Explain why $BeH_{2}$ molecule has a zero dipole moment although the $Be-H$ bonds are polar.
(N/A) The Lewis structure for $BeH_{2}$ is as follows:
$H : Be : H$
There is no lone pair at the central atom $(Be)$ and there are two bond pairs. Hence,$BeH_{2}$ is of the type $AB_{2}$. It has a linear structure.
Dipole moments of each $Be-H$ bond are equal and are in opposite directions. Therefore,they nullify each other. Hence,$BeH_{2}$ molecule has zero dipole moment.
$H : Be : H$
There is no lone pair at the central atom $(Be)$ and there are two bond pairs. Hence,$BeH_{2}$ is of the type $AB_{2}$. It has a linear structure.
Dipole moments of each $Be-H$ bond are equal and are in opposite directions. Therefore,they nullify each other. Hence,$BeH_{2}$ molecule has zero dipole moment.
Explore More
Similar Questions
$A$ body of mass $6 ~kg$ is under a force which causes displacement in it given by $s = \frac{t^2}{4}$ metres,where $t$ is time. The work done by the force in $2$ seconds is: (in $~J$)
If the lines $x^2+2xy-35y^2-4x+44y-12=0$ and $5x+\lambda y-8=0$ are concurrent,then the value of $\lambda$ is
Which of the following has the lowest melting point?
$A$ primary coil and a secondary coil are placed close to each other. $A$ current, which changes at the rate of $25 \, A$ in a millisecond, is present in the primary coil. If the mutual inductance is $92 \times 10^{-6} \, H$, then the value of the induced emf in the secondary coil is:
If $a, b, c$ and $d \in \mathbb{R}$ such that $a^2+b^2=4$ and $c^2+d^2=2$ and if $(a+ib)^2=(c+id)^2(x+iy)$,then $x^2+y^2$ is equal to
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo