WBJEE · Chemistry · Classification of Elements and Periodicity in Properties

Which of the following is correct?

A Radius of \(\mathrm{Ca}^{2+} < \mathrm{Cl}^{-} < \mathrm{S}^{2-}\)
B Radius of \(\mathrm{Cl}^{-} < \mathrm{S}^{2-} < \mathrm{Ca}^{2+}\)
C Radius of \(\mathrm{S}^{2-}=\mathrm{Cl}^{-}=\mathrm{Ca}^{2+}\)
D Radius of \(\mathrm{S}^{2-} < \mathrm{Cl}^{-} < \mathrm{Ca}^{2+}\)

Verified Solution

Answer & Solution

Correct Answer

(A) Radius of \(\mathrm{Ca}^{2+} < \mathrm{Cl}^{-} < \mathrm{S}^{2-}\)

Step-by-step Solution

Detailed explanation

\begin{array}{c|c|c} \hline Species & Atomic number & Number of electrons \\ \hline \mathrm{Ca}^{2+} & 20 & 20-2=18 \\ \mathrm{Cl}^{-} & 17 & 17+1=18 \\ \mathrm{s}^{2-} & 16 & 16+2=18 \\ \hline \end{array} For isoelectronic species, an increase in atomic number, result in…

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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