Identify the correct match.

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

Question

(D) In $S_3O_9$,the structure consists of $S-O-S$ linkages where sulfur atoms are $sp^3$ hybridized. The $\pi$ bonds present are $p\pi - d\pi$ bonds.I

Vedclass · Accepted Answer

$SO_3 - $ both $p\pi - p\pi$ and $p\pi - d\pi$ bonds present

Vedclass · Answer

(D) In $S_3O_9$,the structure consists of $S-O-S$ linkages where sulfur atoms are $sp^3$ hybridized. The $\pi$ bonds present are $p\pi - d\pi$ bonds.In $P_4O_{10}$,there are $16 \ \sigma$ bonds and $4 \ \pi$ bonds. Thus,they are not equal.In $P_4O_6$,there are no $\pi$ bonds as all phosphorus atoms are $sp^3$ hybridized and form only $\sigma$ bonds.In $SO_3$ (cyclic trimer or monomeric gas),the sulfur atom forms $\pi$ bonds with oxygen using its $d$-orbitals. Specifically,in the monomeric $SO_3$,there are $p\pi - p\pi$ and $p\pi - d\pi$ bonds. Thus,option $D$ is correct.

Element	Electronegativity
$W$	$2.7$
$X$	$2.1$
$Y$	$0.8$
$Z$	$3.4$

Similar Questions

Which is the correct electron dot structure of $N_2O$ molecule?

The Lewis structure for $O_3$ molecule is given below. The correct formal charges on oxygen atoms labelled $1$,$2$,$3$ are respectively.

The formal charges of $N_{(1)}$,$N_{(2)}$,and $O$ atoms in the structure $N_{(1)} = N_{(2)} = O$ are respectively:

Element Electronegativity
$W$ $2.7$
$X$ $2.1$
$Y$ $0.8$
$Z$ $3.4$
The incorrect statement regarding the given information is:

$A$ lone pair of electrons in an atom implies

Vedclass Test Series

Exam Paper Generator

Online Exam Module