JEE Main On 8 April 2017 Question 17
Question: Consider the following ionization enthalpies of two elements ‘A’ and ‘B’.
| Element | Ionization enthalpy (kJ/mol) |
|---|---|
| 1st 2nd 3rd | |
| A | 899 1757 14847 |
| B | 737 1450 7731 |
Which of the following statements is correct? [JEE Online 08-04-2017]
Options:
A) Both ‘A’ and ‘B’ belong to group-1 where ‘A’ comes below ‘B’.
B) Both ‘A’ and ‘B’ belong to group-2 where ‘A’ comes below ‘B’.
C) Both ‘A’ and ‘B’ belong to group-1 where ‘B’ comes below ‘A’.
D) Both ‘A’ and ‘B’ belong to group-2 where ‘B’ comes below ‘A’.
Show Answer
Answer:
Correct Answer: D
Solution:
Element A:
- 1st ionization: $899 \mathrm{~kJ} / \mathrm{mol}$
- 2nd ionization: $1757 \mathrm{~kJ} / \mathrm{mol}$
- 3rd ionization: $14847 \mathrm{~kJ} / \mathrm{mol}$ The huge jump from the 2nd to the 3rd ionization indicates that $A$ is losing its third electron from a stable configuration, typical for Group 2.
Element B:
- 1st ionization: $737 \mathrm{~kJ} / \mathrm{mol}$
- 2nd ionization: $1450 \mathrm{~kJ} / \mathrm{mol}$
- 3rd ionization: $7731 \mathrm{~kJ} / \mathrm{mol}$ Similar pattern to A, indicating B is also in Group 2.
Position in the Group: B has a lower 1st ionization energy than $A$, suggesting $B$ is lower in the group. Elements lower in a group have lower ionization energies because their outer electrons are further from the nucleus. These elements typically have two electrons in their outer shell. Removing the third electron (after the first two) requires a lot more energy because it breaks into a stable core.
So, both $A$ and $B$ are in Group 2, and $B$ comes below $A$. This matches statement $D$.
BETA
