GSEB Class 11 Chemistry Solutions Chapter 9 Hydrogen

 

Question 1. Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Answer: Hydrogen is the first element in the periodic table, and it has an electron configuration of \( 1s^1 \). It is similar to alkali metals (ns¹) in Group 1. It also shows similarities with alkali metals found in Group 1 of the periodic table. For these reasons, hydrogen can be placed above the alkali metals in Group 1 of the periodic table.
On the other hand, hydrogen's electron configuration reveals that it needs only one more electron to achieve the nearest noble gas configuration (He), which is \( 1s^2 \). Like halogens, it forms both covalent bonds (e.g., \( H_2, Cl_2, Br_2 \)) and ionic bonds (e.g., \( Na^+H^- \)). It can form an \( H^+ \) ion by giving away one electron and a hydride ion (\( H^- \)) by gaining one electron.
Based on its electron configuration (\( 1s^2 \)), hydrogen is placed with other \( ns^1 \) elements, namely alkali metals in Group 1, as well as in Group 17 of the periodic table. This makes hydrogen's position in the periodic table unusual. Since hydrogen has many distinct properties, it is best placed separately in the periodic table of elements.
In simple words: Hydrogen can fit in two places in the periodic table because it acts like Group 1 metals by losing an electron and like Group 17 halogens by gaining one. This unique behavior means it's often shown in its own special spot.

Exam Tip: To justify hydrogen's position, remember to discuss its dual resemblance to both alkali metals (Group 1) and halogens (Group 17) based on its electronic configuration and bonding behavior.

 

Question 2. Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
Answer: Hydrogen possesses three isotopes:
Protium (\( ^1H \)), deuterium (\( ^2H \) or D), and Tritium (\( ^3H \) or T).
The mass ratio of these isotopes is approximately 1.008 : 2.014 : 3.016, or roughly 1:2:3.
In simple words: Hydrogen has three types: Protium, Deuterium, and Tritium. Their masses are in a ratio of about 1:2:3.

Exam Tip: Always remember the three isotopes of hydrogen and their respective mass numbers. The approximate 1:2:3 ratio is a key detail for this question.

 

Question 3. Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
Answer: The H-H bond dissociation enthalpy is the highest for a single bond between any two atoms of an element. This high energy is why the dissociation of dihydrogen into its individual atoms is very difficult. Only a tiny fraction, about 0.081%, dissociates around 2000 K.
In simple words: Hydrogen stays as two atoms (H2) because it takes a lot of energy to break the bond between them, more than almost any other single bond. This strong bond keeps it together unless it gets extremely hot.

Exam Tip: Focus on the concept of high bond dissociation enthalpy. This is the main reason for hydrogen's diatomic nature under normal conditions.

 

Question 4. How can the production of dihydrogen, obtained from 'coal gasification' be increased ?
Answer: The process of making 'syn-gas' from coal is known as 'coal gasification.' The mixture of carbon monoxide (CO) and hydrogen (\( H_2 \)) is called SYNGAS.
\[ C(s) + H_2O(g) \xrightarrow{1270 K} CO(g) + H_2(g) \] The production of dihydrogen can be increased by reacting the carbon monoxide (CO) from syngas mixtures with steam. This reaction happens in the presence of an iron chromate catalyst.
\[ CO(g) + H_2O(g) \xrightarrow[Catalyst]{673 K} CO_2(g) + H_2(g) \]In simple words: To get more hydrogen from coal gasification, you take the syngas (CO and H2) and react the CO part with steam. This reaction, using a catalyst like iron chromate, makes more hydrogen.

Exam Tip: Remember that 'coal gasification' produces syngas (\( CO + H_2 \)). To increase \( H_2 \) yield, the 'water-gas shift reaction' is crucial, where CO reacts with steam using a catalyst.

 

Question 5. Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process?
Answer: Electrolysis of water that contains an electrolyte (15-25% of acid or alkali) is the best method available to produce dihydrogen when electricity is inexpensive. The presence of an electrolyte helps the solution become conductive, allowing the process to occur efficiently. Platinum electrodes are commonly used in this process.
\[ 2H_2O(l) \xrightarrow{\text{Electrolysis}} 2H_2(g) + O_2(g) \] \[ \text{Traces of acid/base} \]In simple words: Dihydrogen is made in large amounts by passing electricity through water. Adding an electrolyte like acid or alkali makes the water conduct electricity, which is essential for the process.

Exam Tip: When describing the electrolytic method, ensure you mention the use of an electrolyte (acid or alkali) and its role in increasing the conductivity of water, which is essential for efficient electrolysis.

 

Question 6. Complete the following reactions:
(i) \( H_2(g) + M_xO_y(s) \xrightarrow{\Delta} \)
(ii) \( CO(g) + H_2(g) \xrightarrow[Catalyst]{\Delta} \)
(iii) \( C_3H_8(g) + 3H_2O(g) \xrightarrow[Catalyst]{\Delta} \)
(iv) \( Zn(s) + NaOH(aq) \xrightarrow{heat} \)
Answer:
(i) \( H_2(g) + M_xO_y(s) \xrightarrow{\Delta} mM(s) + 0H_2O \)
(ii) \( CO(g) + 2H_2(g) \xrightarrow[Catalyst]{\Delta} CH_3OH(l) \)
(iii) \( C_3H_8(g) + 3H_2O(g) \xrightarrow[Catalyst]{\Delta} 3CO(g) + 7H_2(g) \)
(iv) \( Zn(s) + 2NaOH(aq) \xrightarrow{heat} Na_2ZnO_2 + H_2 \) (Sodium zincate)
In simple words: These are chemical reactions showing how different substances combine or change when heated or mixed with a catalyst. They produce new compounds like methanol, carbon monoxide, hydrogen, or sodium zincate.

Exam Tip: For reaction completion questions, pay close attention to the reactants, conditions (heat, catalyst), and stoichiometry to determine the correct products. Identify common reaction types like reduction, synthesis, or single displacement.

 

Question 7. Discuss the consequences of high enthalpy of H-H bond in terms of chemical reactivity of dihydrogen.
Answer: Due to its high bond dissociation enthalpy (435.88 kJ mol⁻¹), dihydrogen is quite stable. This is the highest enthalpy for a single bond between any two atoms of an element. Because of this factor, the dissociation of dihydrogen into its atoms is only 0.081% at around 2000 K, although it increases to 95.5% at 5000 K. It is also relatively unreactive at room temperature because of its high H-H bond enthalpy.
In simple words: The strong bond between two hydrogen atoms means that hydrogen gas is very stable and doesn't react easily at normal temperatures. It needs a lot of heat to break apart into individual atoms and become more reactive.

Exam Tip: When discussing the chemical reactivity of dihydrogen, emphasize that its high bond dissociation enthalpy makes it highly stable and relatively inert at room temperature, requiring high temperatures to become reactive.

 

Question 8. What do you understand by
(i) electron-deficient
(ii) electron-precise
(iii) electron-rich compounds of hydrogen?
Provide justification with suitable examples.
Answer: Compounds of hydrogen with other elements are known as hydrides. They can be categorized as:

• Electron-deficient
• Electron-precise and
• Electron-rich hydrides.

(i) An electron-deficient hydride has too few electrons to write its conventional Lewis structure. Diborane (\( B_2H_6 \)) is a good example. All elements in Group 13 will form electron-deficient compounds. These compounds act as Lewis acids, meaning they are electron acceptors.
(ii) Electron-precise compounds have the exact number of electrons needed to write their conventional Lewis structures. Methane (\( CH_4 \)) and ethane (\( C_2H_6 \)) are examples of electron-precise compounds. All elements in Group 14 form such compounds, which are tetrahedral in shape.
(iii) Electron-rich hydrides have extra electron(s) that are present as lone pairs. Elements from Group 15 to 17 form such compounds. For example, ammonia (\( NH_3 \)) has one lone pair, water (\( H_2O \)) has two, and hydrogen fluoride (HF) has three lone pairs of electrons. These compounds behave as Lewis bases, meaning they are electron donors.
In simple words: Hydrides are hydrogen compounds grouped by how many electrons they have. Electron-deficient hydrides don't have enough electrons to fill their outer shells, like Diborane. Electron-precise hydrides have just enough electrons, like Methane. Electron-rich hydrides have extra electrons (lone pairs), like Water.

Exam Tip: When classifying hydrides, link each type (electron-deficient, -precise, -rich) to its electron count relative to Lewis structures, its group in the periodic table, and its behavior as a Lewis acid or base, with specific examples.

 

Question 9. What characteristics do you expect from an electron- deficient hydride with respect to its structure and chemical reactions?
Answer: An electron-deficient hydride features an atom of the element that has not completed its octet. For instance, boron (B) in boron trifluoride (\( BF_3 \)) and diborane (\( B_2H_6 \)) is electron-deficient.
Its structure often involves bridging bonds (like in \( B_2H_6 \)) or can be trigonal planar like \( BF_3 \). A 120° trigonal planar molecule of \( BF_3 \) is an example of sp² hybridisation.
\[ \text{B in } BF_3 \text{ has not completed its octet} \] \[ \begin{array}{c} :F: \\ \quad \\ :F:\text{B}:F: \\ \quad \\ :F: \end{array} \] It has a tendency to accept a pair of electrons from a molecule that has one or more lone pairs of electrons. Any substance that accepts a pair of electrons, according to the Lewis concept, is an acid. Therefore, compounds of Group 13 with hydrogen, such as \( BCl_3, BF_3, AlCl_3 \), etc., act as Lewis acids.
For example:
\[ \begin{array}{c} H \\ :N: \\ H \end{array} + \begin{array}{c} F \\ B:F \\ F \end{array} \rightarrow \begin{array}{c} H \quad F \\ H-N \rightarrow B-F \\ H \quad F \end{array} \] Here, ammonia (\( NH_3 \)) is an electron-pair donor (Lewis base), and boron trifluoride (\( BF_3 \)) is an electron-pair acceptor (Lewis acid). A coordinate bond forms between the donor and acceptor.
In simple words: Electron-deficient hydrides have an atom that doesn't have a full set of eight electrons, like boron in \( BF_3 \). Because they are missing electrons, they like to accept electron pairs from other molecules, meaning they act like Lewis acids in chemical reactions.

Exam Tip: When describing electron-deficient hydrides, focus on the incomplete octet, their ability to accept electron pairs (Lewis acid behavior), and classic examples like \( BF_3 \) and \( B_2H_6 \).

 

Question 10. Do you expect the carbon hydrides of the type (\( C_nH_{2n+2} \)) to act as 'Lewis' acid or base ? Justify your answer.
Answer: Carbon hydrides of the type \( C_nH_{2n+2} \), such as \( CH_4 \) and \( C_2H_6 \), have exactly enough electrons to write their conventional Lewis structures. In these compounds, carbon does not have extra electrons nor is it deficient in electrons to complete its octet. Therefore, such compounds with the formula \( C_nH_{2n+2} \) are known as electron-precise compounds. They will act as neither Lewis acids nor Lewis bases.
Here are the Lewis structures for Methane and Ethane:
\[ \begin{array}{cc} H & H \quad H \\ H:C:H & H:C:C:H \\ H & H \quad H \\ \text{Methane} & \text{Ethane} \end{array} \]In simple words: Carbon hydrides like methane (CH4) don't act as Lewis acids or bases. This is because the carbon atoms in these molecules already have a complete outer shell of electrons, so they don't need to gain or lose more.

Exam Tip: For hydrides of the form \( C_nH_{2n+2} \), remember they are "electron-precise" because all atoms achieve a stable octet (or duet for hydrogen), meaning they will not act as Lewis acids (electron acceptors) or bases (electron donors).

 

Question 11. What do you understand by the term "non- stoichiometric hydrides"? Do you expect this type of the hydrides to be formed by alkali metals? Justify your answer.
Answer: Non-stoichiometric hydrides, also known as interstitial hydrides, are compounds with compositions like \( LaH_{2.87}, YbH_{2.55}, TiH_{1.5-1.8}, ZrH_{1.3-1.75}, VH_{0.50} \), etc. These hydrides are deficient in hydrogen. They are formed by many d-block and f-block elements. Alkali metals do not form these types of hydrides. Saline hydrides are stoichiometric, meaning they have fixed integer ratios of elements. D-block or f-block elements have partially filled d or f subshells, which allows hydrogen atoms to occupy the empty spaces (interstices) in their metal lattices.
In simple words: Non-stoichiometric hydrides are compounds where the amount of hydrogen doesn't fit a simple whole-number ratio, like \( TiH_{1.5} \). They happen with d-block and f-block metals because hydrogen atoms can squeeze into the gaps in their structure. Alkali metals don't form these because their structure is different.

Exam Tip: When defining non-stoichiometric hydrides, highlight their variable composition and their formation mainly by d-block and f-block elements due to hydrogen occupying interstitial sites. Remember to state clearly that alkali metals do not form them.

 

Question 12. How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Answer: The property of hydrogen absorption onto transition metals in metallic hydrides is widely utilized in catalytic reduction and hydrogenation reactions for preparing many compounds. Some metals, such as Palladium (Pd) and Platinum (Pt), can accommodate a very large volume of hydrogen. Therefore, these metals can be used as storage media for hydrogen. This property has great potential for hydrogen storage and as an energy source.
In simple words: Metallic hydrides, especially those made with metals like palladium or platinum, can hold a lot of hydrogen. This makes them useful for storing hydrogen safely and compactly, which is important for future energy solutions.

Exam Tip: Focus on the capacity of transition metals (like Pd, Pt) to absorb large volumes of hydrogen within their lattice. This 'storage' ability makes metallic hydrides potential candidates for safe and efficient hydrogen storage and as an energy source.

 

Question 13. How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.
Answer: Atomic hydrogen and oxy-hydrogen torches are commonly used for cutting and welding. In these torches, molecular hydrogen (\( H_2 \)) is dissociated into individual hydrogen atoms with the help of an electric arc:
\[ H_2(g) \xrightarrow{heat} 2H(g) \] These individual hydrogen atoms (H) are then allowed to recombine on the surface of the metal being cut or welded. This recombination process releases a significant amount of heat, generating temperatures of around 4000 K, which is sufficient for cutting and welding purposes.
In simple words: Atomic hydrogen torches work by breaking hydrogen gas into single hydrogen atoms using electricity. When these atoms then rejoin on a metal's surface, they release intense heat, hot enough to cut or weld the metal.

Exam Tip: The key principle of the atomic hydrogen torch is the dissociation of \( H_2 \) into atomic hydrogen (\( H \)) and the subsequent recombination of these atoms on the metal surface, which releases a high amount of heat for welding and cutting.

 

Question 14. Among \( NH_3, H_2O \) and HF, which would you expect to have highest magnitude of hydrogen bonding and why?
Answer: Among \( NH_3, H_2O \), and HF, hydrogen fluoride (HF) will exhibit the highest magnitude of hydrogen bonding. This is because, among nitrogen (N), oxygen (O), and fluorine (F), fluorine possesses the highest electronegativity value.
In simple words: HF will have the strongest hydrogen bonds out of \( NH_3, H_2O \), and HF. This is because fluorine is the most electronegative element among nitrogen, oxygen, and fluorine, making the hydrogen-fluorine bond highly polar and thus stronger in hydrogen bonding.

Exam Tip: The strength of hydrogen bonding is directly related to the electronegativity of the atom bonded to hydrogen. Fluorine is the most electronegative of N, O, and F, leading to the strongest hydrogen bonds in HF.

 

Question 15. Saline hydrides are known to react with water violently producing fire. Can \( CO_2 \), a well known fire extinguisher, be used in this case? Explain.
Answer: Saline hydrides, such as sodium hydride (NaH), react violently with water, liberating hydrogen gas:
\[ NaH(s) + H_2O(l) \rightarrow NaOH(aq) + H_2(g) \] The reaction is exothermic, meaning it releases heat (\( \Delta H = -Q \)). Because of this exothermic nature, the liberated dihydrogen gas (\( H_2 \)) often catches fire. The fire produced by this reaction cannot be extinguished using \( CO_2 \) because the hot metal hydride reduces \( CO_2 \).
In simple words: Saline hydrides react strongly with water, making hydrogen gas that can catch fire because the reaction produces heat. You can't use \( CO_2 \) to put out this fire because the hot metal hydride will react with the \( CO_2 \) itself, making the problem worse.

Exam Tip: Remember that the violent reaction of saline hydrides with water generates flammable hydrogen gas and is exothermic. Crucially, \( CO_2 \) is ineffective as an extinguisher because it can be reduced by the hot metal hydride.

 

Question 16. Arrange the following :
(i) \( CaH_2, BeH_2 \) and \( TiH_2 \) in order of increasing electrical conductance
(ii) \( LiH, NaH \) and \( CsH \) in order of increasing ionic character.
(iii) H-H, D-D and F-F in order of increasing bond dissociation energy.
(iv) \( NaH, MgH_2 \) and \( H_2O \) in order of increasing reducing property.
Answer:
(i) Electrical conductance: \( BeH_2 < CaH_2 < TiH_2 \)
(ii) Ionic character: \( LiH < NaH < CsH \)
(iii) Bond dissociation energy: \( F-F \)
(iv) Reducing property: \( H_2O < MgH_2 < NaH \).
In simple words: The compounds are arranged based on how well they conduct electricity (from less to more), how much they act like ions (from less to more), how much energy is needed to break their bonds (highest listed first if it's F-F), and how good they are at donating electrons (from less to more).

Exam Tip: For arrangement questions, recall the trends within the periodic table for metallic character, electronegativity, and bond strength, as these factors directly influence properties like electrical conductance, ionic character, and reducing ability.

 

Question 17. Compare the structures of \( H_2O \) and \( H_2O_2 \).
Answer: The structure of a \( H_2O \) molecule: Water in its gaseous form is a bent molecule (v-shaped). It has an H-O-H bond angle of 104.5° and an O-H bond length of 95.7 pm. Water is polar in nature.
The structure of \( H_2O_2 \) (hydrogen peroxide): It possesses a non-planar structure. The molecular dimensions in the gas phase and solid phase are shown below. The two oxygen atoms are connected by a single electron pair bond. In the gas phase, the dihedral angle is 111.5°, and in the solid phase at 110 K, the dihedral angle is 90.2°.
In simple words: Water (\( H_2O \)) is a bent, v-shaped molecule with a specific bond angle, while hydrogen peroxide (\( H_2O_2 \)) has a non-planar structure, meaning it's not flat.

Exam Tip: When comparing \( H_2O \) and \( H_2O_2 \), remember that water is bent (V-shaped) and planar, whereas hydrogen peroxide has a non-planar, 'open book' structure with different dihedral angles in gas and solid states.

 

Question 18. What do you understand by the term 'auto-protolysis' of water? What is its significance?
Answer: Water acts both as an acid and as a base, which means it is amphoteric in character. For example, it can behave as an acid with \( NH_3 \) and as a base towards \( H_2S \). Generally, water can act as a base towards acids stronger than itself and as an acid towards a base stronger than itself. Thus, in terms of its amphoteric nature, the auto-protolysis of water can be shown as follows. Two molecules of water react with each other through proton transfer. One \( H_2O \) molecule is converted to \( H_3O^+ \), while the other is converted to \( OH^- \).
\[ H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq) \] \[ \text{acid} \quad \text{base} \quad \text{acid}_2 \quad \text{base}_2 \]In simple words: Auto-protolysis of water means that two water molecules can react with each other. One water molecule gives away a proton (acting as an acid) to another water molecule (acting as a base), forming \( H_3O^+ \) and \( OH^- \). This process shows water's ability to act as both an acid and a base.

Exam Tip: For auto-protolysis, remember that water is amphoteric, meaning it can donate and accept a proton simultaneously from itself. The key products are hydronium (\( H_3O^+ \)) and hydroxide (\( OH^- \)) ions.

 

Question 19. Consider the reaction of water with \( F_2 \) and suggest, in terms of oxidation and reduction which species are oxidised/ reduced.
Answer: When water reacts with fluorine (\( F_2 \)), it can be involved in an oxidation-reduction reaction. The reaction can be:
\[ 2F_2(g) + 2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4F^-(aq) \] In this reaction, water (\( H_2O \)) acts as a reductant (reducing agent), meaning it gets oxidized to either \( O_2 \) or \( O_3 \). On the other hand, \( F_2(g) \) acts as an oxidant (oxidizing agent) and is reduced to \( F^- \).
Alternatively, it can also produce ozone:
\[ 3F_2(g) + 3H_2O(l) \rightarrow O_3(g) + 6F^-(aq) + 6H^+(aq) \]In simple words: When water reacts with fluorine gas, the water loses electrons and becomes oxygen gas (\( O_2 \)) or ozone (\( O_3 \)), which means water is oxidized. The fluorine gas gains electrons and becomes fluoride ions (\( F^- \)), so fluorine is reduced.

Exam Tip: In the reaction of water with \( F_2 \), remember that \( F_2 \) is a powerful oxidizing agent and will always be reduced to \( F^- \). Consequently, water must act as the reducing agent and be oxidized (to \( O_2 \) or \( O_3 \)).

 

Question 20. Complete the following chemical reactions.
(i) \( PbS(s) + H_2O_2(aq) \rightarrow \)
(ii) \( MnO_4^-(aq) + H_2O_2(aq) \rightarrow \)
(iii) \( CaO(s) + H_2O(g) \rightarrow \)
(iv) \( AlCl_3(g) +H_2O (l) \rightarrow \)
(v) \( Ca_3N_2(s) + H_2O (l) \rightarrow \)
Classify the above into (a) hydrolysis (b) redox and (c) hydration reactions.
Answer:
(i) \( PbS(s) + 4H_2O_2(aq) \rightarrow PbSO_4 + 4H_2O \)
Lead sulphide is oxidized to lead sulphate, and \( H_2O_2 \) is reduced to \( H_2O \). This is a **redox reaction**.

(ii) \( 2MnO_4^- + 3H_2O_2 + 2OH^- \rightarrow 2MnO_2 + 3O_2 + 4H_2O \)
In alkaline medium, \( MnO_4^- \) is reduced to \( MnO_2 \) by \( H_2O_2 \). This is a **redox reaction**.

(ii) \( 2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O \)
In acidic medium, \( MnO_4^- \) is reduced to \( Mn^{2+} \) by \( H_2O_2 \). This is a **redox reaction**.

(iii) \( CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq) \)
Calcium oxide is hydrolyzed to calcium hydroxide by water. This is a **hydrolysis reaction**.

(iv) \( AlCl_3(s) + 6H_2O(l) \rightarrow [Al(H_2O)_6]^{3+}(aq) + 3Cl^-(aq) \)
Aluminum chloride is hydrated to \( [Al(H_2O)_6]^{3+} \). This is a **hydration reaction**.

(v) \( Ca_3N_2(s) + 6H_2O(l) \rightarrow 3Ca(OH)_2(aq) + 2NH_3(g) \)
Calcium nitride is hydrolyzed to calcium hydroxide and ammonia. This is a **hydrolysis reaction**.

Thus, (i) and (ii) are redox reactions, (iii) and (v) are hydrolysis reactions, and (iv) is a hydration reaction.
In simple words: The reactions are completed, showing how chemicals interact with hydrogen peroxide or water. They are then sorted into three types: redox (where electrons are transferred), hydrolysis (where water breaks down a compound), and hydration (where water molecules attach to a compound).

Exam Tip: For completing and classifying reactions, pay close attention to the oxidation states of elements, the role of water (as a reactant, solvent, or ligand), and the resulting products. Remember that hydrolysis involves breaking bonds with water, hydration involves forming a complex with water, and redox involves changes in oxidation states.

 

Question 21. Describe the structure of the common form of ice.
Answer: At atmospheric pressure, ice crystallizes in its normal hexagonal form. In this structure, each oxygen atom is tetrahedrally surrounded by four other oxygen atoms. There is a hydrogen atom located between each pair of oxygen atoms. This arrangement gives ice an open, cage-like structure. Each oxygen atom is surrounded by four hydrogen atoms: two by strong covalent bonds (shown by solid lines) and two by weak hydrogen bonds (shown by dotted lines). A number of vacant spaces exist in the crystal lattice, which causes the density of ice to be lower than that of liquid water.
In simple words: Common ice has a hexagonal shape where each oxygen atom connects to four others, creating a spacious, open structure. This open structure, with lots of empty spaces, makes ice less dense than liquid water, which is why ice floats.

Exam Tip: When describing the structure of ice, highlight its hexagonal, open cage-like structure and the tetrahedral arrangement around each oxygen atom. Crucially, explain how the vacant spaces lead to ice being less dense than water.

 

Question 22. What causes the temporary and permanent hardness of water?
Answer: Rainwater is almost pure but can contain some dissolved gases from the atmosphere. As a good solvent, when it flows over the Earth's surface, it dissolves many salts. The presence of calcium and magnesium salts in the form of hydrogen carbonate, chloride, and sulphate in water makes it hard.
The presence of \( Ca(HCO_3)_2 \) and \( Mg(HCO_3)_2 \) is responsible for the temporary hardness of water. Temporary hardness can be removed by boiling.
The presence of calcium and magnesium chlorides and sulphates [e.g., \( CaCl_2, MgCl_2, CaSO_4, MgSO_4 \)] in water is responsible for the permanent hardness of water. Permanent hardness cannot be removed by boiling.
In simple words: Water becomes hard due to dissolved mineral salts. Temporary hardness comes from calcium and magnesium hydrogen carbonates and can be removed by boiling. Permanent hardness comes from calcium and magnesium chlorides and sulphates, and boiling won't remove it.

Exam Tip: Differentiate clearly between temporary and permanent hardness by identifying the specific salts responsible for each. Remember that temporary hardness involves hydrogen carbonates and can be removed by boiling, while permanent hardness involves chlorides and sulfates and requires other methods.

 

Question 23. Discuss the principle and method of softening of hard water by synthetic ion-exchange resins.
Answer: Currently, hard water is softened using synthetic cation exchangers. This method is more effective than the zeolite process. Cation exchangers contain large organic molecules with an \( -SO_3H \) group. An ion exchange resin (\( RSO_3H \)) is converted to its sodium form (RNa) by treating it with sodium chloride (NaCl). This RNa resin then reacts with the metal ions of calcium and magnesium present in hard water, which helps soften the water. Here, R represents the resin anion.
\[ 2RNa(s) + M^{2+}(aq) \rightarrow R_2M(s) + 2Na^+(aq) \] The resin can be regenerated by adding an aqueous NaCl solution. Pure de-mineralized (de-ionized) water, which is free from all soluble mineral salts, is obtained by successively passing water through a cation exchange (in the \( H^+ \) form) and an anion-exchange (in the \( OH^- \) form).
Cation exchange reaction:
\[ 2RH(s) + M^{2+}(aq) \rightleftharpoons MR_2(s) + 2H^+(aq) \] (R is the resin anion)
In this exchange, \( H^+ \) ions are swapped for \( Na^+, Ca^{2+}, Mg^{2+} \), and other cations present in the water. This process makes the water acidic. In the next exchange (anion exchange):
\[ RNH_2(S) + H_2O(l) \rightleftharpoons RNH_3^+ . OH^-(aq) \] \[ RNH_3^+ . OH^-(s) + X^-(aq) \rightleftharpoons RNH_3^+ . X^-(s) + OH^-(aq) \] The \( OH^- \) ions exchange for anions like \( Cl^-, HCO_3^-, SO_4^{2-} \), etc., present in the water. The \( OH^- \) ions then released neutralize the \( H^+ \) ions set free in the first exchange.
\[ H^+(aq) + OH^-(aq) \rightarrow H_2O(l) \] The exhausted cation and anion exchange resin beds are regenerated by treatment with dilute acid and NaOH solution, respectively.
In simple words: Hard water is softened using special resins that swap harmful ions (like calcium and magnesium) with harmless ones (like hydrogen or sodium). First, a resin takes out the metal ions, releasing hydrogen ions. Then, another resin takes out negative ions, releasing hydroxide ions. These combine to make pure water, and the resins can be reused.

Exam Tip: When discussing ion-exchange resins for water softening, explain both the cation and anion exchange processes. Clearly state how hard water ions are replaced and how the resins are regenerated for reuse, highlighting the production of de-ionized water.

 

Question 24. Write chemical reactions to show amphoteric nature of water.
Answer: A substance is termed AMPHOTERIC if it can exhibit both acidic and basic behavior. Water is an amphoteric substance.
It has the ability to act as an acid as well as a base. With ammonia (\( NH_3 \)), which is a base, water acts as an acid:
\[ NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq) \] It acts as a base with hydrogen sulfide (\( H_2S \)), which is a Brønsted acid:
\[ H_2S(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HS^-(aq) \] Water can easily hydrolyze oxides, halides of non-metals. It can also hydrolyze carbides, nitrides, and phosphides of the same metals.
Examples of hydrolysis:
\[ Al_4C_3 + 12H_2O \rightarrow 4Al(OH)_3 + 3CH_4 \] \[ Ca_3N_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2NH_3 \] \[ P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4 \] \[ SiCl_4(l) + 2H_2O \rightarrow SiO_2 + 4HCl \]In simple words: Water is amphoteric, meaning it can be both an acid and a base. It acts as an acid when reacting with a stronger base like ammonia, and it acts as a base when reacting with a stronger acid like hydrogen sulfide. Water also breaks down (hydrolyzes) many other compounds.

Exam Tip: To demonstrate water's amphoteric nature, provide two balanced chemical equations: one where water acts as an acid (e.g., with a base like \( NH_3 \)) and another where it acts as a base (e.g., with an acid like \( H_2S \)).

 

Question 25. Write chemical reactions to justify that hydrogen peroxide can function as an oxidizing as well as reducing agent.
Answer: Hydrogen peroxide (\( H_2O_2 \)) behaves as an oxidizing agent and a reducing agent in both acidic and alkaline mediums. However, \( H_2O_2 \) is primarily a powerful oxidizing agent but a weaker reducing agent.
(a) Oxidizing character: \( H_2O_2 \) acts as an oxidizing agent in both acidic and alkaline medium.
In acidic medium:
\[ H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O \] In basic medium:
\[ H_2O_2 + OH^- + 2e^- \rightarrow 3OH^- \] (i) In acidic medium: It oxidizes \( FeSO_4 \) to \( Fe_2(SO_4)_3 \)
\[ 2FeSO_4 + H_2SO_4 + H_2O_2 \rightarrow Fe_2(SO_4)_3 + 2H_2O \] It liberates iodine from potassium iodide:
\[ 2KI + H_2SO_4 + H_2O_2 \rightarrow K_2SO_4 + I_2 + 2H_2O \] It oxidizes lead sulfide (black) to lead sulfate (white):
\[ PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O \] (ii) In alkaline medium: It oxidizes sulfites to sulfates and nitrites to nitrates:
\[ Na_2SO_3 + H_2O_2 \rightarrow Na_2SO_4 + H_2O \] \[ KNO_2 + H_2O_2 \rightarrow KNO_3 + H_2O \] It oxidizes benzene to phenol in alkaline medium:
\[ C_6H_6 + H_2O_2 \rightarrow C_6H_5-OH + H_2O \] \[ \text{phenol} \]
(b) Reducing Character: In the presence of strong oxidizing agents, \( H_2O_2 \) behaves as a reducing agent in both acidic and basic mediums.
(i) In acidic medium: It reduces acidified \( KMnO_4 \) to \( Mn^{2+} \):
\[ 2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2 \] or
\[ 2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2 \] It reduces acidified \( K_2Cr_2O_7 \) solution (orange color) to chromium salts (green color):
\[ K_2Cr_2O_7 + 3H_2O_2 + 4H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 7H_2O + 3O_2 \] \[ \text{orange} \quad \text{green} \] or
\[ Cr_2O_7^{2-} + 8H^+ + 3H_2O_2 \rightarrow 2Cr^{3+} + 7H_2O + 3O_2 \] \( Cl_2 \) and \( Br_2 \) are reduced to HCl and HBr respectively. This property of \( H_2O_2 \) is called ANTICHLOR:
\[ H_2O_2 + Cl_2 \rightarrow 2HCl + O_2 \] \[ H_2O_2 + Br_2 \rightarrow 2HBr + O_2 \]
(ii) Reducing action of \( H_2O_2 \) in alkaline medium: It oxidizes \( KMnO_4 \) to \( MnO_2 \) in alkaline medium:
\[ 2KMnO_4(aq) + 3H_2O_2(aq) \rightarrow 2MnO_2 + 2KOH + 3O_2 + 2H_2O \] It reduces ferric salts (\( Fe^{3+} \)) to ferrous salts (\( Fe^{2+} \)):
\[ 2Fe^{3+}(aq) + H_2O_2(aq) + 2OH^-(aq) \rightarrow 2Fe^{2+} + O_2(g) + 2H_2O(l) \] It reduces alkaline ferricyanide to ferrocyanide:
\[ 2K_3[Fe(CN)_6] + 2KOH + H_2O_2 \rightarrow 2K_4Fe(CN)_6 + O_2 + H_2O \] or
\[ 2[Fe(CN)_6]^{3-} + 2OH^-(aq) + H_2O_2 \rightarrow 2[Fe(CN)_6]^{4-} + O_2 + H_2O(l) \] It reduces metallic oxides to metal in alkaline medium:
\[ Ag_2O + H_2O_2 \rightarrow 2Ag(s) + H_2O + O_2 \]In simple words: Hydrogen peroxide can both give away and take electrons, acting as an oxidizer or a reducer. It oxidizes other things (like iron or sulfites) when it gets reduced itself. It reduces other strong oxidizers (like permanganate or dichromate) when it gets oxidized itself to oxygen gas.

Exam Tip: To demonstrate \( H_2O_2 \)'s dual nature, provide specific reaction examples for both its oxidizing (e.g., with \( FeSO_4 \), \( PbS \)) and reducing (e.g., with \( KMnO_4 \), \( K_2Cr_2O_7 \)) properties in both acidic and alkaline media.

 

Question 26. What is meant by 'demineralised' water and how can it be obtained?
Answer: Water that does not contain any cations or anions is called 'demineralized' water. It is considered soft water. Demineralized water is obtained similarly to how soft water is produced from hard water. Demineralized or deionized water is produced by first passing hard water through a cation exchange resin (such as RCOOH or \( RSO_3H \)). This resin removes \( Ca^{2+} \) and \( Mg^{2+} \) ions from the hard water by exchanging them with \( H^+ \) ions. Subsequently, the water is passed through an anion exchange resin (like \( RN^+H_3OH^- \)), which removes \( Cl^- \) and \( SO_4^{2-} \) ions present in hard water by exchanging them with \( OH^- \) ions.
In simple words: Demineralized water is water with no dissolved minerals or ions. You get it by sending hard water through two special filters: one that swaps out positive metal ions for hydrogen ions, and another that swaps out negative ions for hydroxide ions, making pure water.

Exam Tip: Define demineralized water as free of both cations and anions. Explain the two-step ion-exchange process: first cation exchange (releasing \( H^+ \)) and then anion exchange (releasing \( OH^- \)), followed by their neutralization to form water.

 

Question 27. Is demineralised or distilled water useful for drinking purposes ? If not, how can it be made useful?
Answer: No, demineralized or distilled water is not suitable for drinking purposes. This is because it lacks essential minerals that are vital for human health and normal bodily functions. Drinking only demineralized water can lead to mineral imbalances in the body. It can be made useful for drinking by treating it chemically to reintroduce small, balanced amounts of beneficial minerals.
In simple words: Demineralized or distilled water is not good for drinking because it's missing important minerals our bodies need. To make it drinkable, you would need to add those minerals back into it.

Exam Tip: Remember that demineralized/distilled water lacks essential minerals, making it unsuitable for drinking. The solution is to re-mineralize it chemically.

 

Question 30. Knowing the properties of H2O and D2O do you think that D₂O can be used for drinking purposes?
Answer: No, heavy water (D2O) despite having the usual characteristics of ordinary water (H2O) is harmful and damaging to people, plants, and animals. This is because it makes the rates of reactions occurring inside them happen more slowly. Therefore, heavy water does not help life as effectively as normal water and is unsuitable for drinking.
In simple words: Heavy water is not good for drinking. It's harmful to living things because it slows down their body processes, making it unable to support life like regular water.

Exam Tip: When comparing substances, always consider their biological effects, not just their chemical similarities, especially for substances meant for consumption.

 

Question 31. What is the difference between the terms 'hydrolysis' and 'hydration'?
Answer: The interaction between cations and anions of a salt with \( \text{OH}^- \) ions and \( \text{H}^+ \) ions provided by water to make the initial acid and base is called hydrolysis.
\( \text{MA} + \text{H}_2\text{O} \implies \text{HA} + \text{MOH} \)
Water can readily break down oxides, halides of non-metals. It can also break down carbides, nitrides, and phosphides of the same metals. For example:
\( \text{Al}_4\text{C}_3 + 12\text{H}_2\text{O} \rightarrow 4\text{Al(OH)}_3 + 3\text{CH}_4 \)
\( \text{Ca}_3\text{N}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{Ca(OH)}_2 + 2\text{NH}_3 \)
\( \text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4 \)
\( \text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl} \)
In hydration processes/reactions, water has a great capacity to make compounds with some metal salts, which are called hydrates.
(i) Water molecules attach themselves to a metal ion in a complex.
(ii) Water fills small gaps within the crystal structure, as seen in \( \text{BaCl}_2 \cdot 2\text{H}_2\text{O} \).
(iii) Water molecules might link with specific oxygen-containing anions using hydrogen bonds, for instance, in \( [\text{Cu(H}_2\text{O)}_4]^{2+} \text{SO}_4^{2-} \cdot \text{H}_2\text{O} \) in \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \).
In simple words: Hydrolysis is when water breaks down a compound into an acid and a base. Hydration is when water molecules simply attach to other substances to form new compounds called hydrates.

Exam Tip: Remember that hydrolysis involves a chemical reaction breaking bonds, while hydration typically involves water molecules forming coordinate bonds or occupying spaces without breaking the original substance apart.

 

Question 32. How can saline hydrides remove traces of water from organic compounds?
Answer: The \( \text{H}^- \) ion found in saline hydrides (\( \text{M}^+\text{H}^- \)) acts as a powerful Bronsted base. This ion reacts with small amounts of water found in organic mixtures to release hydrogen gas:
\( \text{H}_2\text{O} + \text{H}^- \rightarrow \text{H}_2(\text{g}) + \text{OH}^- \)
This reaction effectively removes water traces by converting them into hydrogen gas.
In simple words: Saline hydrides contain H- ions, which are strong bases. These ions react with any water present, turning it into hydrogen gas and removing it from organic compounds.

Exam Tip: Saline hydrides are efficient drying agents because their hydride ion strongly reacts with water, preventing its presence in organic solvents.

 

Question 33. What do you expect the nature of hydrides is, if formed by elements of atomic numbers 15,19,23, and 44 with dihydrogen? Compare their behaviour towards water.
Answer:
For atomic number 15 (Phosphorus): The element with atomic number 15 belongs to the p-block and forms the hydride \( \text{PH}_3 \). This is a molecular or covalent hydride, made up of separate covalent molecules linked by weak van der Waals forces. These hydrides do not react with water but are slightly soluble.
For atomic number 19 (Potassium): The element with atomic number 19 is in the s-block (Potassium). It creates a saline hydride, \( \text{K}^+\text{H}^- \). This reacts vigorously with water, releasing hydrogen gas:
\( \text{KH(s)} + \text{H}_2\text{O(l)} \rightarrow \text{KOH (aq)} + \text{H}_2(\text{g}) \)
Since this reaction produces heat, the hydrogen gas released often catches fire.
For atomic number 23 (Vanadium): The element with atomic number 23 is a transition or d-block element (Vanadium). It forms a metallic or interstitial hydride, like \( \text{VH}_{1.6} \). In these compounds, the proportion of hydrogen atoms to metal atoms is not fixed; it changes based on temperature and pressure. Because of this interstitial hydride creation, these metals can soak up a lot of hydrogen on their surface. Hydrides of transition metals are generally insoluble in water.
For atomic number 44 (Ruthenium): The element with atomic number 44 is in group 8 (Ruthenium). It does not form a hydride. Elements in groups 7 to 9 generally do not form hydrides, a phenomenon known as the hydride gap.
In simple words: Atomic number 15 (P) makes a molecular hydride that doesn't react with water. Atomic number 19 (K) forms a saline hydride that reacts explosively with water, producing fire. Atomic number 23 (V) makes a metallic hydride that can store lots of hydrogen and doesn't dissolve in water. Atomic number 44 (Ru) is in the "hydride gap" and won't form a hydride at all.

Exam Tip: Remember to categorize hydrides by their type (ionic, covalent, metallic) as this directly affects their chemical properties and reactivity, especially with water.

 

Question 34. Do you expect different products in solution when aluminium (III) chloride and potassium chloride treated separately with (i) normal water (ii) acidified water and (iii) alkaline water ? Write equations wherever necessary.
Answer: Potassium chloride (KCl) is a salt made from a strong acid and a strong base. When treated with normal, acidified, or alkaline water, it will only provide \( \text{K}^+(\text{aq}) \) and \( \text{Cl}^-(\text{aq}) \) ions in the solution, without forming different products because it does not hydrolyze significantly. Aluminium(III) chloride, however, will go through hydrolysis and produce various outcomes depending on the water's pH.
(i) With normal water: Aluminium chloride hydrolyzes to form aluminum hydroxide and hydrochloric acid:
\( \text{AlCl}_3(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{Al(OH)}_3(\text{s}) + 3\text{HCl}(\text{aq}) \)
(ii) With acidified water: In acidified water, the hydrolysis of \( \text{AlCl}_3 \) will be suppressed due to the common ion effect from \( \text{H}^+ \) ions. Aluminium ions will largely remain as hydrated complex ions:
\( \text{Al}^{3+}(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) \rightleftharpoons [\text{Al(H}_2\text{O)}_6]^{3+}(\text{aq}) \)
(iii) With alkaline water: In alkaline water, aluminum hydroxide, which is amphoteric, will react with excess \( \text{OH}^- \) to form a soluble complex ion, tetrahydroxoaluminate(III):
\( \text{Al(OH)}_3(\text{s}) + \text{OH}^-(\text{aq}) \rightarrow [\text{Al(OH)}_4]^-(\text{aq}) \)
In simple words: KCl won't change in different waters, just dissolving into ions. But AlCl3 will act differently: in normal water, it forms aluminum hydroxide; in acidic water, it mostly stays as hydrated ions; and in alkaline water, the hydroxide further reacts to make a soluble complex.

Exam Tip: Always consider the amphoteric nature of aluminum compounds when predicting reactions with acids and bases, as this leads to varying products.

 

Question 35. How does H2O2 behave as a bleaching agent?
Answer: Hydrogen peroxide's ability to bleach comes from the reactive oxygen it releases when it breaks down. This process is called decomposition:
\( \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O} \)
This fresh oxygen then reacts with colored substances, making them oxidized. The oxidized coloring matter becomes colorless:
\( \text{Coloring matter} + \text{O} \rightarrow \text{Colorless matter} \)
It helps to whiten delicate items such as ivory, feathers, silk, and wool.
In simple words: Hydrogen peroxide bleaches by breaking down and releasing a powerful, reactive type of oxygen. This oxygen then oxidizes and removes the color from materials, making them lighter.

Exam Tip: The bleaching power of hydrogen peroxide relies on the production of nascent oxygen, a highly reactive form of oxygen that can oxidize pigments and render them colorless.

 

Question 36. What do you understand by the terms (i) hydrogen economy (ii) hydrogenation (iii) 'syngas' (iv) water-gas shift reaction (v) fuel cell ?
Answer:
(i) Hydrogen economy: Hydrogen economy refers to the future possibility of using hydrogen as a clean energy source instead of fossil fuels. When hydrogen burns in oxygen, it creates only water and no pollution, unlike coal or petrol. This idea of using hydrogen for homes, cars, and industries to make electricity is known as the hydrogen economy.
\( 2\text{H}_2(\text{g}) + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \)
(ii) Hydrogenation: Hydrogenation is the process of adding hydrogen (\( \text{H}_2 \)) to compounds that have multiple bonds (like double or triple bonds) in the presence of a catalyst. This converts them into saturated compounds, which have only single bonds. For example, ethene can be converted to ethane, and oils can be changed into fats.
For example:
(a) \( \text{H}_2\text{C} = \text{CH}_2 + \text{H}_2 \xrightarrow{\text{Raney Ni or Pd}} \text{H}_3\text{C}-\text{CH}_3 \)
ethene ethane
(b) \( \text{Oil} + \text{H}_2 \xrightarrow{\text{Powdered Ni}} \text{Fat} \)
(iii) 'Syngas': Syngas, or synthesis gas, is a mixture of carbon monoxide (CO) and hydrogen (\( \text{H}_2 \)). Today, syngas can be made from various materials like sewage, sawdust, waste wood, and old newspapers. The method of producing syngas from coal is called coal gasification.
\( \text{C(s)} + \text{H}_2\text{O(g)} \xrightarrow{1270\text{ K}} \text{CO(g)} + \text{H}_2(\text{g}) \)
(iv) Water-gas shift reaction: The production of hydrogen can be increased by reacting carbon monoxide (CO) from syngas mixtures with steam. This reaction happens in the presence of an iron chromate catalyst and is also known as the water-gas shift reaction.
\( \text{CO(g)} + \text{H}_2\text{O(g)} \xrightarrow{\text{673 K Catalyst}} \text{CO}_2(\text{g}) + \text{H}_2(\text{g}) \)
\( \text{CO}_2 \) is removed by scrubbing with sodium arsenite solution.
(v) Fuel cell: A fuel cell is a device that directly changes the chemical energy from fuel combustion, like hydrogen, into electrical energy. A common example is the hydrogen-oxygen fuel cell. It works by having hydrogen react at the anode and oxygen at the cathode, producing water and electricity. These cells are very efficient, often over 95%.
The following reaction takes place in a fuel cell:
At anode: \( 2\text{H}_2(\text{g}) + 4\text{OH}^-(\text{aq}) \rightarrow 4\text{H}_2\text{O(l)} + 4\text{e}^- \)
At cathode: \( \text{O}_2(\text{g}) + 2\text{H}_2\text{O(l)} + 4\text{e}^- \rightarrow 4\text{OH}^-(\text{aq}) \)
Overall reaction: \( 2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O(l)} \)
In simple words: The "hydrogen economy" is about using hydrogen as a clean future fuel. "Hydrogenation" adds hydrogen to substances to make them saturated. "Syngas" is a mix of carbon monoxide and hydrogen, often made from coal. The "water-gas shift reaction" makes more hydrogen from syngas. A "fuel cell" converts chemical energy directly into electricity.

Exam Tip: For definitions in chemistry, always provide a clear explanation along with a relevant chemical equation or example to illustrate the concept fully.