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AP EAMCET Engineering 18 Sep 2020 Shift 1 Paper

Section: Chemistry

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Question : 160 of 160

Marks: +1, -0

What is the hybridization of

\mathrm{Be}

\mathrm{BeF}_2

molecule?

$\text{dsp}^2$
$\text{sp}^2d$
$\text{sp}$
$\text{sp}^3$

Solution:

Hybridisation state of

\mathrm{Be}

in

\mathrm{BeF}_2

is

\text{s p .}

\mathrm{F} \rightarrow \mathrm{Be} - \mathrm{F}

H = \frac{1}{2}(V+M-C+A)

= \frac{1}{2}(2+2-0+0) = 2(\text{s p})

where,

V=

number of valence electrons of

\mathrm{Be}=2

H=

number of monovalent

\mathrm{F}

atom

=2

C=

number of cationic charge

=0

A=

number of anionic charge

=0

So, the hybridisation of Be in

\mathrm{BeF}_2

molecule is

\text{s p}

.

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