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2nd PUC Chemistry The p-Block Elements NCERT Textbook Questions and Answers

Question 1.
Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy, and electronegativity.
Answer:
The valence shell electronic configuration of group 15 elements is ns²np³. Due to half-filled p-orbitals, these elements have extra stability associated with them.

The common oxidation states of these elements are -3, +3 and +5. The tendency to exhibit -3 oxidation state decreases down the group. The stability of +5 state decreases and that of +3 state increases down the group due to inert pair effect.

The size of group 15 elements increases down the group. There is a considerable increase in covalent radius from N to P. However, from As to Bi only a small increase in covalent radius is observed. This is due to the presence of completely filled d and or f orbitals in heavier members.

Down the group, ionisation enthalpy decreases due to an increase in atomic size. Due to stable half-filled configuration, they have much greater value than group 14 elements.

The electronegativity value, in general, decreases down the group with increasing atomic size. However, amongst the heavier elements, the difference is not that much pronounced.

Question 2
Why does the reactivity of nitrogen differ from phosphorus?
Answer:
Molecular nitrogen exists as a diatomic molecule having a triple bond between the two nitrogen atoms, N = N (due to it stability to form pπ-pπ multiple bonding). The bond dissociation energy is very high (941-4 kJ mol^-1). Thus, under ordinary conditions, nitrogen behaves as an inert gas. On the other hand, white and yellow phosphorus exists as a triatomic molecule (P₄) having single bonds. The dissociation energy of the P-P bond is low (213 kJ mol^-1). Thus, phosphorus is much more reactive than nitrogen.

Question 3
Discuss the trends in chemical reactivity of group 15 elements.
Answer:
General trends in chemical properties of group-15
(i) Reactivity towards hydrogen:
The elements of group 15 react with hydrogen to form hydrides of type EH₃, where E – N, P, As, Sb, or Bi. The stability of hydrides decreases on moving down from NH3 to BiH₃

(ii) Reactivity towards oxygen:
The elements of group 15 form two types of oxides: E₂O₃ arid E₂O₅, where E = N, P, As, Sb or Bi. The oxide with the element in higher oxidation state is more acidic than the other. However, the acidic character decreases on moving down a group.

(iii) Reactivity towards halogens:
The group 15 element react with halogens to form two series of salts: EX₃ and EX₅. However, nitrogen does not form NX₅ as H lacks the d-orbital. All trihalides (except NX₃) are stable.

(iv) Reactivity towards metals:
The group 15 elements react with metals to form binary compounds in which these metals exhibit-3 oxidation states.

Question 4
Why does NH₃ form a hydrogen bond but PH₃ does not?
Answer:
The N – H bond in ammonia is quite polar as nitrogen is highly electronegative in nature. As a result, NH₃ molecules are linked by intermolecular hydrogen bonding. On the other hand, P – H bond is non-polar as both P and H have the same electronegativity. Hence in phosphine, no hydrogen bonding is present.

Question 5.
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.
Answer:
An aqueous solution of ammonium chloride is treated with sodium nitrite
NH₄Cl + NaNO₂(aq) → N₂(g) + 2H₂O(e) + NaCl(aq)
NO, and HNO₃ are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous H₂SO₄ containing potassium dichromate

Question 6
How is ammonia manufactured industrially?
Answer:
Ammonia is prepared on a commercial scale by Haber’s Process from dinitrogen and dihydrogen by the following chemical reaction

Dihydrogen needed for the commercial preparation of ammonia is obtained by the electrolysis of water while dinitrogen is obtained from the liquefied air as fractional distillation. The two gases are purified and also dried.
These are compressed to about 200-atmosphere pressure and are then led into the catalyst chamber packed with the catalyst and the promoter.

Question 7
Illustrate how copper metal can give different products on reaction with HNO₃.
Answer:
On heating with dil HNO₃, copper gives copper nitrate and nitric oxide.

Question 8
Give the resonating structures of NO₂ and N₂O₅.
Answer:
Resonating structures of NO₂ are:

Resonating structures of N₂O₅ are:

Question 9.
The HNH angle value is higher than HPH, HAsH, and HSbH angles. Why?
[Hint: Can be explained on the basis of sp₃ hybridisation in NH₃ and only s-p bonding between hydrogen and other elements of the group].
Answer:
Hydride NH₃, PH₃, AsH₃, SbH₃
H-M-H angle 107°, 92°, 91°, 90°
The central atom (E) in all the hydrides is sp³ hybridised. However, its electronegativity decreases, and atomic size increases down the group. As a result, there is a gradual decrease in the force of repulsion in the shared electron pairs around the central atom. This leads to a decrease in the bond angle.

Question 10.
Why does R₃P = O exist but R₃N = O does not (R = alkyl group)?
Answer:
Nitrogen does not have d-orbitals in its valence shell. Therefore, it can not extend its covalency to five by forming dπ-dπ bonding. As a result, the molecule of R₃N = O does not exist. However, phosphorus has vacant rf-orbitals in the valence shell and can form dx-dx bonding. Thus, a molecule like R₃P = O can exist.

Question 11.
Explain why NH₃ is basic while BiH₃ is only feebly basic.
Answer:
Both NH₃ and PH₃ are Lewis bases due to the presence of lone electron pair on the central atom. However, NH₃ is more basic than PH₃. The atomic size of nitrogen (Atomic radius = 70 pm) is less than that of phosphorus (Atomic radius = 110 pm) As a result, electron density on the nitrogen atom is more than on phosphorus. This means that the electron releasing tendency of ammonia is also more and is, therefore, a stronger base than phosphine.

Question 12.
Nitrogen exists as a diatomic molecule and phosphorus as P₄. Why?
Answer:
Nitrogen because of its small size and high electronegativity is capable of forming pπ-pπ multiple bonding. Therefore, it exists as a diatomic molecule with one σ and two K bonds (N ≡ N). Phosphorus, on the other hand, has a longer size and lower electronegativity and thus is not capable to form pπ-pπ multiple bonds. It prefers to form single bonds hence, it exists as tetrahedral P₄ molecules.

Question 13
Write the main differences between the properties of white phosphorus and red phosphorus.
Answer:

Question 14
Why does nitrogen show catenation properties less than phosphorus?
Answer:
The extent of catenation depends upon the strength of the element-element bond. The N – N bond strength (159 kJ mol^-1 ) is weaker than the P – P bond strength (213 kJ mol^-1 ). Thus, nitrogen shows fewer catenation properties than phosphorus.

Question 15
Give the disproportionation reaction of H₃PO₃.
Answer:
H₃PO₃ on heating undergoes self oxidation-reduction, i.e. disproportionation to form PH, in which P is reduced, and H₃PO₄ in which P is oxidised.

Question 16.
Can PCl₅ act as an oxidising as well as a reducing agent? Justify.
Answer:
Phosphorus has a maximum oxidation state of +5 in PCl₅. It can not increase its oxidation state further and thus it can not act as a reducing agent. However, PCl₅ can act as an oxidising agent as it can itself reduce from +5 to +3 oxidation state.

Question 17.
Justify the placement of Q, S, Se, Te, and Po in the same group of the periodic table in terms of electronic configuration, oxidation state, and hydride formation.
Answer:
The elements of group 16 are collectively called chalcogens.
(i) Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is ns² np⁴, where n varies from 2 to 6.

(ii) oxidation state:
These elements have six valence electrons (ns² np⁴). They should display an oxidation state of -2. However only oxygen predominantly shows the oxidation state of -2 owing to its high electronegativity. It also exhibits an oxidation state of -1 (H₂O₂), Zero (O₂), and +2 (OF₂) However, the stability of the -2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of +2, +4 and +6 due to the availability of d-orbitals.

(iii) Formation of hydrides
These elements form hydrides of formula H₂E, where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type H²E². These hydrides are quite volatile in nature.

Question 18.
Why is dioxygen a gas but sulphur a solid?
Answer:
The oxygen atom has the tendency to form multiple bonds (pπ – pπ interaction) with other oxygen atoms on account of small size while this tendency is missing in sulphur atom. The bond energy of oxygen-oxygen double bond (0 = 0) is quite large (about three times that of an oxygen-oxygen single bond, O – O = 34.9 kcal mol^-1) while sulfur-sulfur double bond (S = S) is not so large (less than double of sulphur-sulphur single bond, S – S = 63.8 kcal mob1). As a result —O—O—O— chains are less stable as compared to O = O molecule while —S—S—S— chains are more stable than S = S molecule. Therefore, at room temperature, while oxygen exists as a diatomic gas molecule, sulphur exists as S₈ solid.

Question 19.
Knowing the electron gain enthalpy values for O → O^– and O → O^2- as -141 and 702 KJ mol^-1 respectively, how can you account for the formation of a large number of oxides having O^2- species and not O^–?
(Hint: Consider lattice energy factor in the formation of compounds).
Answer:
According to available data :
O + e^– → O^–; ∆_(eg) H = – 141 kJ mol^-1
O + 2e^– → O^2-; ∆_(eg) H = + 702 kJ mol ^-1
Although the formation of the divalent anion (O^2-) needs more energy as compared to the monovalent anion (O^–) where energy is actually released, still in a large number of oxides (e.g Na₂O, K₂O, CaO etc.) oxygen is divalent in nature. This is on account of a more stable crystal lattice because of the greater magnitude of electrostatic forces of attraction involving divalent oxygen than the oxides in which oxygen is monovalent in nature.

Question 20
Which aerosols deplete ozone?
Answer:
Freon (CCl₂F₂) (Chlorofluoro carbon).

Question 21.
Describe the manufacture of H₂SO₄ by contact process?
Answer:
H₂SO₄ is manufactured by contact process. It involves the following steps

Step (1)
Sulphur or sulphide ores are burnt in the air to form SO₂.

Step (2)
By a reaction with O₂, SO₂ is converted to SO₃ in the presence of V₂O₅ as a catalyst.

Step (3)
SO₃ produced is absorbed on H₂SO₄ to give H₂S₂O₇(Oleum)

SO₃+ H₂SO₄ → H₂S₂O₇

This oleum is then diluted to obtain H₂SO₄, of the desired concentration. In practice, the plant is operated at 2 bar (pressure) and 720 K (temperature). The H₂SO₄ thus obtained is 96 – 98% pure.

Question 22.
How is SO₂ an air pollutant?
Answer:
(i) Sulphur dioxide released in the atmosphere during the combustion of fuels combines with H20 molecules and oxygen present to form sulphuric acid. The acid being poisonous in
SO₂ + 1/2O₂ + H₂O → H₂SO₄
nature causes pollution. It causes the corrosion of statues and monuments made from marble (CaC03) due to the formation of sulphate.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
(ii) Sulphur dioxide adversely affects the respiratory tract due to its poisonous as well as irritating nature. It causes throat infection as well as irritation in the eyes.
(iii) Even a very low concentration of gas (0·03 ppm) has a very damaging effect on plants and vegetation. This is called chlorosis. It slows down the formation of chlorophyll and the leaves slowly wither.

Question 23.
Why are halogens strong oxidising agents?
Answer:
The halogens are strong oxidising agents due to low bond dissociation enthalpy, high electronegativity, and large negative electron gain enthalpy.

Question 24.
Explain why fluorine forms only one.
Answer:
Fluorine forms only one oxoacid, i.e. HOF because of its high electronegativity and small size.

Question 25
Explain why in spite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Answer:
Both nitrogen (N) and chlorine (Cl) have electronegativity of 3·0. However, only nitrogen is involved in the hydrogen bonds (e.gNH₃) and not chlorine. This is an account of the smaller atomic size of nitrogen (atomic radius = 70 pm) as compared to chlorine (atomic radius = 99 pm). Therefore, N can cause greater polarisation of N—H bond than Cl in case of Cl—H bond. Consequently, the N atom is involved in hydrogen bonding and not chlorine.

Question 26
Write two uses of ClO₂.
Answer:
Two uses of ClO₂:

• It is a powerful oxidising and chlorinating agent.
• It is an excellent bleaching agent for wood pulp, flour for making white bread.

Question 27
Why are halogens coloured?
Answer:
Almost all halogens are coloured. This is because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differs for each halogen, each halogen displays a colour.

Question 28
Write the reactions of F₂ and Cl₂ with water.
Answer:

Question 29
How can you prepare Cl, from HCl and HCl from Cl₂? Write reactions only.
Answer:
Cl₂ can be prepared from HC1 by Deacon’s process

(ii) HCl can be prepared from Cl by treating it with water
Cl₂ + H₂O → HCl + HOCl

Question 30
What inspired N. Bartlett for carrying out a reaction between Xe and PtF6?
Answer:
Neil Bartlett observed that PtF6 reacts with O2 to form an ionic solid O+2 PtF6
O_2(g) + PtF_6(g) → O⁺₂[PtF₆]^–
In this reaction, O₂ gets oxidised to O⁺₂ by PtF₆.
Since the first ionisation energy of xenon is fairly close to that of oxygen. Bartlett thought that PtF₆ should also oxidise Xe to Xe⁺. This prompted Bartlett to carry out the reaction between Xe and PtF₆ and formed the first noble gas compound.

Question 31.
What are the oxidation states of phosphorus in the following:
(i) H₃PO₃
(ii) PCl₃
(iii) Ca₃P₂
(iv) Na₃PO₄
(v) POF₃?
Answer:
Let oxidation state of P be x
(i) H₃PO₃
3 + x + 3 (-2) = 0
3 + x – 6 = 0
x = 3.

(ii) PCl₃
x + 3 (-1) = 0
x – 3 = 0
x = 3

(iii) Ca₃P₂
3(2) + 2x = 0
6 + 2x = 0
x = -3

(iv) Na₃PO₄
3 x 1 +x + 4(-2) = 0
3 + x – 8 = 0
x = 5

(v) POF₃
x + (-2) + 3x (-1) = 0
x – 5 = 0
x = 5.

Question 32.
Write balanced equations for the following:
(i) NaCl is heated with sulphuric acid in the presence of MnO₂.
(ii) Chlorine gas is passed into a solution of Nal in water.
Answer:
(i) Cl₂ is produced
4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂
(ii) Cl₂ being an oxidising agent oxidises Nal to I₂.
Cl_2(g) + 2NaI_(aq) → 2NaCl_(aq) + I_2(s)

Question 33.
How are xenon fluorides XeF₂, XeF₄ and XeF. obtained?
Answer:
XeF₂, X₃F₄ and XeF₆ are obtained by a direct reaction between Xe and F₂. The condition under which the reaction is carried out determines the product.

Question 34
With what neutral molecule’ is ClO^– isoelectronic? Is that molecule a Lewis base?
Answer:
ClO^– has 17 + 8 +1 = 26 electrons.Also, OF₂ has (8 + 2 x 9) = 26 electrons, and ClF has (17 + 9) = 26 electrons.Out of these, ClF can act as Lewis base. The chlorine atom has three lone pair of electrons which it donates to form compounds like ClF₃, ClF₅ and ClF₇

Question 35
How are XeO₃ and XeOF₄ prepared?
Answer:
(i) XeO₃ can be prepared in two ways as shown:
6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂ XeF₆ + 3H₂O → XeO₃ + 6HF
(ii) XeF4 can be prepared using XeF6
XeF₆+ H₂O → XeOF₄ + 2HF

Question 36.
Arrange the following in the order of property indicated for each set :
(i) F₂, Cl₂, Br₂, I₂ – increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI – increasing acid strength.
(iii) NH₃, PH₃, AsH₃, SbH₃, BiH₃ – increasing base strength. (Pb. Board 2009, C.B.S. Sample Paper 2017)
Answer:
(i) I₂ (151-1 kJmol^-1) < F₂ (158-8 kJ mol^-1) < Br₂ (192-8 kJ mol^-1) < Cl₂ (242-6 kJ mol^-1)
(ii) HF < HCl < HBr < HI
(iii) BiH₃ < SbH₃ < ASH₃ < PH₃ < NH₃.

Question 37.
Which one of the following does not exist?
(i) XeOF₄
(ii) NeF₂
(iii) XeF₂
(iv) XeF
Answer:
(ii) NeF₂

Question 38.
Give the formula and describe the structure of a noble gas species which is isostructural with:
(i) ICl₄^–
(ii) IBr₂^–
(iii) BrO₃^–
Answer:
(i) XeF₄ is isoelectronic with ICl₄^– and has square planar geometry

(ii) XeF₂ is isoelectronic to IBr₂^– and has a linear structure

(iii) XeO₃ is isostructural to BrO₃^– and has a pyramidal molecular structure.

Question 39.
Why do noble gases have comparatively large atomic sizes?
Answer:
This is because noble gases have only van der Waal’s radii while others have covalent radii, van der Waal’s radii are larger than covalent radii.

Question 40
List the uses of neon and argon gases.
Answer:
Uses of Neon:

• Neon lamps are used in botanical gardens and also in greenhouses as they are useful in the formation of chlorophyll and thus, stimulate plant growth.
• It is used in filling sodium vapour lamps.
• It is used in safety devices for protecting certain electrical instruments (voltmeters, relays, rectifiers etc.)

Uses of Argon:

• It is used in metal filament electric lamps since it increases the life of the tungsten filament by retarding its vapourisation.
• A mixture of argon and mercury vapours is used in fluorescent tubes.
• It is used to create an inert atmosphere for welding and for carrying certain chemical reactions.

2nd PUC Chemistry The p-Block Elements Additional Questions and Answers

Question 1.
What is the electronic .configuration of p-block elements?
Answer:
ns²np^{0 – 5}

Question 2.
Write three uses of nitrogen?
Answer:

1. Used in the manufacture of nitric acid, ammonia, calcium cyanamide.
2. The inert atmosphere for the iron and steel industry.
3. It is used as a refrigerant.

Question 3.
Write three physical properties of Ammonia?
Answer:

1. It is a colourless gas.
2. It is lighter than air.
3. Extremely soluble in water.

Question 4.
Write three physical properties of Nitric acid?
Answer:

1. Colourless fuming liquid.
2. The melting point is 231.4k.
3. The boiling point is 355.6 k.

Question 5.
What are the three allotropic forms of phosphorous?
Answer:

1.  white phosphorous
2. Red phosphorous
3. Black phosphorous.

Question 6.
Write the chemical formula of Hypo phosphorus acid.
Answer:
H₃PO₂.

Question 7.
Write the chemical formula of orthophosphorous acid.
Answer:
H₃PO₃

Question 8.
Write the chemical formula of pyrophosphoric acid.
Answer:
H₄P₂O₅

Question 9.
Draw the structure of PCl_5.
Answer:

Question 10.
Draw the structure of SF₆.
Answer:

2nd PUC Chemistry Surface Chemistry NCERT Textbook Questions and Answers

Question 1.
Distinguish between the meaning of the terms adsorption and absorption. Give one example of each.
Answer:
The phenomenon of accumulation of the molecules of a substance on a solid or liquid surface resulting in the increased concentration of the molecules on the surface is called adsorption. In absorption, the substance is uniformly distributed throughout the bulk of the solution. A distinction can be made by taking the example of water vapours. Water vapours are absorbed by anhydrous calcium chloride but adsorbed by silica gel.

Question 2.
What is the difference between physisorption and chemisorption? ‘
Answer:

Question 3.
Give a reason why a finely divided substance is more effective as an adsorbent.
Answer:
A finely divided substance (adsorbent) has more absorbing power due to the following reasons:

• The surface area increases and the adsorbate particles get a better opportunity to be adsorbed.
• The number of active sites, also called active centers becomes more and the extent of adsorption increases.

Question 4.
What are the factors which influence the adsorption of a gas on a solid?
Answer:
The adsorption of a gas on a solid surface is an example of physical adsorption. It is influenced by the following factors :

1. Nature of the adsorbate
2. Nature of the adsorbent
3. Surface area of the adsorbent
4. Activation of the adsorbent
5. Effect of pressure
6. Effect of temperature

Factors affecting the adsorption of a gas on solids are :

Nature of the adsorbate: The same gas is adsorbed to different extents by different solids at the same temperature. Also, the greater the surface area of the adsorbent more is the gas adsorbed.

Nature of the adsorbent: Different gases are adsorbed to different extents by different solids at the same temperature. Higher the critical temperature of the gas, the greater is its amount adsorbed.

Surface area of the adsorbent: Surface area available for adsorption per gram of the adsorbent increases the extent of adsorption. Greater the surface area, the higher would be the adsorption therefore, porous or powdered adsorbents are used.

Activation of adsorbent: It means increasing the adsorbing power of an adsorbent by increasing its surface area. It is done by:

• making the adsorbent’s surface rough
• removing gases already adsorbed
• subdividing the adsorbent into smaller pieces.

Pressure: At constant temperature, the adsorption of gas increases with pressure.

Temperature: Since adsorption is an exothermic process, applying Le Chatelier’s principle, we can find out that adsorption decreases with an increase in temperature.

Question 5.
What is an adsorption isotherm? Describe Freundlich adsorption isotherm.
Answer:

Freundlich adsorption isotherm:
Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed by the unit mass of solid adsorbent and pressure at a specific temperature.

From the given plot it is clear that at pressure P_s, \(\frac { x }{ m }\) reaches the maximum value. P_s is called the saturation pressure. Three cases arise from the graph now.

Question 6.
What do you understand by activation of adsorbent? How is it achieved?
Answer:
Activation of an adsorbent means increasing it’s adsorbing power by increasing the surface area of the adsorbent by making its surface rough, by removing already adsorbed gases from it and by subdividing the adsorbent into smaller pieces or grains.

Question 7.
What role does adsorption play in heterogeneous catalysis?
Answer:
Heterogeneous catalysis: A catalytic process in which the catalyst and the reactants are present in different phases is known as heterogeneous catalysis. This heterogeneous catalytic action can be explained in terms of the adsorption theory. The mechanism of catalysis involves the following steps:

1. Adsorption of reactant molecules on the catalyst surface.
2. The occurrence of a chemical reaction through the formation of an intermediate.
3. De-sorption of products from the catalyst surface.
4. Diffusion of products away from the catalyst surface.

In this process, the reactants are usually present in the gaseous state and the catalyst is present in the solid-state. Gaseous molecules are then adsorbed on the surface of the catalyst. As the concentration of reactants on the surface of the catalyst increases, the rate of reaction also increases. In such reactions, the products have less affinity for the catalyst and are quickly desorbed, thereby making the surface free for other reactants.

Case – I at low pressure:
The plot is straight and sloping, indicating that the pressure in directly proportional to
\(\frac { x }{ m }\) i.e., \(\frac { x }{ m }\) ∝ P
\(\frac { x }{ m }\)= kP (k = constant) m

Case – II At high pressure:
When pressure exceeds the saturated pressure,
becomes independent of P values.
\(\frac { x }{ m }\) ∝ P°
\(\frac { x }{ m }\) = kP°

Case – III At intermediate pressure:
At intermediate pressure, \(\frac { x }{ m }\) depends on P raised to the powers between O and ⊥ This relationship is known as the Freundlich adsorption isotherm.
\(\frac { x }{ m }\) ∝ P^1/n
\(\frac { x }{ m }\) = KP^1/n n > ⊥

Now taking log:
log \(\frac { x }{ m }\) = log k + \(\frac { 1 }{ n }\) log P
on plotting the graph between log \(\frac { x }{ m }\) and logP, a straight line is obtained with the slope equal to \(\frac { 1 }{ n }\) and the intercept equal to log k.

Question 8.
Why is adsorption always exothermic?
Answer:
Adsorption is always exothermic. This statement can be explained in two ways:

1. Adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic.
2. ∆H of adsorption is always negative. When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e. As is negative. Now for a process to be spontaneous, ∆G should be negative.
∴ ∆G = ∆n – T ∆s
since ∆s is negative, An has to be negative to make ∆G negative. Hence, adsorption is always exothermic.

Question 9.
How are the colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?
Answer:
One criterion for classifying colloids is the physical state of the dispersed phase and dispersion medium. Depending upon the type of the dispersed phase and dispersion medium (solid, liquid, or gas), there can be eight types of colloidal systems.

Question 10.
Discuss the effect of pressure and temperature on the adsorption of gases on solids.
Answer:

• Adsorption decreases with an increase in, temperature because it is an exothermic process and according to Le Chatelier’s principle the reaction will proceed in a backward direction with an increase in temperature.
•  At a constant temperature, adsorption increases with pressure.

Question 11.
What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Answer:
Lyophilic sols : Those sols in which there are forces of attraction between the dispersed phase and dispersion medium, are called lyophilic sols. Lyophilic means liquid loving. They are directly prepared by mixing substances like gum, gelatine, starch with suitable dispersion medium like water, e.g., starch in water, albumin in water. They are reversible sols.

Lyophobic sols : The word ‘lyophobic’ means liquid hating. Substances like metals, their sulphides etc. when mixed with dispersion medium, do not form sols. They are prepared by special methods, e.g., Gold sol, As2S3 sol etc. In these sols, the particles of disperesed r phase have no affinity for dispersion medium and they are irreversible sols.

Reasons for coagulation of lyophobic sols : Lyophobic sols are easily precipited or coagulated on addition of small amounts of electrolytes, by heating or by shaking because they are not stable due to less force of attraction between dispersed phase and dispersion medium.

Question 12.
What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
Answer:
(i) In multi-molecular colloids, the colloidal particles are an aggregate of atoms or small molecules with a diameter of less than ⊥ nm. The molecules in the aggregate are held together by Van der Waal’s forces of attraction. Examples of such colloids include gold sol and sulphur sol.

(ii) In macro-molecular colloids, the colloidal particles are large molecules having colloidal dimensions. These particles have a high molecular mass. When these particles are dissolved in a liquid, sol is obtained. For example Starch, nylon, cellulose, etc.

(iii) Certain substances tend to behave like normal electrolytes at lower concentrations. However, at higher concentrations, these substances behave as colloidal solutions due to the formation of aggregated particles. Such colloids are called aggregated colloids.

Question 13.
What are enzymes? Write in brief the mechanism of enzyme catalysis.
Answer:
Enzymes are basically protein molecules of high molecular masses. These form colloidal solutions when dissolved in water. These are complex, nitrogenous organic compounds produced by living plants and animals. Enzymes are also called ‘biochemical catalysts’.
E + S ⇌ [E -S] → E + P
Step -1 – Formation of the enzyme-substrate complex
E+S ⇌ E – S
Step -2 – Dissociation of complex
E – S → [EP] → E+P(product)

Mechanism of enzyme catalysis:
On the surface of the enzymes, various cavities are present with characteristics shapes, these cavities process active groups such as NH2, -coon, etc. The reactant molecules having a complementary shape fit into the cavities just like a key fits into a lock. This leads to the formation of an activated complex. This complex then decomposes to give the product.
Step 1: E + S → ES⁺
(Activated complex)
Step 2: ES⁺ → E + P

Question 14.
How are colloids classified on the basis of
(i) physical states of components
(ii) nature of dispersion medium and
(iii) interaction between the dispersed phase and dispersion medium?
Answer:
Colloids can be classified on various bases:
(i) On the basis of the physical state of the components (by components we mean the dispersed phase and dispersion medium). Depending on whether the components are solids, liquids, or gases, we can have eight types of colloids.
(ii) On the basis of the dispersion medium, sols can be divided as

(iii) On the basis of the nature of the interaction between the dispersed phase and dispersion medium, the colloids can be classified as lyophilic (solvent attracting) and lyophobic (solvent repelling).

Question 15.
Explain what is observed
(i) when a beam of light is passed through a colloidal sol.
(ii) an electrolyte, NaCl is added to hydrated ferric oxide sol.
(iii) electric current is passed through a colloidal sol?
Answer:
(i) Scattering of light by the colloidal particles takes place and the path of light becomes visible (Tyndall effect).
(ii) The positively charged colloidal particles of Fe(OH)₃ get coagulated by the oppositely charged Cl^– ions provided by NaCl.
(iii) On passing electric current, colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated.

Question 16.
What are emulsions? What are their different types? Give an example of each type.
Answer:
The colloidal solution in which the dispersed phase and dispersion medium are liquids is called an emulsion.
There are two types of emulsions:
(a) Oil in water type:
Here, oil is the dispersed phase while water is the dispersion medium. For example milk, vanishing cream, etc.
(b) Water in oil type:
Here, water is the dispersed phase while oil is the dispersion medium. For example cold cream, butter, etc.

Question 17.
What is demulsification? Name two demulsifiers.
Answer:
Demulsification is the process of decomposition of an emulsion back to the constituent liquids. It can be achieved by centrifugation or by boiling.

Question 18.
The action of soap is due to emulsification and micelle formation. Comment.
Answer:
The cleansing action of soap is due to emulsification and micelle formation, Soaps are basically sodium and potassium salts of long-chain fatty acids, R-COO^–Na⁺. The end of the molecule to which the sodium is attached is polar in nature, while the alky 1-end is non-polar. Thus, a soap molecule contains a hydrophilic (polar) and a hydrophobic (non-polar) part.

When soap is added to water containing dirt, the soap molecules surround the dirt particles in such a manner that their hydrophobic parts get attached to the dirt molecule and the hydrophilic parts point away from the dirt molecules. This is known as micelle formation. Thus we can say that the polar group dissolves in water while the non-polar group dissolves in the dirt particle. Now, as these micelles are negatively charged, they do not coagulate and a stable emulsion is formed.

Question 19.
Give four examples of heterogeneous catalysis.
Answer:
(i) Oxidation of sulphur dioxide to form sulphur trioxide. In this reaction, Pt acts as a catalyst.

(ii) formation of ammonia by the combination of dinitrogen and dihydrogen in the presence of finely divided iron.

This is called the Haber’s process.

(iii) Oswald’s process: Oxidation of ammonia to nitric oxide in the presence of platinum.

(iv) Hydrogenation of vegetable oil in the presence of Ni Vegetable oil (1) + H₂(g)

Question 20.
What do you mean by activity and selectivity of catalysts?
Answer:
(a) Activity of a catalyst:
The activity of a catalyst is its ability to increase the rate of a particular reaction. Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants on the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.

(b) Selectivity of the catalyst:
The ability of the catalyst to direct a reaction to yield a particular product is refered to as the selectivity of the catalyst. For example, by using different catalysts, we can get different products for the reaction between H₂ and CO.

Question 21.
Describe some features of catalysis by zeolites.
Answer:
(a) Zeolites are hydrated aluminosilicates which have a three-dimensional network structure containing water molecules in their pores.
(b) The pores are made vacant by heating before catalysis.
(c) The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules and also on the pores and cavities in them, e.g., ZSM-5 converts alcohols to hydrocarbons by dehydrating them.
Alcohols \(\underrightarrow { ZSM-5 }\) Hydrocarbons

Question 22.
What is shape-selective catalysis?
Answer:
A catalytic reaction that depends upon the pore structure of the catalyst and on the size of the reactant and the product molecules is called shape-selective catalysis. For example, catalysis by zeolites is shape-selective catalysis. The pore size present in the zeolites ranges from 260-740 pm. Thus, molecules having a pore size more than this cannot enter the zeolite and undergo the reaction.

Question 23.
Explain the following terms:
(i) Peptisation
(ii) Electrophoresis
(iii) Coagulation
(iv) Dialysis
(v) Tyndall effect.
Answer:
(i) Peptisation: The process of converting a freshly prepared precipitate into colloidal form by the addition of a suitable electrolyte in a small amount. Normally a freshly prepared precipitate is preferred because the particles are not so firmly attached to each other and can be easily disintegrated. Please note the electrolyte should not be added in excess because in that case, the oppositely charged ions not involved in peptization, may neutralise the charge on the colloidal particles.

(ii) Electrophoresis: When an electric potential is applied across two platinum electrodes dipped in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied electric potential is called electrophoresis.

(iii) Coagulation: The stability of the lyophobic sols is due to the presence of charge on colloidal particles. If somehow, the charge is removed, the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. The process of settling down colloidal particles is called coagulation.

(iv) Dialysis: It is the process of removing dissolved substances from a colloidal solution by means of diffusion through a suitable membrane. Since particles (ions of smaller molecules) in a true solution can pass through the animal membrane (bladder) or parchment paper or cellophane sheet but not the colloidal particles, the membrane can be used for dialysis. The apparatus used for this purpose is called a dialyzer. A bag of suitable membrane containing the colloidal solution is suspended in a vessel through which freshwater is continuously flowing. The molecules and ions diffuse through the membrane into the outer water and the pure colloidal solution is left behind.

(v) Tyndall effect: When a beam of light is passed through colloidal particles, its path becomes clearly visible and is known as the Tyndall effect. It is due to the scattering of light by colloidal particles. The bright cone of the light is called the Tyndall cone.

Question 24.
Give four uses of emulsions.
Answer:

1. Some of the medicines are effective as emulsions.
2. Paints are emulsions which are used in our daily life.
3. Soaps and detergents act as cleansing agents, the action of which is based on emulsification.
4. Photographic films are coated with an emulsion of AgBr on their surface.

Question 25.
What are micelles? Give an example of a micelle system.
Answer:
Micelle formation is done by substances such as soaps and detergents when dissolved in water. The molecules of such substances contain a hydrophobic and a hydrophilic part when present in water, these substances arrange themselves in spherical structures in such a manner that their hydrophobic parts are present towards the centre, while the hydrophilic parts are pointing towards the outside. This is known as micelles formation.

Question 26.
Explain the following terms with suitable examples. (C.B.S.E. Delhi 2009)

1. Gel
2. Aerosol
3. Hydrosol.

Answer:

1. Gel: Colloidal solution of liquid in a solid, e.g., butter.
2. Aerosol: Colloidal solution of liquid in a gas e.g., fog.
3. Hydrosol: Colloidal solution of solid in water, e.g., gold sol.

Question 27.
Comment on the statement that “colloid is not a substance but a state of substance”.
Answer:
Colloid is not a substance, but a state of substance because the same substance may exist as a colloid or crystalloid under different conditions e.g., sulphur. Colloidal solution of sulphur consists of sulphur molecules dispersed in water. In this state, sulphur atoms combine to form multi molecules whose size lies between 1 nm to 1000 nm and form a colloidal state. Sulfur forms a true solution in carbon disulphide. Similarly, soap is a solution at low concentration but a colloid at higher concentration.

Question 28.
Write any two characteristics of chemisorption.
Answer:

1. Chemisorption is highly specific in nature. It occurs only if there is a possibility of chemical bonding between the adsorbent and the adsorbate.
2. like physisorption, chemisorption also increases with an increase in the surface area of the adsorbent.

Question 29.
Why does physisorption decrease with the increase of temperature?
Answer:
Physisorption is exothermic in nature. Therefore, in accordance with Le-Chateliere’s principle, it decreases with an increase in temperature. This means that physisorption occurs more readily at a lower temperature.

Question 30.
Why are powdered substances more effective adsorbents than their crystalline forms?
Answer:
Powdered substances are more effective adsorbents than their crystalline forms because when a substance is powdered, its surface area increase, and physisorption is directly proportional to the surface area of the adsorbent.

Question 31.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process?
Answer:
It is important to remove CO in the synthesis of ammonia as CO adversely affects the activity of the iron catalyst, used in Haber’s process.

Question 32.
Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?
Answer:
Ester hydrolysis can be represented as: Ester + water → Acid + Alcohol The acid produced in the reaction acts as a catalyst and makes the reaction faster, substances that act as catalysts in the same reaction in which they are obtained as products are known as anticatalysts.

Question 33.
What is the role of desorption in the process of catalysis?
Answer:
The role of desorption in the process of catalysis is to make the surface of the solid catalyst-free for the fresh adsorption of the reactants on the surface.

Question 34.
What modification can you suggest in the Hardy-Schulze law?
Answer:
Hardy-Schulze law states that the greater the valence of the flocculating ion added, the greater is its power to cause precipitation. This law takes into consideration only the charge carried by an ion, not its size. The smaller the size of an ion, the more will be its polarising power. Thus, the Hardy-Schulze law can be modified in terms of the polarising power of the flocculating ion. Thus, the modified Hardy- Schulze law can be stated as the greater the polarising power of the flocculating ion added, the greater is its power to cause precipitation.

Question 35.
Why is it essential to wash the precipitate with water before estimating it quantitatively?
Answer:
When a substance gets precipitated, some ions that combine to form the precipitate get adsorbed on the surface of the precipitate. Therefore, it becomes important to wash the precipitate before estimating it quantitatively in order to remove these adsorbed ions or other such impurities.

2nd PUC Chemistry Surface Chemistry Additional Questions and Answers

Question 1.
What causes Brownian movement in a colloidal solution? (Delhi 2008)
Answer:
The molecules of dispersion medium due to their kinetic motion strike against the colloidal particles (dispersed phase) from all sides with different forces causing them to move.

Question 2.
It is found that when litmus is shaken with animal charcoal adsorption take place

1. Find out the adsorbent and adsorbate in this process.
2. Explain the terms adsorbent and adsorbate.
3. Name the process of removing adsorbate from the adsorbent.
4. Explain the term ‘adsorption’.

Answer:

1. Litmus is the adsorbate and charcoal is the adsorbent.
2. The material providing the surface upon which absorption occurs is known as the adsorbent and the substance adsorbed is called adsorbate.
3. The process of removal of an adsorbed substance from the surface of the adsorbent is called desorption.
4. The phenomenon in which absorption and adsorption occur together is called sorption.

Question 3.
Which has a higher enthalpy of adsorption, physisorption, or chemisorption? (Delhi 2008)
Answer:
Enthalpy of chemisorption is high (80-240 KJ mol^-1) as it involves chemical bond formation.

Question 4.
Explain what is observed when
(i) an electrolyte, KCl, is added to a hydrated ferric oxide solution.
(ii) an electric current is passed through a colloidal solution
(iii) a beam of strong light is passed through a colloidal solution (Delhi 2008, AI2008)
Answer:
(i) When an electrolyte like KCl is added of Fe (OH)3sol, the positively charged colloidal particles of Fe(OH)₃ get coagulated by the oppositely charged Cl ions provided by KCl

(ii) On passing the electric current, colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated.

(iii) When a beam of strong light is passed through a colloidal solution scattering of light by colloidal particles takes place and the path of light becomes visible. This phenomenon is called the Tyndall effect.

Question 5.
(a) What are zeolites? Give any one of its use.
(b) A U-tube is filled with ferric hydroxide solution. Two platinum electrodes are introduced into the two limbs of the U-tube and an electric current is passed through the electrodes. What do you observe? Name the phenomenon.
Answer:
(a) Zeolites are aluminosilicates having a three-dimensional structure. Zeolites are highly porous have cavities of different sizes. The catalytic behaviour of zeolite catalysts depends upon the size of the cavities in them, which usually varies from 260-740 pm. The reaction molecules of a particular shape and size can only enter and get absorbed. Thus Zeolites are called shape-selective catalysts.
(b) Colloidal particles move towards one or another electrode. The phenomenon is called electrophoresis.

Question 6.
Classify the following into homogenous and heterogeneous catalysis
(i) Catalytic decomposition of ozone by chlorine
(ii) Hydrolysis of an organic ester
(iii) Haber’s process.
Answer:

Question 5.
While eating bread and butter, Ramu remembered what he studied yesterday. The teacher taught that butter is a colloidal solution?

1. In which category of colloid, butter belongs?
2. Depending upon the dispersed phase and dispersion medium, how are the colloids classified?
3. Suggest methods to prepare sulphur sol and ferric hydroxide sol?

Answer:
1. Butter is an example of a gel. (Liquid dispersed in solid)

2. The colloids in which the particles of the dispersed phase have a great affinity for the dispersion medium are called lyophilic colloids whereas, the colloids in which the particles of the dispersed phase have no affinity for the dispersion medium are called lyophobic colloids. Glue, gelatin, etc. are examples of lyophilic colloids, and gold sol, Fe(OH₃), sol, etc. are examples of lyophobic colloids.

3. Sulphur sol can be prepared by the oxidation of an aqueous solution of hydrogen sulphide solution with air or SO₂.
2H₂S + SO₂ → 3S + 2H₂O
Ferric hydroxide sol is prepared bv adding a small quantity of ferric chloride to boiling water.
FeCl₃ + 3H₂O → Fe(OH)₃ + 3HCl.

Question 8.
Write any two characteristics of chemisorption.
Answer:

1. Chemisorption is highly specific
2. Chemisorption involves the compound formation and hence is irreversible in nature.

Question 9.
Why is ester hydrolysis slow in the beginning and becomes faster after sometime?
Answer:
When an ester is treated with a mineral acid, gives acid and alcohol. The organic acid undergoes dissociation to give hydrogen ions. These hydrogen ions act as catalysts and hence and the hydrolysis becomes faster.

Question 10.
What is the role of desorption in the process of catalysis?
Answer:
The product formed during catalysis gets detached from the surface at the catalyst by the process of desorption and makes available the catalyst surface for move reaction to occur.