Updated By SaraNextGen

On April 21, 2024, 11:35 AM

Question 9.7:

Discuss the consequences of high enthalpy of H–H bond in terms of chemical reactivity of dihydrogen.

Answer:

The ionization enthalpy of H–H bond is very high (1312 kJ mol^–1). This indicates that hydrogen has a low tendency to form H⁺ ions. Its ionization enthalpy value is comparable to that of halogens. Hence, it forms diatomic molecules (H₂), hydrides with elements, and a large number of covalent bonds.

Since ionization enthalpy is very high, hydrogen does not possess metallic characteristics (lustre, ductility, etc.) like metals.

Question 9.8:

What do you understand by (i) electron-deficient, (ii) electron-precise, and (iii) electron-rich compounds of hydrogen? Provide justification with suitable examples.

Answer:

Molecular hydrides are classified on the basis of the presence of the total number of electrons and bonds in their Lewis structures as:

1. Electron-deficient hydrides

2. Electron-precise hydrides

3. Electron-rich hydrides

An electron-deficient hydride has very few electrons, less than that required for representing its conventional Lewis structure e.g. diborane (B₂H₆). In B₂H₆, there are six bonds in all, out of which only four bonds are regular two centered-two electron bonds. The remaining two bonds are three centered-two electron bonds i.e., two electrons are shared by three atoms. Hence, its conventional Lewis structure cannot be drawn.

An electron-precise hydride has a sufficient number of electrons to be represented by its conventional Lewis structure e.g. CH₄. The Lewis structure can be written as:

Four regular bonds are formed where two electrons are shared by two atoms.

An electron-rich hydride contains excess electrons as lone pairs e.g. NH_3.

There are three regular bonds in all with a lone pair of electrons on the nitrogen atom.

Question 9.9:

What characteristics do you expect from an electron-deficient hydride with respect to its structure and chemical reactions?

Answer:

An electron-deficient hydride does not have sufficient electrons to form a regular bond in which two electrons are shared by two atoms e.g., B₂H₆, Al₂H₆ etc.

These hydrides cannot be represented by conventional Lewis structures. B₂H₆, for example, contains four regular bonds and two three centered-two electron bond. Its structure can be represented as:

Since these hydrides are electron-deficient, they have a tendency to accept electrons. Hence, they act as Lewis acids.

Question 9.10:

Do you expect the carbon hydrides of the type (CnH₂n _{+ 2}) to act as ‘Lewis’ acid or base? Justify your Answer.

Answer:

For carbon hydrides of type CnH₂n _{+ 2}, the following hydrides are possible for

For a hydride to act as a Lewis acid i.e., electron accepting, it should be electron-deficient. Also, for it to act as a Lewis base i.e., electron donating, it should be electron-rich.

Taking C₂H₆ as an example, the total number of electrons are 14 and the total covalent bonds are seven. Hence, the bonds are regular 2e^–-2 centered bonds.

Hence, hydride C₂H₆ has sufficient electrons to be represented by a conventional Lewis structure. Therefore, it is an electron-precise hydride, having all atoms with complete octets. Thus, it can neither donate nor accept electrons to act as a Lewis acid or Lewis base.

Question 9.11:

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 are hydrogen-deficient compounds formed by the reaction of dihydrogen with d-block and f-block elements. These hydrides do not follow the law of constant composition. For example: LaH_2.87, YbH_2.55, TiH_{1.5 – 1.8} etc.

Alkali metals form stoichiometric hydrides. These hydrides are ionic in nature. Hydride ions have comparable sizes (208 pm) with alkali metal ions. Hence, strong binding forces exist between the constituting metal and hydride ion. As a result, stoichiometric hydrides are formed.

Alkali metals will not form non-stoichiometric hydrides.

Question 9.12:

How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.

Answer:

Metallic hydrides are hydrogen deficient, i.e., they do not hold the law of constant composition. It has been established that in the hydrides of Ni, Pd, Ce, and Ac, hydrogen occupies the interstitial position in lattices allowing further absorption of hydrogen on these metals. Metals like Pd, Pt, etc. have the capacity to accommodate a large volume of hydrogen. Therefore, they are used for the storage of hydrogen and serve as a source of energy.

Question 9.13:

How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.

Answer:

Atomic hydrogen atoms are produced by the dissociation of dihydrogen with the help of an electric arc. This releases a huge amount of energy (435.88 kJ mol^–1). This energy can be used to generate a temperature of 4000 K, which is ideal for welding and cutting metals. Hence, atomic hydrogen or oxy-hydrogen torches are used for these purposes. For this reason, atomic hydrogen is allowed to recombine on the surface to be welded to generate the desired temperature.

Question 9.14:

Among NH₃, H₂O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?

Answer:

The extent of hydrogen bonding depends upon electronegativity and the number of hydrogen atoms available for bonding. Among nitrogen, fluorine, and oxygen, the increasing order of their electronegativities are N < O < F.

Hence, the expected order of the extent of hydrogen bonding is HF > H₂O > NH₃.

But, the actual order is H₂O > HF > NH₃.

Although fluorine is more electronegative than oxygen, the extent of hydrogen bonding is higher in water. There is a shortage of hydrogens in HF, whereas there are exactly the right numbers of hydrogens in water. As a result, only straight chain bonding takes place. On the other hand, oxygen forms a huge ring-like structure through its high ability of hydrogen bonding.

In case of ammonia, the extent of hydrogen bonding is limited because nitrogen has only one lone pair. Therefore, it cannot satisfy all hydrogens.

Question 9.15:

Saline hydrides are known to react with water violently producing fire. Can CO₂, a well known fire extinguisher, be used in this case? Explain.

Answer:

Saline hydrides (i.e., NaH, LiH, etc.) react with water to form a base and hydrogen gas. The chemical equation used to represent the reaction can be written as:

The reaction is violent and produces fire.

CO₂ is heavier than dioxygen. It is used as a fire extinguisher because it covers the fire as a blanket and inhibits the supply of dioxygen, thereby dousing the fire.

CO₂ can be used in the present case as well. It is heavier than dihydrogen and will be effective in isolating the burning surface from dihydrogen and dioxygen.

Question 9.16:

Arrange the following

(i) CaH₂, BeH₂ and TiH₂ 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 enthalpy.

(iv) NaH, MgH₂ and H₂O in order of increasing reducing property.

Answer:

(i) The electrical conductance of a molecule depends upon its ionic or covalent nature. Ionic compounds conduct, whereas covalent compounds do not.

BeH₂ is a covalent hydride. Hence, it does not conduct. CaH₂ is an ionic hydride, which conducts electricity in the molten state. Titanium hydride, TiH₂ is metallic in nature and conducts electricity at room temperature. Hence, the increasing order of electrical conductance is as follows:

BeH₂ < CaH₂ < TiH₂

(ii) The ionic character of a bond is dependent on the electronegativities of the atoms involved. The higher the difference between the electronegativities of atoms, the smaller is the ionic character.

Electronegativity decreases down the group from Lithium to Caesium. Hence, the ionic character of their hydrides will increase (as shown below).

LiH < NaH < CsH

(iii) Bond dissociation energy depends upon the bond strength of a molecule, which in turn depends upon the attractive and repulsive forces present in a molecule.

The bond pair in D–D bond is more strongly attracted by the nucleus than the bond pair in H–H bond. This is because of the higher nuclear mass of D₂. The stronger the attraction, the greater will be the bond strength and the higher is the bond dissociation enthalpy. Hence, the bond dissociation enthalpy of D–D is higher than H–H.

However, bond dissociation enthalpy is the minimum in the case of F–F. The bond pair experiences strong repulsion from the lone pairs present on each F-centre.

Therefore, the increasing order of bond dissociation enthalpy is as follows:

F–F < H–H < D–D

(iv) Ionic hydrides are strong reducing agents. NaH can easily donate its electrons. Hence, it is most reducing in nature.

Both, MgH₂ and H₂O are covalent hydrides. H₂O is less reducing than MgH₂ since the bond dissociation energy of H₂O is higher than MgH₂.

Hence, the increasing order of the reducing property is H₂O < MgH₂ < NaH.

Question 9.17:

Compare the structures of H₂O and H₂O₂.

Answer:

In gaseous phase, water molecule has a bent form with a bond angle of 104.5°. The O–H bond length is 95.7 pm. The structure can be shown as:

Hydrogen peroxide has a non-planar structure both in gas and solid phase. The dihedral angle in gas and solid phase is 111.5° and 90.2° respectively.

Question 9.18:

What do you understand by the term ’auto-protolysis’ of water? What is its significance?

Answer:

Auto-protolysis (self-ionization) of water is a chemical reaction in which two water molecules react to produce a hydroxide ion (OH^–) and a hydronium ion (H₃O⁺).

The reaction involved can be represented as:

Auto-protolysis of water indicates its amphoteric nature i.e., its ability to act as an acid as well as a base.

The acid-base reaction can be written as:

Question 9.19:

Consider the reaction of water with F₂ and suggest, in terms of oxidation and reduction, which species are oxidized/reduced.

Answer:

The reaction between fluorine and water can be represented as:

This is an example of a redox reaction as water is getting oxidized to oxygen, while fluorine is being reduced to fluoride ion.

The oxidation numbers of various species can be represented as:

Fluorine is reduced from zero to (– 1) oxidation state. A decrease in oxidation state indicates the reduction of fluorine.

Water is oxidized from (– 2) to zero oxidation state. An increase in oxidation state indicates oxidation of water.

Question 9.20:

Complete the following chemical reactions.

(i) 

(ii) 

(iii) 

(iv) 

(v) 

Classify the above into (a) hydrolysis, (b) redox and (c) hydration reactions.

Answer:

(i) 

H₂O₂ is acting as an oxidizing agent in the reaction. Hence, it is a redox reaction.

(ii) 

is acting as a reducing agent in the acidic medium, thereby oxidizing  . Hence, the given reaction is a redox reaction.

(iii) 

The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction is hydrolysis.

(iv) 

The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction represents hydrolysis of AlCl_3.

(v) 

The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction represents hydrolysis of Ca₃N₂.

Question 9.21:

Describe the structure of the common form of ice.

Answer:

Ice is the crystalline form of water. It takes a hexagonal form if crystallized at atmospheric pressure, but condenses to cubic form if the temperature is very low.

The three-dimensional structure of ice is represented as:

The structure is highly ordered and has hydrogen bonding. Each oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of 276 pm. The structure also contains wide holes that can hold molecules of appropriate sizes interstitially.

Question 9.22:

What causes the temporary and permanent hardness of water?

Answer:

Temporary hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of hydrogen carbonates (MHCO₃, where M = Mg, Ca) in water.

Permanent hardness of water is because of the presence of soluble salts of calcium and magnesium in the form of chlorides in water.

Question 9.23:

Discuss the principle and method of softening of hard water by synthetic ion-exchange resins.

Answer:

The process of treating permanent hardness of water using synthetic resins is based on the exchange of cations (e.g., Na⁺, Ca²⁺, Mg²⁺ etc) and anions (e.g., Cl^–, SO₄^2–, HCO₃^– etc) present in water by H⁺ and OH^– ions respectively.

Synthetic resins are of two types:

1) Cation exchange resins

2) Anion exchange resins

Cation exchange resins are large organic molecules that contain the –SO₃H group. The resin is firstly changed to RNa (from RSO₃H) by treating it with NaCl. This resin then exchanges Na⁺ ions with Ca²⁺ and Mg²⁺ ions, thereby making the water soft.

There are cation exchange resins in H⁺ form. The resins exchange H⁺ ions for Na⁺, Ca²⁺, and Mg²⁺ ions.

Anion exchange resins exchange OH^– ions for anions like Cl^–, , and SO₄^2– present in water.

During the complete process, water first passes through the cation exchange process. The water obtained after this process is free from mineral cations and is acidic in nature.

This acidic water is then passed through the anion exchange process where OH^– ions neutralize the H⁺ ions and de-ionize the water obtained.

Question 9.24:

Write chemical reactions to show the amphoteric nature of water.

Answer:

The amphoteric nature of water can be described on the basis of the following reactions:

1) Reaction with H2S

The reaction takes place as:

In the forward reaction,   accepts a proton from . Hence, it acts as a Lewis base.

2) Reaction with NH3

The reaction takes place as:

In the forward reaction,   denotes its proton to . Hence, it acts as a Lewis acid.

3) Self-ionization of water

In the reaction, two water molecules react as:

Question 9.25:

Write chemical reactions to justify that hydrogen peroxide can function as an oxidizing as well as reducing agent.

Answer:

Hydrogen peroxide, H₂O₂ acts as an oxidizing as well as a reducing agent in both acidic and alkaline media.

Reactions involving oxidizing actions are:

1) 

2) 

3) 

4) 

Reactions involving reduction actions are:

1) 

2) 

3) 

4)