Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Match List-I with List-II:
| List-I Species | List-II Geometry/Shape | ||
|---|---|---|---|
| A. | H$$_3$$O$$^+$$ | I. | Tetrahedral |
| B. | Acetylide anion | II. | Linear |
| C. | NH$$_4^+$$ | III. | Pyramidal |
| D. | ClO$$_2^-$$ | IV. | Bent |
We need to determine the geometry/shape of each species using VSEPR theory.
A. H$$_3$$O$$^+$$:
Central atom O has 3 bond pairs + 1 lone pair = 4 electron pairs (sp$$^3$$ hybridized).
Shape: Pyramidal (III)
B. Acetylide anion (C$$_2^{2-}$$):
The acetylide anion has a triple bond between two carbon atoms: $$[:C \equiv C:]^{2-}$$
Shape: Linear (II)
C. NH$$_4^+$$:
Central atom N has 4 bond pairs + 0 lone pairs = 4 electron pairs (sp$$^3$$ hybridized).
Shape: Tetrahedral (I)
D. ClO$$_2^-$$:
Central atom Cl has 2 bond pairs + 2 lone pairs = 4 electron pairs (sp$$^3$$ hybridized).
Shape: Bent (IV)
Therefore: A → III, B → II, C → I, D → IV
Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.