Chemical Bonding & Molecular Structure — Page 11 | JEE Chemistry

Q101

Arrange the bonds in order of increasing ionic character in the molecules.

\mathrm{LiF}

,

\mathrm{K}_2 \mathrm{O}, \mathrm{N}_2, \mathrm{SO}_2

and

\mathrm{ClF}_3

:

A

\mathrm{N}_2<\mathrm{SO}_2<\mathrm{ClF}_3<\mathrm{K}_2 \mathrm{O}<\mathrm{LiF}

B

\mathrm{ClF}_3<\mathrm{N}_2<\mathrm{SO}_2<\mathrm{K}_2 \mathrm{O}<\mathrm{LiF}

C

\mathrm{LiF}<\mathrm{K}_2 \mathrm{O}<\mathrm{ClF}_3<\mathrm{SO}_2<\mathrm{N}_2

D

\mathrm{N}_2<\mathrm{ClF}_3<\mathrm{SO}_2<\mathrm{K}_2 \mathrm{O}<\mathrm{LiF}

Correct Answer

Option A

Solution

To determine the order of increasing ionic character in the bonds of the given molecules, we should consider the concept of electronegativity, which is a measure of the tendency of an atom to attract a bonding pair of electrons.

The larger the difference in electronegativity between two atoms, the more ionic the bond is.

Covalent bonds, on the other hand, have smaller differences in electronegativity.

We can qualitatively assess the ionic character of the bonds in each molecule as follows : - $\mathrm{N_2}$ : Both nitrogen atoms have the same electronegativity since it is a diatomic molecule of the same element, therefore the bond is purely covalent with no ionic character. - $\mathrm{SO_2}$ : Sulfur and oxygen have different electronegativities, with oxygen being more electronegative than sulfur, but not as drastically different as in ionic compounds.

Therefore, the bond has some ionic character but is primarily covalent. - $\mathrm{ClF_3}$ : Chlorine and fluorine have different electronegativities, with fluorine being more electronegative.

However, the difference is not as large as between metals and nonmetals, so while this bond has ionic character, it is not as ionic as a bond between a metal and a nonmetal. - $\mathrm{K_2O}$ : Potassium is a metal with low electronegativity, and oxygen is a nonmetal with high electronegativity.

Therefore, bonds between potassium and oxygen will have a high degree of ionic character. - $\mathrm{LiF}$ : Lithium is a metal with low electronegativity, and fluorine is a nonmetal with very high electronegativity.

The difference in electronegativity is very large, indicating a very high ionic character, even more so than in $\mathrm{K_2O}$ because fluorine is more electronegative than oxygen.

Given these considerations, the correct order from least to most ionic character is as follows :

\mathrm{N_2}Therefore, the correct option is : Option A :

\mathrm{N_2}

Q102

Choose the polar molecule from the following:

A

\mathrm{CCl}_4

B

\mathrm{CO}_2

C

\mathrm{CH}_2=\mathrm{CH}_2

D

\mathrm{CHCl}_3

Correct Answer

Option D

Solution

\mu \neq 0

\mathrm{CHCl}_3

is polar molecule and rest all molecules are non-polar.

Q103

Which of the following is least ionic?

A CoCl

_2

B KCl

C BaCl

_2

D AgCl

Correct Answer

Option D

Solution

\mathrm{AgCl} Reason:

\mathrm{Ag}^{+}$$ has pseudo inert gas configuration.

Q104

The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals A. have the same energy B. have the minimum overlap C. have same symmetry about the molecular axis D. have different symmetry about the molecular axis Choose the most appropriate from the options given below:

A B, C, D only

B A, B, C only

C B and D only

D A and C only

Correct Answer

Option D

Solution

* Molecular orbital should have maximum overlap * Symmetry about the molecular axis should be similar

Q105

The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is

A electromeric energy

B resonance energy

C ionization energy

D hyperconjugation energy

Correct Answer

Option B

Solution

The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is known as resonance energy.

Q106

The molecule / ion with square pyramidal shape is

A

\mathrm{PCl}_5

B

\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}

C

\mathrm{PF}_5

D

\mathrm{BrF}_5

Correct Answer

Option D

Solution

\mathrm{BrF_5}

Q107

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : There is a considerable increase in covalent radius from

\mathrm{N}

to

\mathrm{P}

. However from As to Bi only a small increase in covalent radius is observed. Reason (R) : Covalent and ionic radii in a particular oxidation state increases down the group. In the light of the above statements, choose the most appropriate answer from the options given below:

A (A) is false but (R) is true

B Both (A) and (R) are true and (R) is the correct explanation of (A)

C Both (A) and (R) are true but (R) is not the correct explanation of (A)

D (A) is true but (R) is false

Correct Answer

Option C

Solution

According to NCERT, Statement-I : Factual data, Statement-II is true.

But correct explanation is presence of completely filled

d

and

f

-orbitals of heavier members

Q108

When

\psi_{\mathrm{A}}

and

\psi_{\mathrm{B}}

are the wave functions of atomic orbitals, then

\sigma^*

is represented by :

A

\psi_A+2 \psi_B

B

\psi_{\mathrm{A}}-\psi_{\mathrm{B}}

C

\psi_A-2 \psi_B

D

\psi_{\mathrm{A}}+\psi_{\mathrm{B}}

Correct Answer

Option B

Solution

In Molecular Orbital Theory, molecular orbitals are formed by the linear combination of atomic orbitals (LCAO).

The wave functions of atomic orbitals $\psi_A$ and $\psi_B$ can combine in two ways: Constructive Interference: This leads to the formation of a bonding molecular orbital ( $\sigma$ ): $\sigma = \psi_A + \psi_B$ Destructive Interference: This leads to the formation of an antibonding molecular orbital ( $\sigma^*$ ): $\sigma^* = \psi_A - \psi_B$ Conclusion The antibonding molecular orbital $\sigma^*$ is represented by: $\sigma^* = \psi_A - \psi_B$ Therefore, the correct answer is: Option B: $\psi_{\mathrm{A}} - \psi_{\mathrm{B}}$

Q109

Consider ' n ' is the number of lone pair of electrons present in the equatorial position of the most stable structure of

\mathrm{ClF}_3

. The ions from the following with ' n ' number of unpaired electrons are A.

\mathrm{V}^{3+}

B.

\mathrm{Ti}^{3+}

C.

\mathrm{Cu}^{2+}

D.

\mathrm{Ni}^{2+}

E.

\mathrm{Ti}^{2+}

Choose the correct answer from the options given below:

A B and D Only

B A, D and E Only

C B and C Only

D A and C Only

Correct Answer

Option B

Solution

$\mathrm{n}=2($ No of lone pair present in equitorial plane)

\begin{array}{lc} & \text{ (Unpaired } \mathrm{e}^{-} \text{) } \\ \text{ (A) } \mathrm{V}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^2 & 2 \\ \text{ (B) } \mathrm{Ti}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^1 & 1 \\ \text{ (C) } \mathrm{Cu}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^9 & 1 \\ \text{ (D) } \mathrm{Ni}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^8 & 2 \\ \text{ (E) } \mathrm{Ti}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^2 & 2 \end{array}

Q110

Number of

\sigma

and

\pi

bonds present in ethylene molecule is respectively :

A 3 and 1

B 5 and 1

C 4 and 1

D 5 and 2

Correct Answer

Option B

Solution

Let's analyze the bond structure in an ethylene molecule (C2H4) to determine the count of $\sigma$ and $\pi$ bonds.

Ethylene consists of two carbon atoms double-bonded to each other, with each carbon atom also bonded to two hydrogen atoms.

Each single bond is a $\sigma$ bond.

The double bond between the carbon atoms is made up of one $\sigma$ bond and one $\pi$ bond.

Here's how the bonds break down: Each carbon-hydrogen bond is a $\sigma$ bond.

Since there are four C-H bonds, we have 4 $\sigma$ bonds from C-H bonding.

The carbon-carbon double bond consists of one $\sigma$ bond and one $\pi$ bond.

Thus, we have 1 additional $\sigma$ bond from the C=C double bond.

Total $\sigma$ bonds = 4 (from C-H) + 1 (from C=C) = 5.

There is 1 $\pi$ bond present in the carbon-carbon double bond.

Therefore, the correct option is: Option B: 5 and 1