Identify the metal and complexes from the data

We are given:

• $X = MCl_2\cdot 6H_2O$ is pink in aqueous solution.
• Spin-only magnetic moment of $X$ is $3.87\,\text{B.M.}$
• On adding excess aqueous ammonia in presence of air and $NH_4Cl$, it gives an octahedral complex $Y$.
• In aqueous solution, $Y$ behaves as a $1:3$ electrolyte.
• $\mu$ for $Y$ is zero.
• With excess $HCl$ at room temperature, $X$ gives a blue complex $Z$.
• $\mu$ of $Z$ is also $3.87\,\text{B.M.}$

Since $\mu = \sqrt{n(n+2)}$ and $\mu = 3.87\,\text{B.M.}$ corresponds to $n=3$ unpaired electrons.

A common ion showing:

• pink hydrated complex,
• blue chloro complex,
• 3 unpaired electrons in both,
• diamagnetic ammine complex after oxidation by air,

is cobalt.

So, $X \equiv [Co(H_2O)_6]Cl_2$ i.e. aqueous cobalt(II) chloride hexahydrate.

Check magnetic moment of $X$

In $X$, cobalt is in +2 oxidation state: $Co^{2+} : 3d^7$ For octahedral high-spin $d^7$, number of unpaired electrons $=3$.

Hence, $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\,\text{B.M.}$ This matches.

Determine complex $Y$

Excess $NH_3$ in presence of air oxidizes $Co^{2+}$ to $Co^{3+}$ and forms the hexaammine complex: $[Co(NH_3)_6]^{3+}$

Since the solution behaves as a $1:3$ electrolyte, the compound must be: $Y = [Co(NH_3)_6]Cl_3$

In solution: $[Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^-$ so it is indeed a $1:3$ electrolyte.

For $Co^{3+}$: $Co^{3+} : 3d^6$ In the strong field of $NH_3$, the octahedral complex is low spin: $t_{2g}^6e_g^0$ hence all electrons are paired and magnetic moment is zero.

Determine complex $Z$

Excess $HCl$ converts the pink aqua complex into the blue chloro complex of cobalt(II): $[Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O$

So, $Z = [CoCl_4]^{2-}$ (formed in solution, e.g. as $H_2[CoCl_4]$-type environment / chloride-rich medium).

For $Co^{2+}$, $d^7$ tetrahedral complexes are high spin with 3 unpaired electrons, so $\mu \approx 3.87\,\text{B.M.}$ This matches.

Evaluate each option

Option A

Addition of silver nitrate to $Y$ gives only two equivalents of silver chloride

We found: $Y = [Co(NH_3)_6]Cl_3$ All three chloride ions are outside the coordination sphere. Therefore with $AgNO_3$: $[Co(NH_3)_6]Cl_3 + 3AgNO_3 \rightarrow [Co(NH_3)_6](NO_3)_3 + 3AgCl \downarrow$ So three equivalents of $AgCl$ are obtained, not two.

Hence, A is incorrect.

Option B

The hybridization of the central metal ion in $Y$ is $d^2sp^3$

In $Y = [Co(NH_3)_6]^{3+}$, cobalt is $d^6$ low spin. The complex is inner-orbital octahedral.

Therefore hybridization is: $d^2sp^3$

Hence, B is correct.

Option C

$Z$ is a tetrahedral complex

We identified: $Z = [CoCl_4]^{2-}$ and this complex is well-known to be tetrahedral and blue.

Hence, C is correct.

Option D

When $X$ and $Z$ are in equilibrium at $0^\circ C$, the color of the solution is pink

The equilibrium is: $[Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O$

• $[Co(H_2O)_6]^{2+}$ is pink
• $[CoCl_4]^{2-}$ is blue

Formation of $[CoCl_4]^{2-}$ is favored at higher temperature, while lower temperature favors the pink aqua complex.

Therefore at $0^\circ C$, equilibrium shifts toward $X$ (pink form), so the solution appears pink.

Hence, D is correct.

Final answer

Correct options are: $\boxed{B,\ C,\ D}$

Comparison with stored correct answer

Stored correct answer: B, C, D

This matches exactly.