Identify the molecule (X) with the maximum nu | vedclass

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

Question

(C) Let us calculate the total number of lone pairs for each molecule:$HNO_{3}$: $N$ has $0$, $O$ atoms have $2+2+3 = 7$ lone pairs. Total = $7$.$H_{2

Vedclass · Accepted Answer

$102$

Vedclass · Answer

(C) Let us calculate the total number of lone pairs for each molecule:$HNO_{3}$: $N$ has $0$, $O$ atoms have $2+2+3 = 7$ lone pairs. Total = $7$.$H_{2}SO_{4}$: $S$ has $0$, $O$ atoms have $2+2+2+2 = 8$ lone pairs. Total = $8$.$NF_{3}$: $N$ has $1$, each $F$ has $3$. Total = $1 + 3(3) = 10$ lone pairs.$O_{3}$: Central $O$ has $1$, terminal $O$ atoms have $2$ and $3$. Total = $6$ lone pairs.Thus, $NF_{3}$ has the maximum number of lone pairs $(10)$.The hybridization of $N$ in $NF_{3}$ is $sp^{3}$. Due to the high electronegativity of $F$ atoms, the bond angle is compressed to approximately $102^{\circ}$.

Identify the molecule $(X)$ with the maximum number of lone pairs of electrons (obtained using Lewis dot structure) among $HNO_{3}$, $H_{2}SO_{4}$, $NF_{3}$, and $O_{3}$. Choose the correct bond angle made by the central atom of the molecule $(X)$. (in $^{\circ}$)

Similar Questions

Amongst $ClF_3$,$BF_3$ and $NH_3$ molecules,the one with non-planar geometry is:

The geometry of $XeOF_4$ by $VSEPR$ theory is

Predict the correct order of repulsion among the following:

Assertion : Lone pair-lone pair repulsive interactions are greater than lone pair-bond pair and bond pair-bond pair interactions.
Reason : The space occupied by lone pair electrons is more as compared to bond pair electrons.

Which of the following molecules has a pyramidal shape?

Vedclass Test Series

Exam Paper Generator

Online Exam Module