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

Step 1: Check the electron count.
Count electrons in each ion: $O^{2-}$ has $10$, $F^-$ has $10$, $Na^+$ has $10$, $Mg^{2+}$ has $10$. They are all isoelectronic with $10$ electrons.

Step 2: Pick the deciding factor.
When electron number is the same, size is decided by nuclear charge. More protons pull the same electron cloud in tighter, giving a smaller ion.

Step 3: List the nuclear charges.
$O^{2-}$ has $Z = 8$, $F^-$ has $Z = 9$, $Na^+$ has $Z = 11$, $Mg^{2+}$ has $Z = 12$.

Step 4: Find the largest ion.
Smallest nuclear charge means weakest pull, so $O^{2-}$ with $Z = 8$ is the biggest.

Step 5: Find the smallest ion.
Largest nuclear charge means strongest pull, so $Mg^{2+}$ with $Z = 12$ is the smallest.

Step 6: State the pair.
Highest radius is $O^{2-}$ and lowest is $Mg^{2+}$.
\[ \boxed{O^{2-},\ Mg^{2+}} \]