Chemistry Chapter 3 Short Questions For Class 11 (with Answers)

The most important chemistry chapter 3 short questions for class 11. The 1st-year chapter 3 of chemistry is related to gases. These questions are for the Punjab Textbook Board and can be used within all of Punjab where this syllabus is taught.

Students are advised to prepare these questions in order to perform the best in the board examination.

Chemistry Chapter 3 Short Questions for Class 11

Q.1: What is chemical bond?
Ans: Chemical bond:
“A chemical bond is the force, which holds together two or more atoms or ions together to form a large variety of compounds”.
The forces which are responsible for such bonding and the shapes of the molecules formed, are as a result of chemical combination.
Examples:
₁₁Na (2, 8, 1) —lose e⁻→ Na⁺¹(2, 8) + 1e⁻
₁₇Cl (2, 8, 7) + 1e⁻ —Gain e⁻→ Cl⁻¹(2, 8, 8)
Na⁺(g) + Cl⁻(g) ———→ NaCl(s) (ionic bond)

Q.2: What is octet rule? Give an example.
Ans: Octet rule:
“The tendency of atoms to attain a maximum of eight electrons in the valence shell is known as the octet rule”.
All the elements have inherent tendency to stabilize themselves. They got their stabilization by losing, gaining or sharing electrons to attain the nearest noble gas configuration.
Examples:
₁₁Na (2, 8, 1) —lose e⁻→ Na⁺¹(2, 8) + 1e⁻ (Neon = 2,8)
₁₇Cl (2, 8, 7) + 1e⁻ —Gain e⁻→ Cl⁻¹(2, 8, 8) (Argon = 2,8,8)

Q.3: What is an ionic bond? Give an example.
Ans: Ionic bond:
According to the Lewis theory,
“An ionic bond is formed by the complete transfer of electron or electrons from an atom having low ionization energy to another atom having high electron affinity.”
Example:
₁₁Na (2, 8, 1) —lose e⁻→ Na⁺¹(2, 8) + 1e⁻
₁₇Cl (2, 8, 7) + 1e⁻ —Gain e⁻→ Cl⁻¹(2, 8, 8)
Na⁺(g) + Cl⁻(g) ———→ NaCl(s) (ionic bond)

Q.4: What is a covalent bond? Give an example.
Ans: Covalent Bond:
According to Lewis and Kossel,
“A covalent bond is formed by the mutual sharing of electrons between two atoms”
While sharing, each atom completes its valence shell and attains the nearest inert gas configuration.
Example:
:Cl̈: + :Cl̈: or Cl—Cl (Single covalent bond)
Ö + Ö or O=Ö (Double covalent bond)

Q.5: Differentiate between a polar and non-polar covalent bond.
Ans:
(1) Non-Polar Covalent Bond: The covalent bond in which, the bonding electron pair is equally shared between the bonded atoms are called as non-polar covalent bond. e.g. H-H, Cl-Cl
(2) Polar Covalent Bond: The covalent bond in which the bonding electron pair is not equally shared and are displaced towards one of the bonded atoms is called polar covalent bond. e.g. H-Cl, etc.
(3) Non-polar bond is generally a weaker bond. Polar bond is generally a stronger bond.

Q.6: What is a coordinate covalent bond? Give an example.
Ans: Coordinate covalent Bond:
“A coordinate covalent bond is formed between two atoms when the shared pair of electrons is donated by one of the bonded atoms”.
Examples:
NH₃ has three covalent bonds and there is a lone pair of electrons on nitrogen atom. On the other hand, boron atom in BF₃ is deficient in electrons. Therefore, nitrogen can donate the pair of electrons to the acceptor BF₃ and this results in the formation of a coordinate covalent bond.
(Diagram: H₃N -> BF₃)
Donor Acceptor Complex

Q.7: Differentiate between covalent and coordinate covalent bond.
Ans:
(1) Covalent Bond: The bond which is formed by the mutual sharing of electrons between two atoms is called as covalent bond. e.g. H-H
(2) Coordinate covalent bond: is formed between two atoms when the shared pair of electrons is donated by one of the bonded atoms. e.g. H₃N -> BF₃
(3) Single covalent bond is represented by a single line, double covalent bond by two lines and triple covalent bond by three lines.
(4) Coordinate covalent bond is represented by an arrow pointing from the donor atom to the acceptor atom.

Q.8: Why NH₃ and H₂O, give co-ordinate covalent bonds with H⁺?
Ans: Co-ordinate covalent bonds:
NH₃ and H₂O, both have lone pair of electrons. Thus they can donate an electron pair to the H⁺ ion, and can make co-ordinate covalent bond with them.
Examples:
NH₃ donate is electron pair to H⁺ and form NH₄⁺ ion.
(Diagram: H₃N + H⁺ -> NH₄⁺)

Q.10: The distinction between a co-ordinate covalent bond and a covalent bond vanishes after bond formation in NH₄⁺, H₃O⁺ and CH₃NH₃⁺.
Ans: Distinction between a co-ordinate and a covalent bond:
(a) NH₄⁺ ion:
NH₃ have three covalent bonds. It donates its lone pair to H⁺ ion to form NH₄⁺ ion.
All the four bonds in NH₄⁺ ion behave alike and each bond have 25% co-ordinate and 75% covalent character.
(b) H₃O⁺ ion:
H₂O molecule have two covalent bonds. It forms one co-ordinate covalent bond with one H⁺ ion.
In H₃O⁺ ion, each bond has 33% co-ordinate covalent and 66.6% covalent character.
(c) CH₃N⁺H₃ ion
Methylamines have three covalent bonds. It forms a co-ordinate covalent bond with H⁺ ion.
All the bonds between N and H are alike in all respect.

Q.11: How the criteria of electronegativity help us to understand the nature of a bond? / How the difference of electronegativity decides the nature of a chemical bond?
Ans: Nature of Chemical Bond:
The difference in the electronegativity value of the bonded atoms determines the nature of the bond.
(1) If the difference is 0 to 0.4, the bond between the two atoms is non-polar. e.g., H₂, Cl₂, Br₂ etc.
(2) If the difference is less than 1.8 but greater than 0.4, the bond between the atoms is polar covalent e.g., HCl, HBr etc.
(3) A difference of 1.8 units shows equal contribution of ionic and covalent bond e.g., AlCl₃.
(4) If the difference is greater than 1.8 which, the bond will be ionic in nature e.g., NaCl KBr etc.

Q.12: What is expanded octet rule? Give an example.
Ans: Expanded Octet rule:
In some polyatomic ions, the central atom violates the octet rule by expanding its electron density to the higher orbitals. These are said to have expanded octets. Some prominent examples are SO₄²⁻, ClO₃¹⁻, PO₄³⁻ etc.
Example:
From the Lewis structures of SO₄²⁻, we can calculate the number of electrons around the central S atom.
The number of electrons in the valence shell of S atom can be calculated as:
No of valence electrons = 2 x (double bond electrons) + 2 x (single bond electrons)
= 2 (4) + 2 (2) = 12
Thus, S has 12 electrons in its valence shell and it exceeds the octet by 4 electrons.

Q.13: CO₂ is non-polar molecules although its bonds are polar. Why?
Ans: CO₂ is non-polar:
CO₂ is a triatomic molecule in which carbon atom makes two double bonds with two oxygen atoms. Both the bonds are polar in nature due to electronegativity difference of 1.0 unit (3.5 – 2.5 = 1.0).
The geometry of CO₂ is linear.
Due to linear geometry of CO₂ molecule, both the dipoles cancel each other and net dipole moment of CO₂ is zero debye. Therefore due to zero dipole moment CO₂ is a non-polar molecule.

Q.14: The molecule of water is not linear. How dipole moment justify this statement?
Ans: Dipole moment of H₂O:
The dipole moment of water is 1.85 D which is directed from the end having two hydrogen atoms to the end with the oxygen atom as in the structure shown in Figure. A linear H₂O molecule (H-O-H) would have zero dipole moment. The non-zero dipole moment value shows that water is a non-linear molecule. Experiments reveal that the water molecule has a v-shaped structure.

Q.15: The dipole moments of CO₂ and CS₂ are zero? justify
Ans: Dipole moments of CO₂ and CS₂:
CO₂ and CS₂ have linear geometries. The individual bonds are polar, and there are two dipoles in each molecule. These dipoles are equal in magnitude and opposite in direction.
Therefore these dipoles cancel each other and CO₂ and CS₂ are non-polar in nature having a zero dipole moments.

Q.16: The dipole moments of SO₂ is 1.61D. comment
Ans: Dipole moments of SO₂:
SO₂ have a angular geometry due to presence of lone pair of electrons. SO₂ molecules have two dipoles. These dipoles cannot cancel each other due to non-linear geometry. Hence SO₂ is a highly polar molecule with a dipole moment of 1.61 D.

Q.17: Why molecules like BF₃ and CCl₄ are non-polar?
Ans: BF₃ and CCl₄ are non-polar:
Molecules having symmetrical geometries have zero dipole moment due to cancellation of individual bond moments. BF₃ and CCl₄ show symmetrical geometries as all the bonded atoms are similar as well as have same bond angles.
So due to cancellation of all bond moments these molecules are non-polar.

Q.18: Define bond energy with an example.
Ans: Bond energy:
The bond energy is the average amount of energy required to break all the bonds of a particular type in one mole of a substance.
e.g. H – H(g) ———→ H(g) + H(g) ΔH = +436 Kj mol⁻¹
The bond energy is expressed in Kjmol⁻¹.

It is the energy required to break 6.02 x 10²³ bonds.
It is also the amount of energy released when 6.02 x 10²³ bonds are formed.
It is the measure of the strength of a covalent bond.

Q.19: Discuss the factors affecting the bond energy.
Ans: Factors affecting the bond energies:
The bond energies of the molecules depend upon the following factors.

Number of bonds. (bond order)
Bond length.
Bond polarity.

Q.20: Define bond length? Give an example.
Ans: Bond length:
“The distance between the nuclei of two atoms forming a covalent bond is called the bond length.”
Determination of bond length:
The bond lengths are experimentally determined by electron diffraction, X-ray diffraction or spectral studies.
The covalent bond length between two atoms is often but not always independent of the nature of the molecules.
For instance, in most of the aliphatic hydrocarbons, the C-C bond length is very close to 154 pm. The C-C bond length is also found to be the same in diamond.

Q.21: The abnormality of bond length and bond strength in HI is less prominent than HCl.
Ans: Abnormality of bond length and bond energy:
The bond length and bond energy of a molecule is affected by the bond polarity. Highly polar molecules show greater deviation in the properties like bond length and bond energy.
For example:
HI is a weakly polar molecule (ΔE.N = 2.5 – 2.1 = 0.4) as compared to HCl. (ΔE.N = 3.0 – 2.1 = 0.8).
Therefore, due to small electronegativity difference in HI molecule it is less polar than HCl and hence the abnormality in the bond length and bond energy is less prominent in HI than HCl.

Ans: Atomic orbital hybridization:
“According to this theory, atomic orbitals differing slightly in energy and shape intermix to form new orbitals, which are called hybrid atomic orbitals. These orbitals are similar in shape and energy with each other but differ from their parent atomic orbitals in shape and possess specific geometry”.
₆C = (excited state) = 1s 2s 2px 2py 2pz
₆C = (hybridized state) = 1s sp³ sp³ sp³ sp³
Hybridization leads to entirely new shape and orientation of the valence orbitals of an atom.

Q.31: Write main assumptions of the hybridization theory.
Ans: Main assumptions of hybridization theory:
The atomic orbital hybridization gives a satisfactory explanation for the valency of the elements.

In some cases, the electrons belonging to the ground state are promoted to the excited state as a result of which there is an increase in the number of unpaired electrons.
These excited orbitals undergo hybridization simultaneously, which leads to entirely new shape and orientation of the valence orbitals of an atom.

Q.32: The bond angles of H₂O and NH₃ are not 109.5° like that of CH₄. Although O and N atoms are sp³ hybridized.
Ans: Bond angles of H₂O and NH₃:
The geometry of a molecule depends upon the hybridization state of the central atom. In CH₄, NH₃ and H₂O, the central atoms (C,N,O) are sp³ hybridized. Therefore, it is expected that the bond angles in CH₄, NH₃ and H₂O are 109.5°.
In CH₄ all the electron pairs are bonding, so the geometry of the molecule is perfect tetrahedron with a bond angle of 109.5°. But in NH₃ there is one lone pair, and in H₂O there are two lone pairs. There lone pairs cause greater repulsion and hence the bond angles reduced from 109.5° to 107.5° and 104.5° respectively.

Q.33: What is the basic assumption of VSEPR theory?
Ans: Basic Assumption:
The valence electron pairs (lone pairs and the bond pairs) are arranged around the central atom to remain at a maximum distance apart to keep repulsions at a minimum.
Example:
Molecules which have two electrons pairs around the central atom are arranged at farther distance apart at an angle of 180°, in order to minimize repulsions between them. Thus, they form a linear geometry.

Q.34: What are AB₂ type molecules according to the VSEPR theory? Give an example. / Explain the shape of BeCl₂ according to the VSEPR theory?
Ans: Shape of BeCl₂:
In AB₂ molecules two electrons pairs around the central atom are arranged at farther distance apart at an angle of 180°, in order to minimize repulsions between them. Thus, they form a linear geometry.
Examples:
Beryllium chloride is a typical linear molecule, which contains two electrons pairs.

Q.35: Why lone pair of electrons occupy more space than the bond pairs?
Ans: Lone pair occupies more space:
A non bonding electron pair occupy more space than the lone pair on the central atom.
Reason:
A bonding electron pair is attracted by both the nuclei, while non-bonding by only one nucleus. So a lone pair experiences less nuclear attraction, due to which its electronic cloud is spread out more in space than that for the bonding pair. As a result, the non-bonding electron pairs exert greater repulsive forces on bonding electron pairs and thus tend to compress the bond pairs.
The magnitude of repulsions between the electron pairs in a given molecule decreases in the following order.
Lone pair – lone pair > Lone pair – bond pair > bond pair – bond pair

Q.36: What are AB₃ type molecules according to the VSEPR theory? Give an example. / Explain the shape of BH₃ according to the VSEPR theory?
Ans: Shape of BH₃:
BH₃ is a typical example of AB₃ molecule.
The boron atom in BH₃ is surrounded by three charge clouds, which remain farthest apart in one plane, each pointing towards the corners of an equilateral triangle. Thus, BH₃ molecules has a trigonal planar geometry, with each H-B-H bond angles of 120°.

Q.37: What are AB₄ type molecules according to the VSEPR theory? Give an example. / Explain the shape of CH₄ according to the VSEPR theory?
Ans: Shape of CH₄:
The four electron pairs are directed from the center towards the corners of a regular tetrahedron, with each apex representing a hydrogen nucleus. The arrangement permits a non-planar arrangement of electron pairs.
Each H-C-H bond is perfectly 109.5°.

Q.38: Explain the shape of NH₃ according to the VSEPR theory?
Ans: Shape of NH₃:
Ammonia, NH₃ is a typical example of AB₄ molecule with one lone pair.
The non-bonding electron in 2s orbital takes up more space and exerts a strong repulsive force on the bonding electron pairs.
Consequently, to avoid a larger repulsion, the bonding electron pairs move closer that reduces the ideal bond angle from 109.5° to 107.5°. This effect compels ammonia to assume a triangular pyramidal geometry instead of tetrahedral.

Q.39: Why bond angle in NF₃ is less than NH₃?
Ans: Bond angle of NF₃:
Substitution of hydrogen with electronegative atoms like F or Cl further reduces the bond angle. In NF₃, the strong polarity of N-F bond pulls the lone pair of N atom closer to its nucleus, which in turn exerts a stronger repulsion over bonding electrons. Thus, the angle further shrinks to 102°.
Moreover, the bond pairs N-F bonds are more close to F atoms than N atoms. The increased distances in these bond pairs makes their repulsions less operative.

Q.40: Explain the shape of H₂O molecule according to the VSEPR theory?
Ans: Shape of H₂O:
Water, H₂O is a typical example of AB₄ molecule with two lone pairs.
Two of the corners of a tetrahedron are occupied by each of the two lone pairs and remaining by bond pairs. But owing to spatial arrangement of lone pairs and their repulsive action among themselves and on bond pairs, the bond angle is further reduced to 104.5°. H₂S, H₂Se, H₂Te form similar geometries.

Q.41: Differentiate between bonding molecular orbitals and anti-bonding molecular orbital?
Ans:

Bonding Molecular Orbital	Anti-Bonding Molecular Orbital
(i) A molecular orbital is formed due to constructive interference of electronic waves.	(i) An anti-bonding orbital is formed due to destructive interference of electronic waves.
(ii) In bonding molecular orbital, the area of maximum electron density is present between the bond axis.	(ii) In anti-bonding molecular orbital, the area of maximum electron density is away from the bond axis.
(iii) Bonding molecular orbital is denoted by σ or σ*.	(iii) Anti-bonding molecular orbital is denoted by π or π*.

Q.42: Sketch the molecular orbital pictures of σ (1s) and σ (1s)*
Ans: σ (1s) and σ (1s):*
When two 1s orbitals of two atoms approach along the same axis (i.e.’x’ axis). This combination of the atomic orbitals gives rise to σ (1s) bonding and σ* (1s) anti-bonding molecular orbitals.
Both are symmetrical about the nuclear axis.

Q.43: Sketch the molecular orbital pictures of σ (2px) and σ (2px)*
Ans: σ (2px) and σ (2px):*
When two 2px -orbitals of the two atoms approach along the same axis (i.e.’x’ axis). This combination of the atomic orbitals gives rise to σ (2px) bonding and σ* (2px) anti-bonding molecular orbitals.
Both are symmetrical about the nuclear axis.

Q.44: Sketch the molecular orbital pictures of π(2py) and π*(2py).
Ans: π(2py) and π*(2py)
2py and 2py orbitals overlaps by a sideways approach to form two molecular orbitals.
One is low energy bonding molecule orbitals called as π(2py). The other is high energy anti-molecule orbital called as π*(2py).

Q.45: Explain the term bond order. Giving one example.
Ans: Bond order:
“The number of bonds formed between two atoms after the atomic orbitals overlap, is called the bond order and is taken as half of the difference between the number of bonding electrons and anti-bonding electrons.”
Bond order = (a – b) / 2
For, H₂ molecule, the number of electron in bonding orbital σ(1s) is two, while the anti-bonding σ*(1s) orbital is empty. Hence its bond can be calculate as,
Bond order = (2 – 0) / 2 = 1

Q.46: He₂ molecule does not exist. How MOT justify this statement. / The bond order of He₂ molecule is zero. Justify with example.
Ans: He₂ molecule:
The electronic configuration of He is 1s².
₂He = 1s²
The 1s orbitals of He-atoms combine to form one bonding σ(1s) and one anti-bonding σ* (1s) orbitals. Each He-atom contributes two electrons. Two electrons enter bonding molecular orbital σ (1s) and the remaining two go to antibonding σ* (1s) molecular orbital.
Bond order of He₂ molecule can be calculated as
Bond order = (2 – 2) / 2 = 0
Since bond order is zero, the He₂ molecule is not formed.

Q.47: O₂ molecule is paramagnetic in nature. How MOT justify this statement.
Ans: Paramagnetic nature of O₂:
Oxygen molecule is paramagnetic in nature, which means it is attracted by a magnetic field. MOT successfully explains the paramagnetic behavior of oxygen molecule.
The MO diagram of oxygen shows the presence of two unpaired electrons, one in π(2py) and π(2pz) each.
Due to the presence of these unpaired electrons, oxygen molecule has a net magnetic field, which interacts with the external magnetic field.