QJEEM19Ch2127: JEE Main 2019 Chemistry - Class 12 JEEM19Ch2127 Solution: The crystal field stabilization energy (CFSE) of [Fe(H 2 O) 6…

CN of [Fe(H₂O)₆]Cl₂ = 6

For CN = 6, CFSE = $$\left[\frac{3}{5}\times n-\frac{2}{5}\times n_1\right]{\mathrm{\Delta }}_0$$

Here, n = number of electron in e_g and
n₁ = number of electron in t_2g

Electronic configuration of Fe⁺² according to crystal field theory = $$t_{2g}^4{e}_g^2$$

$$\therefore$$ CFSE = $$\left[\frac{3}{5}\times 2-\frac{2}{5}\times 4\right]{\mathrm{\Delta }}_0$$ = - 0.4 $${\mathrm{\Delta }}_0$$

CN of K₂[NiCl₄] = 4

For CN = 4, CFSE = $$\left[\frac{2}{5}\times n-\frac{3}{5}\times n_1\right]{\mathrm{\Delta }}_t$$

Here, n = number of electron in t_2g and
n₁ = number of electron in e_g

Electronic configuration of Ni⁺² according to crystal field theory = $$e_g^4{t}_{2g}^4$$

$$\therefore$$ CFSE = $$\left[\frac{2}{5}\times 4-\frac{3}{5}\times 4\right]{\mathrm{\Delta }}_t$$ = - 0.8 $${\mathrm{\Delta }}_t$$