Which of the following consists of an equal n | vedclass

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(B) In $SO_3$, the sulfur atom is $sp^2$ hybridized. It forms three $S=O$ double bonds. Each double bond consists of one $\sigma$ bond and one $\pi$ b

Vedclass · Accepted Answer

$SO_2$

Vedclass · Answer

(B) In $SO_3$, the sulfur atom is $sp^2$ hybridized. It forms three $S=O$ double bonds. Each double bond consists of one $\sigma$ bond and one $\pi$ bond. The $\pi$ bonds are formed by the overlap of $p$-orbitals of oxygen with $p$ and $d$ orbitals of sulfur. Specifically, $SO_3$ has one $p\pi - p\pi$ bond and two $p\pi - d\pi$ bonds. In $SO_2$, the sulfur atom is $sp^2$ hybridized. It has one lone pair and forms two $S=O$ bonds. One $\pi$ bond is $p\pi - p\pi$ and the other is $p\pi - d\pi$. Thus, $SO_2$ has an equal number ($1$ each) of $p\pi - p\pi$ and $p\pi - d\pi$ bonds. In $CO_2$, there are two $p\pi - p\pi$ bonds and no $p\pi - d\pi$ bonds. In $SO_4^{2-}$, there are two $p\pi - p\pi$ bonds and two $p\pi - d\pi$ bonds, but the question typically refers to neutral molecules like $SO_2$ in this context. Therefore, $SO_2$ is the correct answer.

Which of the following consists of an equal number of $p\pi - p\pi$ and $p\pi - d\pi$ bonds?

Similar Questions

The order of the average bond length of the given bonds is

Select the correct order:
$(a)$ $CO < CO_2 < CO_3^{2-}$ $\Rightarrow$ Bond length
$(b)$ $O_2 < O_3 < O_2^{2-}$ $\Rightarrow$ Bond length
$(c)$ $N_2 < N_2^+$ $\Rightarrow$ Bond energy

Which of the following statements is correct?

The correct order of the $O-O$ bond length in $O_2$,$H_2O_2$,and $O_3$ is:

What are the similarities and differences in the Lewis structures of $H_2SO_4$ and $SO_4^{2-}$?

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