Which of the following molecules contains 25 | vedclass

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

Question

(C) The percentage of $s$-character in a hybrid orbital is calculated as: $s$-character $= (\frac{1}{n}) 	imes 100$,where $n$ is the number of hybrid

Vedclass · Accepted Answer

Methane

Vedclass · Answer

(C) The percentage of $s$-character in a hybrid orbital is calculated as: $s$-character $= (\frac{1}{n}) 	imes 100$,where $n$ is the number of hybrid orbitals. For $sp^{3}$ hybridization,there are $4$ hybrid orbitals,so $s$-character $= (\frac{1}{4}) 	imes 100 = 25 \%$. In $CH_{4}$ (Methane),the carbon atom is $sp^{3}$-hybridized. In $C_{2}H_{4}$ (Ethylene),carbon is $sp^{2}$-hybridized ($33.3 \% \,s$-character). In $C_{2}H_{2}$ (Acetylene),carbon is $sp$-hybridized ($50 \% \,s$-character). In $C_{6}H_{6}$ (Benzene),carbon is $sp^{2}$-hybridized ($33.3 \% \,s$-character). Therefore,methane contains $25 \% \,s$-character.

List-$I$ (molecule)	List-$II$ (hybridization,shape)
$A. XeO_3$	$I. sp^3$; pyramidal
$B. XeF_2$	$II. sp^3d$; linear
$C. XeOF_4$	$III. sp^3d^2$; square pyramidal
$D. XeF_6$	$IV. sp^3d^3$; distorted octahedral

Which of the following molecules contains $25 \% \,s$-character of the carbon atom in its hybrid state?

Similar Questions

What is the shape of the molecule $CH_2 = CH - CH = CH_2$?

Match List-$I$ with List-$II$.

List-$I$ (molecule) List-$II$ (hybridization,shape)

$A. XeO_3$ $I. sp^3$; pyramidal

$B. XeF_2$ $II. sp^3d$; linear

$C. XeOF_4$ $III. sp^3d^2$; square pyramidal

$D. XeF_6$ $IV. sp^3d^3$; distorted octahedral

Choose the correct answer from the options given below.

The carbon atoms in an acetylene molecule are in which state of hybridisation?

The shape of $CH_3^+$ species is

Explain $sp^{3}d$ hybridization by a suitable example.

Vedclass Test Series

Exam Paper Generator

Online Exam Module