QJEEM21Ch9925: JEE Main 2021 Chemistry - Class 12 JEEM21Ch9925 Solution: The Crystal Field Stabilization Energy (CFSE) and magnetic…

Co²⁺ = [Ar] 3d⁷4s^o

in [Co(H₂O)₆]²⁺, H₂O will behave as weak field ligand

Co²⁺ = t_2g⁵ , e_g²


CFSE = (–0.4 × 5 + 2 × 0.6) Δ₀ = –0.8 Δ₀

Magnetic moment (spin only) can be calculated as :

$$\mu =\sqrt{n\left(n+2\right)}$$

where, $$\mu$$ = spin only magnetic moment and

n = number of unpaired electrons = 3

= $$\sqrt{3\left(3+2\right)}=\sqrt{15}$$ = 3.87 BM