JEE Chemistry 2026 Question with Solution

Given: A $p$-block element $E$ forms a binary cation $\left(EH_x\right)^+$, and $EH_3$ gives basic mercury(II) amido-iodide with $K_2HgI_4$ in alkaline medium.

Find: The correct first ionisation enthalpy value for element $E$.

The formation of a protonated binary cation indicates ammonium-type behaviour, which is characteristic of group $15$ hydrides.

The test with $K_2HgI_4$ in alkaline medium is a confirmatory test for ammonia or ammonia-like hydrides, so $EH_3$ behaves like $NH_3$. Therefore, element $E$ belongs to group $15$.

Now match the given first ionisation enthalpy values of the first elements of groups $13$ to $16$:

$$\text{Group 13: } 801,\quad \text{Group 14: } 1086,\quad \text{Group 15: } 1402,\quad \text{Group 16: } 1312$$

Hence, for group $15$, the first ionisation enthalpy is

$$1402 \, \text{kJ mol}^{-1}$$

Therefore, the correct option is A.

Given: $EH_3$ gives a precipitate of basic mercury(II) amido-iodide with $K_2HgI_4$ in alkaline medium.

Find: Which first ionisation enthalpy corresponds to element $E$.

This reagent test is associated with ammonia-like behaviour. Among the first elements of groups $13$, $14$, $15$ and $16$, the element whose hydride most closely matches this behaviour is the group $15$ element.

So the relevant first element is nitrogen, whose first ionisation enthalpy is

$$1402 \, \text{kJ mol}^{-1}$$

Thus the correct option is A.