JEE MainChemistryp Block Elements (Group 15, 16, 17 & 18)
Four elements of Group 15 form hydrides denoted as EH ₃ . The E-H bond lengths for these four hydrides are given as L₁ < L₂ < L₃ < L₄ . Which of the following correctly describes the properties of the hydride with the bond length L₄ among the four?
Options
- AIt has the lowest basicity and the highest reducing power.
- BIt has the highest basicity and the lowest reducing power.
- CIt has the lowest basicity and the lowest reducing power.
- DIt has the highest basicity and the highest reducing power.
Correct answer
A. It has the lowest basicity and the highest reducing power.
Step-by-step solution
The E-H bond length in Group 15 hydrides increases down the group as the atomic size of the central atom E increases. Therefore, the hydride with the longest bond length, L₄ , corresponds to the heaviest element among the four. As we move down the group, the electron density of the lone pair is dispersed over a larger volume due to the increasing size of the central atom, which leads to a decrease in Lewis basicity. Thus, the heaviest hydride has the lowest basicity. Furthermore, the increase in bond length results