The Crystal Field Stabilization Energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion M Z + are - 0 . 8 Δ 0 and 3 . 87 BM , respectively. Identify M Z + :

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CFSE for octahedral complex = - 0 . 4 n t 2 g + 0 . 6 n eg Δ 0 + n ( P ) As μ = 3 . 87 B M so number unpaired electron = 3 . Metal ion configuration CFSE 27 Co 2 + = 3 d 7 t 2 g 2 , 2 , 1 , e g 1 , 1 CFSE = - 0 . 8 Δ o 25 Mn 4 + = 3 d 3 t 2 g 1 , 1 , 1 , e g 0 , 0 CFSE = - 1 . 2 Δ 0 V 23 2 + = 3 d 3 t 2 g 1 , 1 , 1 , e g 0 , 0 CFSE = - 1 . 2 Δ 0 Cr 24 3 + = 3 d 3 t 2 g 1 , 1 , 1 , e g 0 , 0 CFSE = - 1 . 2 Δo

Metal ion	configuration	$CFSE$
$27 {Co}^{2 +} = 3 d^{7}$	$t_{2 g}^{2, 2, 1}, e_{g}^{1, 1}$	$CFSE = - 0.8 Δ o$
$25 {Mn}^{4 +} = 3 d^{3}$	$t_{2 g}^{1, 1, 1}, e_{g}^{0, 0}$	$CFSE = - 1.2 Δ_{0}$
${{}_{23}V}^{2 +} = 3 d^{3}$	$t_{2 g}^{1, 1, 1}, e_{g}^{0, 0}$	$CFSE = - 1.2 Δ_{0}$
${}_{24}{Cr}^{3 +} = 3 d^{3}$	$t_{2 g}^{1, 1, 1}, e_{g}^{0, 0}$	$CFSE = - 1.2 Δo$

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