Given below are h¥o statements:
Statement I: Hybridisation, shape and spin only magnetic moment of $$K_{3}[Co(CO_{3})_{3}]$$ is $$sp^{3}d^{2}$$, octahedral and 4.9 BM respectively.
Statement II: Geometty, hybridisation and spin only magnetic moment values (BM) of the ions $$[Ni(CN)_{4}]^{2-},[MnBr_{4}]^{2-}\text{ and }[CoF_{6}]^{3-}$$ respectively are square planar, tetrahedral, octahedral;$$dsp^{2},sp^{3},sp^{3}d^{2}$$ and 0, 5.9, 4.9.
In the light of the above statements, choose the correct answer from the options given below

We need to evaluate two statements about coordination compounds.

Statement I: $$K_3[Co(CO_3)_3]$$ has hybridisation $$sp^3d^2$$, octahedral shape and spin-only magnetic moment of 4.9 BM.

In $$K_3[Co(CO_3)_3]$$:

The charge on the complex ion $$[Co(CO_3)_3]^{3-}$$: Since $$CO_3^{2-}$$ is bidentate, each carbonate acts as a bidentate ligand. With 3 carbonate ions giving a charge of -6, and the complex having a charge of -3, Co has an oxidation state of +3.

Co³⁺ has configuration: $$[Ar] 3d^6$$

$$CO_3^{2-}$$ is a weak field ligand, so the electrons remain unpaired.

With 6 donor atoms (3 bidentate carbonates), the geometry is octahedral.

For weak field: 4 unpaired electrons, $$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.9$$ BM

Hybridisation: $$sp^3d^2$$ (outer orbital complex for weak field ligand)

Statement I is TRUE.

Statement II:

$$[Ni(CN)_4]^{2-}$$: Ni²⁺ (3d⁸), CN⁻ is a strong field ligand, forces pairing. Hybridisation: $$dsp^2$$, square planar, 0 unpaired electrons, $$\mu = 0$$ BM. ✓

$$[MnBr_4]^{2-}$$: Mn²⁺ (3d⁵), Br⁻ is a weak field ligand. Hybridisation: $$sp^3$$, tetrahedral, 5 unpaired electrons, $$\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.9$$ BM. ✓

$$[CoF_6]^{3-}$$: Co³⁺ (3d⁶), F⁻ is a weak field ligand. Hybridisation: $$sp^3d^2$$, octahedral, 4 unpaired electrons, $$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.9$$ BM. ✓

Statement II is TRUE.

Both statements are true, which corresponds to Option 2.