1. Identify the complex that gives 2 moles of $AgCl$ with excess $AgNO_3$

Excess $AgNO_3$ precipitates only ionisable chloride ions (chloride outside the coordination sphere).

We examine the given compounds using Werner formulations:

• $CoCl_3.4NH_3$ can be written as either:

  • $[Co(NH_3)_4Cl_2]Cl$ giving 1 mole $AgCl$, or
  • $[Co(NH_3)_4Cl]Cl_2$ giving 2 moles $AgCl$

• $NiCl_2.6H_2O$ is typically $[Ni(H_2O)_6]Cl_2$, so it gives 2 moles of $AgCl$.

• $PtCl_4.2HCl$ is $H_2[PtCl_6]$, where chloride is coordinated / present as complex acid form, not giving 2 moles of free $Cl^-$ in the usual Werner sense.

Now the question asks for an octahedral complex among these that gives 2 moles of $AgCl$.

Among the candidates giving 2 moles $AgCl$:

• $[Co(NH_3)_4Cl]Cl_2$ is 5-coordinate, not octahedral.
• $[Ni(H_2O)_6]Cl_2$ contains octahedral $[Ni(H_2O)_6]^{2+}$.

Hence the required complex is: $[Ni(H_2O)_6]Cl_2$ with octahedral ion $[Ni(H_2O)_6]^{2+}$.

2. Find the electronic configuration of $Ni^{2+}$

Nickel: $Z=28$

Neutral Ni: $[Ar]3d^8 4s^2$

For $Ni^{2+}$, remove two electrons from $4s$: $Ni^{2+} : [Ar]3d^8$

3. Determine number of unpaired electrons in octahedral $d^8$

For octahedral field: $t_{2g}^6 e_g^2$

This has 2 unpaired electrons. So, $n=2$

4. Calculate spin-only magnetic moment

Spin-only formula: $\mu = \sqrt{n(n+2)}\, \text{B.M.}$

Substitute $n=2$: $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83\, \text{B.M.}$

Nearest integer: $\boxed{3}$

5. Compare with stored answer

Derived answer = $3$

Stored correct answer = $3$

So they agree.