Coordination Compounds 2 Question 47
49. Amongst the following, the lowest degree of paramagnetism per mole of the compound at $298 K$ will be shown by
(a) $MnSO _4 \cdot 4 H _2 O$
(b) $CuSO _4 \cdot 5 H _2 O$
(c) $FeSO _4 \cdot 6 H _2 O$
(d) $NiSO _4 \cdot 6 H _2 O$
$(1988,2 M)$
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Answer:
Correct Answer: 49. (b)
Solution:
- Salt with least number of unpaired electrons in $d$ - orbital of central metal will show lowest degree of paramagnetism
$Mn^{2+}\left(3 d^{5}, 5\right.$ unpaired electrons)
$Cu^{2+}\left(3 d^{9}, 1\right.$ unpaired electrons)
$Fe^{2+}\left(3 d^{6}, 4\right.$ unpaired electrons $)$
$Ni^{2+}$ ( $3 d^{8}, 2$ unpaired electrons)
Hence, $CuSO _4 \cdot 5 H _2 O$ has lowest degree of paramagnetism.
BETA
