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JEE CHEMISTRY MCQ: CHEMICAL BONDING & MOLECULAR STRUCTURE

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Chlorine atom, in its third excited state, reacts with fluorine to form a compound X. The formula and shape of X are

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Correct option: D

\mathrm{Cl}{\mathrm{F}_7}

pentagonal bipyramidal

Option D (

\mathrm{Cl}{\mathrm{F}_7}

pentagonal bipyramidal) is correct because in the third excited state, chlorine can form 7 bonds with fluorine, resulting in a pentagonal bipyramidal shape.

More Questions from CHEMICAL BONDING & MOLECULAR STRUCTURE

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The hybridization in sulphur dioxide is

\mathrm{C}{\mathrm{O}_2}

has the same geometry as:

\left( A \right)\mathrm{HgC}{\mathrm{l}_2},\left( B \right)\mathrm{N}{\mathrm{O}_2},\left( C \right)\mathrm{SnC}{\mathrm{l}_4},\left( D \right){\mathrm{C}_2}{\mathrm{H}_2}

The pair of molecules forming strongest hydrogen bonds are Paramagnetism is shown by Which statement is correct? Assertion (A) :

{BF}_3

is a weaker Lewis acid than

{BCl}_3

.

Reason (R) : The planar

{BF}_3

molecule is stabilised to a greater extent than

{BCl}_3

{B}-X \pi

-bonding.