JEE Main 2019 — D And F Block Elements Question with Solution

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⁺