Question

Find the correct statements related to group 15 hydrides:
A. Reducing nature increases from NH₃ to BiH₃
B. Tendency to donate lone pair of electrons decreases from NH₃ to BiH₃
C. The stability of hydrides decreases from NH₃ to BiH₃
D. HEH bond angle decreases from NH₃ to SbH₃

• A. A and B only
• B. B and C only
• C. A, B, C and D
• D. A, C and D Only

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We evaluate the physical and chemical trends of Group 15 hydrides ($EH_3$), moving down the group from Nitrogen to Bismuth. Trends in bond length, bond dissociation enthalpy, electron density, and orbital hybridization shape these properties.
Step 2: Key Formula or Approach:
As we go down the group:
- Atomic size of the central atom E increases.
- E-H bond length increases, so E-H bond strength/dissociation enthalpy decreases.
- Electronegativity of central atom decreases.
Step 3: Detailed Explanation:
Let's analyze each statement sequentially:
A. Reducing nature depends on the ease of breaking the E-H bond to liberate hydrogen. Since the atomic size increases down the group ($N<P<As<Sb<Bi$), the E-H bond length increases and bond dissociation enthalpy drops rapidly. Thus, it's easier to give off hydrogen, making $BiH_3$ the strongest reducing agent. Statement A is correct.
B. The Lewis basicity (tendency to donate a lone pair) depends on electron density. While all have one lone pair, the volume of the central atom increases drastically down the group. The lone pair electron density gets highly diffused over a larger volume, lowering the capacity to effectively donate it. So, basicity drops from $NH_3$ to $BiH_3$. Statement B is correct.
C. Thermal stability depends entirely on the E-H bond strength. As highlighted earlier, increasing bond length means decreasing bond strength. $BiH_3$ is highly unstable compared to $NH_3$. Statement C is correct.
D. Bond angles: For $NH_3$, $sp^3$ hybridization with high electronegativity causes bond pairs to repel closer to the central atom, pushing the angle to $107.8^\circ$. Moving down the group, electronegativity drops, bond pairs move further from the central atom towards H, lowering repulsion. Drago's Rule indicates that for $PH_3$ and below, hybridization barely occurs, and nearly pure $p$-orbitals are used, causing bond angles to drop close to $90^\circ$ ($PH_3 \approx 93.6^\circ, SbH_3 \approx 91.3^\circ$). Statement D is correct.
Since all four statements accurately describe the chemical behaviors, the correct choice includes all of them.
Step 4: Final Answer:
Statements A, B, C, and D are all correct.