JEE MAIN - Chemistry (2019 - 10th April Morning Slot - No. 17)
Consider the hydrated ions of Ti2+, V2+, Ti3+, and Sc3+. The correct order of their spin-only magnetic
moments is :
Sc3+ < Ti3+ < V2+ < Ti2+
Sc3+ < Ti3+ < Ti2+ < V2+
Ti3+ < Ti2+ < Sc3+ < V2+
V2+ < Ti2+ < Ti3+ < Sc3+
Explanation
As we know that
$$\mu = \sqrt {n\left( {n + 2} \right)} $$ where n = no. of unpaired electrons i.e. greater the no. of unpaired electron more will be the spin-only magnetic moments.
. Electronic configuration of the given transition metal ions are
Sc3+ (Z = 21) = 1s22s22p63s23p6
$$ \therefore $$ n = 0
Ti2+ (Z = 22) = 1s22s22p63s23p63d2
$$ \therefore $$ n = 2
Ti3+ (Z = 22) = 1s22s22p63s23p63d1
$$ \therefore $$ n = 1
V2+ (Z = 23) = 1s22s22p63s23p63d3
$$ \therefore $$ n = 3
So the correct increasing order of magnetic moment is
Sc3+ < Ti3+ < Ti2+ < V2+
$$\mu = \sqrt {n\left( {n + 2} \right)} $$ where n = no. of unpaired electrons i.e. greater the no. of unpaired electron more will be the spin-only magnetic moments.
. Electronic configuration of the given transition metal ions are
Sc3+ (Z = 21) = 1s22s22p63s23p6
$$ \therefore $$ n = 0
Ti2+ (Z = 22) = 1s22s22p63s23p63d2
$$ \therefore $$ n = 2
Ti3+ (Z = 22) = 1s22s22p63s23p63d1
$$ \therefore $$ n = 1
V2+ (Z = 23) = 1s22s22p63s23p63d3
$$ \therefore $$ n = 3
So the correct increasing order of magnetic moment is
Sc3+ < Ti3+ < Ti2+ < V2+
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