JEE Chemistry 2026 Question with Solution

Given: Two statements about atomic/ionic radii and electron gain enthalpy are to be checked.

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

Statement I: Across Period 3, atomic radius decreases from left to right. Therefore,

$$Mg > Al$$

Also, $Mg^{2+}$ and $Al^{3+}$ are isoelectronic. For isoelectronic species, higher positive charge means smaller radius. Hence,

$$Mg^{2+} > Al^{3+}$$

So the correct overall order is

$$Mg > Al > Mg^{2+} > Al^{3+}$$

Therefore, Statement I is false.

Statement II: The magnitude of electron gain enthalpy is higher for halogens than for Group 16 elements. Also, chlorine has higher magnitude than bromine, and sulfur has higher magnitude than oxygen because of greater interelectronic repulsion in the small oxygen atom.

$$Cl > Br > S > O$$

Therefore, Statement II is true.

Hence, Statement I is false but Statement II is true. The correct option is C.

Given:

• A radius-order statement involving $Al$, $Mg$, $Mg^{2+}$, and $Al^{3+}$
• An electron gain enthalpy order involving $Cl$, $Br$, $S$, and $O$

Find: Evaluate each statement using periodic properties.

1. Atomic size trend: Atomic radius decreases across a period because effective nuclear charge increases.
2. Since $Mg$ lies to the left of $Al$ in Period 3,

$$Mg > Al$$

3. Cation size trend: Cations are smaller than their parent atoms.
4. Isoelectronic ion comparison: $Mg^{2+}$ and $Al^{3+}$ have the same number of electrons. Among isoelectronic species, radius decreases as nuclear charge increases. Therefore,

$$Mg^{2+} > Al^{3+}$$

5. Combining these results,

$$Mg > Al > Mg^{2+} > Al^{3+}$$

So the proposed order in Statement I is incorrect, hence false.

For Statement II, electron gain enthalpy generally becomes more negative across a period and less negative down a group, with known exceptions.

• Among halogens,

$$Cl > Br$$

in magnitude.

• Between $S$ and $O$, sulfur has greater magnitude than oxygen because the very small size of oxygen causes stronger electron-electron repulsion in the compact $2p$ subshell.

$$S > O$$

Thus,

$$Cl > Br > S > O$$

This order is correct in magnitude, so Statement II is true.

Therefore, the final conclusion is: Statement I is false but Statement II is true, so the correct option is C.