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


11 Questions MCQ Test Chemistry for JEE Advanced | Retro (Past 13 Years) IIT-JEE Advanced (Chemical Bonding)


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This mock test of Retro (Past 13 Years) IIT-JEE Advanced (Chemical Bonding) for JEE helps you for every JEE entrance exam. This contains 11 Multiple Choice Questions for JEE Retro (Past 13 Years) IIT-JEE Advanced (Chemical Bonding) (mcq) to study with solutions a complete question bank. The solved questions answers in this Retro (Past 13 Years) IIT-JEE Advanced (Chemical Bonding) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Retro (Past 13 Years) IIT-JEE Advanced (Chemical Bonding) exercise for a better result in the exam. You can find other Retro (Past 13 Years) IIT-JEE Advanced (Chemical Bonding) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Assuming 2s-2p mixing is not operative, the paramagnetic species among the following is

(2014 Adv., Single Option Correct Type)

Solution:

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

*Multiple options can be correct
QUESTION: 2

Hydrogen bonding plays a central role in the following phenomena

(2014 Adv., One or More than Option Correct Type)

Solution:

(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.

QUESTION: 3

Match the orbital overlap figures shown in Column I with the description given in Column II and select the correct answer using the codes given below the columns.

Solution:

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




QUESTION: 4

Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is

(2010, Single Option Correct Type)

Solution:

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


Where, π2py and π2pz 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.

QUESTION: 5

The species having pyramidal shape is

(2010, Single Option Correct Type)

Solution:


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

QUESTION: 6

Match each of the diatomic molecules in Column I with its property/properties in Column II.

 

Solution:



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.

QUESTION: 7

Hyperconjugation involves overlap of the following orbitals

(2008, Single Option Correct Type)

Solution:


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

QUESTION: 8

The species having bond order different from that in CO is

(2007, Single Option Correct Type)

Solution:

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.

QUESTION: 9

Among the following, the paramagnetic compound is

(2007, Single Option Correct Type)

Solution:

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

QUESTION: 10

Match the reactions in Column I with nature of the reactions/type of the products in Column II.

Solution:

(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.

QUESTION: 11

Which of the following contains maximum number of lone pairs on the central atom?

(2005, Single Option Correct Type)

Solution: