Chemical Bonding and Atomic Structure MCQs for NEET — Chemistry Questions with Answers

As covalent character increase, melting point decreases, thus order of melting point is 

CaF₂ > CaCl₂ > CaBr₂ > Cal₂ 

Bond order of O₂⁺   = (10-5)/2=2.5

Bond order of O₂^-    = (10-7)/2=1.5

Bond order of O₂^2-  = (10-8)/2=1

Bond order of O₂     = (10-6)/2=2

Maximum bond order = minimum bond length.

Bond length is minimum for O₂⁺ 

(b) As the size of halogen atom increases, the acidic strength of boron halides increases. Thus, BF₃ is the weakest Lewis acid. This is because of the p$\mathrm{\pi }$ - p$\mathrm{\pi }$ back bonding between the fully-filled unutilised 2p orbitals of F and vacant 2p orbitals of boron which makes BF₃ less electron deficient. Such back donation is not possible in case of ₃ or BBr₃ due to larger energy difference between their orbitals. Thus, these are more electron deficient. Since on moving down the group the energy difference increases, the Lewis acid character also increases. Thus, the tendency to behave as Lewis acid follows the order

BBr₃> BCl₃ > BF₃

The dominant intermolecular force in liquid CH3OH (methanol) is hydrogen bonding due to the presence of the O-H bond. This hydrogen bonding must be overcome to convert liquid methanol into the gaseous state.

Key Idea : As the number of lone pair of electrons increases, bond angle decreases.

NO₂⁺ ion is isoelectronic with CO₂ molecule. It is a linear ion and its central atom (N+) undergoes sp-hybridisation, hence bond angle is 180°.

In NO₂^- ion, N-atom undergoes sp²-hybridisation. The angle between hybrid orbital should be 120° but one lone pair of electrons is lying on N-atom, hence bond angle decreases to 115°.

In NO₂ molecule, N-atom has one unpaired electron in sp²-hybrid orbital. The bond angle should be 120° but actually it is 132°. It maybe due to one unpaired electron in sp²-hybrid orbital.

Therefore, the increasing order of bond angles is:

NO₂^-  < NO₂  < NO₂⁺

(115°) (132°) (180°)

Key Idea:

  -C-C- sp3

 -C=C- sp2

-C$\equiv$C- sp3

=C=C= sp

 sp³  sp²  sp² sp³  sp   sp

CH₃-CH=CH=CH₂-C$\equiv$CH

6      5     4    3      2    1

Hence, the state of hybridisation of carbons 1,3 and 5 are sp, sp³ and sp² respectively.

(b) The correct order regarding the electronegativity of hybrid orbitals of carbon is sp>sp²>sp³ because in sp, sp² and sp³ hybrid orbitals s-orbital character is 50%, 33.33% and 25% respectively and due to higher s-orbital character electron attraction tendency i.e, electronegativity increases.

(c) $\stackrel{+}{\mathrm{N}}{\mathrm{O}}_2$ has linear shape due to sp-hybridisation of N of $\stackrel{+}{\mathrm{N}}{\mathrm{O}}_2$

(c) In Al₂O₃, Sb₂O₃, Bi₂O₃ and SeO₂.Bi₂O₃ is most basic oxide due to higher reactivity with acid.

${\mathrm{Bi}}_2{\mathrm{O}}_3 + 6\mathrm{HCl} \stackrel{ }{\to }2{\mathrm{BiCl}}_3 + 3{\mathrm{H}}_2\mathrm{O}$