JEE Advance - Chemistry (1997 - No. 5)
A compound of vanadium has a magnetic moment of 1.73 BM. Work out the electronic configuration of the vanadium ion in the compound.
1s2 2s2 2p6 3s2 3p6 3d5
1s2 2s2 2p6 3s2 3p6 3d3 4s2
1s2 2s2 2p6 3s2 3p6 3d2 4s1
1s2 2s2 2p6 3s2 3p6 3d1
1s2 2s2 2p6 3s2 3p6 3d4 4s1
Giải thích
We know that, magnetic moment, $$\mu = \sqrt {n(n + 2)} $$
where [n = number of unpaired electrons]
$$1.73 = \sqrt {n(n + 2)} $$
$$\therefore$$ n = 1 [After solving we get n = 1]
So, vanadium ion contains only one unpaired electron.
But $$_{23}V = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}4{s^2}$$
$$\therefore$$ $${V^{4 + }} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^2}3{d^1}$$
(will have one unpaired electron).