The correct order of decreasing ionic radii among the following isoelectronic species
\(K^+, \, Ca^{2+}, \, Cl^-, \, S^{2-}\)
\(K^+, \, Cl^-, \, Ca^{2+}, \, S^{2-}\)
\(S^{2-}, \, Ca^{2+}, \, Cl^-, \, K^+\)
\(S^{2-}, \, Cl^-, \, K^+, \, Ca^{2+}\)
Choose the correct answer from the options given below:

Question

Consider the following sequence of reactions:

4-nitrotoluene
Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce_______ mg of B. (Given molar mass in g mol⁻¹ H: 1, C: 12, N: 14, O: 16, Br: 80)}

Answer 1

In an isoelectronic series, a greater nuclear charge results in a smaller ionic radius. Consequently, the ionic radii of the given isoelectronic species decrease in the following order, corresponding to increasing nuclear charge: \[ \text{S}^{2-} > \text{Cl}^- > \text{K}^+ > \text{Ca}^{2+}. \] \( \text{S}^{2-} \) possesses the smallest nuclear charge, while \( \text{Ca}^{2+} \) has the largest.

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The Correct Option is D

Solution and Explanation

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The correct order of decreasing ionic radii among the following isoelectronic species\(K^+, \, Ca^{2+}, \, Cl^-, \, S^{2-}\)\(K^+, \, Cl^-, \, Ca^{2+}, \, S^{2-}\)\(S^{2-}, \, Ca^{2+}, \, Cl^-, \, K^+\)\(S^{2-}, \, Cl^-, \, K^+, \, Ca^{2+}\)Choose the correct answer from the options given below:

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The Correct Option is D

Solution and Explanation

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Questions Asked in CUET (PG) exam