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JEE CHEMISTRY MCQ: CHEMICAL BONDING & MOLECULAR STRUCTURE

1 Easy

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Analyzing +4 | -1

s{p^3}{d^2}

hybridization is not displayed by

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Correct option: D

P{F_5}

Option B is correct because the equilibrium constant for the reaction D ⇌ 4A is found by manipulating the given equilibrium constants: K = Z / (X² * Y).

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.