QJEEM20Chaec5: JEE Main 2020 Chemistry - Class 12 JEEM20Chaec5 Solution: The correct order of the calculated spin-only magnetic moments…

(A) Ni(CO)₄

Ni = 3d⁸4s²

CO is strong field ligand. So pairing of elections happens.

$$\therefore$$ Number of unpaired electrons = 0

$$\therefore$$ $$\mu$$_spin = 0

(B) [Ni(H₂O)₆]Cl₂

Ni⁺² = 3d⁸4s⁰

H₂O is weak field ligand. So no pairing of electrons happens.

Number of unpaired electron = 2

$$\therefore$$ $$\mu$$_spin = $$\sqrt{n\left(n+2\right)}$$ = $$\sqrt{2\left(2+2\right)}$$ = $$\sqrt{8}$$ B.M

(C) Na₂[Ni(CN)₄]

Ni⁺² = 3d⁸4s⁰

CN^- is strong field ligand. So pairing of electrons happens.

Number of unpaired electron = 0

$$\therefore$$ $$\mu$$_spin = 0

(D) PdCl₂(PPh₃)₂

Pd²⁺ = 4d⁸

This is dsp² complex. And shape is square planar.

$$\therefore$$ $$\mu$$_spin = 0