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The spin-only magnetic moments of [Mn(CN...

The spin-only magnetic moments of `[Mn(CN)_(6)]^(4-)` and `[MnBr_(4)]^(2-)` in Bohr Magnetons, respectively, are

A

5.92 and 5.92

B

4.89 and 1.73

C

1.73 and 5.92

D

1.73 and 1.73

Text Solution

Verified by Experts

The correct Answer is:
C
`[Mn^(+2)(CN)_(6)]^(-4)`
`Mn^(+2) rarr 3d^(5)4s^(0) 4p`
`CN^(-)` is strong ligands so creates back paring effect of `(n-1) d` orbitals configuration

So, unpaired `e^(-)=1`
`mu=sqrt(n(n+2))B.M`
`mu=1.73 B.M`
And in `[MnBr_(4)]^(-2)`
Br is a weak ligands so no back pairing effect on `(n-1)` d orbital so, unpaired `e^(-)` is `= 5`
`mu=sqrt(5(5+2))=sqrt(35)=5.92 B.M`
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