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Q.

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

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a

$[CoCl]^{2 -}, {[Fe {(H_{2} O)}_{6}]}^{2 +}$

b

${[Mn {(H_{2} O)}_{6}]}^{2 +} \cdot {[Cr {(H_{2} O)}_{6}]}^{2 +}$

c

${[Cr {(H_{2} O)}_{6}]}^{2 +}, {[{CoCl}_{4}]}^{2 -}$

d

${[Cr {(H_{2} O)}_{6}]}^{2 +}, {[Fe {(H_{2} O)}_{6}]}^{2 +}$

answer is B.

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Detailed Solution

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[Cr(H₂O)₆]²⁺=Cr²⁺=3d⁴= four unpaired electrons
[CoCl₄]²⁻=Co²⁺=3d⁷=3 unpaired electrons
[Fe(H₂O)₆]²⁺=Fe²⁺=3d⁶=4 unpaired electrons
[Mn(H₂O)₆]²⁺=Mn²⁺=3d⁵=5 unpaired electrons
So, [Cr(H₂O)₆]²⁺ and [Fe(H₂O)₆]²⁺ has same magnetic moment.

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