Problem 82 - A Complete Resource Book in Chemistry for JEE Main

Question

If the anions (b) form hexagonal close packing and cations (a) occupy only \(2 / 3\) octahedral holes in it, then the general formula of the compound is (a) \(\mathrm{AB}_{2}\) (b) \(\mathrm{A}_{2} \mathrm{~B}_{3}\) (c) \(\mathrm{A}_{3} \mathrm{~B}_{3}\) (d) \(\mathrm{A}_{2} \mathrm{~B}_{2}\)

Step-by-Step Solution

Answer

The correct formula is option (b) \( \mathrm{A}_{2} \mathrm{~B}_{3} \).

Step 1: Understand the Problem

In this exercise, we are given that anions form a hexagonal close-packing (hcp) structure and cations occupy only \( \frac{2}{3} \) of the octahedral holes. We need to determine the general chemical formula of the resulting compound.

Step 2: Determine the Ratio in HCP

In hexagonal close packing (hcp), let's assume the number of anions (\( B \)) in one full unit cell is \( x \). There are as many octahedral holes as there are anions, so there are \( x \) octahedral holes.

Step 3: Calculate Cations in the Structure

It's given that cations (\( A \)) occupy \( \frac{2}{3} \) of the available octahedral holes. Thus, the number of \( A \) ions is \( \frac{2}{3}x \).

Step 4: Establish Formula Ratio

The ratio of cations to anions is \( \frac{2}{3}x : x \), which simplifies to \( \frac{2}{3} : 1 \) or equivalently \( 2 : 3 \). Therefore, the chemical formula should reflect this ratio.

Step 5: Select the Correct Option

From the options provided, \( \text{A}_2 \text{~B}_3 \) matches the ratio \( 2:3 \) of cations to anions, as determined in step 4.

Key Concepts

Octahedral Holes

In crystal structures, elements arrange in specific patterns to achieve maximum stability. In the hexagonal close packing (hcp) arrangement, we come across a fascinating concept known as "octahedral holes." These are void spaces formed between the layers of packed spheres (typically atoms or ions). Octahedral holes get their name because they are situated in locations where an octahedron can fit. Each hole is surrounded by six spheres: three from the layer below and three from the layer above, creating a space that can be occupied by smaller particles, usually cations.

• In a complete hexagonal close packing cell, there are just as many octahedral holes as there are anions.
• These octahedral holes provide the accommodation necessary for cations to reside within the structure without disrupting the close packing of the anions.

Understanding how these holes accommodate cations helps in predicting the stoichiometry and the resultant chemical formula of a compound package.

Chemical Formula

A chemical formula succinctly denotes the types and numbers of atoms in a compound. It paints a clear picture of the constituent elements and their ratios. When dealing with compounds formed in crystalline structures like hexagonal close packing (hcp), chemical formulas become crucial to identify the ideal proportions of each element involved.To derive the chemical formula, we need to understand how the anions and cations interact. In the problem scenario:

• Anions form the hexagonal close packed base structure.
• Cations occupy the available octahedral holes, but not completely. They fill only \(\frac{2}{3}\) of these holes.

This limited occupation leads to a formula that must accommodate more anions than cations. By calculating the exact number of moles for these ions, we achieve a ratio that guides the chemical formula, reflecting the elemental incorporation within the compound, as revealed upon final determination of the ratio to match \(\text{A}_2 \text{B}_3\).

Cation to Anion Ratio

The cation to anion ratio is a crucial factor in defining the stoichiometry of ionic compounds. This ratio indicates how many cations there are for every anion in the compound's structure.In our exercise, an important step was to find that cations filled only \(\frac{2}{3}\) of the octahedral holes. Consequently, for every unit, if there are \(x\) anions, there are \(\frac{2}{3}x\) cations.

• The original ratio is expressed as \(\frac{2}{3} : 1\), showcasing the cations are fewer relative to the anions.
• When adjusted to whole numbers for practical use in chemical formulas, this translates to \(2 : 3\).

This ratio defines the general formula. Hence, the formula \(\text{A}_2 \text{B}_3\) is concluded, aligning perfectly with the available options, as it accurately represents the cation (A) and anion (B) mixing within the structure.