Question

The geometry and the type of hybrid orbital present about the central atom in $$B{F_3}$$  is

A. linear, $$sp$$
B. trigonal planar, $$s{p^2}$$  
C. tetrahedral, $$s{p^3}$$
D. pyramidal, $$s{p^3}$$
Answer :   trigonal planar, $$s{p^2}$$
Solution :
$$H = \frac{1}{2}\left( {3 + 3 + 0 - 0} \right) = 3$$
∴ Boron, in $$B{F_3},$$  is $$s{p^2}$$  hybridised leading to trigonal planar shape.

Releted MCQ Question on
Inorganic Chemistry >> Chemical Bonding and Molecular Structure

Releted Question 1

The compound which contains both ionic and covalent bonds is

A. $$C{H_4}$$
B. $${H_2}$$
C. $$KCN$$
D. $$KCl$$
Releted Question 2

The octet rule is not valid for the molecule

A. $$C{O_2}$$
B. $${H_2}O$$
C. $${O_2}$$
D. $$CO$$
Releted Question 3

Element $$X$$ is strongly electropositive and element $$Y$$ is strongly electronegative. Both are univalent. The compound formed would be

A. $${X^ + }{Y^ - }$$
B. $${X^ - }{Y^{ + \,}}$$
C. $$X - Y$$
D. $$X \to Y$$
Releted Question 4

Which of the following compounds are covalent?

A. $${H_2}$$
B. $$CaO$$
C. $$KCl$$
D. $$N{a_2}S$$

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Chemical Bonding and Molecular Structure


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