NEET Re-2022 Q19: Bond Angle Trends — Full solution + Why they asked this

PYQ + Trends

1. TITLE

NEET Re-2022 Q19: Bond Angle Trends — Full solution + Why they asked this

2. THE ACTUAL NEET QUESTION

NEET Re-2022 — Question
The correct order of bond angles in the following compounds/ species is :

A. $CO_{2} < NH_{3} < H_{2}O < NH_{4}^{+}$
B. $H_{2}O < NH_{3} < NH_{4}^{+} < CO_{2}$
C. $H_{2}O < NH_{4}^{+} < NH_{3} < CO_{2}$
D. $H_{2}O < NH_{4}^{+} = NH_{3} < CO_{2}$

3. OFFICIAL ANSWER

NEET Answer Key: (B)

4. COMPLETE SOLUTION

Step-by-Step Analysis:

To solve this, we must determine the hybridization and the number of lone pairs for each species. According to VSEPR theory, lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion, which squeezes the bond angle.

1. $CO_{2}$ (Carbon Dioxide):
* Hybridization: The Carbon has 2 sigma bonds and 0 lone pairs. Hybridization is sp.
* Geometry: Linear.
* Bond Angle: 180°.

2. $NH_{4}^{+}$ (Ammonium Ion):
* Hybridization: Nitrogen has 5 valence electrons. +1 charge means 4 electrons. It forms 4 sigma bonds with H.
* Pairs: 4 Bond Pairs (bp), 0 Lone Pairs (lp).
* Hybridization: $sp^3$.
* Geometry: Tetrahedral (Ideal).
* Bond Angle: 109.5°.

3. $NH_{3}$ (Ammonia):
* Hybridization: Nitrogen has 5 valence electrons. Forms 3 bonds, 2 electrons remain.
* Pairs: 3 Bond Pairs (bp), 1 Lone Pair (lp).
* Hybridization: $sp^3$.
* Geometry: Trigonal Pyramidal.
* Deviation: The lone pair pushes the bonds down.
* Bond Angle: 107°.

4. $H_{2}O$ (Water):
* Hybridization: Oxygen has 6 valence electrons. Forms 2 bonds, 4 electrons remain.
* Pairs: 2 Bond Pairs (bp), 2 Lone Pairs (lp).
* Hybridization: $sp^3$.
* Geometry: Bent / V-shaped.
* Deviation: Two lone pairs create maximum repulsion ($lp-lp > lp-bp$).
* Bond Angle: 104.5°.

Final Ordering:
$H_{2}O$ ($104.5^{\circ}$) < $NH_{3}$ ($107^{\circ}$) < $NH_{4}^{+}$ ($109.5^{\circ}$) < $CO_{2}$ ($180^{\circ}$)

This matches Option B.

5. WHY NEET ASKED THIS

The examiner is testing your mastery of VSEPR Theory beyond just memorizing shapes.
* Concept Tested: The hierarchy of repulsion: Lone Pair-Lone Pair > Lone Pair-Bond Pair > Bond Pair-Bond Pair.
* The Trap: Students often assume all $sp^3$ hybridized molecules have the same angle ($109.5^{\circ}$). You must adjust for the number of lone pairs.
* Intent: To verify if you can integrate hybridization ($sp$ vs $sp^3$) with VSEPR distortions (Lone pairs reducing angles).

6. TREND ANALYSIS

PATTERN ALERT:
Questions on bond parameters (Angle, Length, Moment) involving Hydrides ($NH_3, H_2O$) and simple p-block molecules appear consistently.

Prediction for NEET 2026:
Expect a shift to Group 15/16 Vertical Trends (Drago's Rule context).
* Likely Question: Compare bond angles of $NH_3$ vs $PH_3$ vs $AsH_3$.
* Concept: While $NH_3$ is $107^{\circ}$, $PH_3$ drops drastically to $\sim93^{\circ}$ because lower group elements (P, As, Sb, S, Se) tend to use pure p-orbitals for bonding rather than hybrid orbitals (Drago's Rule condition), keeping angles near $90^{\circ}$.

7. SIMILAR QUESTIONS TO PRACTICE

"If you understood this, you can solve:"
* NEET 2016 (Phase I) Q43: "Consider the molecules $CH_{4}, NH_{3}$ and $H_{2}O$. Which of the given statements is false?" (Direct comparison of angles).
* NEET 2015 Cancelled Q47: Maximum bond angle in Nitrogen species ($NO_2, NO_2^+, NO_2^-$).
* Practice Variation: Arrange bond angles: $CH_4, NH_3, PH_3, H_2O$. (Hint: $PH_3$ is the smallest, near $93^{\circ}$).

8. QUICK REVISION FOR THIS TYPE

For Bond Angle questions, always follow this priority:

1. Check Hybridization:
   • $sp$ ($180^{\circ}$) > $sp^2$ ($120^{\circ}$) > $sp^3$ ($109.5^{\circ}$).
2. Check Lone Pairs (if hybridization is same):
   • More Lone Pairs = Smaller Angle.
   • Example: $CH_4$ (0 lp, $109^{\circ}$) > $NH_3$ (1 lp, $107^{\circ}$) > $H_2O$ (2 lp, $104.5^{\circ}$).
3. Watch out for Drago's Rule (Down the Group):
   • If central atom is Period 3 or below (P, S, As) and surrounds are less electronegative (like H): Angle drops to $\sim90^{\circ}$.
   • Example: $H_2O$ ($104.5^{\circ}$) >>> $H_2S$ ($92^{\circ}$) > $H_2Se$ ($91^{\circ}$).

9. NCERT BASE

This question tests: NCERT Class 11, Chapter 4 (Chemical Bonding), Page 116, Table 4.8 (Shapes of Molecules containing Bond Pair and Lone Pair).

10. CTA

All PYQs from this chapter → Chemical Bonding Guide
Practice similar questions → Mock exams