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JEE CHEMISTRY MCQ: CHEMICAL BONDING & MOLECULAR STRUCTURE

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Analyzing +4 | -1

Correct order of stability of the given species is

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Correct option: B

\mathrm{N}_2 > \mathrm{N}_2^{+} > \mathrm{N}_2^{-}

Option B (

\mathrm{N}_2 > \mathrm{N}_2^{+} > \mathrm{N}_2^{-}

) is correct because higher bond order implies greater stability, and the number of antibonding electrons affects stability when bond orders are equal.

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.