Which one of the following complexes will consume more equivalents of aqueous solution of $$AgNO_3$$?

Our aim is to find, for every given salt, how many chloride ions lie outside the coordination sphere, because only those free chloride ions will react with an aqueous solution of $$AgNO_3$$ to give a precipitate of $$AgCl$$. One mole of such chloride needs one equivalent of $$AgNO_3$$, so the salt that gives the greatest number of outside chloride ions will consume the largest number of equivalents of $$AgNO_3$$.

We recall the charge-balance idea: in a neutral salt, the algebraic sum of the charges of the coordination entity and of the counter-ions must be zero.

Now we study every option one by one, carefully writing the charge equations.

Option A : $$Na_2[CrCl_5(H_2O)]$$

Let the oxidation state of chromium be $$x$$.

Inside the coordination sphere we have five $$Cl^-$$ ligands and one neutral $$H_2O$$ ligand, so their total charge is $$5(-1)+0=-5$$.

The complex carries some overall negative charge, which must balance the two $$Na^+$$ ions. Writing the neutrality condition,

$$2(+1) \;+\; \text{charge of complex} \;=\;0,$$

so the complex charge is $$-2$$. Hence

$$x-5=-2 \;\Longrightarrow\; x=+3.$$

All five chlorides are inside the coordination sphere, and there is no chloride outside. Therefore no $$AgCl$$ can be formed.

Equivalents of $$AgNO_3$$ consumed  =  $$0$$.

Option B : $$Na_3[CrCl_6]$$

Let chromium have oxidation state $$x$$. Inside the sphere are six $$Cl^-$$ ligands, total charge $$6(-1)=-6$$.

The complex must balance three $$Na^+$$ ions:

$$3(+1)+\text{charge of complex}=0 \;\Longrightarrow\; \text{charge of complex}=-3.$$

Then $$x-6=-3 \;\Longrightarrow\; x=+3.$$

All six chlorides are inside; none are outside, so again

Equivalents of $$AgNO_3$$ consumed  =  $$0$$.

Option C : $$[Cr(H_2O)_5Cl]Cl_2$$

Let chromium have oxidation state $$x$$. Inside we have one $$Cl^-$$ and five neutral $$H_2O$$ molecules, so the ligand charge is $$-1$$.

The entire salt contains two outside $$Cl^-$$ ions. Hence the neutrality condition is

$$\bigl[\text{charge of complex}\bigr] \;+\; 2(-1)=0 \;\Longrightarrow\; \text{charge of complex}=+2.$$

Writing the oxidation-state equation,

$$x + (-1)=+2 \;\Longrightarrow\; x=+3.$$

Therefore exactly two chloride ions lie outside the square brackets and will react with $$AgNO_3$$.

Equivalents of $$AgNO_3$$ consumed  =  $$2$$.

Option D : $$[Cr(H_2O)_6]Cl_3$$

All six ligands inside the sphere are neutral $$H_2O$$ molecules; their total charge is zero.

The salt has three outside $$Cl^-$$ ions. For electrical neutrality,

$$\bigl[\text{charge of complex}\bigr] \;+\; 3(-1)=0 \;\Longrightarrow\; \text{charge of complex}=+3.$$

Consequently, chromium is in the $$+3$$ oxidation state, and all three chloride ions remain outside the coordination sphere.

Equivalents of $$AgNO_3$$ consumed  =  $$3$$.

Comparing the numbers obtained,

$$0 \;(\text{A}) \;,\; 0 \;(\text{B}) \;,\; 2 \;(\text{C}) \;,\; 3 \;(\text{D})$$

we see that Option D provides the maximum of three ionisable chloride ions and therefore requires the greatest number of equivalents of $$AgNO_3$$.

Hence, the correct answer is Option D.