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JEE CHEMISTRY MCQ: CHEMICAL BONDING & MOLECULAR STRUCTURE

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Geometry of ammonia molecule and the hybridisation of nitrogen involved in it are
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Correct option: B
sp3s{p^3} hyridisation and distorted tetrahedral geometry
Option B (sp3s{p^3} hyridisation and distorted tetrahedral geometry) is correct because ammonia has sp3s{p^3} hybridization with a lone pair, resulting in a distorted tetrahedral (pyramidal) geometry.

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The hybridization in sulphur dioxide is CO2\mathrm{C}{\mathrm{O}_2} has the same geometry as: (A)HgCl2,(B)NO2,(C)SnCl4,(D)C2H2\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) : BF3{BF}_3 is a weaker Lewis acid than BCl3{BCl}_3.

Reason (R) : The planar BF3{BF}_3 molecule is stabilised to a greater extent than BCl3{BCl}_3 by B−Xπ{B}-X \pi-bonding.
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