JEE Chemistry 2026 Question with Solution

Given: Two statements about second ionisation enthalpy and ionic radius are to be checked.

Find: Which option correctly describes the truth values of Statement I and Statement II.

For Statement I:

$$IE_2(\text{Na})$$

involves removing an electron from the stable

$$2p^6$$

noble gas configuration of

$$\text{Na}^+$$

whereas

$$IE_2(\text{Mg})$$

involves removing the

$$3s^1$$

electron from

$$\text{Mg}^+$$

Therefore,

$$IE_2(\text{Na}) > IE_2(\text{Mg})$$

So Statement I is true.

For Statement II:

$$O^{2-}$$

and

$$F^-$$

are isoelectronic species with $10$ electrons. For isoelectronic species, ionic radius decreases as effective nuclear charge increases. Since

$$Z_{\text{eff}}(O^{2-}) < Z_{\text{eff}}(F^-)$$

we get

$$r(O^{2-}) > r(F^-)$$

So Statement II is true.

Therefore, both Statement I and Statement II are true. The correct option is C.

Given: One statement compares $IE_2$ of Na and Mg, and the other compares the ionic radii of $O^{2-}$ and $F^-$.

Find: Evaluate each statement separately.

1. For Na:

$$\text{Na} = 1s^2 2s^2 2p^6 3s^1$$

After losing one electron,

$$\text{Na}^+ = 1s^2 2s^2 2p^6$$

This is a noble gas configuration, so removing the next electron requires very high energy.

2. For Mg:

$$\text{Mg} = 1s^2 2s^2 2p^6 3s^2$$

After losing one electron,

$$\text{Mg}^+ = 1s^2 2s^2 2p^6 3s^1$$

The second ionisation removes a

$$3s^1$$

electron, which is easier than removing an electron from the noble gas core of

$$\text{Na}^+$$

Hence,

$$IE_2(\text{Na}) > IE_2(\text{Mg})$$

So Statement I is true.

3. Now compare

$$O^{2-}$$

and

$$F^-$$

Both have $10$ electrons, so they are isoelectronic.

4. In an isoelectronic series, the species with smaller nuclear charge has larger radius. Here,

$$Z(O^{2-}) = 8, \quad Z(F^-) = 9$$

Therefore,

$$O^{2-}$$

has the larger ionic radius.

So Statement II is true.

Therefore, both statements are true, so the correct option is C.