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Question

Match coloum - I with coloum - II and select the correct answer using the codes given below the coloums.
ColumnI ColumnII (Metal ion) (Magnetic moment (BM))(a)Cr3+(p)35(b)Fe2+(q)30(c)Ni2+(r)24(d)Mn2+(s)15 (t)8

A
(a) s; (b) r; (c) t; (d) p
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B
(a) r; (b) s; (c) t; (d) p
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C
(a) s; (b) r; (c) p; (d) t
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D
(a) p; (b) r; (c) t; (d) s
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Solution

The correct option is A (a) s; (b) r; (c) t; (d) p
Cr3+ has 3 unpaired electrons in the [Ar]3d34s0 configuration after losing 3 electrons from the [Ar]3d54s1 configuration.

The magnetic moment in (BM) is calculated as = n(n+2) where n is the number of unpaired electrons.

For Cr3+ it comes out to 15.
Likewise, for Fe2+ we get 24 as n =4 in the 3d4 configuration.

Ni2+ only has n = 2 unpaired electrons in the [Ar]3d84s0 configuration and thus it has a magnetic moment of 8.

Lastly, Mn2+ has n = 5 unpaired electrons and thus it has trhe highest magnetic moment of 35

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