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(C) To determine the hybridisation state,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,where $V$ is the

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$NH_4^+, H_3O^+, OF_2$

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(C) To determine the hybridisation state,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For option $C$:$1$. $NH_4^+$: $SN = \frac{1}{2} (5 + 4 - 1 + 0) = 4$ ($sp^3$ hybridisation). Shape: Tetrahedral.$2$. $H_3O^+$: $SN = \frac{1}{2} (6 + 3 - 1 + 0) = 4$ ($sp^3$ hybridisation). Shape: Pyramidal.$3$. $OF_2$: $SN = \frac{1}{2} (6 + 2 - 0 + 0) = 4$ ($sp^3$ hybridisation). Shape: Bent.All three species have $sp^3$ hybridisation but different shapes due to the presence of different numbers of lone pairs on the central atom.

Which of the following sets of species have the same hybridisation state but different shapes?

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