Question:medium

Which among the following pairs are not having same number of total electrons?

Show Hint

Always calculate the actual number of electrons after adding or removing electrons due to ionic charge.
Updated On: Jun 5, 2026
  • $\mathrm{Na}^+$ and $\mathrm{Al}^{3+}$
  • $\mathrm{O}^{2-}$ and $\mathrm{F}^{-}$
  • $\mathrm{Mg}^{2+}$ and $\mathrm{Ar}$
  • $\mathrm{P}^{3-}$ and $\mathrm{Ar}$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understand the question.
We must find the pair that does NOT have the same number of electrons. Species with equal electron counts are called isoelectronic. So we just count electrons in each species and look for the odd pair out.

Step 2: Recall the counting rule.
For an atom, electrons equal the atomic number $Z$. For a positive ion, subtract the charge; for a negative ion, add the charge. This is because removing electrons gives a positive ion and adding electrons gives a negative ion.

Step 3: Check $\mathrm{Na}^+$ and $\mathrm{Al}^{3+}$.
Sodium ion: $11-1=10$. Aluminium ion: $13-3=10$. Both have $10$, so they match.

Step 4: Check $\mathrm{O}^{2-}$ and $\mathrm{F}^{-}$.
Oxide ion: $8+2=10$. Fluoride ion: $9+1=10$. Both have $10$, so they match too.

Step 5: Check $\mathrm{Mg}^{2+}$ and $\mathrm{Ar}$.
Magnesium ion: $12-2=10$. Argon atom: $18$. These are $10$ and $18$, which are different. This is our odd pair.

Step 6: Confirm with $\mathrm{P}^{3-}$ and $\mathrm{Ar}$.
Phosphide ion: $15+3=18$. Argon: $18$. These match, so they are not the answer. Hence the pair that is not equal is $\mathrm{Mg}^{2+}$ and $\mathrm{Ar}$, option 3. \[ \boxed{\mathrm{Mg}^{2+}\ \text{and}\ \mathrm{Ar}} \]
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