Solveeit Logo
Login

Question

Chemistry Question on Quantum Mechanical Model of Atom

The correct order of magnetic moments (spin only values in B.M.) is:

A

[Fe(CN)6]4−>[MnCl4]2−>[CoCl4]2−[Fe(CN)_6 ]^{4-} > [MnCl_4 ]^{2-} > [CoCl_4]^{2-}

B

[MnCl4]2−>[Fe(CN)6]4−>[CoCl4]2−[MnCl_4 ]^{2-} > [Fe(CN)_6 ]^{4-} > [CoCl_4]^{2-}

C

[MnCl4]2−>[CoCl4]2−>[Fe(CN)6]4−[MnCl_4 ]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6 ]^{4-}

D

[Fe(CN)6]4−>[CoCl4]2−>[MnCl4]2−[Fe(CN)_6 ]^{4-}> [CoCl_4]^{2-} > [MnCl_4 ]^{2-}

Answer

[MnCl4]2−>[CoCl4]2−>[Fe(CN)6]4−[MnCl_4 ]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6 ]^{4-}

Explanation

Solution

Magnetic moment of [MnCl4]2−\left[ MnCl _{4}\right]^{2-} is 5.9 BM5.9\, BM.
MnMn is in +2+2 oxidation state having configuration of [Ar]3d5[ Ar ] 3 d ^{5}.
It has 55 unpaired electrons.
μ=n(n+2)\mu= \sqrt{ n ( n +2)}
=(5(5+2)=\sqrt{(5(5+2)}
=5.9 BM=5.9 \,BM
Magnetic moment of [CoC14]2−\left[ CoC 1_{4}\right]^{2-} is 3.9 BM3.9 \,BM.
CoCo is in +2+2 oxidation state having configuration of [Ar] 3d 7^{7}.
It has 33 unpaired electrons.
μ=n(n+2)\mu= \sqrt{ n ( n +2)}
=(3(3+2)=\sqrt{(3(3+2)}
=3.9 BM=3.9 \,BM
Magnetic moment of [FeCN4]2−\left[ FeCN _{4}\right]^{2-} is 0.0BM0.0 BM.
FeFe is in +2+2 oxidation state having configuration of [Ar]3d6[ Ar ] 3 d ^{6}. In this complex CN−CN ^{-}ion is a strong ligand and all electrons are paired and there are no unpaired electrons.
μ=n(n+2)\mu= \sqrt{ n ( n +2)}
=(0(0+2)=\sqrt{(0(0+2)}
=0 BM=0 \,BM
The decreasing order of the magnetic moments is:
[MnCl4]2−>[CoCl4]2−>Fe(CN)6]4−\left.\left[ MnCl _{4}\right]^{2-}>\left[ CoCl _{4}\right]^{2-}> Fe ( CN )_{6}\right]^{4-}