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The ratio of magnetic moment (spin only value) between `[FeF_(6)]^(3-)` and `[Fe(CN)_(6)]^(3-)` is approximatelyA. 4B. 2C. 5D. 3 |
Answer» Correct Answer - 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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