Retro (Past 13 Years) IIT-JEE Advanced (Chemical Bonding) - JEE MCQ

If (2s-2p) mixing is not operative, then

(a) When liquid water changes to ice, there is an air gap inside the cage formed due to intermolecular H-bonding, Thus, volume of ice > volume of water.

Hence, density (ice) < density (water).
Thus ice floats on water - hydrogen bonding plays its role.
(b) In gaseous phase, basicity of 1° < 2° < 3° amines due to increase of electron density at N-atom

But in aqueous (H₂O) solvent, RNH₂ is stabilised by solvation. Thus, 1° amine is more

basic than 3° amine (3° < T < 2°) amines, 

Dimer due to intermolecular H- bonding.

Plan This problem includes basic concept of bonding. It can be solved by using the concept of molecular orbital theory.

Any orbital has two phase +ve and -ve. In the following diagram +ve phase is shown by darkening the lobes and -ve by without darkening the lobes.

When two same phase overlap with each other it forms bonding molecular orbital otherwise anti-bonding.

On the basis of above two concepts correct matching can be done as shown below

For molecules lighter than O₂, the increasing ord er order of energies of molecular orbitals is :

Where, π²p_y and π²p_z are degenerate molecular orbitals, first singly occupied and then pairing starts if Hund’s rule is obeyed. If Hund’s rule is violated in B₂, electronic arrangement would be :


No unpaired electron-diamagnetic.

SO₃ is planar (S is sp² hybridised), BrF₃ is T-shaped and is planar (Si is sp² hybridised).

Bond is formed by mixing of s and p orbitals. B2 undergoes both oxidation and reduction as



In N₂, bonds are formed by mixing of s and p orbitals.


Paramagnetic with bond-order undergoes both oxidation and reduction and bond involves mixing of s and p-orbitals.

Paramagnetic with bond order = 2.
O₂ undergoes reduction and the bond involves mixing of s and p-orbitals.

I and II are hyperconjugation structures of propene and involves σ-electrons of C—H bond and p-orbitals of pi-bond in delocalisation.

The bond order of CO = 3. NO⁺, CN^- and N₂ are isoelectronic with CO, have the same bond orders as CO. NO^- (16e^-) has bond order of 2.

O^-₂ in KO₂ has 17 electrons, species with odd electrons are always paramagnetic.

(i) In the reaction
Oxygen on reactant side is in   oxidation state. In product side, one of the oxygen is in zero oxidation state, i.e. oxidised while the other oxygen is in -1 oxidation state, i.e, reduced. Hence, in the above reaction, oxygen (O^-1/2) is simultaneously oxidised and reduced (disproportionated),

(ii) In acid medium,  is converted into  which is a dimeric, bridged tetrahedral :


The above is a redox reaction and a product NO₃ has trigonal planar structure.

The above is a redox reaction.