JEE Questions for Chemistry Chemical Bonding And Molecular Structure Quiz 14 - MCQExams.com

Statement (a) is not correct as hybridisation is not a prior phenomenon.

Bond order cannot be negative because number of bonding electrons is always greater than anti bonding electrons.

o-Nitrophenol undergoes intramolecular H –bonding and hence exists as a monomer but p- nitrophenol undergoes intermolecular H – bonding and hence exists as an associated molecule. Therefore, o-nitrophenol is more volatile than p- nitrophenol.

Configuration (c) is correct.

C₂H₄ has sp² - hybridization.

[NF₃ and H₃O⁺] are pyramidal while [NO₃^- and BF₃] are planar. Hence answer (c) is correct.

F^- forms H- Bond with HF, therefore, the species [ H…F- H]- or HF₂⁻ exists and hence option (c) is correct.

In H - O - O - H, O - H bonds are polar but O –O bond is non –polar.

Only SO₂ has sp² hybridization while all the remaining molecules, i.e. CO₂,N₂O and CO have sp- hybridization.

H –bonding interactions.

Since the element X forms XCl₃, X₂O₅ and Ca₃X₂, therefore, it must be N or P. Since it does form MCl5, therefore, it must be N since it has no d- orbitals to expand its covalency from 3 to 5. Therefore, X must be N.

Metallic lustre is due to oscillation of loose electrons.

Each N atom in N₂ molecule shares three electrons, i.e., :N::N:
 

SF₆ has six bond pairs and hence according to VSEPR theory, it should be octahedral.

Since MX₄ is tetrahedral, therefore, total number of ∠XMX is six.

Strength of H - bond depends upon the electronegativity of the atom carrying the H - atom. Since O is more electronegative than N, therefore, H- bonding is maximum in ethanol.

Only BCl₃ is electron deficient.

sp - Hybridized (O = C= C= C =O).

All are gaseous molecules and do not conform to the octet rule. Thus, option (d) is correct.

Only statement I is correct.

H₂O has highest boiling point due to hydrogen bonding.

Hybridization of N in NH₃ = sp³, that of Pt²⁺ in [PtCl₄]²- = dsp², that of P in PCl₅ = dsp³and that of B in BCl₃ = sp²