2nd PUC Chemistry Question Bank Chapter 7 The p-Block Elements

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:

White phosphorous	Red phosphorous
It is a soft and waxy solid. It possesses a garlic smell.	It is hard and crystalline solid.
It is poisonous	without any smell
It is insoluble in water but soluble in carbon disulphide	It is non-poisonous It is insoluble in both water and carbon
It undergoes spontaneous combustion in air	disulphide It is relatively
In both solid and vapour states, it exists as a P4 molecule	less reactive It exists as a chain of tetrahedral P4.

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: