Table of Contents
The development of a chemical connection between at least two iotas, atoms, or particles to bring about a chemical compound is known as chemical holding. These chemical bonds hold the iotas together in the subsequent compound. The appealing power which keeps different constituents intact and balances out them by the general loss of energy is known as chemical bonding. Subsequently, it very well may be perceived that chemical mixtures are dependent on the strength of the chemical connections between their constituents. The more grounded the holding between the constituents, the more steady the subsequent compound would be.
IIT JEE aspirants can find Chemical bonding JEE Main Chemistry previous year solved questions here. This arrangement of questions assists aspirants with understanding the kind of questions asked in the JEE tests.
JEE Main Previous Year Solved Questions on Chemical Bonding
1. The bond separation energy of B-F in BF3 is 646 kJ mol-1 while that of C-F in CF4 is 515 kJ mol-1. The right justification for higher B-F bond separation energy when contrasted with that of C-F is
(1) Significant pπ – pπ collaboration among B and F in BF3 though there is no chance of such communication among C and F in CF4.
(2) Lower level of pπ – pπ collaboration among B and F in BF3 than that
among C and F in CF4
(3) Smaller size of B-molecule when contrasted with that of C-particle
(4) Stronger bond among B and F in BF3 when contrasted with that among C and F in CF4.
Solution:
In light of pπ – pπ back bonding in BF3 atom, all B-F bonds have fractional twofold bond character.
Thus choice (1) is the answer.
2. Among the accompanying species which two have a three-sided bipyramidal shape?
(1) NI3 (2) I3-(3) SO32-(4) NO3-
(1) II and III
(2) III and IV
(3) I and IV
(4) I and III
Solution:
Allow us to track down the hybridization (H) and state of given species.
(1) For NI3, H = ½ (5+3) = 8/2 = 4 → sp3 hybridized state. It is three-sided pyramidal in shape.
(2) For I3-, H = ½ (7+2+1) = 10/2 = 5 → sp3d hybridized state. It is straight in shape.
(3) For SO32-, H = ½ (6+2) = 8/2 = 4 → sp3 hybridized state. It is three-sided pyramidal in shape.
(4) For NO3-, H = ½ (5+1) = 6/2 = 3 → sp2 hybridized state. It is three-sided planar in shape.
Henceforth choice (4) is the answer.
3. Utilizing the MO hypothesis, anticipate which of the accompanying species has the most limited bond length?
(1) O2-
(2) O22-
(3) O22+
(4) O2+
Solution:
Chemical species O2- O22- O22+ O2+
Bond order 1.5 1 3 2.5
In this way bond length request O22->O2->O2+ >O22+
Henceforth choice (3) is the answer.
4. Among the accompanying, the species having the littlest bond is :
(1) NO
(2) NO+
(3) O2
(4) NO-
Solution:
Bigger the bond request, the more modest the bond length. NO+ has bond request 3.
Consequently choice (2) is the answer.
5. The hybridisation of orbitals of N particle in NO3-, NO2+, NH4+ are individual:
(1) sp2, sp3, sp
(2) sp, sp3, sp2
(3) sp, sp2, sp3
(4) sp2, sp, sp3
Solution:
In NO3, the focal N particle has 3 bonding spaces and no solitary sets of electrons.
In NO2, the focal N iota has 2 bonding spaces and no solitary sets of electrons.
In NH4, the focal N particle has 4 bonding spaces and no solitary sets of electrons.
The Hybridization of N iota in NO3-, NO2+, NH4+ are sp2, sp, sp3 individually.
Subsequently choice (4) is the answer.
6. In view of grid energy and different contemplations, which one of the accompanying antacid metal chlorides is relied upon to have the most noteworthy softening point?
(1) RbCl
(2) LiCl
(3) KCl
(4) NaCl
Solution:
NaCl has the most noteworthy dissolving point.
Subsequently choice (4) is the answer.
7. The structure of IF7 is :
(1) octahedral
(2) pentagonal bipyramid
(3) square pyramid
(4) three-sided bipyramid
Solution:
For IF7, hybridisation – sp3d3. The shape is pentagonal bipyramidal.
Henceforth choice (2) is the answer.
8. Which of the accompanying has the square planar structure :
(1) NH4+
(2) CCl4
(3) XeF4
(4) BF4-
Solution:
Hybridization of XeF4 sp3d2
It has a square planar shape.
Consequently choice (3) is the answer.
9. Among the accompanying the most extreme covalent person is shown by the compound :
(1) AlCl3
(2) MgCl2
(3) FeCl2
(4) SnCl2
Solution:
Al+3 is having the most elevated polarizing power than different compounds having a more noteworthy covalent person.
Subsequently choice (1) is the answer.
10. The compound of Xenon with zero dipoles second is :
(1) XeO3
(2) XeO2
(3) XeF4
(4) XeOF4
Solution:
XeF4 has a dipole in second zero.
Henceforth choice (3) is the answer.
FAQs
What are the significant topics in chemical bonding for JEE mains?
The significant topics in Chemical bonding are: For JEE Main: Kossel - Lewis way to deal with the chemical bond arrangement, the concept of ionic and covalent bonds. Ionic Bonding: Formation of ionic bonds, factors influencing the arrangement of ionic bonds, computation of grid enthalpy.
Which chemistry has more weightage in JEE mains?
Physical Chemistry bears the most weightage in JEE Main Chemistry. High-weightage sections from this area, like Electrochemistry, Atomic Structure, and Equilibrium, are to be concentrated.
Is chemical bonding a significant topic for JEE?
Yes, Chemical bonding is very important for JEE as 2-3 questions are asked, and this is a moderate topic that you can prepare by solving past papers and scoring good marks through these.
