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:

Question

5.33 gram of $CrCl_3 \cdot 6H_2O$ (1:3 electrolyte) is passed through cation exchanger. The resulting solution is then treated with an excess of $AgNO_3$, leading to formation of 8.61 gran of precipitate. Calculate :
$\frac{\text{number of moles of complex reacted}}{\text{number of moles of AgCl precipitated}} \times 100$

Answer 1

Let's analyze each statement given in the question to determine their truthfulness.

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.

The coordination compound $\mathrm{K_3[Co(CO_3)_3]}$ involves cobalt in the +3 oxidation state. The ligand $\mathrm{CO_3^{2-}}$ is a weak field ligand, which generally does not cause pairing of electrons.
$\mathrm{Co^{3+}}$ has an electronic configuration of $[Ar]\,3d^6$. With weak field ligands, the electrons do not pair up, leading to a high-spin complex. This results in four unpaired electrons.
The hybridization and shape for a high-spin octahedral complex should be $d^2sp^3$ or $sp^3d^2$ (with six ligands), but here the correct assignment should be $d^2sp^3$, and the magnetic moment for four unpaired electrons is calculated as:

$\mu = \sqrt{n(n+2)}\ \text{BM}$

Where $n$ is the number of unpaired electrons. For four unpaired electrons, $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9\ \text{BM}$.
However, the hybridization mentioned as $sp^3d^2$ is incorrect for the given structure because it should ideally be $d^2sp^3$.

Conclusion for Statement I: The hybridization is not correct; hence, Statement I is false.

Statement II: Geometry, hybridization, 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.

$\mathrm{[Ni(CN)_4]^{2-}}$ - CN^- is a strong field ligand, causing pairing and resulting in a square planar geometry with $dsp^2$ hybridization and no unpaired electrons, hence a magnetic moment of 0 BM.
$\mathrm{[MnBr_4]^{2-}}$ - Br^- is a weak field ligand, resulting in a tetrahedral geometry with $sp^3$ hybridization. Mn²⁺ has five unpaired electrons, making the magnetic moment approximately 5.9 BM.
$\mathrm{[CoF_6]^{3-}}$ - F^- is also a weak field ligand, leading to an octahedral geometry with $sp^3d^2$ hybridization. Co³⁺ has four unpaired electrons, hence approximately 4.9 BM.

Conclusion for Statement II: The hybridizations, geometries, and magnetic moment values are correct; hence, Statement II is true.

Final Conclusion: Based on the above analysis, the correct option is: Statement I is false but Statement II is true.

Column-I (Complex compound)	Column-II ($\Delta_0$ (CFSE) $\text{cm}^{-1}$)
(i) $[Cr(CN)_6]^{3-}$	(P) 17000
(ii) $[Cr(H_2O)_6]^{3+}$	(Q) 15000
(iii) $[Cr(en)_3]^{3+}$	(R) 12000
(iv) $[CrF_6]^{3-}$	(S) 20000

Show Hint

The Correct Option is A

Solution and Explanation

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