Question

An ionic compound is made up of A cations and B anions. If A cations are present at the alternate corners and B anions are present on the body of the diagonal, what is the formula of the ionic compound?

Answer 1

Hint: To solve the problems, firstly we have to calculate the total number of atoms which will contribute to the alternate corners and the body diagonal. Then the ratio which will come will be the formula of the compound.

Complete step by step Answer:
- In the given question, we have to find the molecular formula of the ionic solid from the given data.
- It is given in the question that the cation A is present on the alternate corners and we know that there are a total of 8 corners in the cube so the total number of cations will be 4.
- As each corner will contribute $\text{1/8}$ to each atom, so the total number of an atom will be:
$\dfrac{1}{8}\text{ }\times \,\text{ 4 = 0}\text{.5}$
- Now, as we know that cation B is present at the body diagonal so there is only one atom which is present at the centre of the cube.
- So, total no. of the anion B is 1.
- Hence, the ratio of cation A to the cation B will be 0.5:1 or the whole number ratio will be 2: 1.
- So, the molecular formula will be ${{\text{A}}_{2}}\text{B}$.

Therefore, ${{\text{A}}_{2}}\text{B}$ is the correct answer.

Note: When the atoms are present on the corner of the cube and on the body diagonal then such type of unit cell is known as body-centred cubic or BCC. Whereas FCC or face Centred Cubic is the unit cell in the atoms are present on the corners and at the centre of the face of the cube.