The pair in which both the species have the same magnetic moment (spin only) is:

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No. unpaired e - Cr H 2 O 6 2 + ⇒ Cr 2 + ⇒ 3 d 4 ⇒ t 2 g 1 , 1 , 1 , e g 1 , 0 4 Fe H 2 O 6 2 + ⇒ Fe 2 + ⇒ 3 d 6 ⇒ t 2 g 1 , 1 , 1 , e g 1 , 1 4 Fe NH 3 6 2 + ⇒ Fe 2 + ⇒ 3 d 6 ⇒ t 2 g 2 , 1 , 1 , e g 1 , 1 4 Co OH 4 2 - ⇒ Co 2 + ⇒ 3 d 7 ⇒ e g 2 , 2 , t 2 1 , 1 , 1 3 CoCI 4 2 - ⇒ Co 2 + ⇒ 3 d 7 ⇒ e 2 , 2 , t 2 1 , 1 , 1 3 Mn H 2 O 6 2 + ⇒ MN 2 + ⇒ 3 d 5 ⇒ t 2 g 1 , 1 , 1 , e g 1 , 1 5 So Cr H 2 O 6 2 + and Fe H 2 O 6 2 + have same magnetic moment (spin only)

	No. unpaired $e^{-}$
${[Cr {(H_{2} O)}_{6}]}^{2 +} \Rightarrow {Cr}^{2 +} \Rightarrow 3 d^{4} \Rightarrow {t_{2 g}}^{1, 1, 1}, {e_{g}}^{1, 0}$	$4$
${[Fe {(H_{2} O)}_{6}]}^{2 +} \Rightarrow {Fe}^{2 +} \Rightarrow 3 d^{6} \Rightarrow {t_{2 g}}^{1, 1, 1}, {e_{g}}^{1, 1}$	$4$
${[Fe {({NH}_{3})}_{6}]}^{2 +} \Rightarrow {Fe}^{2 +} \Rightarrow 3 d^{6} \Rightarrow {t_{2 g}}^{2, 1, 1}, {e_{g}}^{1, 1}$	$4$
${[Co {(OH)}_{4}]}^{2 -} \Rightarrow {Co}^{2 +} \Rightarrow 3 d^{7} \Rightarrow {e_{g}}^{2, 2}, {t_{2}}^{1, 1, 1}$	$3$
${[{CoCI}_{4}]}^{2 -} \Rightarrow {Co}^{2 +} \Rightarrow 3 d^{7} \Rightarrow e^{2, 2}, {t_{2}}^{1, 1, 1}$	$3$
${[Mn {(H_{2} O)}_{6}]}^{2 +} \Rightarrow {MN}^{2 +} \Rightarrow 3 d^{5} \Rightarrow {t_{2 g}}^{1, 1, 1}, {e_{g}}^{1, 1}$	$5$
So ${[Cr {(H_{2} O)}_{6}]}^{2 +}$ and ${[Fe {(H_{2} O)}_{6}]}^{2 +}$	have same magnetic moment (spin only)

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