The incorrect order of bond angle is:

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(D) $XeF_2$ has a linear geometry with a bond angle of $180^{\circ}$.$BeF_3^-$,$BF_3$,and $CH_3^+$ are all $sp^2$ hybridized with $0$ lone pairs,resul

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$PH_3 > PCl_3$

Vedclass · Answer

(D) $XeF_2$ has a linear geometry with a bond angle of $180^{\circ}$.$BeF_3^-$,$BF_3$,and $CH_3^+$ are all $sp^2$ hybridized with $0$ lone pairs,resulting in a bond angle of $120^{\circ}$.$CH_4$,$SiCl_4$,and $BF_4^-$ are all $sp^3$ hybridized with $0$ lone pairs,resulting in a bond angle of $109.5^{\circ}$.$CCl_4$ has a bond angle of $109.5^{\circ}$ and $H_2O$ has a bond angle of $104.5^{\circ}$.In $PH_3$ and $PCl_3$,the bond angle depends on the electronegativity of the central atom and the surrounding atoms. $PH_3$ has a bond angle of $\approx 93^{\circ}$ and $PCl_3$ has a bond angle of $\approx 100^{\circ}$.Therefore,the order $PH_3 > PCl_3$ is incorrect.

Similar Questions

Given below are two statements:
Statement $I:$ $SO_2$ and $H_2O$ both possess $V$-shaped structure.
Statement $II:$ The bond angle of $SO_2$ is less than that of $H_2O$.
In the light of the above statements,choose the most appropriate answer from the options given below:

The correct structure of $PCl_3F_2$ is

Which of the following represents pairs of isostructural species?

Match the following:
$($Molecule$/$Ion$)$ $($Shape$)$
$A. I_3^- \rightarrow 4.$ Linear
$B. ClF_3 \rightarrow 1.$ $T-$shaped
$C. H_2O \rightarrow 3.$ Angular
$D. SF_4 \rightarrow 2.$ See$-$Saw

Which of the following compounds has the smallest bond angle?

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