Chemistry of Oxygen (O) - Practice Questions

Multiple Choice Questions (MCQs)

Question 1

Which of the following statements about oxygen and ozone is incorrect?

A. Ozone is diamagnetic, whereas oxygen is paramagnetic. B. The O-O bond length in ozone is shorter than that in oxygen molecule. C. Ozone acts as a more powerful oxidizing agent than oxygen. D. Oxygen can be obtained by heating potassium chlorate in the presence of manganese dioxide.

Solution:

The correct answer is B.

Explanation:

• A. Ozone is diamagnetic, whereas oxygen is paramagnetic.
  • Oxygen (O₂) has two unpaired electrons in its π* antibonding molecular orbitals, making it paramagnetic.
  • Ozone (O₃) has no unpaired electrons, making it diamagnetic. This statement is correct.
• B. The O-O bond length in ozone is shorter than that in oxygen molecule.
  • In oxygen (O₂), the O=O bond is a double bond, with a bond length of approximately 121 pm.
  • In ozone (O₃), there is resonance, leading to delocalized bonding, and the O-O bond length is approximately 128 pm. This bond length is intermediate between a single bond and a double bond, but it is longer than the O=O double bond in O₂. Therefore, this statement is incorrect.
• C. Ozone acts as a more powerful oxidizing agent than oxygen.
  • Ozone readily decomposes to form dioxygen and nascent oxygen ($O_3 \rightarrow O_2 + [O]$), which is a very strong oxidizing agent. Oxygen is a comparatively weaker oxidizing agent. This statement is correct.
• D. Oxygen can be obtained by heating potassium chlorate in the presence of manganese dioxide.
  • This is a standard laboratory preparation of oxygen: $2KClO_3(s) \xrightarrow{MnO_2, \Delta} 2KCl(s) + 3O_2(g)$. Manganese dioxide ($MnO_2$) acts as a catalyst. This statement is correct.

Question 2

Ozone reacts with moist potassium iodide solution to produce:

A. $KIO_3$ B. $I_2$ C. $KI_3$ D. $K_2O$

Solution:

The correct answer is B.

Explanation: Ozone is a powerful oxidizing agent. It oxidizes iodide ions ($I^-$) to iodine ($I_2$) in the presence of water. The reaction is: $O_3(g) + 2KI(aq) + H_2O(l) \rightarrow 2KOH(aq) + I_2(s) + O_2(g)$ The liberated iodine can then be titrated against a standard sodium thiosulphate solution (hypo) to estimate ozone quantitatively.

Question 3

Which of the following species has the highest bond order?

A. $O_2^+$ B. $O_2$ C. $O_2^-$ D. $O_2^{2-}$

Solution:

The correct answer is A.

Explanation: To determine the bond order, we can use Molecular Orbital Theory (MOT). Bond Order = $\frac{1}{2}$ (Number of electrons in bonding MOs - Number of electrons in antibonding MOs)

• $O_2$: Total electrons = 16 Electronic configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{1} \pi_{2p_y}^{1}$ Bonding electrons ($N_b$) = 10 (2 from $\sigma_{2s}$, 2 from $\sigma_{2p_z}$, 4 from $\pi_{2p_x}$, $\pi_{2p_y}$) Antibonding electrons ($N_a$) = 6 (2 from $\sigma_{2s}^$, 2 from $\pi_{2p_x}^$, 2 from $\pi_{2p_y}^$) Bond order = $\frac{1}{2}(10 - 6) = 2$

• $O_2^+$: Total electrons = 15 (one electron removed from $O_2$, typically from the highest energy antibonding orbital) Electronic configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1}$ Bonding electrons ($N_b$) = 10 Antibonding electrons ($N_a$) = 5 Bond order = $\frac{1}{2}(10 - 5) = 2.5$

• $O_2^-$: Total electrons = 17 (one electron added to $O_2$, to the highest energy antibonding orbital) Electronic configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*2} \pi_{2p_y}^{*1}$ Bonding electrons ($N_b$) = 10 Antibonding electrons ($N_a$) = 7 Bond order = $\frac{1}{2}(10 - 7) = 1.5$

• $O_2^{2-}$: Total electrons = 18 (two electrons added to $O_2$, filling the highest energy antibonding orbitals) Electronic configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*2} \pi_{2p_y}^{*2}$ Bonding electrons ($N_b$) = 10 Antibonding electrons ($N_a$) = 8 Bond order = $\frac{1}{2}(10 - 8) = 1$

Comparing the bond orders: $O_2^+$ (2.5) > $O_2$ (2) > $O_2^-$ (1.5) > $O_2^{2-}$ (1). Therefore, $O_2^+$ has the highest bond order.

Assertion-Reason Questions

Question 1

Assertion (A): Ozone acts as a powerful oxidizing agent. Reason (R): Ozone readily decomposes to form dioxygen and nascent oxygen.

A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, but R is not the correct explanation of A. C. A is true, but R is false. D. A is false, but R is true. E. Both A and R are false.

Solution:

The correct answer is A.

Explanation:

• Assertion (A) is true. Ozone is indeed a very strong oxidizing agent, stronger than oxygen.
• Reason (R) is also true. Ozone readily decomposes, particularly at elevated temperatures or in the presence of certain catalysts, into a molecule of dioxygen ($O_2$) and an atom of nascent oxygen ($[O]$): $O_3(g) \rightarrow O_2(g) + [O]$
• The nascent oxygen is extremely reactive and is responsible for ozone’s strong oxidizing power. It readily accepts electrons or abstract atoms from other substances, thereby oxidizing them. Thus, the reason correctly explains why ozone acts as a powerful oxidizing agent.

Question 2

Assertion (A): Dioxygen is a gas while sulphur is a solid at room temperature. Reason (R): Oxygen forms pπ-pπ multiple bonds, while sulphur prefers to form S-S single bonds.

A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, but R is not the correct explanation of A. C. A is true, but R is false. D. A is false, but R is true. E. Both A and R are false.

Solution:

The correct answer is A.

Explanation:

• Assertion (A) is true. Dioxygen ($O_2$) exists as a diatomic gas at room temperature, while sulphur typically exists as $S_8$ puckered rings in solid form (e.g., rhombic or monoclinic sulphur) at room temperature.
• Reason (R) is also true and is the correct explanation for Assertion (A).
  • Oxygen has a small atomic size and high electronegativity. These properties favour effective lateral overlap of its p-orbitals to form strong pπ-pπ multiple bonds. Thus, oxygen exists as discrete $O_2$ molecules with a double bond. These $O_2$ molecules are held together by weak van der Waals forces, resulting in a low boiling point and gaseous state.
  • Sulphur, being larger, has a reduced tendency to form pπ-pπ bonds due to less effective orbital overlap. Instead, sulphur prefers to form strong S-S single bonds. These single bonds allow sulphur atoms to link together to form large polymeric structures, such as $S_8$ rings or long chains. The strong covalent S-S bonds within these large molecules, and stronger intermolecular forces between them compared to $O_2$, lead to sulphur being a solid at room temperature.

Short Answer Questions

Question 1

How is ozone quantitatively estimated? Explain the chemical principle involved.

Model Answer:

Ozone is quantitatively estimated by passing a known volume of ozonised oxygen through a neutral solution of potassium iodide (KI).

Chemical Principle and Reaction: Ozone acts as a powerful oxidizing agent and oxidizes iodide ions ($I^-$) to iodine ($I_2$) in the presence of water (or neutral/acidic medium). Oxygen gas is also formed in the process. The reaction is: $O_3(g) + 2KI(aq) + H_2O(l) \rightarrow 2KOH(aq) + I_2(s) + O_2(g)$

The liberated iodine ($I_2$) is then titrated against a standard solution of sodium thiosulphate ($Na_2S_2O_3$), commonly known as “hypo” solution, using starch as an indicator. Starch forms a blue complex with iodine, which disappears at the endpoint.

The titration reaction is: $I_2(s) + 2Na_2S_2O_3(aq) \rightarrow Na_2S_4O_6(aq) + 2NaI(aq)$ (Tetrathionate)

From the stoichiometry of the reactions and the volume of sodium thiosulphate consumed, the amount of iodine liberated can be calculated, which in turn allows for the quantitative estimation of ozone. This method is called iodometric titration.

Question 2

Explain why dioxygen ($O_2$) is paramagnetic despite having an even number of electrons.

Model Answer:

Dioxygen ($O_2$) is paramagnetic because its molecular orbital configuration contains unpaired electrons, as predicted by Molecular Orbital Theory (MOT).

1. Total Electrons: An oxygen atom has 8 electrons, so a dioxygen molecule ($O_2$) has a total of 16 electrons.
2. Expected Behavior by Valence Bond Theory: According to simple Valence Bond Theory (VBT), $O_2$ would be represented with a double bond ($O=O$), implying all electrons are paired. This would predict diamagnetism. However, experimental evidence shows $O_2$ is paramagnetic.
3. Molecular Orbital Theory (MOT) Explanation:
   • MOT describes the formation of molecular orbitals (MOs) by the combination of atomic orbitals (AOs).
   • The electron configuration of $O_2$ according to MOT is: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1} \pi_{2p_y}^{*1}$
   • In this configuration, the two electrons entering the degenerate $\pi_{2p_x}^$ and $\pi_{2p_y}^$ antibonding molecular orbitals occupy these orbitals individually with parallel spins, according to Hund’s Rule of Maximum Multiplicity.
   • These two unpaired electrons in the $\pi^*$ antibonding orbitals are responsible for the paramagnetic nature of dioxygen.
4. Conclusion: The presence of two unpaired electrons makes the $O_2$ molecule attracted to an external magnetic field, thus confirming its paramagnetic property.

High-Order Thinking Skills (HOTS) Question

Explain the mechanism of ozone layer depletion by chlorofluorocarbons (CFCs) in the stratosphere, highlighting the catalytic nature of the process.

Detailed Chemical Explanation:

The depletion of the ozone layer by chlorofluorocarbons (CFCs) is a critical environmental issue. CFCs are highly stable compounds that, when released into the atmosphere, can persist for many decades, eventually reaching the stratosphere where they wreak havoc on the ozone layer. The process is a chain reaction involving radical species, primarily chlorine atoms, and is catalytic in nature.

1. Photolytic Decomposition of CFCs: In the lower atmosphere (troposphere), CFCs (e.g., $CCl_2F_2$, $CCl_3F$) are inert. However, when they diffuse into the stratosphere, they are exposed to high-energy ultraviolet (UV) radiation (specifically UV-C, $\lambda < 240 nm$). This UV radiation provides enough energy to break the weakest carbon-chlorine bonds in CFC molecules, leading to the homolytic cleavage of the C-Cl bond and the generation of highly reactive chlorine free radicals ($Cl \cdot$). Example: $CCl_2F_2(g) \xrightarrow{UV light} \cdot Cl(g) + \cdot CClF_2(g)$

2. Ozone Destruction by Chlorine Radicals (Catalytic Cycle): The released chlorine radicals are extremely reactive towards ozone ($O_3$). They initiate a catalytic cycle that efficiently destroys ozone molecules.

   • Step 1: Ozone Reaction with Chlorine Radical A chlorine radical reacts with an ozone molecule, abstracting an oxygen atom to form chlorine monoxide radical ($ClO \cdot$) and a molecule of dioxygen ($O_2$). $\cdot Cl(g) + O_3(g) \rightarrow ClO \cdot (g) + O_2(g)$
   • Step 2: Regeneration of Chlorine Radical The chlorine monoxide radical then reacts with an atomic oxygen ($O \cdot$) (which is naturally present in the stratosphere due to the photolysis of $O_2$ by UV light: $O_2 \xrightarrow{UV} O \cdot + O \cdot$). This reaction regenerates the chlorine radical and produces another molecule of dioxygen. $ClO \cdot (g) + O \cdot (g) \rightarrow \cdot Cl(g) + O_2(g)$

3. Overall Reaction and Catalytic Nature: Adding the two steps, the net reaction for the ozone destruction cycle is: $O_3(g) + O \cdot (g) \rightarrow 2O_2(g)$ Notice that the chlorine radical ($\cdot Cl$) consumed in the first step is regenerated in the second step. This means that a single chlorine radical can participate in destroying thousands of ozone molecules before it is eventually removed from the stratosphere by other termination reactions (e.g., forming HCl or $ClONO_2$). This makes the process highly catalytic and extremely efficient in depleting stratospheric ozone.

4. Role of Heterogeneous Catalysis (Polar Stratospheric Clouds - PSCs): While the gas-phase reactions described above are significant, the depletion is dramatically enhanced during polar winters due to the formation of Polar Stratospheric Clouds (PSCs).

   • At very low temperatures (below -78°C) in the polar regions, nitric acid and water condense to form PSCs.
   • These ice crystal surfaces provide sites for heterogeneous reactions that convert relatively stable chlorine reservoir compounds, such as chlorine nitrate ($ClONO_2$) and hydrogen chloride ($HCl$), into more reactive species like $Cl_2$ and $HOCl$. $ClONO_2(g) + HCl(g) \xrightarrow{PSC surface} Cl_2(g) + HNO_3(s)$ $ClONO_2(g) + H_2O(g) \xrightarrow{PSC surface} HOCl(g) + HNO_3(s)$
   • When sunlight returns in the polar spring, these $Cl_2$ and $HOCl$ molecules are rapidly photolyzed to produce a large burst of active chlorine radicals ($\cdot Cl$), leading to massive ozone depletion, commonly known as the “ozone hole.” $Cl_2(g) \xrightarrow{UV light} 2 \cdot Cl(g)$ $HOCl(g) \xrightarrow{UV light} \cdot OH(g) + \cdot Cl(g)$

In summary, CFCs release chlorine radicals in the stratosphere upon photolysis. These radicals then catalytically destroy ozone, with the process being significantly exacerbated by heterogeneous reactions on polar stratospheric clouds.