Coordination Compounds Part 4

Concept:Valence Bond Theory (VBT) explains the hybridisation and magnetic nature of coordination complexes based on ligand field strength and electron pairing.Explanation:Check each complex using VBT and ligand field strength from NCERT.A. $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$Co oxidation state: $+3$, so $\mathrm{Co}^{3+}$ has $3d^6$ configuration.$\mathrm{NH}_3$ is a strong field ligand causing pairing.It forms an inner orbital (low spin) octahedral complex with $d^2sp^3$ hybridisation.✅ A is correct.B. $[\mathrm{MnCl}_6]^{3-}$Mn oxidation state: $+3$, so $\mathrm{Mn}^{3+}$ has $3d^4$ configuration.$\mathrm{Cl}^-$ is weak field, no pairing, high spin outer orbital complex.Hybridisation: $sp^3d^2$.✅ B is correct.C. $[\mathrm{CoF}_6]^{3-}$Co oxidation state: $+3$, so $\mathrm{Co}^{3+}$ has $3d^6$ configuration.$\mathrm{F}^-$ is weak field, no pairing, high spin outer orbital complex.Hybridisation should be $sp^3d^2$, not $d^2sp^3$.❌ C is incorrect.D. $[\mathrm{FeF}_6]^{3-}$Fe oxidation state: $+3$, so $\mathrm{Fe}^{3+}$ has $3d^5$ configuration.$\mathrm{F}^-$ is weak field, no pairing, high spin outer orbital octahedral complex.Hybridisation: $sp^3d^2$.✅ D is correct.E. $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$Ni oxidation state: $+2$, so $\mathrm{Ni}^{2+}$ has $3d^8$ configuration.$\mathrm{CN}^-$ is strong field, causes pairing, square planar geometry.Hybridisation for square planar is $dsp^2$, not $sp^3$.❌ E is incorrect.Answer:Correct set: A, B and D only → Option D.