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Chemistry Question on coordination compounds

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

A

[Cr(H2O)6]2+,[CoCl4]2−\left[ Cr \left( H _{2} O \right)_{6}\right]^{2+},\left[ CoCl _{4}\right]^{2-}

B

[Cr(H2O)6]2+,[Fe(H2O)6]2+\left[ Cr \left( H _{2} O \right)_{6}\right]^{2+},\left[ Fe \left( H _{2} O \right)_{6}\right]^{2+}

C

[Mn(H2O)6]2+,[Cr(H2O)6]2+\left[ Mn \left( H _{2} O \right)_{6}\right]^{2+},\left[ Cr \left( H _{2} O \right)_{6}\right]^{2+}

D

[CoCl4]2−∗,[Fe(H2O)6]2+\left[ CoCl _{4}\right]^{2-*},\left[ Fe \left( H _{2} O \right)_{6}\right]^{2+}

Answer

[Cr(H2O)6]2+,[Fe(H2O)6]2+\left[ Cr \left( H _{2} O \right)_{6}\right]^{2+},\left[ Fe \left( H _{2} O \right)_{6}\right]^{2+}

Explanation

Solution

[Cr(H2O)6]2+=Cr2+=3d4=\left[ Cr \left( H _{2} O \right)_{6}\right]^{2+}= Cr ^{2+}=3 d ^{4}= four unpaired electrons [CoCl4]2−=Co2+=3d7=3\left[ CoCl _{4}\right]^{2-}= Co ^{2+}=3 d ^{7}=3 unpaired electrons [Fe(H2O)6]2+=Fe2+=3d6=4\left[ Fe \left( H _{2} O \right)_{6}\right]^{2+}= Fe ^{2+}=3 d ^{6}=4 unpaired electrons [Mn(H2O)6]2+=Mn2+=3d5=5\left[ Mn \left( H _{2} O \right)_{6}\right]^{2+}= Mn ^{2+}=3 d ^{5}=5 unpaired electrons So, [Cr(H2O)6]2+\left[ Cr \left( H _{2} O \right)_{6}\right]^{2+} and [Fe(H2O)6]2+\left[ Fe \left( H _{2} O \right)_{6}\right]^{2+} has same magnetic moment.