Neet Solved Paper 2015 Question 12
Question: Magnetic moment 2.84 BM is given by (At. no. $ Ni=28,,Ti=22,,Cr=24,,Co=27 $ )
Options:
A) $ N{i^{2+}} $
B) $ T{i^{3+}} $
C) $ C{r^{3+}} $
D) $ C{O^{2+}} $
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Answer:
Correct Answer: A
Solution:
Magnetic moment, $ \mu =\sqrt{n(n+2)} $ BM where, n = number of unpaired electrons $ \mu =2.84(given) $
$ \therefore $ $ 284=\sqrt{n(n+2)}B.M $ $ {{(2.84)}^{2}}=n(n+2) $ $ 8=n^{2}+2n $ $ n^{2}+2n-8=0 $ $ n^{2}+4n-2n-8=0 $ $ n(n+4)-2(n+4)=0 $
$ n=2 $ $ N{i^{2+}}=[Ar]3d^{8}4s^{0} $ (two unpaired electrons)
$ T{i^{3+}}=[Ar]3d^{1}4s^{0} $ (one unpaired electrons)
$ C{r^{3+}}=[Ar]3d^{3} $ , (three unpaired electrons)
$ C{o^{2+}}=[Ar],3d^{7},4s^{0} $ (three unpaired electrons)
So, only $ N{i^{2+}} $ has 2 unpaired electrons.
