Complex (A) has a composition of H12O6Cl3Cr. If the complex on treatment with conc. H2SO4 loses 13.5

Q: Why does conc. $\text{H}_2\text{SO}_4$ only remove water from outside the sphere?

Water molecules inside the coordination sphere are coordinately bonded to the central metal ion, making them much harder to remove. Concentrated sulfuric acid acts as a dehydrating agent and primarily targets the loosely held water of crystallization outside the sphere.

Q: How do we know the coordination number of Chromium is 6?

Chromium in its +3 oxidation state almost exclusively forms octahedral complexes, which inherently require a coordination number of 6. This is a standard fact in coordination chemistry.

Q: Could the lost mass be chlorine gas instead of water?

No, concentrated $\text{H}_2\text{SO}_4$ acts primarily as a dehydrating agent in this context, specifically targeting water. It doesn't oxidize chloride to chlorine gas under these mild conditions without heating or the presence of other strong oxidizing agents.

Q: What does the dot ($\cdot$) in the formula mean?

The dot represents water of crystallization. These are water molecules that are incorporated into the crystal lattice of the solid complex but are not directly bonded to the central metal ion.

Q: Why are there two chlorines inside and one outside?

To satisfy the coordination number of 6 for Chromium. With 4 water molecules inside, we need 2 more ligands, which are provided by the chloride ions. The remaining chloride acts as a counter ion outside the sphere to balance the overall charge.

Question

Complex (A) has a composition of $	ext{H}_{12}	ext{O}_6	ext{Cl}_3	ext{Cr}$. If the complex on treatment with conc. $	ext{H}_2	ext{SO}_4$ loses $13.5\%$ of its original mass, the correct molecular formula of (A) is
[Given : atomic mass of $	ext{Cr} = 52	ext{ amu}$ and $	ext{Cl} = 35	ext{ amu}$]

Options:

(A) $[	ext{Cr}(	ext{H}_2	ext{O})_4	ext{Cl}_2]	ext{Cl}\cdot 2	ext{H}_2	ext{O}$

(B) $[	ext{Cr}(	ext{H}_2	ext{O})_6]	ext{Cl}_3$

(C) $[	ext{Cr}(	ext{H}_2	ext{O})_5	ext{Cl}]	ext{Cl}_2\cdot 	ext{H}_2	ext{O}$

(D) $[	ext{Cr}(	ext{H}_2	ext{O})_3	ext{Cl}_3]\cdot 3	ext{H}_2	ext{O}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let's embark on a fascinating journey into the world of coordination chemistry. We are given a complex with the empirical formula $	ext{H}_{12}	ext{O}_6	ext{Cl}_3	ext{Cr}$. Our first task is to understand the total mass we are dealing with. By summing up the atomic masses of all the constituent atoms, we can find the molar mass of the complex.

The molar mass $M$ is calculated as:
$$M = 52 + 3(35.5) + 6(18) = 266.5 	ext{ g/mol}$$

This $266.5 	ext{ g/mol}$ represents the total mass of one mole of our starting material.

### The Role of the Reagent

The problem introduces a classic chemical actor: concentrated sulfuric acid ($	ext{H}_2	ext{SO}_4$). What is its role here? Concentrated $	ext{H}_2	ext{SO}_4$ is renowned for its powerful dehydrating properties. When it interacts with a coordination complex, it acts like a sponge, specifically targeting and absorbing water molecules.

However, there is a catch! It can only easily remove the water molecules that are loosely held outside the coordination sphere—these are known as the **water of crystallization**. The water molecules tightly bound directly to the central metal ion (inside the square brackets) are safe from its grasp.

### Calculating the Lost Mass

We are told that the complex loses $13.5\%$ of its original mass upon treatment. Let's translate this percentage into a tangible mass in grams for one mole of the complex.

$$	ext{Mass lost} = \frac{13.5}{100} 	imes 266.5 \approx 36 	ext{ g}$$

This $36 	ext{ g}$ is the exact mass of the water molecules that were stripped away by the acid.

### Deducing the Molecular Formula

Now, we need to figure out how many water molecules correspond to this $36 	ext{ g}$. Since the molar mass of a single water molecule ($	ext{H}_2	ext{O}$) is $18 	ext{ g/mol}$, we can easily find the number of moles lost:

$$	ext{Moles of } 	ext{H}_2	ext{O lost} = \frac{36}{18} = 2 	ext{ moles}$$

This is our breakthrough! Losing exactly $2 	ext{ moles}$ of water means that there are **two water molecules residing outside the coordination sphere**.

Given that the empirical formula contains a total of $6$ water molecules, the remaining $4$ must be inside the coordination sphere, acting as ligands directly bonded to the Chromium ion. Chromium(III) typically exhibits a coordination number of $6$, forming octahedral complexes. To satisfy this coordination number, we need $2$ more ligands inside the sphere, which will be provided by the chloride ions.

This leaves one chloride ion outside the sphere to act as a counter ion. Putting it all together, the correct structural formula is:

$$[	ext{Cr}(	ext{H}_2	ext{O})_4	ext{Cl}_2]	ext{Cl} \cdot 2	ext{H}_2	ext{O}$$

This perfectly matches option (a).

Select Answer:

Visualized Solution

\text{Molar Mass of Complex}

\text{Role of Conc. } \text{H}_2\text{SO}_4

\text{Mass Lost Setup}

\text{Evaluating Lost Mass}

\text{Moles of Water Lost}

\text{Deducing the Formula}

\text{The Way Forward}

The Sigma Insight: Nomenclature, Isomerism, Importance and Werner's Theory

Analyzing the Setup

The Role of the Reagent

Calculating the Lost Mass

Deducing the Molecular Formula

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