JEE Main 2020 — Coordination Compounds Question with Solution

[Ti(H₂O)₆]³⁺

Ti = [Ar]3d²4s²

Ti⁺³ = [Ar]3d¹


For octahedral

Crystal field stabilization energy (CFSE)

= (-0.4$$\Delta$$₀) $$\times$$ n_t2g + (+0.6$$\Delta$$₀) $$\times$$ n_eg + np

n_t2g = number of electrons in t_2g orbital

n_eg = number of electrons in e_g orbital

n = number of extra pairs

p = Pairing energy

Here n_t2g = 1

n_eg = 0

n = 0 ( For weak field ligands n always zero. Here H₂O is weak field ligand)

$$\therefore$$ Crystal field stabilization energy (CFSE)

= (-0.4$$\Delta$$₀) $$\times$$ 1 + (+0.6$$\Delta$$₀) $$\times$$ 0 + 0$$\times$$P

= -0.4$$\Delta$$₀ = –0.4 × 20300 = –8120 cm^–1

CFSE (in kJ) = $$\frac{8120}{83.7}$$ = 97 kJ/mol