The magnetic moment of K3[Fe(CN)6] is found to be 1.7 BM. How many unpaired electron (s) is/are present per molecule?
Answers
Answered by
17
Final Answer : 1
Steps:
1) The spin only magnetic moment is given by
√n(n+2) .
Since,
Spin only magnetic moment is given by 1.7BM
=> √n(n+2) = 1.7
=> n = 1 since n is whole number.
=> CN-is strong field Ligand.
=> This is Low spin complex.
Steps:
1) The spin only magnetic moment is given by
√n(n+2) .
Since,
Spin only magnetic moment is given by 1.7BM
=> √n(n+2) = 1.7
=> n = 1 since n is whole number.
=> CN-is strong field Ligand.
=> This is Low spin complex.
Answered by
3
the right answer is 1....
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