Chemical Bonding JEE Mains Questions: Important MCQs, PYQs & Solutions

Chemical Bonding JEE Mains Weightage

For the syllabus, you can go through the Chemical Bonding JEE Mains Weightage Syllabus. The total Chemical Bonding JEE Mains weightage is given below:

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Chemical Bonding PYQ JEE mains

Some of the previously asked chemical bonding jee mains questions and answers are given below. These chemical bonding PYQ JEE mains will help you get an idea of what kind of questions you need to practice to score good marks in JEE Mains 2026. All the questions below are given accordingly latest JEE Mains syllabus:

Question 1:Given below are two statements:
Statement (I) : for $C \ell \mathrm{F}_3$, all three possible structures may be drawn as follows.

Statement (II) : Structure III is most stable, as the orbitals having the lone pairs are axial, where the $\ell \mathrm{p}-\mathrm{bp}$ repulsion is minimum.

In light of the above statements, choose the most appropriate answer from the options given below:

1) Statement I is incorrect but Statement II is correct.

2) Statement I is correct but Statement II is incorrect.

3) Both Statement I and Statement II are correct.

4) Both Statement I and Statement II are incorrect.

Solution:

Statement 1 is correct.
Statement 2 is incorrect since in $\mathrm{sp}^3 \mathrm{~d}$ hybridization, a lone pair cannot occupy an axial position due to lone pair-bond pair repulsion. The lone pairs will be at the equatorial position for maximum stability.

Hence, the correct answer is option (2).

Question 2: $\mathrm{sp}^3 \mathrm{~d}^2$ hybridization is not displayed by :

1) $\mathrm{BrF}_5$

2) $\mathrm{SF}_6$

3) $\left[\mathrm{CrF}_6\right]^{3-}$

4) $\mathrm{PF}_5$

Solution:

$\mathrm{sp}^3 \mathrm{~d}^2$ hybridization is not displayed by $\mathrm{PF}_5$. The central P atom has 5 sets of electron pairs (all of which are bonding) hence it shows $\mathrm{sp}^3 \mathrm{~d}$ hybridization. $\mathrm{sp}^3 \mathrm{~d}^2$ hybridization corresponds to 6 sets of electron pairs.

Rest all other options display $\mathrm{sp}^3 \mathrm{~d}^2$ hybridization.

Hence, the answer is option (4).

Question 3: The incorrect geometry is represented by :

1)$B F_3$ - trigonal planar

2) $\mathrm{H}_2 \mathrm{O}$ - bent

3) $N F_3$ - trigonal planar

4) $A s F_5$ - trigonal bipyramidal

Solution:

As we have learned

Determination of the shape of molecules using VSEPR Theory -

calculate $X$

$X=($ No. of valence electronsof central atom $)+($ No.of other atom $)+$ (Negative charge on the molecule) - (Positive charge on the molecule)

The geometry of $N F_3$ is pyramidal.

Hence, the answer is option (3).

Q. 4 Among the following species, the diamagnetic molecule is :

1) NO

2)$C O$

3)$B_2$

4) $O_2$

Solution:

If the molecular orbitals have unpaired electrons then the molecule will be paramagnetic.

or if all are paired then the molecule will be diamagnetic.

In CO, molecular orbital configuration

$\sigma_{1 s}^2, \sigma_{1 s}^{* 2}, \sigma_{2 s}^2, \sigma_{2 s}^{* 2}, \pi_{2 p_x}^2=\pi_{2 p_y}^2, \sigma_{2 p_z}^2$

There are no unpaired electrons thus CO is a diamagnetic.

Hence, the answer is the option (2).

Question 5: During the change of $\mathrm{O}_2$ to $\mathrm{O}_2^{-}$, the incoming electron goes to the orbital:

1) $\pi 2 p_y$

2) $\sigma^* 2$

3) $\pi^* 2 p_x$

4) $\pi 2 p_x$

Solution:

For $O_2$ molecule, the molecule orbital sequence follows the given order :

$\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^1=\pi^* 2 p_y^1$

Now, from $O_2$ to $O_2^{-}$, the incoming electron will go into ( $\pi^* 2 p_x$ ) orbital and thus

The final sequence will be

$\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^2=\pi^* 2 p_y^1$

Hence, the answer is option (3).

Question 6: The total number of electrons in all bonding molecular orbitals of $\mathrm{O}_2^{2-}$ is $\_\_\_\_$ (Round off to the nearest integer)

Solution:

The orbital configuration of $\mathrm{O}_2^{2-}$ is

$\sigma_{1 S}^2 \sigma_{1 s}^{* 2} \sigma_{2 s}^2 \sigma_{2 s}^{* 2} \sigma_{2 p z}^2 \pi_{2 p x}^2=\pi_{2 p_y}^2 \pi_{2 p x}^2=\pi_{2 p y}^{* 2}$

Thus, there are a total of 10 electrons present in the bonding molecular orbitals.

Hence, the answer is (10).

Question 7: In $\mathrm{SO}_2, \mathrm{NO}_2^{-}$and $\mathrm{N}_3^{-}$the hybridizations at the central atom are respectively :

1) $\mathrm{sp}^2, \mathrm{sp}^2$ and sp

2) $\mathrm{sp}^2, \mathrm{sp}$ and sp

3) $\mathrm{sp}^2, \mathrm{sp}^2$ and $\mathrm{sp}^2$

4) $\mathrm{sp}, \mathrm{sp}^2$ and sp

Solution:

$\mathrm{SO}_2 \Rightarrow 2 \sigma$ bond +1 l.p. $\Rightarrow s p^2$ hybridisation

$\mathrm{NO}_2^{-} \Rightarrow 2 \sigma$ bond +1 l.p. $\Rightarrow s p^2$ hybridisation

$\mathrm{N}_3^{-} \Rightarrow 2 \sigma$ bond $\Rightarrow s p$ hybridisation

Hence, the correct answer is option (1).

Question 8: The bond order and the magnetic characteristics of $\mathrm{CN}^{-}$are:

1) 3, paramagnetic

2) $2 \frac{1}{2}$, diamagne

3) 3, diamagnetic

4) $2 \frac{1}{2}$, paramagnetic

Solution:

The bond order of $\mathrm{CN}^{-}$is 3, and all the electrons are paired in this ion, thus it is diamagnetic.

Hence, the answer is option(3).

Question 9: The interaction energy of London forces between two particles is proportional to $r^x$, where $r$ is the distance between the particles. The value of $x$ is :

1) 6

2) -6

3) 3

4) -3

Solution:

As we have learned, the interaction energy of London Forces between two particles $\alpha \frac{1}{r^6}$

Thus, the value of $x=-6$

Hence, the answer is option (2).

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Chemical Bonding JEE Mains questions 2025: Important Concepts

Some of the most important concepts asked in Chemical Bonding JEE Mains questions with solutions are listed below: