Which of the following sets of species have t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$NH_4^+, H_3O^+, OF_2$

Vedclass · Answer

(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?

Similar Questions

Why is the bond angle in $CH_4$ not $90^{\circ}$,even though the outermost $p$-orbitals of carbon are at $90^{\circ}$ to each other?

In the following molecule,determine the hybridisation of carbon atoms $a, b$ and $c$ respectively:

What are the hybridization states of carbon in diamond,graphite,and acetylene,respectively?

Which one of the following is a correct set?

In which one of the following species does the central atom have a type of hybridization that is not the same as that present in the other three?

Vedclass Test Series

Exam Paper Generator

Online Exam Module