The calculated spin–only magnetic moments(BM) of the anionic and cationic species of [Fe(H₂O)₆]₂ and [Fe(CN)₆], respectively are

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2.84 and 5.92

0 and 4.9

0 and 5.92

4.9 and 0

answer is A.

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Detailed Solution

Oxidation state of Fe in the (B) is 0 since water is a neutral ligand, having a configuration of t_2g⁴ and e_g², thus accounting for four numbers of unpaired electrons.
Magnetic moment = $\sqrt{n (n + 2)}$ = $\sqrt{4 (4 + 2)}$ = 4.92 BM, where n is the number of unpaired electrons.
Oxidation state of Fe in the (A) is +2 taking to account that CN has a negative donation per itself, having a configuration of t_2g⁶, thus accounting for zero numbers of unpaired electrons.
Magnetic moment = $\sqrt{n (n + 2)}$ = $\sqrt{0 (0 + 2)}$ = 0 BM, where n is the number of unpaired electrons.
Hence, the correct option is option (A) 4.9 and 0.