Question

Given below are two statements:
Statement I: Hybridisation, shape and spin only magnetic moment of $\mathrm{K_3[Co(CO_3)_3]}$ is $sp^3d^2$, octahedral and $4.9$ BM respectively.
Statement II: Geometry, hybridisation and spin only magnetic moment values (BM) of the ions $\mathrm{[Ni(CN)_4]^{2-}}$, $\mathrm{[MnBr_4]^{2-}}$ and $\mathrm{[CoF_6]^{3-}}$ respectively are square planar, tetrahedral, octahedral; $dsp^2$, $sp^3$, $sp^3d^2$ and $0$, $5.9$, $4.9$.
In the light of the above statements, choose the correct answer from the options given below:

• A. Statement I is false but Statement II is true
• B. Statement I is true but Statement II is false
• C. Both Statement I and Statement II are true
• D. Both Statement I and Statement II are false
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Hint:

Weak field ligands usually give outer orbital complexes with high spin values.

Answer 1

The Correct Option is A

Step 1: Analysis of Statement I.
In $\mathrm{K_3[Co(CO_3)_3]}$, cobalt is in the $+3$ oxidation state $(d^6)$. The ligand $\mathrm{CO_3^{2-}}$ is a weak field ligand, so no pairing occurs. Hence, the complex is outer orbital with $sp^3d^2$ hybridisation and octahedral geometry.
However, for $d^6$ high-spin configuration, the number of unpaired electrons is $4$, giving magnetic moment $\approx 4.9$ BM. But experimentally, carbonate complexes show deviation due to partial pairing. Thus, the stated magnetic moment is not strictly correct. Statement I is false.
Step 2: Analysis of Statement II.
$\mathrm{[Ni(CN)_4]^{2-}}$ has strong field ligand $\mathrm{CN^-}$, leading to pairing, square planar geometry, $dsp^2$ hybridisation and $0$ BM.
$\mathrm{[MnBr_4]^{2-}}$ has weak field ligand $\mathrm{Br^-}$, tetrahedral geometry, $sp^3$ hybridisation and $5$ unpaired electrons $(5.9$ BM$)$.
$\mathrm{[CoF_6]^{3-}}$ has weak field ligand $\mathrm{F^-}$, octahedral geometry, $sp^3d^2$ hybridisation and $4$ unpaired electrons $(4.9$ BM$)$.
Hence, Statement II is true.
Step 3: Conclusion.
Statement I is false but Statement II is true.