kalakadu.com: The P-block Elements Class 12th Chemistry Part I CBSE Solution

The P-block Elements Class 12th Chemistry Part I CBSE Solution

Class 12^th Chemistry Part I CBSE Solution

Intext Questions Pg-169
Why are pentahalides more covalent than trihalides ?
Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements?…
Intext Questions Pg-189
What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?…
Comment on the nature of two S-O bonds formed in SO2 molecule. Are the two S-O bonds in…
How is the presence of SO2 detected?
Intext Questions Pg-191
Mention three areas in which H2SO4 plays an important role.
Write the conditions to maximise the yield of H2SO4 by contact process.…
Why is for H2SO4in water?
Intext Questions Pg-196
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy, and…
Give two examples to show the anomalous behaviour of fluorine.
Sea is the greatest source of some halogens. Comment.
Intext Questions Pg-198
Give the reason for bleaching action of Cl2.
Name two poisonous gases which can be prepared with chlorine gas.…
Intext Questions Pg-202
Why is ICl more reactive than I2 ?
Intext Questions Pg-205
Why is helium used in diving apparatus?
Balance the following equation: XeF6 + H2O → XeO2F2 +HF.
Why has it been difficult to study the chemistry of radon?
Exercises
Discuss the general characteristics of Group 15 elements with reference to their…
Why does the reactivity of nitrogen differ from phosphorus?
Discuss the trends in chemical reactivity of group 15 elements.
Why does NH3 form hydrogen bond but PH3 does not?
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions…
How is ammonia manufactured industrially?
Illustrate how copper metal can give different products on reaction with HNO3.…
Give the resonating structures of NO2 and N2O5.
The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?…
Why does R3P = O exist but R3N = O does not (R = alkyl group)?
Explain why NH3 is basic while BiH3 is only feebly basic.
Nitrogen exists as diatomic molecule and phosphorus as P4. Why?
Write main differences between the properties of white phosphorus and red phosphorus.…
Why does nitrogen show catenation properties less than phosphorus?…
Give the disproportionate reaction of H3PO3.
Can PCl5 act as an oxidizing as well as a reducing agent? Justify.…
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in…
Why is dioxygen a gas but sulphur a solid?
Knowing the electron gain enthalpy values for O → O- and O → O^2 - as -141 and 702 kJ…
Which aerosols deplete ozone?
Describe the manufacture of H2SO4 by contact process?
How is SO2 an air pollutant?
Why are halogens strong oxidising agents?
Explain why fluorine forms only one oxoacid, HOF.
Explain why inspite of nearly the same Electronegativity, nitrogen forms hydrogen bonding…
Write two uses of ClO2.
Why are halogens coloured?
Write the reactions of F2 and Cl2 with water.
How can you prepare Cl2 from HCl and HCl from Cl2? Write reactions only.…
What inspired N. Bartlett for carrying out the reaction between Xe and PtF6?…
What are the oxidation states of phosphorus in the following: (i) H3PO3 (ii) PCl3 (iii)…
Write balanced equations for the following: (i) NaCl is heated with sulphuric acid in the…
How are xenon fluorides XeF2, XeF4 and XeF6 obtained?
With what neutral molecule is ClO- isoelectronic? Is that molecule a Lewis base?…
How are XeO3 and XeOF4 prepared?
Arrange the following in the order of property indicated for each set: (i) F2, Cl2, Br2,…
Which one of the following does not exist? (i) XeOF4 (ii) NeF2 (iii) XeF2 (iv) XeF6…
Give the formula and describe the structure of a noble gas species which is isostructural…
Why do noble gases have comparatively large atomic sizes?
List the uses of neon and argon gases.
Intext Questions Pg-180
What is the basicity of H3PO4?
What happens when H3PO3 is heated?
Intext Questions Pg-170
Why is N2 less reactive at room temperature?
Intext Questions Pg-172
Mention the conditions required to maximise the yield of ammonia.…
How does ammonia react with a solution of Cu2+?
What happens when white phosphorus is heated with concentrated NaOH solution in an inert…
Intext Questions Pg-173
What is the covalence of nitrogen in N2O5 ?
Intext Questions Pg-177
Bond angle in PH4+ is higher than that in PH3. Why?
Intext Questions Pg-178
What happens when PCl5 is heated?
Write a balanced equation for the hydrolytic reaction of PCl5 in heavy water.…
Intext Questions Pg-183
List the important sources of sulphur.
Write the order of thermal stability of the hydrides of Group 16 elements.…
Why is H2O a liquid and H2S a gas?
Intext Questions Pg-185
Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe…
Complete the following reactions: (i) C2H4 + O2→ (ii) 4Al + 3 O2→…
Intext Questions Pg-187
Why does O3 act as a powerful oxidizing agent?
How is O3 estimated quantitatively?

Intext Questions Pg-169

Question 1.
Why are pentahalides more covalent than trihalides ?

Answer:
Pentahalides (like MX₅) means the compounds in which metal is bonded with five halogen atoms. Thus, the oxidation state of metal here is +5. Similarly, trihalides (like MX₃) means the compounds in which metal is bonded with three halogen atoms. Thus, the oxidation state of metal here is +3. Now, as the polarizing power is directly proportional to the charge, the metals with the higher charge will have higher polarizing power.
Now,we all know that according to Fazan's rule more the polarising power, more will be its covalency. Hence, Pentahalides are more covalent than trihalides.
Note: Polarizing power is the ability of a cation to distort an anion.

Question 2.
Why is BiH₃ the strongest reducing agent amongst all the hydrides of Group 15 elements?

Answer:
Atomic size increases down the group and thus the stability of the hydrides of Group-15 elements decreases. Now, the least stable hydride is the one which is more reactive and possessing higher reducing strength. The reducing character of hydrides increases on moving from NH₃ to BiH₃. Hence, BiH₃ is the strongest reducing agent among hydrides of Group-15 elements.

Intext Questions Pg-189

Question 1.
What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?

Answer:
SO₂ acts as a reducing agent when passed through an aqueous solution containing Fe(III) salt.

It reduces Fe(III) to Fe(II) i.e., ferric ions to ferrous ions.

2Fe³⁺ + SO₂ + 2H₂O ⇒ 2Fe²⁺ + SO₄^2- + 4H⁺

Question 2.
Comment on the nature of two S–O bonds formed in SO₂ molecule. Are the two S–O bonds in this molecule equal?

Answer:
The electronic configuration of S is 1s² 2s² 2p⁶ 3s² 3p⁴.

One electron from 3p orbital goes to the 3d orbital and S undergoes sp² hybridization, during the formation of SO₂ molecule. Two of these orbitals form sigma bonds with two oxygen atoms and the third contains a lone pair of electrons. The p-orbital and d-orbital contain an unpaired electron each. An electron in the p-orbital forms p π- p π bond with one oxygen atom and the other electron forms p π- d π bond with the other oxygen. This is the reason SO₂ has a bent structure. Also, it is a resonance hybrid of structures I and II, with equal bond lengths of 143pm.

Question 3.
How is the presence of SO₂ detected?

Answer:
When SO₂ is passed through acidified potassium permanganate(VII) solution it decolourises the solution. This can be used for the detection of the SO₂ gas. The below reaction represents the test,

5SO₂ + 2MnO₄^- + 2H₂O → 5SO₄^2- + 4H⁺ + 2Mn²⁺

Intext Questions Pg-191

Question 1.
Mention three areas in which H₂SO₄ plays an important role.

Answer:
H₂SO₄ is one the most important and widely used chemicals for many purposes. It has following uses:
(i) As an electrolyte sulphuric acid dissociates in aqueous solution to form ions. These ions are responsible for the conduction of electricity. As a result, sulphuric acid is used in the manufacture of the lead storage battery.

(ii) As a dehydrating agent (to remove water) because it has a great affinity for water. It doesn’t get dissolved in water but absorb it, so it is used as a drying agent.

(iii) For removing layers of basic oxides from the metal surfaces like iron, copper, etc. before the metals are galvanized, electroplated. Because when the metals are dissolved in sulphuric acid the basic oxide layer formed (due to reaction with air, water) will be dissolved in it as it is acidic in nature.

Question 2.
Write the conditions to maximise the yield of H₂SO₄ by

contact process.

Answer:
(i) Contact process consists of three steps. First is the burning of sulphide ores to form SO₂ gas. The second step (key step) is SO₂ is
oxidised to SO₃ by heating the mixture on the heterogeneous catalyst V₂O₅.

Now, this step is very crucial for maximising the yield of H₂SO₄

It can be done by controlling the temperature and pressure of reaction.

(ii) The forward reaction is exothermic and there is a decrease in a number of moles and volume. As Le-Chatelier principle states that when a system experiences a disturbance (concentration, pressure changes) it will respond to nullify the changes produced to bring initial condition.

So, when we decrease temperature (BEING EXOTHERMIC),

it will try to restore original equilibrium, by experiencing a forward reaction where the temperature is high.

Now in case of pressure, moles of product is less than that of reactant and we know that no. of moles is directly proportional to volume (pressure is inversely proportional to volume). So volume of reactant is more than product if we increase pressure (decrease volume) the reaction will shift to forward direction.

NOTE: The optimum conditions to maximise the yield of sulphuric acid is the pressure of 2-3 atm and temperature around 723K.

Question 3.
Why is for H₂SO₄in water?

Answer:
H₂SO₄ is dibasic acid (two replacable H⁺ ions) . In aqueous solution it dissociates in two steps as follows:

The formula for dissociation constants are:

The neutral H₂SO₄ molecule has more tendency to lose a proton (H⁺) than the Lowry-Bronsted base (which can donate electron i.e. accept a proton H⁺) HSO₄^-.This is because a neutral has a much higher tendency to lose a proton than the negatively charged. Thus, the former is a much stronger acid than the latter.

Intext Questions Pg-196

Question 1.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy, and hydration enthalpy, compare the oxidizing power of F₂ and Cl₂.

Answer:
The oxidizing power of an element depends on three factors.
1. Bond dissociation energy

2. Electron gain enthalpy

3. Hydration enthalpy

Fluorine is considered strongest oxidizing agent compared to all known elements due to its small size (being the first member of 17 group and ionic radii decrease downwards), low bond enthalpy of F-F (due to high electron-electron repulsion). High negative value of electron gain enthalpy (highest electronegativity ) and very high hydration energy of F^- ions (hydration energy is the energy released when a compound is dissolved in water . In case of fluorine it is more due to the strong interaction between F^- ions and water, as it is electro-negative). The comparison of both the elements makes this clear that F₂ is a strong oxidizing power as compared to Cl_2.

The electron gain enthalpy of chlorine (-349 KJ mol^-1) is more negative than that of fluorine (-333 KJ mol^-1). However, the bond dissociation energy of fluorine (158.8 KJ mol^-1) is much lesser than that of chlorine (242.6 KJ mol^-1). Also, because of its small size, the hydration energy of fluorine (515 KJ mol^-1) is much higher than that of chlorine(381 KJ mol^-1). Therefore, the latter two factors more than compensate for the less negative electron gain enthalpy of fluorine. Thus, fluorine is a much stronger oxidizing agent than chlorine.

Question 2.
Give two examples to show the anomalous behaviour of fluorine.

Answer:
The reasons for the anomalous behaviour of fluorine are as follows:
(i) The smallest size of fluorine

(ii) The highest electronegativity

(iii) Low bond dissociation enthalpy of F-F bond

(iv) Non-availability of d-orbitals in its valence shell

The anomalous properties are as follows;

(i) Fluorine shows only one oxidation state-1 while all other halogens show variable oxidation states like -1, +1, +3, +5, +7(due to small size, high electronegativity)

(ii) Fluorine forms strong hydrogen bonding in its hydrides unlike other halogens. (due to high hydration energy than other elements)

(iii) The compounds of fluorine have higher ionic character than other halogens. (due to high electronegativity forming polar interactions)

(iv) Fluorine has no tendency to form polyhalide ion whereas other halogen form polyhalide ion(due to non-availability of d orbitals in valence shell)

Question 3.
Sea is the greatest source of some halogens. Comment.

Answer:
Sea water is a treasure of many elements and halogens are one of them. It contains many compounds of halogen including bromides, iodides, and chlorides. The solid deposits from sea water contain NaCl and Carnalite. It primarily contains COMMON SALT (2.5% by mass). The deposits of dried up sea bed also contain sodium chloride and Carnallite, KCl.MgCl₂.6H₂O . The Marine life also contains iodine in their systems. For example, sea weed may contain 0.5% iodine as sodium iodide(NaI) and Chile saltpetre (sodium compound) contains about 0.3% of sodium iodate.
Thus, the sea is the greatest source of halogens.

Intext Questions Pg-198

Question 1.
Give the reason for bleaching action of Cl₂.

Answer:
(i) Moist chlorine is a good reducing agent because it can accept electrons from other species as it is very electronegative.
(iii) The bleaching action of chlorine is due to oxidation by nascent oxygen produced. This nascent oxygen can be produced by

Chlorine dissolves in water in absence of sunlight and forms hydrochloric acid and hypochlorous acid(HOCl)

Since hypochlorous acid is not much stable it decomposes giving nascent oxygen.

Cl₂ + H₂O ⇒ HCl + HOCl

HOCl ⇒ HCl + [O]

This [O] is called as nascent oxygen. This nascent oxygen is responsible for bleaching action.

NOTE:Since chlorine has the ability to kill harmful micro-organism it also acts as a good disinfecting agent.

Question 2.
Name two poisonous gases which can be prepared with chlorine gas.

Answer:
Two poisonous gases that can be prepared with chlorine gas are:
(i) Phosgene(COCl₂)

This is how phosphene is formed:

CO + Cl₂→ COCl₂(in presence of sunlight)

(ii) Mustard gas(Cl-C₂H₄-S-C₂H₄-Cl)

(iii) Tear gas(CCl₃NO₂)

Intext Questions Pg-202

Question 1.
Why is ICl more reactive than I₂ ?

Answer:
(i) In interhalogen compound ICl and I₂ the atoms are bonded by covalent bonds.
Reactivity refers to the rate at which a chemical species will undergo reaction in time. For reaction to take place the compound needs to be broken into separate elements first, then the individual elements react with other elements to form new compounds. So any compound which can easily break into their individual elements can react faster.

(ii) The covalent bonds between dissimilar atoms I and Cl atoms in ICl are weaker than between similar atoms in I₂. Therefore the bond between ICL will break easily so the I and Cl atom will be easily available to form another compound as compared to the strong bond between I_2.

(iii) Therefore bond dissociation enthalpy of all bond is less than that of I₂ bond.

Therefore ICl is more reactive than I₂.

Intext Questions Pg-205

Question 1.
Why is helium used in diving apparatus?

Answer:
Helium mixed with oxygen under pressure is given to sea divers for respiration. Because pure oxygen can be toxic at a great concentration at the depth. Therefore oxygen can be mixed with helium to reduce oxygen concentration while eliminating nitrogen. The main reason for adding helium to the breathing mix is to reduce the proportion of nitrogen and oxygen below those of air, to allow the gas mix to be breathed safely on deep dives. A low proportion of nitrogen is required to reduce nitrogen narcosis and other physiological effects of gas at depth.

Question 2.
Balance the following equation: XeF₆ + H₂O → XeO₂F₂ +HF.

Answer:
Balanced equation: XeF₆ + 2H₂O → XeO₂F₂ +4HF.
The steps for balancing the reaction are as follows:

1. The main element is Xe. First, check if the atoms of Xe on both sides are balanced. YES, they are.

2. Then take another element i.e. F. It is not balanced on the right side, there are 3 atoms of F missing. So, to balance it multiply HF by 4.

3. Now check the other secondary atoms which are oxygen and hydrogen. Oxygen is not balanced on the left side as there are 2 O atoms so multiply H₂O by 2.

4. Now check for the hydrogen atom. They are balanced on both sides.

Question 3.
Why has it been difficult to study the chemistry of radon?

Answer:
It is difficult to study the chemistry of radon because it is a radioactive substance having a half-life(the time period to decompose the substance half to its initial concentration) of only 3.82 days.
Radon belongs to the 18 group elements with chemical formula as Rn. In general, the elements at the bottom o periodic table are radioactive, they are very dangerous to study as they emit harmful radiations. Also, compounds of radon such as RnF₂ have not been isolated, they are still in the phase of discovery. They have only been identified by radiotracer technique and no any further properties have been determined.

Exercises

Question 1.
Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionization enthalpy and electronegativity.

Answer:
The general characteristics of Group 15 elements are:
(i)Electronic configuration: All Group 15 elements have 5 electrons in their valence shell. The general electronic configuration of these elements is ns² np³.

(ii)Oxidation state: Group 15 elements have 5 valence electrons and they require 3 more electrons to complete their octet. However, the gaining of 3 electrons is difficult

(iii)Atomic size: Atomic size increases as we move down the group due to increase in the number of shells.

(iv)Ionisation enthalpy: Ionisation enthalpy decreases as we move down the group because of increase in atomic sizes.

(v)Electronegativity: Electronegativity decreases on moving down the group due to increase in atomic radius.

Question 2.
Why does the reactivity of nitrogen differ from phosphorus?

Answer:
Nitrogen atom can bond with another nitrogen atom by strong pπ–pπ overlap resulting in N≡N. The triple bond in N₂ has high bond strength resulting in high bond dissociation energy. Phosphorous do not show this property of pπ–pπ overlap. Hence, nitrogen is less reactive than phosphorous.

The p-orbitals which interact to form two π bonds in N₂ molecule are shown by joining them with a curve line.

Question 3.
Discuss the trends in chemical reactivity of group 15 elements.

Answer:
Chemical reactivity of group 15 elements towards hydrogen, oxygen, halogens, and metals are discussed below.
(i)Reactivity towards hydrogen: Group 15 elements react with H to form hydrides of type EH₃ where E=N, P, As, Sb or Bi. On moving down from NH₃ to BiH₃, the stability of the hydrides decreases. For example, the P-H bond in PH₃ is less stable than the N-H bond in NH₃. The strength of the E-H bond gets weaker as the size of the central atom increases.

Stability order of E-H bond (where E is group 15 elements) can be represented as

N-H > P-H > As-H > Sb-H > Bi-H

(ii)Reactivity towards oxygen: Group 15 elements react with O to form oxides of the type E₂O₃ and E₂O₅ where E = N, P, As, Sb or Bi. The oxide of the element with higher oxidation state is more acidic than those with lower oxidation states. Down the group, the acidic character of the oxides decreases.

(iii)Reactivity towards halogens: Group 15 elements form halides of the type EX₃ and EX₅ where E = N, P, As, Sb or Bi. However, N does not form pentahalides EX₅ as it lacks d-orbital. All group 15 elements form trihalides but trihalides of N are unstable.

(iv)Reactivity towards metals: Group 15 elements react with metals to form compounds of general formula M₃E₂ where E= N, P, As, Sb or Bi. The binary compounds formed by the reaction of group 15 elements with metal show oxidation state -3.

Question 4.
Why does NH₃ form hydrogen bond but PH₃ does not?

Answer:
The electronegativity of N (3.0) is higher than that of P (2.1). Due to this N-H bond is more polar than a P-H bond. Also, P and H have the same electronegativity of 2.1 i.e., the P-H bond is non-polar. Therefore, PH₃ does not form hydrogen form.
The structure of NH₃ and PH₃ with their electronegativity is represented below.

Question 5.
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.

Answer:
Nitrogen is prepared in the laboratory by heating aqueous ammonium chloride (NH₄Cl) with sodium nitrite (NaNO₂) to form ammonium nitrite (NH₄NO₂), which is unstable. Ammonium nitrite breaks down to form nitrogen and water.
NH₄Cl(aq) + NaNO₂(aq) NH₄NO₂ + NaCl(aq)

NH₄NO₂N₂(g)+ H₂O(l)

Small amounts of NO and HNO₃ are also produced which can be removed by passing nitrogen gas through aqueous sulphuric acid containing potassium dichromate.

Question 6.
How is ammonia manufactured industrially?

Answer:
Ammonia is manufactured industrially by Haber’s process. Nitrogen from gas is combined with hydrogen derived from natural gas (methane) in the ratio 1:3 giving rise to ammonia. The reaction is reversible and exothermic.
N₂(g)+ 3H₂(g)2NH₃(g)

Fig: Haber’s Process

The optimum conditions for the manufacturing of ammonia are pressure 200×10⁵ Pa, the temperature of 4700 K and iron oxide catalyst with a small amount of Al₂O₃ and K₂O.

Question 7.
Illustrate how copper metal can give different products on reaction with HNO₃.

Answer:
Copper metal on reaction with HNO₃ gets oxidized and give different by-products depending on the temperature, concentration of the acid and the copper metal undergoing oxidation.
The reaction of copper with concentrated and dilute HNO₃ is as shown below:

3Cu + 8HNO₃(did.) ⇒ 3Cu(NO₃)₂ + 2NO +4H₂O

Cu + 4HNO₃(conc.) ⇒ Cu(NO₃)₂ + 2NO₂ +2H₂O

Question 8.
Give the resonating structures of NO₂ and N₂O₅.

Answer:
To draw the resonating structure of NO₂
No. of valence electrons for N = 5

No. of valence electrons for O = 2 × 6 = 12

Total no. of valence electrons = 5 + 12 = 17

The N and O atoms are arranged in such a way that the less electronegative atom N is placed as the central atom.

Electron pairs are placed between bonds and distributed around the atoms to form octet as shown. Both the O atoms have their complete octet but N has only 5 electrons.

A pair of electrons is transferred from O to the bond between O and N. Both the O atoms have their complete octets and N has 7 electrons.

The resonating structures of NO₂ can be shown as:

To draw the resonating structure of N₂O₅

No. of valence electrons for N = 2 × 5 = 10

No. of valence electrons for O = 5× 6 = 30

Total no. of valence electrons = 10 + 30 = 40

Just like in NO_2, the valence electrons are distributed between bonds and then around atoms to complete their octet.

(Formal charge of each atom is enclosed in bracket)

When the valence electrons around the atoms, the formal charge of N atoms are found to be +1 while that of the center O is +2. But N is less electronegative than O. So, the electrons can be re-arranged as

The centre O now has formal charge 0 and the terminal O atoms have formal charge -1 which the N atoms have +1 which is more appropriate since O is more electronegative than O.

The resonance structures of N₂O₅ can be shown as:

Question 9.
The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?

Answer:
Among the group 15 elements, N has the highest electronegativity, because of which there is high electron density around N. This causes repulsion between the electron pairs around causing high HNH angle value. As we go down the group, the electronegativity of the elements decease and the bond angle also decreases due to lesser repulsion.

Question 10.
Why does R₃P = O exist but R₃N = O does not (R = alkyl group)?

Answer:
Nitrogen does not have d-orbital to expand its octet. So, it cannot have coordination number greater than 4. But, phosphorous has d-orbital and can extend its octet and form R₃P = O. Therefore, R₃P = O exist but R₃N = O does not.

Question 11.
Explain why NH₃ is basic while BiH₃ is only feebly basic.

Answer:
Nitrogen has a smaller size than Bismuth, because on going down the group the atomic size increases i.e. N<P<As<Sb<Bi, as shown below:

Due to smaller size, there is very high electron density around Nitrogen as compared to Bismuth. Therefore, nitrogen atom can easily release electrons. And, we know higher the electron donating tendency, higher is the basic strength. Due to which NH3 is more basic than BiH₃.

Note: The basic strength in group 15 is in the following pattern:

NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃

Question 12.
Nitrogen exists as diatomic molecule and phosphorus as P₄. Why?

Answer:
Nitrogen owing to its small size has a tendency to form pπ – pπ multiple bonds with itself. Nitrogen thus forms a very stable diatomic molecule, N₂.

On moving down, a group, the tendency to form pπ–pπ bonds decreases [because of the large size of heavier elements]. With the increase in the size of atoms, the strength pπ– pπ bonds decreases. Thus pπ– pπ bonds are weaker for heavier atoms. Therefore, phosphorus [like other heavier metals] exists in the P₄ state, as shown below.

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

Answer:
Question 14.
Why does nitrogen show catenation properties less than phosphorus?

Answer:
Catenation is the bonding of atoms of the same element into a series to form a chain. Catenation is much more common in phosphorous compounds than in nitrogen compounds.
This is because of the relative weakness of the N−N single bond as compared to the P−P single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N−N single bond. This is the reason why Nitrogen shows catenation properties less than phosphorous.

Phosphorous has larger atomic size because of which it has less repulsion of electron density of two phosphorous atoms and hence P-P single bond is stronger.

Question 15.
Give the disproportionate reaction of H₃PO₃.

Answer:
Disproportionation is a chemical reaction typically a redox reaction where a molecule is transformed into two or more dissimilar products.
On heating, orthophosphorus acid (H₃PO₃) disproportionate to give orthophosphoric acid (H₃PO₄) and phosphine (PH₃). The oxidation states of Phosphorous in various species involved in the reaction are mentioned below.

Question 16.
Can PCl₅ act as an oxidizing as well as a reducing agent? Justify.

Answer:
Phosphorous Pentachloride [PCl₅] cannot act as reducing agent because in the PCl₅, the oxidation state of chlorine is +5 and it is not possible for Chlorine to increase its oxidation state beyond +5 but it can very easily act as an oxidizing agent because it can easily reduce its oxidation state from +5 to +3.

Question 17.
Justify the placement of O, 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: As these elements have six valence electrons [ns²np⁴], 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 the 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 H₂E_2. These hydrides are quite volatile in nature.

Question 18.
Why is dioxygen a gas but sulphur a solid?

Answer:
Oxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively form pπ –pπ bonds and form O₂[ O = = O] molecule. Also, the intermolecular forces in oxygen are weak van der Wall’s, which cause it to exist as a gas.
On the other hand, sulphur does not form M₂ molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid.

Question 19.
Knowing the electron gain enthalpy values for O → O– and O → O²^– 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:
Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be. Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving O^2- ion is much more than the oxide involving O^- ion. Hence, the oxide having O^2-ions are more stable than oxides having O^-. Hence, we can say that formation of O^2- is energetically more favorable that formation of O^-.

Question 20.
Which aerosols deplete ozone?

Answer:
Freons or chlorofluorocarbons [CFCs] are aerosols that accelerate the depletion of ozone. In presence of ultraviolet radiation [UV radiation], these chlorofluorocarbons break down to give chlorine ions which combine with the ozone atoms in the atmosphere to give oxygen atoms.

Question 21.
Describe the manufacture of H₂SO₄ by contact process?

Answer:
Theory: This process involves the catalytic oxidation of sulphur dioxide into sulphur trioxide by atmospheric air.
2SO₂ + O₂ → 2 SO_3; A_H = - 196.6 kJ

The reaction is reversible, exothermic and involves a decrease in the number of moles. Therefore, according to the Le-Chatelier’s principle, the favourable conditions for the maximum yield of sulphur trioxide are as follows.

[i]. Low temperature: A decrease in temperature would favour the forward reaction. The optimum temperature is experimentally found to be 670-720 K.

[ii]. High pressure: An increase in pressure should favour the forward reaction because the reaction involves a decrease in a number of moles. In practice, about 2-atmosphere pressure is maintained.

[iii]. Excess of oxygen: An increase in the concentration of reactants favours the forward reaction. Hence, oxygen [air] is used in excess.

[iv]. Use of catalyst:

1] Even at low temperature and high pressure, the yield of sulphur trioxide is not very satisfactory.

2] Hence, a suitable catalyst must be used to increase the rate of formation of sulphur trioxide.

3] Several catalysts such as platinized asbestos, platinized silica gel, vanadium pentoxide, etc., may be used for the purpose.

4] Nowadays, vanadium pentoxide is extensively used as a catalyst in this process. It gives good results at 67-72 K and is less susceptible to poisoning.

The sulphur trioxide obtained as above is dissolved in concentrated sulphuric acid to get oleum. Sulphuric acid of desired concentration can be obtained by diluting it with water.

Plant Used and the Procedure:

The plant used in the process consists of following parts.

[i]. Sulphur burners: In these burners, sulphur is burnt to produce sulphur dioxide.

S + O₂ → SO₂

Iron pyrites may also be used in place of sulphur.

4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

Sulphur dioxide thus produced is mixed with air and is taken to the purification unit.

[ii]. Purification unit: The impurities present in the mixture of sulphur dioxide and air coming from sulphur burners are removed in this unit. The unit consists of following parts.

[a] Dust precipitator: It makes the dust particles to settle down.

[b] Washing and cooling tower: The dust-free gases are washed with water to remove the impurities and to cool the gases.

[c] Drying Tower: In this tower, the incoming gases are dried by a spray of conc. H₂SO₄.

[d] Arsenic purifier: It consists of gelatinous ferric hydroxide and absorbs any impurity of arsenic oxide present in the gases.

[e] Testing box: This box detects the presence of solid particles in the incoming gases. When the gases are found free from solid particles, they are allowed to move onwards to the contact tower.

[iii] Contact tower:

1] It consists of iron pipes packed with vanadium pentoxide.

2] The gases coming out of the testing box are heated to 725 K in a pre-heater and then allowed to enter into this tower.

3] As the mixture of sulphur dioxide and air passed through the pipes containing the catalyst, sulphur dioxide gets oxidized to sulphur trioxide.

4] The conversion of SO₂ into SO₃is exothermic and heat generated in the reaction is sufficient to maintain the temperature of the catalyst.

5] Therefore, once the process has started, the gases are not heated in the pre-heater and are passed directly into the contact tower.

[iv]. Absorption tower:

1] The sulphur trioxide formed in the contact tower is now led to the absorption tower where it is brought in contact with conc. H₂SO₄which is sprayed from the top of the tower.

2] Sulphur trioxide gets dissolved in conc.Sulphuric acid to give oleum. This can be converted into sulphuric acid of desired concentration by diluting it with a calculated amount of water.

3] Sulphur trioxide is not absorbed in water directly because it results in the formation of the dense fog of sulphuric acid particles which makes the operation difficult to handle.

4] This is why it is preferred to dissolve SO₃ into conc. Sulphuric acid to obtain oleum.

Question 22.
How is SO₂ an air pollutant?

Answer:
SO₂ is a highly irritating gas and causes serious respiratory problems, and may cause a fit of coughing.
1. It reacts with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil [soil become more acidic], plants, and buildings get corroded, especially those made of marble.

In the air, SO₂ is oxidized to SO₃ which is also an irritant.

2SO₂ + O₂ → 2SO₃

SO₃ + H₂O → H₂SO₄

2. Even in very low concentrations, SO₂ causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.

3. It is extremely harmful to plants. Plants exposed to sulphur dioxide for a prolong time lose colour from their leaves. This condition is known as chlorosis. This is because the formation of chlorophyll is retarded in presence of sulphur dioxide.

Question 23.
Why are halogens strong oxidising agents?

Answer:
The general electronic configuration of halogens is np⁵, where n = 2-6. Thus, halogens needonly one more electron to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron.
Whenever an atom accepts an electron from another atom, the atom accepting the electron is getting reduced and the atom donating the electron is oxidized. But the electron accepting atom acts as an oxidizing agent while the electron donating atom acts as reducing agent.

Hence, they act as strong oxidizing agents.

Question 24.
Explain why fluorine forms only one oxoacid, HOF.

Answer:
Fluorine forms only one oxoacid i.e., HOF because of its high Electronegativity and small size.

Question 25.
Explain why inspite of nearly the same Electronegativity, nitrogen forms hydrogen bonding while chlorine does not.

Answer:
Both chlorine and nitrogen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, nitrogen has a smaller size and as a result, a higher electron density per unit volume. Hence nitrogen forms hydrogen bonding more readily.
Chlorine has larger atomic size as compared to nitrogen and so has lower electron density per unit volume. Hence chlorine does not readily form hydrogen bonding.

Question 26.
Write two uses of ClO₂.

Answer:
ClO₂ which is called as Chlorine Dioxide has following uses:
1)It is used as a bleaching agent in paper pulp and textile industries.

2)It is used as disinfectant in sewage and for purification of drinking water.

3) Chlorine dioxide is used to control tastes and odours associated with algae and decaying vegetation.

Question 27.
Why are halogens coloured?

Answer:
Most of the halogens are coloured because they absorb radiations which are in the visible region which in turn excite the valence electrons to higher energy levels. As the amount of energy which is required to excite the electrons to a higher level is different for different halogens, each halogen has a different colour.

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

Answer:
Chlorine gas reacts with water to give Hydrochloric acid and Hypochlorous acid

Cl₂ + H₂O → HCl (Hydrochloric Acid) + HOCl (Hypochlorous acid)

Fluorine gas reacts with water to give Hydrogen ions, Fluorine ions, Oxygen gas and Hydrofluoric acid.

2F₂ + 2H₂O → 4H⁺ + 4F^- + O₂ + 4HF

Question 29.
How can you prepare Cl₂ from HCl and HCl from Cl₂? Write reactions only.

Answer:
1) By Deacon’s process Cl₂ can be prepared from HCl in

presence of CuCl₂

4HCl + O₂ → 2Cl₂ + 2H₂O

2) Cl₂ on treating with water gives HCl

Cl₂ + H₂O → HCl + HOCl

Question 30.
What inspired N. Bartlett for carrying out the reaction between Xe and PtF₆?

Answer:
Neil Bartlett first performed an experiment in which reaction between oxygen and PtF₆ was carried out which lead to the formation of a red coloured compound O₂⁺[PtF₆]^-.

He observed that the first ionization energy of Oxygen and Xenon is almost same (~1170 kJ/mol). So, he tried to react Xe and PtF₆ in which he was successful to obtain a red coloured compound Xe⁺[PtF₆]^-.

Question 31.
What are the oxidation states of phosphorus in the following:

(i) H₃PO₃ (ii) PCl₃ (iii) Ca₃P₂ (iv) Na₃PO₄ (v) POF₃?

Answer:
Let the oxidation state of P be x:

(i)

(ii)

(iii)

(iv)

(v)

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 NaI in water.

Answer:
1) 4NaCl + MnO₂ + 4H₂SO₄→ MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂

Manganese(IV) oxide reacts with sodium chloride and sulfuric acid to produce manganese(II) chloride, chlorine, sodium bisulfate and water.

This reaction takes place at a temperature near 100°C.

2) Cl₂ + NaI → 2NaCl + I₂

Chlorine reacts with sodium iodide to produce sodium chloride and iodine.

Chlorine - diluted solution.

Sodium iodide - cold solution.

Question 33.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained?

Answer:
Xe and F₂ combine under different conditions to produce XeF₂,XeF₄,XeF₆ as follows:

Question 34.
With what neutral molecule is ClO^– isoelectronic? Is that molecule a Lewis base?

Answer:
ClO^- isisoelectronic to ClF as both the compounds contain 26 electrons in all.

ClO^- : 17+8+1 = 26

ClF : 17+9 = 26

Yes, ClF Molecule is a Lewis base as it accepts electrons from F to form ClF₃.

Question 35.
How are XeO₃ and XeOF₄ prepared?

Answer:
1) XeO₃ can be produced by hydrolysis of XeF₄ and XeF₆ under controlled pH of the medium in which reaction is taking place as shown below:

6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂

XeF₆ + 3H₂O → XeO₃ + 6HF

2) XeOF₄ can be obtained on partial hydrolysis of XeF₆ as shown below:

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.

Answer:
1) As Bond dissociation energy generally decreases on moving down the group as the atomic size of the element increases. However, among halogens, the bond dissociation energy of F₂ is lower than that of Cl₂ and F₂ due to the small atomic size of fluorine.

Thus increasing order for bond dissociation energy among halogens is as follows:

I₂<F₂<Br₂<Cl₂

2) As Bond dissociation energy of H-X molecules where X is the halogen decreases with increase in the atomic size. HI is the strongest acid as it loses H atom easily due to weak bonding between H and I.

So Increasing acid strength is as follows:

HF<HCl<HBr<HI

3) Basic strength decreases as we move from Nitrogen to Bismuth down the group as the size of the atom increases which is electron density of the atom decreases.

So Basic Strength is as follows:

BiH₃<SbH₃<AsH₃<PH₃<NH₃

Question 37.
Which one of the following does not exist?

(i) XeOF₄ (ii) NeF₂ (iii) XeF₂ (iv) XeF₆

Answer:
NeF₂does not exist as it would require d orbital for bonding which Ne does not have as it belongs to period 2 and moreover a lot of energy would be used to excite the electrons to a higher energy level as the valence electrons are very close to the nucleus thus they experience strong electrostatic attraction.

Question 38.
Give the formula and describe the structure of a noble gas species which is isostructural with:

(i) ICl^–₄

(ii) IBr^–₂

(iii) BrO^–₃

Answer:
1) XeF₄ is isoelectronic with ICl₄^-.

By isoelectronic, we mean that the two have the same number of valence electrons.

It has square planar geometry as shown below:

2) XeF₂ is isoelectronic with IBr₂^-. And the geometry is linear structure.

3) XeF₃ is isoelectronic with BrO₃^-. And the geometry is pyramidal structure.

Question 39.
Why do noble gases have comparatively large atomic sizes?

Answer:
Noble gases have comparatively large atomic sizes as their shells are completely filled with electrons and they have a stable configuration due to this completely filled electronic configuration

Electrons tend to be away from each other so as to reduce the electronic repulsion leading to the larger size of the noble gases.

The atomic radius of the noble gases is determined by vader waals radius which is comparatively greater than covalent or ionic radii.

Question 40.
List the uses of neon and argon gases.

Answer:
Uses of Argon:

1) Argon is used in gas-filled electric lamps due to its inert property.

2) It is used in laboratories to handle air-sensitive substances.

3) It is used to provide an inert temperature in a high metallurgical process.

Uses of Neon:

1) It is used in beacon lights

2) It is filled in discharge tubes with characteristic colours.

3) It is mixed with helium to protect electrical appliances from high voltage.

Intext Questions Pg-180

Question 1.
What is the basicity of H₃PO₄?

Answer:
The basicity of a compound is defined as the number of acidic hydrogen atoms present in the compound. For a hydrogen atom to be acidic, it should be bonded to a strongly electronegative element like fluorine or oxygen atom. As shown in the figure, 3 hydrogen atoms (green circled) are attached to an oxygen atom, which makes them acidic. Hence, the basicity of H₃PO₄ is 3.

Question 2.
What happens when H₃PO₃ is heated?

Answer:
H₃PO₃ on heating dissociates into orthophosphoric acid(H₃PO₄) and phosphine (PH₃). Since, the oxidation states of orthophosphorous acid (H₃PO₃), orthophosphoric acid (H₃PO₄) and phosphine (PH₃) are +3,+5 and -3 respectively, this reaction is known as a disproportionate reaction (since the oxidation states of one of the product are positive and the other is negative). The reaction is as follows.

Intext Questions Pg-170

Question 1.
Why is N₂ less reactive at room temperature?

Answer:
N₂ (Dinitrogen0 is formed by sharing three electron pairs between two nitrogen atoms, which are joined by a triple bond as shown below:

Due to a complete octet of the nitrogen atoms and bonded by strong triple bonds, due to which its bond dissociation energy is very high. Due to this reason, N₂ is very less reactive at room temperature.

Intext Questions Pg-172

Question 1.
Mention the conditions required to maximise the yield of ammonia.

Answer:
Ammonia (NH₃) is generally prepared through a many numbers of processes out of which Haber’s process is the most important one. Following the Le Chatlier’s principle and taking into considerations the reaction conditions of the Haber’s process, the yield of ammonia can be maximized by
(i) High pressure (~ 200 atm)

(ii) High temperature (~ 700K)

(iii) A mixture of Iron oxide with small amounts of K₂O and Al₂O₃ used as a catalyst. (This mixture acts as a positive catalyst in the Haber’s process)

Question 2.
How does ammonia react with a solution of Cu²⁺?

Answer:
Ammonia on reacting with Cu²⁺ acts as a Lewis base and donates its electron pair to the metal ion and forms a linkage with the metal ion.

Here, colour of Cu²⁺ is blue whereas the colour of [Cu(NH₃)₄]²⁺ is deep blue.

Question 3.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂?

Answer:
White phosphorous when heated with concentrated NaOH solution (in a CO₂ atmosphere) gives phosphine, PH₃. Another product is Sodium hypophosphite, NaH₂PO₂.

Intext Questions Pg-173

Question 1.
What is the covalence of nitrogen in N₂O₅ ?

Answer:
Diagram of N₂O₅:

Covalence of nitrogen in N₂O₅ is 4 (as can be said from the above diagram).

Note: Covalence is the number of electron pairs that an atom can share with other atoms.

Intext Questions Pg-177

Question 1.
Bond angle in PH₄⁺ is higher than that in PH₃. Why?

Answer:
Hybridisation of P in PH₃ is sp³. Out of the four sp³ hybrid orbitals, three are involved in bonding with the three H-atoms and the fourth one has a lone pair in it.
PH₃ when combines with a proton, it forms PH₄⁺ in which the lone pair is absent. So, here all the four hybrid orbitals are involved in bonding.

Now; the lone pair- bond-pair repulsion is stronger than the corresponding bond pair-bond pair repulsion; thus, the tetrahedral shape associated with sp³ bonding in PH₃ is changed to pyramidal and due to the absence of lone pair in PH₄⁺, there is no lone pair-bond pair repulsion. Hence; the bond angle in PH₄⁺ is higher than the bond angle in PH₃.

Intext Questions Pg-178

Question 1.
What happens when PCl₅ is heated?

Answer:
Generally, in CCl₄, all the C-Cl bonds are equal. But that is not in the case of PCl₅. All the bonds in PCl₅ are not similar. It has three equatorial and two axial bonds. Therefore, when PCl₅ is heated strongly, the axial bonds breakdown and thus PCl₅ decomposes to PCl₃.

Question 2.
Write a balanced equation for the hydrolytic reaction of PCl₅ in heavy water.

Answer:
The balanced equation for the hydrolytic reaction of PCl₅ in heavy water(D₂O):

Therefore, on adding the above two equation we can say that the net reaction is:

Intext Questions Pg-183

Question 1.
List the important sources of sulphur.

Answer:
Sulphur generally exists in the combined form in the following sulphate minerals like gypsum (CaSO₄.2H₂O), Epsom salt (MgSO₄.7H₂O), barite (BaSO₄) and sulphide minerals such as galena (PbS), zinc blende (ZnS), copper pyrites (CuFeS₂). Traces of sulphur are present in volcanoes as hydrogen sulphide. Sulphur is also present in organic materials such as eggs, proteins, garlic, onion, mustard, hair, and wool.

Question 2.
Write the order of thermal stability of the hydrides of Group 16 elements.

Answer:
The enthalpy of dissociation of H-E bond (E= O, S, Se, Te, Po) in the hydrides decreases down the group. Hence, the thermal stability of the hydrides decreases down the group. The thermal stability is in the order H₂O > H₂S> H₂Se> H₂Te >H₂Po.

Question 3.
Why is H₂O a liquid and H₂S a gas?

Answer:
The oxygen atom has a smaller size and greater electronegativity than sulphur atom, hence intermolecular hydrogen bonding is possible in water (H₂O). But, only weak Van der Waal’s force holds H₂S molecules, hence water is liquid and H₂S is a gas.

Intext Questions Pg-185

Question 1.
Which of the following does not react with oxygen directly?

Zn, Ti, Pt, Fe

Answer:
Pt is a noble metal (inert) and does not react with oxygen, whereas Zn, Ti, and Fe react with quickly with oxygen.

Question 2.
Complete the following reactions:

(i) C₂H₄ + O₂→

(ii) 4Al + 3 O₂→

Answer:
(i) C₂H₄ + O₂→ 2CO₂ + 2H₂O

Ethane in reaction with oxygen gives carbon dioxide and water.

(ii) 4Al + 3 O₂→ 2Al₂O₃

Aluminium on reaction with oxygen gives alumina.

Intext Questions Pg-187

Question 1.
Why does O₃ act as a powerful oxidizing agent?

Answer:
Since ozone is not very stable at room temperature; it decomposes to give oxygen molecule and nascent oxygen on heating. The nascent oxygen, being a free radical is very reactive. Hence, it acts as a powerful oxidizing agent. The equation for the above reaction is given by,

O₃→ O₂ + [O]

Question 2.
How is O₃ estimated quantitatively?

Answer:
One of the quantitative method for estimating O₃ gas is when ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This iodine can be titrated against a standard solution of sodium thiosulphate.

2I^–(aq) + H₂O(l) + O₃(g) → 2OH^–(aq) + I₂(s) + O₂(g)