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

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).