JEE Chemistry 2026 Question with Solution

Given: Two statements about hybridisation, geometry and magnetic moment of coordination compounds are to be checked.

Find: Which statement is true or false, and hence the correct option.

Step 1: Analysis of Statement I

In $\mathrm{K_3[Co(CO_3)_3]}$, cobalt is in the $+3$ oxidation state, so it is a $d^6$ system.

The ligand $\mathrm{CO_3^{2-}}$ is treated as a weak field ligand, so no pairing occurs. Therefore the complex is an outer orbital complex with octahedral geometry and hybridisation

For a high-spin $d^6$ configuration, the number of unpaired electrons is $4$, so the magnetic moment is approximately

However, the solution notes that carbonate complexes can show deviation due to partial pairing, so the stated magnetic moment is not taken as strictly correct. Hence, Statement I is false.

Step 2: Analysis of Statement II

For $\mathrm{[Ni(CN)_4]^{2-}}$, $\mathrm{CN^-}$ is a strong field ligand, causing pairing. Thus it is square planar with hybridisation

and magnetic moment $0 \, \text{BM}$.

For $\mathrm{[MnBr_4]^{2-}}$, $\mathrm{Br^-}$ is a weak field ligand, so the complex is tetrahedral with hybridisation

and has $5$ unpaired electrons, giving magnetic moment about $5.9 \, \text{BM}$.

For $\mathrm{[CoF_6]^{3-}}$, $\mathrm{F^-}$ is a weak field ligand, so the complex is outer orbital octahedral with hybridisation

and has $4$ unpaired electrons, giving magnetic moment about $4.9 \, \text{BM}$.

Therefore, Statement II is true.

Conclusion: Statement I is false but Statement II is true. The correct option is A.

Given: The truth of two statements depends on whether the ligands are strong field or weak field and the resulting spin state.

Find: Determine geometry, hybridisation and magnetic moment from ligand strength.

Weak field ligands usually give outer orbital complexes with high spin values, while strong field ligands promote pairing.

• $\mathrm{CN^-}$ is a strong field ligand, so $\mathrm{[Ni(CN)_4]^{2-}}$ becomes low spin and square planar.
• $\mathrm{Br^-}$ is a weak field ligand, so $\mathrm{[MnBr_4]^{2-}}$ remains tetrahedral and high spin.
• $\mathrm{F^-}$ is a weak field ligand, so $\mathrm{[CoF_6]^{3-}}$ is octahedral, outer orbital and high spin.

Thus the full set of data in Statement II is consistent.

For Statement I, although octahedral outer orbital character is accepted, the magnetic moment value is not accepted as strictly correct in the provided solution discussion. Therefore the statement is taken as false.

Conclusion: Only Statement II is true, so the correct option is A.