Question:

A forms $hcp$ lattice and B are occupying $1/3^{rd}$ of tetrahedral voids, then the formula of compound is

Updated On: Jul 5, 2022
  • $AB$
  • $A_3B_2$
  • $A_2B_3$
  • $AB_4$
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The Correct Option is B

Solution and Explanation

Let no. of atoms of A used in close packing = n Number of tetrahedral voids = 2n Number of atoms of $B=\frac{1}{3}\times2n=\frac{2}{3}n$ $A : B=n : \frac{2}{3}n=3 : 2$ Formula of the compound =$ A_{3}B_{2}$
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Concepts Used:

Solid State

Solids are substances that are featured by a definite shape, volume, and high density. In the solid-state, the composed particles are arranged in several manners. Solid-state, in simple terms, means "no moving parts." Thus solid-state electronic devices are the ones inclusive of solid components that don’t change their position. Solid is a state of matter where the composed particles are arranged close to each other. The composed particles can be either atoms, molecules, or ions. 

Solid State

Types of Solids:

Based on the nature of the order that is present in the arrangement of their constituent particles solids can be divided into two types;

  • Amorphous solids behave the same as super cool liquids due to the arrangement of constituent particles in short-range order. They are isotropic and have a broad melting point (range is about greater than 5°C).
  • Crystalline solids have a fixed shape and the constituent particles are arranged in a long-range order.