Cu2+=[Ar]3d19,4s0CuSO _{4} \Rightarrow Cu ^{2+}=[ Ar ] 3 d^{\frac{1}{9}}, 4 s^{0} (one unpaired electron) (b) CuCl2Cu2+=[Ar]3d9,4s0CuCl _{2} \Rightarrow Cu ^{2+}=[ Ar ] 3 d^{9}, 4 s^{0} (one unpaired electron) TiCl3Ti3+=[Ar]3d1,4sTiCl _{3} \Rightarrow Ti ^{3+}=[ Ar ] 3 d^{1}, 4 s (one unpaired electron) CuCl2\therefore CuCl _{2} and TiCl3TiCl _{3} have same magnetic moment. μ=1(1+2)=3=1.73BM\mu=\sqrt{1(1+2)}=\sqrt{3}=1.73 BM (c) TiO2Ti4+=[Ar]3d0,4s0TiO _{2} \Rightarrow Ti ^{4+}=[ Ar ] 3 d^{0}, 4 s^{0} (no unpaired electron) CuSO4Cu2+=[Ar]3d9,4s0CuSO _{4} \Rightarrow Cu ^{2+}=[ Ar ] 3 d^{9}, 4 s^{0} (one unpaired electron) (d) TiCl3Ti3+=[Ar]3d1,4s0TiCl _{3} \Rightarrow Ti ^{3+}=[ Ar ] 3 d^{1}, 4 s^{0} (one unpaired electron) NiCl2Ni2+=[Ar]3d8,4s0NiCl _{2} \Rightarrow Ni ^{2+}=[ Ar ] 3 d^{8}, 4 s^{0} (two unpaired electrons)