Do you expect the carbon hydrides of the type $(C_{n}H_{2n+2})$ to act as 'Lewis' acid or base? Justify your answer.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) For carbon hydrides of type $C_{n}H_{2n+2},$ the following hydrides are possible:
$n=1 \Rightarrow CH_{4}$
$n=2 \Rightarrow C_{2}H_{6}$
$n=3 \Rightarrow C_{3}H_{8}$
For a hydride to act as a Lewis acid,it must be electron-deficient (electron acceptor). For it to act as a Lewis base,it must be electron-rich (electron donor).
Taking $C_{2}H_{6}$ as an example,the total number of valence electrons is $14$ and the total number of covalent bonds is $7$. Hence,the bonds are regular $2e^{-}-2$ centered bonds.
Thus,$C_{2}H_{6}$ has sufficient electrons to be represented by a conventional Lewis structure. It is an electron-precise hydride,where all atoms have complete octets (or duplets for hydrogen). Therefore,it can neither donate nor accept electrons to act as a Lewis acid or Lewis base.

Explore More

Similar Questions

If $f: R \rightarrow R$ is defined by $f(x) = x - [x]$,where $[x]$ is the greatest integer not exceeding $x$,then the set of points of discontinuity of $f$ is

The lungs are covered by which of the following?

$A$ ray of light strikes a plane mirror $M$ at an angle of $45^o$ as shown in the figure. After reflection,the ray passes through a prism of refractive index $1.5$ whose apex angle is $4^o$. The total angle through which the ray is deviated is......$^o$

Identify the correct order in which the covalent radius of the following elements increases: $(I)$ $Ti$,$(II)$ $Ca$,$(III)$ $Sc$.

If $A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$ and $hA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$,then the values of $h, a, b$ are respectively

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