Question

In a cubic unit cell of an ionic compound, the eight corners are occupied by anions and cations at the center of the cube. Calculate the volume of the unit cell if density of the unit cell is \(4\text{ g cm}^{-3}\).
[Molar mass of compound \(= 168.6\) g mol\(^{-1}\) and \(N_A = 6.022\times 10^{23}\)]

• A. \(5.0\times 10^{-23}\text{ cm}^3\)
• B. \(6.0\times 10^{-23}\text{ cm}^3\)
• C. \(7.0\times 10^{-23}\text{ cm}^3\)
• D. \(8.0\times 10^{-23}\text{ cm}^3\)

Hint:

Count $Z$ first: corners give $8\times\frac18=1$ anion, center gives 1 cation.

Answer 1

The Correct Option is C

Step 1: Count formula units
Anions at 8 corners give \(8\times\frac18=1\) anion. One cation at the centre. So the cell holds \(Z=1\) formula unit (like CsCl).

Step 2: Use the density formula
\[ \rho=\frac{ZM}{N_AV}\;\Rightarrow\;V=\frac{ZM}{\rho N_A} \]

Step 3: Substitute
\[ V=\frac{1\times168.6}{4\times6.022\times10^{23}}=\frac{168.6}{2.409\times10^{24}}=7.0\times10^{-23}\text{ cm}^3 \]

Step 4: Check
If one used \(Z=2\) the answer would double, which is not an option. So option (C).

Final Answer:
\(V=7.0\times10^{-23}\text{ cm}^3\), option (C). \[ \boxed{\text{(C)}} \]