KCET · Chemistry · Classification of Elements and Periodicity in Properties

Volume occupied by single \( \mathrm{CsCl} \) ion pair in a crystal is \( 7.014 \times 10^{-23} \mathrm{~cm}^{3} \). The smallest \( \mathrm{Cs}- \)
Cs inter-nuclear distance is equal to length of the side of the cube corresponding to volume of
one \( \mathrm{CsCl} \) ion pair. The smallest \( \mathrm{Cs}-\mathrm{Cs} \) inter-nuclear distance is nearly

A \( 4.3 Å \)
B \( 4.5 Å \)
C \( 4.4 Å \)
D None of the above

Verified Solution

Answer & Solution

Correct Answer

(D) None of the above

Step-by-step Solution

Detailed explanation

Volume of the unit cell \(=7.014 \times 10^{-23}\)
\(\Rightarrow a^{3}=7.014 \times 10^{-23}\) (a is the smallest Cs-Cs distance)
\(a=\sqrt{7.014 \times 10^{-23}}\)
\(a=4.12 Å\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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