Question

Identify the two ions having the highest and lowest length of ionic radius respectively out of the following: \[ F^{-},\quad Mg^{2+},\quad Na^{+},\quad O^{2-} \]

• A. \(O^{2-},\; Mg^{2+}\)
• B. \(O^{2-},\; Na^{+}\)
• C. \(F^{-},\; Mg^{2+}\)
• D. \(F^{-},\; Na^{+}\)

Hint:

For an isoelectronic series: \[ \text{Radius} \propto \frac{1}{\text{Nuclear Charge}} \] Thus, \[ O^{2-} \gt F^{-} \gt Na^{+} \gt Mg^{2+} \]

Answer 1

The Correct Option is A

Concept: All the given ions are isoelectronic. \[\begin{aligned} O^{2-},\; F^{-},\; Na^{+},\; Mg^{2+} \end{aligned}\] each contains \[\begin{aligned} 10 \text{ electrons} \end{aligned}\] For an isoelectronic series, ionic radius decreases as nuclear charge increases.

Step 1: Determine the nuclear charges. \[ \begin{aligned} O^{2-} &: \quad Z=8,\qquad \text{Electrons}=10 \\ F^{-} &: \quad Z=9,\qquad \text{Electrons}=10 \\ Na^{+} &: \quad Z=11,\qquad \text{Electrons}=10 \\ Mg^{2+} &: \quad Z=12,\qquad \text{Electrons}=10 \end{aligned} \]

Step 2: Apply the isoelectronic rule. Higher nuclear charge pulls electrons more strongly. Therefore, \[\begin{aligned} O^{2-} \gt F^{-} \gt Na^{+} \gt Mg^{2+} \end{aligned}\]

Step 3: Identify highest and lowest ionic radii. \[\begin{aligned} \text{Highest radius} &= O^{2-} \\ \text{Lowest radius} &= Mg^{2+} \end{aligned}\] \[\begin{aligned} \boxed{ O^{2-},\; Mg^{2+} } \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.