The H-N-H angle in ammonia is 107.6^{circ} wh | vedclass

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

Question

(A) Key point: As the percentage of $s$-character in a bond increases, the bond angle also increases. According to Bent's Rule, the lone pair occupies

Vedclass · Accepted Answer

Less

Vedclass · Answer

(A) Key point: As the percentage of $s$-character in a bond increases, the bond angle also increases. According to Bent's Rule, the lone pair occupies an orbital with more $s$-character as the electronegativity of the central atom increases. In $NH_3$, the $N$ atom is more electronegative than $P$ in $PH_3$. Since the bond angle in $NH_3$ $(107.6^{\circ})$ is greater than in $PH_3$ $(93.5^{\circ})$, the bonding orbitals in $NH_3$ have more $s$-character. Consequently, the lone pair orbital in $NH_3$ has more $s$-character and less $p$-character compared to the lone pair in $PH_3$. Thus, the $p$-character of the lone pair on ammonia is less than that of phosphine. Hence, option $(A)$ is the correct answer.

The $H-N-H$ angle in ammonia is $107.6^{\circ}$ while the $H-P-H$ angle in phosphine is $93.5^{\circ}.$ Relative to phosphine, the $p$-character of the lone-pair on ammonia is expected to be

Similar Questions

The correct increasing order of the number of lone pairs of electrons on the central atom of $SnCl_2$,$XeF_2$,$ClF_3$,and $SO_3$ is:

The increasing order of the number of lone pairs of electrons on the central atom of the following molecules is:
$I) \ ClF_3$
$II) \ XeF_2$
$III) \ SF_4$
$IV) \ SiH_4$

The number of molecules/ions that show linear geometry among the following is $.........$
$SO_2, BeCl_2, CO_2, N_3^{-}, NO_2, F_2O, XeF_2, NO_2^{+}, I_3^{-}, O_3$

According to Lewis theory, the total number of bond-pairs and lone pairs of electrons around the central atom of $XeO_6^{4-}$ ion is . . . . . . .

Which of the following has a square planar structure?

Vedclass Test Series

Exam Paper Generator

Online Exam Module