Aqueous solution of which of the following salts will not contain ions with the electronic configuration $$1s^2 2s^2 2p^6 3s^2 3p^6$$?

First, let us write clearly the electronic configuration given in the question:

$$1s^2\,2s^2\,2p^6\,3s^2\,3p^6$$

This configuration contains a total of $$2+2+6+2+6 = 18$$ electrons. Thus any ion that possesses exactly eighteen electrons will have this noble-gas (argon) configuration.

To decide which salt does not give any ion with 18 electrons, we must examine every ion produced by each salt on dissolution in water and count its electrons one by one. We recall the simple fact:

Number of electrons in an ion $$=$$ atomic number of the element $$\pm$$ charge on the ion.

If the ion is positive (cation), we subtract electrons; if it is negative (anion), we add electrons.

Salt A  $$\bigl({\rm NaF}\bigr)$$

On dissolving $$NaF$$ we obtain $$Na^+$$ and $$F^-$$.

For $$Na^+$$: atomic number of Na is $$11$$, charge $$+1$$ ⇒ electrons $$= 11-1 = 10$$.

For $$F^-$$: atomic number of F is $$9$$, charge $$-1$$ ⇒ electrons $$= 9+1 = 10$$.

Neither ion possesses $$18$$ electrons.

Salt B  $$\bigl({\rm KBr}\bigr)$$

Dissolution gives $$K^+$$ and $$Br^-$$.

For $$K^+$$: atomic number $$19$$, charge $$+1$$ ⇒ electrons $$= 19-1 = 18$$.

We already obtain an 18-electron ion here, so $$KBr$$ does furnish the required configuration.

Salt C  $$\bigl({\rm NaCl}\bigr)$$

Dissolution gives $$Na^+$$ and $$Cl^-$$.

For $$Cl^-$$: atomic number $$17$$, charge $$-1$$ ⇒ electrons $$= 17+1 = 18$$.

Thus $$NaCl$$ also supplies an ion with the argon configuration.

Salt D  $$\bigl({\rm CaI_2}\bigr)$$

Dissolution gives $$Ca^{2+}$$ and two $$I^-$$ ions.

For $$Ca^{2+}$$: atomic number $$20$$, charge $$+2$$ ⇒ electrons $$= 20-2 = 18$$.

So $$CaI_2$$ likewise produces an 18-electron ion.

Let us summarise:

$$\begin{array}{lcl} NaF &:& \;10\;{\rm e^-,\;}10\;{\rm e^-} \\ KBr &:& \;18\;{\rm e^-} \\ NaCl &:& \;18\;{\rm e^-} \\ CaI_2 &:& \;18\;{\rm e^-} \end{array}$$

Only the aqueous solution of $$NaF$$ fails to contain any ion with $$18$$ electrons.

Hence, the correct answer is Option A.