Question

In the basic $CsCl$ crystal structure, ${Cs}^{+}$ and ${Cl}^{-}$ ions are arranged in a bcc configuration as shown in the figure. The net electrostatic force exerted by the eight ${Cs}^{+}$ ions on the ${Cl}^{-}$ ion is :

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{4e^2}{3a^2}$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{16e^2}{3a^2}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{32e^2}{3a^2}$
• D. $zero$

Answer 1

The correct option is D $zero$

$\text{Step 1: Net force on}$ $C{l}^{-}$ $\text{ion at the centre}$

From Figure, Position of $C{l}^{-}$ is symmetric at the centre, and distances of all $C{s}^{+}$ ions from center is same, so magnitude of all attraction forces on $C{l}^{-}$ will also be same.

Therefore, all $C{s}^{+}$ ions will attract $C{l}^{-}$ at center with equal forces in opposite directions. So, all their forces will cancel out.

Hence Net force on $C{l}^{-}$ ion will be zero.

Therefore correct answer is $D$

$\text{Step 2: Detailed Explanation of forces on}$ $C{l}^{-}$
The forces on $C{l}^{-}$ will be get cancelled due to equal and opposite forces.

Forces due to $1$ and $7$ will cancel

Forces due to $2$ and $8$ will cancel

Forces due to $3$ and $5$ will cancel

Forces due to $4$ and $6$ will cancel

So, Net force on $C{l}^{-}$ will be zero

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