The lowest degree of paramagnetism per mole of the compound at 298 K will be shown by :

F e S O_{4} .6 H_{2} O

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N i S O_{4} .6 H_{2} O

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M n S O_{4} .4 H_{2} O

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C u S O_{4} .5 H_{2} O

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Solution

The correct option is D $C u S O_{4} .5 H_{2} O$
$F e^{2 +}$ , $N i^{2 +}$ , $M n^{2 +}$ and $C u^{2 +}$ ions have outer electronic configuration of $3 d^{6}$ , $3 d^{8}$ , $3 d^{5}$ and $3 d^{9}$ respectively. They contain 4, 2, 5 and 1 unpaired electron respectively.

Hence, the lowest degree of paramagnetism per mole of the compound at 298 K will be shown by, $C u S O_{4} .5 H_{2} O$

Hence the correct option is D.

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Similar questions

Among the following the lowest degree of paramagnetism per mole of the compound at 298 K will be shown by:

The increasing order of paramagnetism of the following compound is:

(I) $M n S O_{4} .4 H_{2} O$

(II) $F e S O_{4} .7 H_{2} O$

(III) $N i S O_{4} .6 H_{2} O$

(IV) $C u S O_{4} .5 H_{2} O$

M n S O_{4} .4 H_{2} O

F e S O_{4} .7 H_{2} O

N i S O_{4} .6 H_{2} O

C u S O_{4} .5 H_{2} O

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