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The magnetic moment of K(3)[Fe(CN)(6)] i...

The magnetic moment of `K_(3)[Fe(CN)_(6)]` is found to be 1.7 B.M. how many unpaired electron (s) is/are present per molecule

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
A
`K_(3)[Fe(CN)_(6)]`
`Fe_(26)=4s^(2)3d^(6)`
`Fe^(3+) =3d^(5)4s^(0)`
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