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The ratio of magnetic moment (spin only ...

The ratio of magnetic moment (spin only value) between `[FeF_(6)]^(3-)` and `[Fe(CN)_(6)]^(3-)` is approximately

A

4

B

2

C

5

D

3

Text Solution

Verified by Experts

The correct Answer is:
D
`mu` = `sqrt n(n+2)` B.M
For `[Fe(CN)_6]^(3-)` , n = 1 , `mu` `sqrt 1(1+2)` = 173 B.M
For `[FeF_6]^(3-)` , n = 5
`mu` = `sqrt 5(5 + 2)` = 5.92 B.M
Ratio = `5.92 / 1.73 = 3.`
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