Chemistry of Nitrogen (N) Practice Questions

Multiple Choice Questions (MCQs)

1. Which of the following statements best explains the inertness of dinitrogen (N$_2$) at room temperature?

A) Nitrogen has a small atomic size. B) Nitrogen has high electronegativity. C) The N≡N bond in N$_2$ has a very high bond dissociation enthalpy. D) Nitrogen is a non-metal.

Solution: C Explanation: Dinitrogen (N$_2$) contains a triple bond (N≡N), which is very strong. The bond dissociation enthalpy for the N≡N bond is approximately 946 kJ/mol, one of the highest known for a diatomic molecule. This high energy required to break the bond makes N$_2$ chemically inert and unreactive at room temperature.

2. What is the oxidation state of Nitrogen in ammonium nitrate (NH$_4$NO$_3$)?

A) +5 and -3 B) +3 and -5 C) +1 and -1 D) +4 and -4

Solution: A Explanation: Ammonium nitrate is an ionic compound composed of the ammonium ion (NH$_4$$^+$) and the nitrate ion (NO$_3$$^-$). In NH$_4$$^+$: Let the oxidation state of N be ‘x’. x + 4(+1) = +1 => x + 4 = +1 => x = -3. In NO$_3$$^-$: Let the oxidation state of N be ‘y’. y + 3(-2) = -1 => y - 6 = -1 => y = +5. Therefore, nitrogen exhibits two different oxidation states in NH$_4$NO$_3$: -3 in the ammonium ion and +5 in the nitrate ion.

3. Which of the following is the correct set of conditions for the Haber’s process for the manufacture of ammonia?

A) High temperature, low pressure, presence of catalyst. B) Low temperature, high pressure, absence of catalyst. C) High temperature, high pressure, presence of catalyst. D) Low temperature, low pressure, presence of catalyst.

Solution: C Explanation: The Haber’s process (N$_2$(g) + 3H$_2$(g) ⇌ 2NH$_3$(g), ΔH = -92.4 kJ/mol) is an exothermic reaction that produces a decrease in the number of moles. According to Le Chatelier’s principle:

Temperature: Since the reaction is exothermic, a lower temperature favors ammonia formation. However, to achieve an economically viable reaction rate, a compromise temperature of 673-773 K (400-500 °C) is used.
Pressure: The reaction involves a decrease in the number of moles (4 moles of reactants to 2 moles of product). Therefore, high pressure favors ammonia formation. Typically, pressures of 200-300 atm are used.
Catalyst: An iron catalyst (often with molybdenum as a promoter) is used to increase the rate of reaction, helping to achieve equilibrium faster without affecting the position of equilibrium.

Thus, the correct set of conditions is high temperature (relative to kinetic favorability), high pressure, and the presence of a catalyst.

Assertion-Reason Questions

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option. 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.

1. Assertion (A): Dinitrogen (N$_2$) is less reactive than white phosphorus (P$_4$).

Reason (R): Nitrogen has a strong N≡N bond and lacks d-orbitals for bond expansion, unlike phosphorus.

Solution: A Explanation:

Assertion (A): Dinitrogen is indeed less reactive than white phosphorus. N$_2$ is quite inert at room temperature, while white phosphorus is highly reactive and spontaneously ignites in air.
Reason (R): The N≡N triple bond in dinitrogen has an exceptionally high bond dissociation enthalpy (946 kJ/mol), making it difficult to break. This is the primary reason for its inertness. Furthermore, nitrogen lacks accessible d-orbitals in its valence shell, preventing it from expanding its octet and forming more than four bonds. Phosphorus, on the other hand, has vacant d-orbitals, allowing it to form more than four bonds (e.g., PCl$_5$) and engage in a wider range of reactions. The P-P single bond in P$_4$ is also weaker than the N≡N triple bond. Both A and R are true, and R correctly explains why N$_2$ is less reactive.

2. Assertion (A): Concentrated nitric acid acts as a strong oxidizing agent.

Reason (R): In concentrated nitric acid, nitrogen is in its highest possible oxidation state of +5.

Solution: A Explanation:

Assertion (A): Concentrated nitric acid (HNO$_3$) is a powerful oxidizing agent, capable of oxidizing many metals (except noble metals like Au, Pt) and non-metals.
Reason (R): In HNO$_3$, the oxidation state of nitrogen is calculated as: x + 1(+1) + 3(-2) = 0 => x - 5 = 0 => x = +5. This is the highest possible oxidation state for nitrogen. Since nitrogen is already in its highest oxidation state, it can only decrease its oxidation state, meaning it can only be reduced. Species that get reduced act as oxidizing agents. Both A and R are true, and R provides the correct explanation for A.

Short Answer Questions

1. Explain why ammonia (NH$_3$) acts as a Lewis base.

Model Answer: Ammonia (NH$_3$) acts as a Lewis base because of the presence of a lone pair of electrons on the nitrogen atom. According to the Lewis theory, a Lewis base is a species that can donate a pair of electrons. The nitrogen atom in ammonia has five valence electrons; three are shared with hydrogen atoms to form covalent bonds, leaving one lone pair. This readily available lone pair can be donated to an electron-deficient species (a Lewis acid), forming a coordinate covalent bond. For example: NH$_3$ + H$^+$ → NH$_4$$^+$ (Ammonia donates its lone pair to H$^+$) NH$_3$ + BF$_3$ → H$_3$N → BF$_3$ (Ammonia donates its lone pair to electron-deficient BF$_3$)

2. Describe the main steps involved in the Ostwald process for the manufacture of nitric acid.

Model Answer: The Ostwald process is an industrial method for the manufacture of nitric acid (HNO$_3$) from ammonia (NH$_3$). It involves three main steps:

Catalytic Oxidation of Ammonia: Ammonia gas, mixed with air (or oxygen), is passed over a heated platinum-rhodium gauze catalyst at about 800-950 °C and 9 bar pressure. Ammonia is oxidized to nitric oxide (NO) and water. 4NH$_3$(g) + 5O$_2$(g) $\xrightarrow{\text{Pt/Rh gauze, 800-950°C}}$ 4NO(g) + 6H$_2$O(g)
Oxidation of Nitric Oxide: The nitric oxide (NO) produced is cooled and then further oxidized by reacting with excess air (or oxygen) to form nitrogen dioxide (NO$_2$). This reaction is exothermic and proceeds spontaneously. 2NO(g) + O$_2$(g) $\rightarrow$ 2NO$_2$(g)
Absorption of Nitrogen Dioxide in Water: Nitrogen dioxide (NO$_2$) is then absorbed in water in an absorption tower, usually counter-current to water flow. This reaction produces nitric acid and regenerates some nitric oxide. The nitric oxide is then recycled back to step 2. 3NO$_2$(g) + H$_2$O(l) $\rightarrow$ 2HNO$_3$(aq) + NO(g) The aqueous nitric acid obtained is typically around 68% by mass.

High-Order Thinking Skills (HOTS) Question

Explain why BiH$_3$ is the strongest reducing agent among the Group 15 hydrides (NH$_3$, PH$_3$, AsH$_3$, SbH$_3$, BiH$_3$), while NH$_3$ is the weakest.

Model Answer: The reducing character of the Group 15 hydrides (MH$_3$) increases down the group from NH$_3$ to BiH$_3$. This trend is explained by considering the bond strength of the M-H bond and the size of the central atom.

Bond Dissociation Enthalpy (M-H Bond Strength): As we move down Group 15 from nitrogen to bismuth, the atomic size of the central atom (M) increases significantly. Consequently, the M-H bond length increases (e.g., N-H bond is shorter than Bi-H bond). A longer bond generally means a weaker bond, as the overlap between the valence orbitals of the central atom and hydrogen becomes less effective. Therefore, the bond dissociation enthalpy of the M-H bond decreases progressively from NH$_3$ to BiH$_3$.
Ease of Hydrogen Release: Reducing agents are substances that donate electrons or provide hydrogen atoms for reduction (get oxidized themselves). In the case of hydrides, their reducing strength is related to the ease with which they can release hydrogen atoms. Since the M-H bond strength decreases down the group, it becomes progressively easier to break the M-H bond and release hydrogen.
Stability of Hydrides: The thermal stability of these hydrides also follows the same trend, decreasing from NH$_3$ to BiH$_3$. NH$_3$ is very stable, while BiH$_3$ is highly unstable and readily decomposes to bismuth and hydrogen upon heating.

Conclusion:

NH$_3$ (Ammonia): Has the strongest N-H bond due to the small size of nitrogen and effective orbital overlap. This makes it difficult to release hydrogen, hence it is the weakest reducing agent (or most stable).
BiH$_3$ (Bismuthine): Has the weakest Bi-H bond due to the large size of bismuth and poor orbital overlap. This makes it very easy to release hydrogen, thus making it the strongest reducing agent (or least stable).

The reducing power order is: NH$_3$ < PH$_3$ < AsH$_3$ < SbH$_3$ < BiH$_3$.