Identification and Properties of Cobalt Coordination Complexes

To determine the correct statements, we analyze the given experimental observations step-by-step:

1. Identification of Metal Complex $\mathbf{X}$: The formula of the metal salt is given as $\text{MCl}_2 \cdot 6\text{H}_2\text{O}$, containing the metal ion $\text{M}^{2+}$. The calculated spin-only magnetic moment ($\mu$) for $\mathbf{X}$ is $3.87 \text{ B.M.}$ Using the spin-only magnetic moment formula: $\mu = \sqrt{n(n+2)} = 3.87 \text{ B.M.}$ $\implies n(n+2) = 15 \implies n = 3 \text{ unpaired electrons}$ A $+2$ oxidation state transition metal ion with $3$ unpaired electrons in a high-spin octahedral geometry corresponds to a $d^7$ configuration ($t_{2g}^5 e_g^2$). Thus, the metal ion $\text{M}^{2+}$ is $\text{Co}^{2+}$.

   • Complex $\mathbf{X}$ is the octahedral pink-colored complex $[\text{Co}(\text{H}_2\text{O})_6]\text{Cl}_2$.

2. Identification and Hybridization of Complex $\mathbf{Y}$: When $\mathbf{X}$ ($[\text{Co}(\text{H}_2\text{O})_6]\text{Cl}_2$) is treated with excess aqueous ammonia in the presence of air ($\text{O}_2$), $\text{Co}^{2+}$ is oxidized to $\text{Co}^{3+}$: $4[\text{Co}(\text{H}_2\text{O})_6]^{2+} + 24\text{NH}_3 + \text{O}_2 + 4\text{H}^+ \rightarrow 4[\text{Co}(\text{NH}_3)_6]^{3+} + 26\text{H}_2\text{O}$ Given that $\mathbf{Y}$ acts as a $1:3$ electrolyte, it dissociates into one complex cation and three chloride anions: $\mathbf{Y} = [\text{Co}(\text{NH}_3)_6]\text{Cl}_3 \rightleftharpoons [\text{Co}(\text{NH}_3)_6]^{3+} + 3\text{Cl}^-$

   • In $[\text{Co}(\text{NH}_3)_6]^{3+}$, the oxidation state of $\text{Co}$ is $+3$ ($d^6$ configuration).
   • Ammonia ($\text{NH}_3$) acts as a strong field ligand for $\text{Co}^{3+}$, causing complete pairing of $d$-electrons to give a low-spin configuration ($t_{2g}^6 e_g^0$) with $n = 0$ unpaired electrons ($\mu = 0 \text{ B.M.}$).
   • The hybridization of $\text{Co}^{3+}$ in $[\text{Co}(\text{NH}_3)_6]^{3+}$ involves two inner $3d$ orbitals, one $4s$ orbital, and three $4p$ orbitals, giving $d^2sp^3$ hybridization.
   • Since $\mathbf{Y}$ contains three ionizable chloride ions, reaction with excess $\text{AgNO}_3$ yields three equivalents of $\text{AgCl}$, not two.

3. Identification and Geometry of Complex $\mathbf{Z}$: Treating $\mathbf{X}$ with excess $\text{HCl}$ converts the octahedral aquo complex into a tetrachloro complex: $[\text{Co}(\text{H}_2\text{O})_6]^{2+} \text{ (pink)} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} \text{ (blue)} + 6\text{H}_2\text{O}$

   • Complex $\mathbf{Z}$ is $[\text{CoCl}_4]^{2-}$.
   • For $\text{Co}^{2+}$ ($d^7$), weak-field ligand $\text{Cl}^-$ forms a tetrahedral complex ($e^4 t_2^3$) with $n = 3$ unpaired electrons ($\mu = 3.87 \text{ B.M.}$).

4. Effect of Temperature on the Equilibrium ($\mathbf{X} \rightleftharpoons \mathbf{Z}$): The equilibrium reaction: $[\text{Co}(\text{H}_2\text{O})_6]^{2+} \text{ (pink)} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} \text{ (blue)} + 6\text{H}_2\text{O}$ is endothermic ($\Delta H > 0$) in the forward direction. According to Le Chatelier's principle, lowering the temperature to $0^\circ\text{C}$ shifts the equilibrium in the reverse (exothermic) direction towards the pink-colored $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$ complex. Thus, at $0^\circ\text{C}$, the solution appears pink.

Evaluation of the Options:

• Option A: Incorrect. Complex $\mathbf{Y}$ ($[\text{Co}(\text{NH}_3)_6]\text{Cl}_3$) produces 3 equivalents of $\text{AgCl}$, not 2.
• Option B: Correct. The hybridization of $\text{Co}^{3+}$ in $\mathbf{Y}$ is $d^2sp^3$.
• Option C: Correct. $\mathbf{Z}$ ($[\text{CoCl}_4]^{2-}$) is a tetrahedral complex.
• Option D: Correct. At $0^\circ\text{C}$, the equilibrium shifts backward, making the solution pink.

Hence, the correct options are B, C, and D.