Consider the complexes (I)[MnCl6]3–,(II)[FeF6]3– and (III)[CoF6]3–. Then the correct order of their magnetic moment (spin only) is correctly represented in
A
I=III<II
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B
I=II=III
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C
I=II>III
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D
I=III<II
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Solution
The correct option is AI=III<II [MnCl6]2–n=4 [FeF6]3–n=5 [CoF6]3–n=4 So, I=III<II