Question

rank the following atoms/ions in order of atomic radius from smallest to largest: se²⁻ / sr²⁺ / rb⁺ / br⁻

Explanation:

Step1: Analyze the isoelectronic nature

All these species (\(Se^{2 - },Sr^{2+},Rb^{+},Br^{-}\)) are isoelectronic (have the same number of electrons, \(36\) electrons).

Step2: Consider the nuclear charge

For isoelectronic species, the greater the nuclear charge (\(Z\), number of protons in the nucleus), the smaller the ionic radius.

• The nuclear charge of \(Sr^{2+}\): \(Z = 38\)
• The nuclear charge of \(Rb^{+}\): \(Z = 37\)
• The nuclear charge of \(Br^{-}\): \(Z = 35\)
• The nuclear charge of \(Se^{2 - }\): \(Z = 34\)

Since \(Z(Sr^{2+})>Z(Rb^{+})>Z(Br^{-})>Z(Se^{2 - })\)

Answer:

\(Sr^{2+}