What is the proportion of $s$-character in $sp^3$,$sp^2$,and $sp$ hybrid orbitals? What is the order of electronegativity in it?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The $s$-character in hybrid orbitals is calculated as follows:
$sp^3$: $1/4 = 25 \%$,$sp^2$: $1/3 = 33.3 \%$,$sp$: $1/2 = 50 \%$.
Electronegativity is directly proportional to the $s$-character because $s$-orbitals are closer to the nucleus.
Therefore,the order of electronegativity is $sp > sp^2 > sp^3$.
Hybrid Orbital $s$-character
$sp^3$ $25 \%$
$sp^2$ $33.3 \%$
$sp$ $50 \%$

Explore More

Similar Questions

Which of the following molecules contains a bond formed by the overlap of $sp$ and $sp^3$ hybrid orbitals?

What type of covalent bond is formed by hybrid orbitals?

What is the decreasing order of carbon-carbon single bond length in the following compounds?
$(I) CH_2 = CH - C \equiv CH$
$(II) CH \equiv C - C \equiv CH$
$(III) CH_3 - CH = CH_2$
$(IV) CH_2 = CH - CH = CH_2$

Difficult
View Solution

The linear shape of $CO_2$ is due to

Bond $X$ is formed by the overlap of which type of hybridized orbitals?

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo