Chemistry of Hydrogen: Practice Questions

Multiple Choice Questions (MCQs)

Question 1

Which of the following statements regarding the isotopes of hydrogen is incorrect?

A) Protium is the most abundant isotope of hydrogen. B) Deuterium is also known as heavy hydrogen. C) Tritium is radioactive and emits beta particles. D) All isotopes of hydrogen have the same number of neutrons.

Solution:

The correct answer is D).

Explanation:

• Protium ($\text{^1H}$ or $\text{H}$): Contains 1 proton and 0 neutrons. It is the most abundant isotope (approx. 99.98%).
• Deuterium ($\text{^2H}$ or $\text{D}$): Contains 1 proton and 1 neutron. It is known as heavy hydrogen.
• Tritium ($\text{^3H}$ or $\text{T}$): Contains 1 proton and 2 neutrons. It is radioactive with a half-life of 12.33 years and emits low energy beta particles.

Since Protium has 0 neutrons, Deuterium has 1 neutron, and Tritium has 2 neutrons, they do not have the same number of neutrons.

Question 2

Which of the following compounds is an interstitial hydride?

A) $\text{NaH}$ B) $\text{CH}4$ C) $\text{TiH}{1.7}$ D) $\text{HCl}$

Solution:

The correct answer is C).

Explanation:

• Interstitial hydrides (also known as metallic hydrides) are formed by many d-block and f-block elements. In these hydrides, hydrogen occupies interstitial sites in the metal lattice. They are typically non-stoichiometric (e.g., $\text{TiH}{1.5-1.8}$, $\text{LaH}{2.87}$). $\text{TiH}_{1.7}$ is an example of an interstitial hydride.
• $\text{NaH}$ is an example of a saline (ionic) hydride, formed by highly electropositive s-block elements.
• $\text{CH}_4$ is an example of a covalent (molecular) hydride, formed by p-block elements.
• $\text{HCl}$ is also a covalent (molecular) hydride.

Question 3

In the industrial preparation of dihydrogen, the water-gas shift reaction is often employed. Which of the following statements correctly describes the role of steam in this reaction?

$\text{CO(g)} + \text{H}_2\text{O(g)} \xrightarrow{\text{Catalyst, High temp.}} \text{CO}_2\text{(g)} + \text{H}_2\text{(g)}$

A) Steam acts as a reducing agent, converting $\text{CO}$ to $\text{CO}_2$. B) Steam acts as an oxidizing agent, converting $\text{CO}$ to $\text{CO}_2$. C) Steam acts as a catalyst, speeding up the reaction. D) Steam provides the hydrogen atoms for the formation of $\text{H}_2$.

Solution:

The correct answer is B).

Explanation:

In the water-gas shift reaction: $\text{CO(g)} + \text{H}_2\text{O(g)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{(g)}$

• The oxidation state of carbon in $\text{CO}$ is +2, and in $\text{CO}_2$ it is +4. Carbon is oxidized.
• The oxidation state of hydrogen in $\text{H}_2\text{O}$ is +1, and in $\text{H}_2$ it is 0. Hydrogen is reduced.
• Therefore, $\text{CO}$ acts as a reducing agent, and $\text{H}_2\text{O}$ (steam) acts as an oxidizing agent by accepting electrons (its hydrogen atoms get reduced from +1 to 0).

Assertion-Reason Questions

Question 1

Assertion (A): Hydrogen exhibits properties similar to both alkali metals and halogens. Reason (R): Hydrogen has only one electron in its outermost shell and can either lose an electron to form $\text{H}^+$ or gain an electron to form $\text{H}^-$.

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. Hydrogen has an electronic configuration of $1s^1$.
  • Like alkali metals (Group 1, $ns^1$), it can lose its valence electron to form a unipositive ion ($\text{H}^+$), though this is rare in condensed phases.
  • Like halogens (Group 17, $ns^2np^5$), it can gain one electron to achieve a stable duplet configuration ($1s^2$) and form a uninegative ion ($\text{H}^-$).
• Reason (R) is true and correctly explains the assertion. The ability to both lose and gain an electron due to its $1s^1$ configuration is precisely why hydrogen shows dual resemblance to both alkali metals and halogens.

Question 2

Assertion (A): Dihydrogen is less reactive at room temperature. Reason (R): The $\text{H}-\text{H}$ bond dissociation enthalpy is very high.

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. Dihydrogen ($\text{H}_2$) is relatively inert at room temperature. It typically reacts at higher temperatures or in the presence of catalysts.
• Reason (R) is true. The $\text{H}-\text{H}$ bond in the $\text{H}_2$ molecule is very strong, with a bond dissociation enthalpy of 435.88 kJ/mol. This high bond dissociation enthalpy means that a significant amount of energy is required to break the $\text{H}-\text{H}$ bond, which is the initial step for most reactions of dihydrogen.
• The high bond dissociation enthalpy directly explains why dihydrogen is less reactive at room temperature, as enough energy is not available to break the strong bond. Thus, Reason (R) is the correct explanation of Assertion (A).

Short Answer Questions

Question 1

Explain the terms ‘ortho’ and ‘para’ hydrogen. How do they differ in properties?

Model Answer:

Dihydrogen ($\text{H}_2$) exists in two forms based on the relative spin of the nuclei of the two hydrogen atoms:

1. Ortho hydrogen: In this form, the spins of the two hydrogen nuclei (protons) are in the same direction (parallel spins).
2. Para hydrogen: In this form, the spins of the two hydrogen nuclei (protons) are in the opposite direction (antiparallel spins).

Differences in properties:

• Energy: Para hydrogen has lower energy and is more stable than ortho hydrogen.
• Specific Heat: Ortho hydrogen has a higher specific heat capacity than para hydrogen.
• Thermal Conductivity: Ortho hydrogen has higher thermal conductivity than para hydrogen.
• Magnetic Properties: Ortho hydrogen has a net nuclear magnetic moment, while para hydrogen has zero net nuclear magnetic moment.
• Equilibrium: At room temperature (25°C), the equilibrium mixture consists of approximately 75% ortho hydrogen and 25% para hydrogen. As the temperature decreases, the percentage of para hydrogen increases, reaching nearly 100% at absolute zero. The conversion between the two forms is slow but can be catalyzed by activated charcoal, atomic hydrogen, or paramagnetic substances.

Question 2

Describe the water-gas shift reaction, including the catalyst used and its significance in hydrogen production.

Model Answer:

The water-gas shift reaction is a chemical reaction used to produce additional dihydrogen ($\text{H}_2$) from carbon monoxide ($\text{CO}$) and steam ($\text{H}_2\text{O}$). It is represented by the following equilibrium:

$\text{CO(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g)} + \text{H}_2\text{(g)}$

Process and Catalyst: The reaction is carried out by passing carbon monoxide (obtained from the steam reforming of hydrocarbons or from the gasification of coal, forming “water gas” which is a mixture of CO and H2) over a hot catalyst.

• Catalyst: Typically, iron chromate ($\text{FeCr}_2\text{O}_4$) or a mixture of iron oxide and chromium oxide ($\text{Fe}_2\text{O}_3/\text{Cr}_2\text{O}_3$) is used at high temperatures ($\sim 400-500^\circ\text{C}$). For low-temperature shift, a copper-zinc oxide catalyst is used.
• Conditions: High temperature is required to achieve a reasonable reaction rate.

Significance in Hydrogen Production:

1. Enhances $\text{H}_2$ yield: It allows for the conversion of the $\text{CO}$ component of water gas (a mixture of $\text{CO}$ and $\text{H}_2$) into more $\text{H}_2$, thereby increasing the overall yield of dihydrogen from processes like coal gasification or steam reforming of natural gas.
2. Removes $\text{CO}$: The reaction converts $\text{CO}$ into $\text{CO}_2$. $\text{CO}$ is a poison for many catalysts, especially those used in ammonia synthesis (Haber process) or fuel cells. Removing $\text{CO}$ ensures the purity of hydrogen required for these applications. The produced $\text{CO}_2$ can then be easily removed by absorption in solutions of potassium carbonate or other absorbents.

High-Order Thinking Skills (HOTS) Question

Despite having an electronic configuration similar to alkali metals ($1s^1$), hydrogen does not typically form $\text{H}^+$ ions in aqueous solutions and its chemistry is distinct. Explain why hydrogen’s behavior is unique and not entirely analogous to alkali metals or halogens.

Detailed Chemical Explanation:

Hydrogen’s unique position in the periodic table and its distinct chemical behavior stem primarily from its extremely small size, high ionization enthalpy, and relatively low electron gain enthalpy compared to halogens.

1. High Ionization Enthalpy:

   • For hydrogen, the single electron in the $1s$ orbital is very close to the nucleus, resulting in strong nuclear attraction. Its ionization enthalpy (1312 kJ/mol) is much higher than that of alkali metals (e.g., Li: 520 kJ/mol, Na: 496 kJ/mol).
   • This high ionization enthalpy means that forming a bare $\text{H}^+$ ion (a proton) by losing an electron is energetically very unfavorable. A bare proton has an extremely high charge density and is never found free in aqueous solutions. Instead, it readily associates with other molecules, particularly water, to form hydronium ions ($\text{H}_3\text{O}^+$) or other protonated species. This distinguishes it significantly from alkali metals, which readily form stable, hydrated $\text{M}^+$ ions.

2. Electron Gain Enthalpy:

   • Hydrogen’s electron gain enthalpy is slightly negative (-73 kJ/mol), which is much less negative (less exothermic) than that of halogens (e.g., F: -328 kJ/mol, Cl: -349 kJ/mol).
   • While hydrogen can gain an electron to form a hydride ion ($\text{H}^-$), the stability of $\text{H}^-$ is generally lower than that of halide ions ($\text{X}^-$). The formation of $\text{H}^-$ requires significant energy input when reacting with less electropositive elements. This behavior is primarily observed with highly electropositive elements like alkali and alkaline earth metals, forming ionic hydrides.

3. Covalent Bond Formation:

   • Given its intermediate electronegativity (2.20 on the Pauling scale), hydrogen predominantly forms stable covalent bonds by sharing its electron. This is in stark contrast to alkali metals, which primarily form ionic compounds, and halogens, which can also form significant covalent bonds but typically aim for an octet.
   • The formation of a single covalent bond allows hydrogen to achieve the stable duplet configuration (like helium), which is its preferred state. This propensity for covalent bonding is the basis for the vast number of organic compounds and many inorganic hydrides.

4. Absence of d-orbitals:

   • Unlike many p-block elements that can expand their octet due to the availability of empty d-orbitals, hydrogen has no d-orbitals. It can only form one bond, whether covalent or ionic. This limits its coordination number and overall chemical versatility compared to elements further down its potential groups.

In summary, hydrogen’s unique size, ionization energy, and electron affinity place it in a chemical category of its own. It’s a bridge element that can mimic certain aspects of both Group 1 and Group 17, but fundamentally, its small size and tendency to achieve a stable duplet by covalent bonding dictate its distinct and rich chemistry.