G11 C3 Chemical Bonding Short Question answers

Chapter 3 Chemical Bonding

Short Answer Questions

2. The table shows the atomic number and boiling points of some noble gases.

a. Explain this trend in boiling points.

The boiling points of noble gases increase down the group from helium to xenon. This is because the atomic size and number of electrons increase. This leads to stronger instantaneous dipole-induced dipole forces (London dispersion forces) between the atoms, which require more energy to overcome, thus raising the boiling point.

b. Xenon forms a number of covalently bonded compounds with fluorine.

i. Draw a dot-and-cross diagram for xenon tetrafluoride, XeF$_4$.

In XeF$_4$, xenon is the central atom with 8 valence electrons. Each of the four fluorine atoms shares one electron with xenon to form a single covalent bond. This uses 4 of xenon's valence electrons, leaving 4 electrons as 2 lone pairs.

ii. Suggest a shape for XeF$_4$. Explain why you chose this shape.

The shape of XeF$_4$ is **square planar**. The central xenon atom has four bonding pairs and two lone pairs of electrons. According to the VSEPR theory, these six electron pairs arrange themselves in an octahedral electron geometry to minimize repulsion. The lone pairs occupy positions opposite to each other (axial positions) to further minimize repulsion, forcing the four bonding pairs into a flat, square planar arrangement.

3. Aluminium chloride, AlCl$_3$, and ammonia, NH$_3$, are both covalent molecules.

a. Draw a diagram of an ammonia molecule, showing its shape. Show any lone pairs of electrons. Also State the bond angle HNH in the ammonia molecule.

The ammonia molecule ($\text{NH}_3$) has a trigonal pyramidal shape. The central nitrogen atom is bonded to three hydrogen atoms and has one lone pair of electrons. The lone pair repels the bonding pairs more strongly than the bonding pairs repel each other, compressing the bond angle from the ideal tetrahedral angle of 109.5° to approximately **107°**.

b. What type of forces are present in ammonia molecule. Draw diagram to show forces between ammonia molecules.

Within the ammonia molecule itself, the bonds are **covalent**. Between ammonia molecules, the primary intermolecular force is **hydrogen bonding**. This occurs because the highly electronegative nitrogen atom is bonded to hydrogen, creating a strong dipole. This allows the lone pair on one nitrogen to form a strong electrostatic attraction with the partially positive hydrogen of a neighboring ammonia molecule.

c. An ammonia molecule and an aluminium chloride molecule can join together by forming a co-ordinate bond.

i. Explain how a co-ordinate bond is formed.

A coordinate bond, also known as a dative covalent bond, is a type of covalent bond where both of the shared electrons in the bond are donated by one of the atoms. In this case, the nitrogen atom in ammonia donates its lone pair of electrons to the electron-deficient aluminium atom in aluminium chloride.

ii. Draw a dot-and-cross diagram to show the bonding in the compound formed between ammonia and Aluminium chloride, H$_3$NAlCl$_3$.

The diagram shows the nitrogen atom sharing its lone pair with the aluminium atom. This forms a dative bond, where the electrons are shared but originate solely from the nitrogen atom. This fills the valence shell of the aluminium atom.

4. Electronegativity values can be used to predict the polarity of bonds.

a. Explain the term electronegativity.

Electronegativity is a measure of the tendency of an atom to attract a shared pair of electrons towards itself in a chemical bond. The higher the electronegativity value, the more strongly an atom will pull the shared electrons, resulting in a polar bond.

b. The electronegativity values for some atoms are given below: H = 2.1, C = 2.5, F = 4.0, Cl = 3.0, I = 2.5. Use these values to predict the polarity of each of the following bonds.

i. H—I

Electronegativity difference: |2.1 - 2.5| = 0.4. This is a very small difference, indicating the bond is **non-polar to slightly polar**. Polarity can be shown with partial charges: H$^{δ+}$—I$^{δ-}$.

ii. F—I

Electronegativity difference: |4.0 - 2.5| = 1.5. This is a significant difference, indicating the bond is **polar covalent**. Polarity can be shown with partial charges: F$^{δ-}$—I$^{δ+}$.

iii. C—Cl

Electronegativity difference: |2.5 - 3.0| = 0.5. This is a small difference, making the bond **slightly polar**. Polarity can be shown with partial charges: C$^{δ+}$—Cl$^{δ-}$.

c. Describe the shape of the ICl$_3$ molecule. Also mention bond angle in it.

The central iodine atom in ICl$_3$ has 7 valence electrons, forming 3 single bonds with chlorine atoms and having 2 lone pairs. This results in a trigonal bipyramidal electron geometry. However, the molecular shape is a **T-shape** due to the lone pairs occupying equatorial positions to minimize repulsion. The bond angles are approximately **90°** and **180°**.

d. The boiling points of the hydrogen halides are shown in the table.

i. Explain the trend in boiling points from HCl to HI.

The boiling points increase from HCl to HI. This is due to the increase in the number of electrons and atomic size from chlorine to iodine. This results in stronger London dispersion forces between the molecules, requiring more energy to overcome, thus increasing the boiling point.

ii. Explain why the boiling point of HF is so much higher than the boiling point of HCl.

The boiling point of HF is anomalously high because of the strong **hydrogen bonding** between its molecules. Fluorine is highly electronegative and its small size allows it to form strong hydrogen bonds with neighboring HF molecules. HCl only has weak dipole-dipole forces, as chlorine is not electronegative enough to form hydrogen bonds.

e. Tetrachloromethane, CCl$_4$, is a non-polar molecule. Draw a diagram to show the shape of this molecule. Explain why this molecule is non-polar.

The CCl$_4$ molecule has a **tetrahedral** shape with a central carbon atom bonded to four chlorine atoms. The bond angles are 109.5°. Even though the C-Cl bonds are individually polar, the molecule as a whole is non-polar. This is because the symmetrical tetrahedral shape ensures that the individual bond dipole moments cancel each other out, resulting in a net dipole moment of zero.

Question 5

a. Hydrogen sulphide, H₂S, is a covalent compound. Explain the type of hybridization also write bond angle in HSH. Also show on your diagram the partial charges on each atom as δ+ or δ- and an arrow showing the exact direction of the dipole in the molecule as a whole.

The central sulfur atom in H₂S is **sp³ hybridized**. It has two bonding pairs and two lone pairs of electrons. The ideal bond angle for sp³ hybridization is 109.5°, but due to the repulsion from the lone pairs, the H-S-H bond angle is compressed to approximately **92°**. This is a V-shaped or bent molecular geometry. The sulfur atom is more electronegative than hydrogen, so it will have a partial negative charge (δ-) and the hydrogen atoms will have a partial positive charge (δ+). The dipole moments of the S-H bonds do not cancel out, resulting in a net dipole moment for the molecule. The overall dipole moment points towards the more electronegative sulfur atom.

b. Oxygen, O, sulphur, S, and selenium, Se, are in the same group in the Periodic Table.

i. Explain why hydrogen selenide, H₂Se, has a higher boiling point than hydrogen sulphide, H₂S.

H₂Se has a higher boiling point than H₂S because of its **larger molar mass**. As you move down a group, the size and mass of the atoms increase. This leads to stronger **London dispersion forces** (a type of van der Waals force) between the molecules. More energy is required to overcome these stronger intermolecular forces, resulting in a higher boiling point.

ii. Explain why the boiling point of water is so much higher than the boiling point of hydrogen sulphide.

Water (H₂O) has a much higher boiling point than H₂S primarily because water molecules can form **hydrogen bonds**. Hydrogen bonding is a strong type of intermolecular force that occurs when hydrogen is bonded to a highly electronegative atom like oxygen, nitrogen, or fluorine. H₂S does not form hydrogen bonds as sulfur is not electronegative enough. Therefore, the intermolecular forces in H₂S are weaker (primarily dipole-dipole interactions and London dispersion forces), and less energy is needed to separate the molecules, leading to a lower boiling point.

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Question 6

a. Describe the shape of the carbon dioxide molecule.

The carbon dioxide ($$CO_2$$) molecule has a **linear shape**. The central carbon atom is sp hybridized and has no lone pairs of electrons. The two oxygen atoms are bonded to the carbon atom with double bonds, and the bond angle is **180°**.

b. Bromine is a liquid at room temperature. Weak van der Waals' forces hold the bromine molecules together. Describe how van der Waals' forces arise.

**Van der Waals' forces** are the general term for attractive forces between neutral molecules. They arise from temporary fluctuations in electron distribution. As electrons move around the nucleus, a temporary dipole can form in one molecule. This temporary dipole can then induce a corresponding dipole in a nearby molecule, leading to a weak, short-lived electrostatic attraction. These forces are present in all molecules but are the dominant intermolecular force in nonpolar molecules like bromine ($$Br_2$$). The larger the molecule, the more electrons it has, and the more significant these temporary dipoles and resulting forces become.

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Question 7

a. Water is extensively hydrogen bonded. This gives it anomalous (peculiar) properties. Explain why ice is less dense than liquid water. Also state two other anomalous properties of water.

Ice is less dense than liquid water because of **hydrogen bonding**. In liquid water, molecules are constantly forming and breaking hydrogen bonds, allowing them to be closely packed. However, as water freezes, the hydrogen bonds lock the water molecules into a rigid, open, and ordered crystalline lattice structure. This lattice structure holds the molecules further apart than they are in the liquid state, increasing the volume and thus decreasing the density. This is why ice floats.

Two other anomalous properties of water are:

• **High specific heat capacity:** Water requires a large amount of energy to change its temperature. This is because a significant amount of the added energy is used to break the hydrogen bonds before the kinetic energy of the molecules can increase.
• **High surface tension:** The strong cohesive forces created by hydrogen bonds at the surface of water give it a high surface tension, allowing some insects to walk on its surface.

b. Propanone has the structure shown below.

      CH₃
      |
      C=O
      |
      CH₃
    

When propanone dissolves in water, it forms a hydrogen bond with water. Draw a diagram to show a propanone molecule and a water molecule forming a hydrogen bond.

c.

i. Propanone has a double bond. One of the bonds is a σ bond (sigma bond). The other is a π bond (pi bond). Explain the difference between a σ bond and a π bond in terms of how they are formed.

A **sigma (σ) bond** is a type of covalent bond formed by the **direct, head-on overlap** of atomic orbitals along the internuclear axis. This overlap is very strong and allows for free rotation around the bond. A **pi (π) bond** is formed by the **sideways overlap** of unhybridized p orbitals. This overlap occurs above and below the plane of the sigma bond. Pi bonds are generally weaker than sigma bonds and prevent free rotation around the double or triple bond.

ii. Copy the diagram, then complete it to show the shapes of the electron clouds in the σ bond and the π bond between the carbon atoms in ethene. Label your diagram.

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Question 8

Energies of orbitals can be explained by molecular orbital theory. It has been observed that in case of Nitrogen molecule σ2pₓ is higher in energy than π 2pᵧ and π 2p₂.

a. Draw molecular orbital energy diagram for nitrogen molecule.

b. Give reason why the σ 2pₓ energy is greater than π 2pᵧ and π 2p₂.

The σ 2pₓ orbital has a higher energy than the π 2pᵧ and π 2p₂ orbitals in molecules like N₂ due to the **mixing of the 2s and 2p orbitals**. In atoms with a small energy difference between their 2s and 2p atomic orbitals (like nitrogen), a phenomenon called **s-p mixing** occurs. This mixing causes the σ2s and σ*2s orbitals to decrease in energy and the σ2p and σ*2p orbitals to increase in energy. This upward shift in the energy of the σ2p orbital causes it to be higher in energy than the π2p orbitals.

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Question 9

Carbon can make a bond with hydrogen to form ethyne. Bond energy of C-H is same although 2s and 2p orbitals are involved which have difference in energies. Explain the formation of ethyne molecule on the basis of hybridization with the help of diagram.

The formation of the ethyne ($$C_2H_2$$) molecule can be explained by **sp hybridization** of the carbon atoms. Each carbon atom undergoes sp hybridization, where one 2s orbital mixes with one 2p orbital to form two sp hybrid orbitals. The remaining two 2p orbitals (2pᵧ and 2p₂) remain unhybridized.

One sp hybrid orbital from each carbon atom overlaps head-on to form a strong **carbon-carbon sigma bond**. The other sp hybrid orbital on each carbon atom overlaps with the 1s orbital of a hydrogen atom to form a strong **carbon-hydrogen sigma bond**. The two unhybridized 2p orbitals on each carbon atom then overlap sideways to form two separate **pi bonds**. This results in a **linear molecule** with a triple bond (one sigma and two pi bonds) between the carbon atoms and single bonds to the hydrogen atoms. The C-C-H bond angle is 180°.

The reason the C-H bond energies are the same is that the hybrid orbitals (sp) that form the bonds are identical in energy and shape, despite being formed from parent orbitals with different energies. Hybridization results in new orbitals of equal energy that are optimized for bonding.

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Question 10

Explain the magnetic properties of O₂, O₂⁻ and N₂ by applying molecular orbital theory.

Molecular orbital (MO) theory can predict the magnetic properties of molecules by examining the number of unpaired electrons in their molecular orbitals. Molecules with one or more unpaired electrons are **paramagnetic** and are attracted to a magnetic field. Molecules with all paired electrons are **diamagnetic** and are weakly repelled by a magnetic field.

• **Oxygen ($$O_2$$):** According to its MO diagram, the two highest energy electrons are located in the degenerate $$π*2p$$ antibonding orbitals. These electrons occupy these orbitals singly, according to Hund's rule. Since there are **two unpaired electrons**, $$O_2$$ is **paramagnetic**.
• **Superoxide ion ($$O_2⁻$$):** The superoxide ion has one more electron than $$O_2$$. This extra electron goes into one of the $$π*2p$$ antibonding orbitals, pairing up with one of the existing electrons. This leaves **one unpaired electron**. Therefore, $$O_2⁻$$ is also **paramagnetic**, but less so than $$O_2$$.
• **Nitrogen ($$N_2$$):** The MO diagram for $$N_2$$ shows that all electrons occupy molecular orbitals in pairs. There are **no unpaired electrons**. Therefore, $$N_2$$ is **diamagnetic**.

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Question 11

Differentiate between a sigma bond and a pi bond.

Feature	Sigma ($$σ$$) Bond	Pi ($$π$$) Bond
Formation	Formed by the direct, head-on overlap of atomic orbitals.	Formed by the sideways overlap of unhybridized p orbitals.
Overlap Region	Electron density is concentrated along the internuclear axis.	Electron density is concentrated above and below the internuclear axis.
Strength	Stronger than a pi bond due to greater orbital overlap.	Weaker than a sigma bond.
Rotation	Free rotation is possible around the sigma bond.	Rotation is restricted or prevented.
Hybridization	Involves hybrid orbitals (e.g., sp, sp², sp³) or s orbitals.	Involves only unhybridized p orbitals.
Presence	Present in single, double, and triple bonds. A single bond is a sigma bond.	Present only in double (one π) and triple (two π) bonds.

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