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

Step2: Consider the nuclear charge

For isoelectronic species, the greater the nuclear charge (number of protons in the nucleus), the more the electrons are attracted towards the nucleus, and the smaller the ionic radius.

• The atomic number of \(Sr\) is \(38\), so \(Sr^{2+}\) has \(38 - 2=36\) electrons and \(Z = 38\) (nuclear charge).
• The atomic number of \(Rb\) is \(37\), so \(Rb^{+}\) has \(37- 1 = 36\) electrons and \(Z=37\).
• The atomic number of \(Br\) is \(35\), so \(Br^{-}\) has \(35 + 1=36\) electrons and \(Z = 35\).
• The atomic number of \(Se\) is \(34\), so \(Se^{2-}\) has \(34+2 = 36\) electrons and \(Z = 34\).

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

Answer:

\(Sr^{2+}