Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements are part of Plus Two Chemistry Chapter Wise Questions and Answers. Here we have given Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements

Plus Two Chemistry The p Block Elements One Mark Questions and Answers

Question 1.
The gas obtained by the thermal decomposition of barium azide is _________.
Answer:
Nitrogen

Question 2.
The number of replaceable hydrogen atoms in ortho phosphorous acid (H₃PO₃) is
(a) 3
(b) 2
(c) 1
(d) 0
Answer:
(b) 2

Question 3.
The stable allotrope of sulphur at room temperature is ___________ .
Answer:
Rhombic sulphur

Question 4.
Which element would readily replace oxygen from an oxide?
(a) N
(b) S
(c) F
(d) Cl
(Answer:
(c) F

Question 5.
The geometry of XeF₄ is.
Answer:
Square planar

Question 6.
The shape and hybridisation of XeF₄ molecule
Answer:
Square planar and sp³d²

Question 7.
The least stable hydride of 15^th group elements is.
Answer:
(BiH₃)

Question 8.
Choose the weak monobasic acid among the following.
(a) H₃BO₃
(b) H₃PO₂
(c) H₃PO₄
(d) HNO₃
Answer:
(a) H₃BO₃

Question 9.
The bond enthalpy is highest for ________
Answer:
H₂

Question 10.
Magnetic moment of an atom with atomic no. 24 in aqeous solution is ____________.
Answer:
4.90 B.M.

Plus Two Chemistry The p Block Elements Two Mark Questions and Answers

Question 1.
In the class, a student argued that “heterogenous” catalysis and Le-Chatlier’s principles are applied in contact process.”

1. Do you agree with this statement?
2. Justify.

Answer:
1. Yes

The reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are the favourable conditions for maximum yield according to Le Chatelier’s principle.

2. Contact process is an example of heterogeneous catalysis since the catalyst V₂O₅ is solid while the reactants are gases.

Question 2.
ClF₃ exists, but FCl₃ doesn’t, why?
Answer:
In the valence shell of Cl vacant 3d orbitals are available. Hence it can form ClF₃. But, F has no vacant d orbital to show higher oxidation state and FCl₃ is not possible.

Question 3.
In a classroom discussion, a student argued that carbon has the maximum catenation property.

1. Do you agree?
2. What do you understand by the term catenation?
3. List out some other elements which can show catenation?

Answer:

1. Yes.
2. Catenation is the self linking property of an atom. It is maximum for carbon, because C-C bond is very strong.
3. Sulphur and Phosphorus.

Question 4.
Some elements and their ores are given in table. Arrange them correctly.

Answer:

Question 5.
Analyse the statement: ‘Ionisation enthalpy of halogens decreases with a decrease in atomic size’. Is it true? Justify your answer.
Answer:
No. On moving from top to bottom in a given group, the size of the atom increases and hence ionisation enthalpy decreases.

Question 6.
Hydrides of group 15 are Lewis bases.

1. Give reason.
2. Arrange the group 15 hydrides in the decreasing order of basic strength.

Answer:

1. Due to the presence of lone pair on the central atom which is available for donation.
2. NH₃ > PH₃ > ASH₃ > SbH₃ ≥ BiH₃.

Question 7.
According to VSEPR theory assign structure to XeOF₄.
Answer:
XeOF₄ possess square pyramidal shape.

Question 8.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:

1. Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.
2. Nitrogen does not contain vacant d-orbitals in its valence shell whereas phosphorus contains vacant d-orbitals in its valence shell.
3. Nitrogen has the ability to form triple bond ( N ≡ N) as a result of which its bond enthalpy (941.4 kJ mol^-1) is very high making it less reactive.

Question 9.
Why does NH₃ form hydrogen bond but PH₃ does not?
Answer:
In NH₃, the nitrogen atom forms hydrogen bond because of the following reasons:

• Small size of nitrogen
• High electronegativity (3.0) of nitrogen

N-H bond is polar forming hydrogen bond.
P-H bond is almost purely covalent due to larger size and lesser electronegativity.

Question 10.
The HNH angle is higher than HPH, HAsH and HSbH angles. Why?
Answer:
Because in NH₃ is sp³ hybridised. Due to lone pair of electrons the bond angle contracts from 109° 28’ to 106.5°. The sp³ hybridisation becomes less and less distinct with increasing size of the central atom. Thus, the bond angle of the hydrides of group 15 decreases as.

Question 11.
Can PCl₃ act as an oxidising as well as a reducing agent? Justify.
Answer:
This is because in PCl₃ phosphorus is in the intermediate oxidising state of -3.
1. As reducing agent:
The following reactions support the reducing behaviour of PCl₃.

• PCl₃ + SO₂Cl₂ → PCl₅ + SO₂
• PCl₃ + SO₃ → POCl₃ + SO₂

2. As an oxidising agent:
It oxidises metals to their respective chlorides.

• 12Ag + 4PCl₃ → 12AgCl + P₄
• 6Na + PCl₃ → 3NaCl + Na₃P

Question 12.
Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.
Answer:
Due to its larger size (99 pm) as compared to oxygen (66 pm).

Question 13.
Ozone is a strong oxidising agent. Why?
Answer:
O₃ undergoes dissociation to form nascent oxygen. This nascent oxygen is responsible for oxidising property of ozone.

O₃ → O₂ + [O]

Question 14.
White phosphorus is more reactive than other solid phases of phosphorus. Give reason.
Answer:
This is beacuse of angular strain in the P₄ molecule where the angles are only 60°.

Question 15.
What happens when

1. Concentrated H₂SO₄ is added to CaF₂?
2. SO₃ is passed through water?

Answer:
1. It form hydrogen fluoride.
CaF₂ + H₂SO₄ → CaSO₄ + 2HF

2. It dissolves SO₃ to give H₂SO₄.
SO₃ + H₂O → H₂SO₄

Question 16.
Give two important fluorides of Xenone. Predict their structure.
Answer:

• XeF₂ – linear
• XeF₄ – Square planar
• XeF₆ – Distorted octahedral

Question 17.
Why is N₂ less reactive at room temperature?
Answer:
Due to the presence of triple bond between N atoms N₂ has high bond dissociation energy and is less reactive.

Question 18.
Arrange the following in the increasing order of thermal stability.
ASH₃, NH₃, PH₃
Answer:
Thermal stability of hydrides, decreases from nitrogen to bismuth: NH₃ > PH₃ > AsH₃.

Question 19.
The 3 allotropic forms of Pare white, red and black.

1. Name the thermodynamically most stable allotrope.
2. Which allotrope of P is stored in water?

Answer:

1. Black phosphorus
2. White phosphorus

Question 20.
Nitrogen and Phosphorus are in the same group. PCI₅ is known but NCI₅ is not known. Why?
Answer:
In phosphorus vacant d-orbitals are present. It can form pentahalides. But due to absence of d-orbitals nitrogen cannot form NCI₅.

Question 21.
Why is helium used in diving apparatus?
Answer:
Helium is used in diving apparatus because of its very low solubility in blood.

Question 22.
Why has it been difficult to study the chemistry of radon?
Answer:
Radon is radioactive with very short half-life which makes the study difficult.

Question 23.
Account for the following:
SO₃ is more covalent than SO₂.
Answer:
Due to high charge and small size of sulphur in +6 oxidation state in SO₃ it is more covalent than SO₂, in which sulphur is in +4 oxidation state.

Question 24.
I₂ is more soluble in KI than in water.
Answer:
I₂ combines with KI to form the soluble complex, KI₃.
KI + I₂ → KI₃

Question 25.
H₃PO₃ is dibasic while H₃PO₄ is tribasic.
Answer:
There are three ‘P-OH’ bonds that are ionisable in H₃PO₄. So, it is tribasic. In H₃PO₃ there are only 2 ionisable ‘P-OH’ bonds. Hence it is dibasic.

Plus Two Chemistry The p Block Elements Three Mark Questions and Answers

Question 1.
Consider the given reaction:
2A + 6SiO₂ → 6CaSiO₃ + B
B + 10C → D + 10CO

1. Identify A, B and D.
2. D is stored underwater. Why?
3. Give two examples of oxoacids of D and compounds of D.

Answer:
1. A – Ca₃(PO₄)₂
B – P₄O₁₀
D – P₄

2. Because P₄ readily catches fire in air.

3. Two examples of oxoacids of D and compounds of D

• Oxoacids of D:
  Oxoacids – Phosphorous acid(H₃PO₃), Phosphoric acid (H₃PO₄).
• Compounds of D:
  phosphine (PH₃), PCI₅, PCI₄, P₄O₆

Question 2.

1. Suggest a method for the preparation of PCI₃.
2. Why does PCI₃ fume in moist air?

Answer:
1. By passing dry chlorine overheated white phosphorus.
P₄ + 6CI₂ → 4PCI₃

2. PCI₃ hydrolyses in the presence of moisture giving fumes of HCI.
PCI₃ + 3H₂O → H₃PO₃ + 3HCI

Question 3.
A student argued that electronegativity of p-block elements decrease along period and increases down the group.

1. Do you agree with this? Explain.
2. Write about the metallic character of p-block elements.
3. Arrange the following of p-block elements in the order of decreasing oxidising power.

F₂/F-( E° =+2.85 V), Br/Br^–( E° =+1.07 V), CI₂/CI^– (E° =+1.36V), I₂/I^–( E° =+0.57V).
Answer:

1. No. electronegativity of p-block elements increases along a period and decreases down a group.
2. Most of the p-block elements are non-metallic in nature. Among p-block elements the metallic character decreases along a period and increases down a group.
3. F₂ > CI₂ > Br₂ > I₂, because the standard electrode potential values decreases in the same order.

Question 4.
The shape and hybridisation of some interhalogen compounds are given below in wrong order. Match them correctly.

Answer:

Question 5.
Account for the following:

1. H₂O is a liquid while H₂S is a gas.
2. H₂S is more acidic than H₂O
3. SF₆ is known while SH₆ is not known.

Answer:

1. There is intermolecular hydrogen bonding in H₂O molecule but there is no hydrogen bonding in H₂S .
2. The S-H bond is weaker than O-H bond because size of S atom is greater than that of O atom. Hence H₂S can dissociate to give H⁺ ions in aqueous solution.
3. In the higher oxidation state, S can combine only with highly electronegative elements like F.

Question 6.
Sulphur forms many allotropes such as a – sulphur, β -sulphur etc.

1. What do you mean by allotropy?
2. How can you convert a – sulphur to b-sulphur?

Answer:

1. Certain elements can exist in different forms with different physical property and same chemical properties.
2. b -sulphur is prepared by melting a -sulphur in a dish and cooling, till crust is obtained. Holes are then pierced into the crust, and the liquid is taken out. On removing the crust, needle shaped β -sulphur is obtained.

Question 7.
Account for the following:

1. PH₃ has lower boiling point than NH₃.
2. Pentahalides are more covalent than trihalides.
3. ICI is more reactive than I₂.

Answer:

1. NH₃, PH₃ molecules are not associated through intermolecular hydrogen bonding in liquid state.
2. Higher the positive oxidation state of central atom, more will be its polarising power which, in turn, increases the covalent character.
3. Bond energy of I-CI bond is less than that fo I-I bond.

Plus Two Chemistry The p Block Elements Four Mark Questions and Answers

Question 1.
My name is ‘X’. I am a poisonous colourless gas with the smell of rotten fish.

1. Identify ‘X’.
2. Explain the laboratory preparation of X.

Answer:

1. Phosphine(PH₃).
2. By heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO₂.

Question 2.

1. How is bleaching powder prepared?
2. Give the composition of bleaching powder.

Answer:
1. By treating Cl₂ with dry slaked lime.
2Ca(OH)₂ + 2CI₂ → Ca(OCI)₂ + CaCI₂ + 2H₂O

2. Ca(OCI)₂ CaCI₂.Ca(OH)₂.2H₂O.

Question 3.

1. Why do noble gases form compounds with fluorine and oxygen only?
2. Does the hydrolysis of XeF₆ lead to a redox reaction?

Answer:
1. Fluorine and oxygen are small atoms with high value of electronegativity. Fluorine is also highly reactive in nature. It is for this reason they form compounds with noble gases.

2. No, the products of hydrolysis are XeOF₄ and XeO₂F₂ where the oxidation states of all the elements remain the same as it was in the reacting state.

Question 4.

1. Interhalogen compounds are more reactive than halogens. Why?
2. Is there any exception to the above generalisation. Explain.

Answer:

1. This is because the bond in the interhalogen (X – X’) is weaker than X – X bond in the halogens.
2. Yes. F₂ is more reactive than interhalogen compounds. This is because the F – F bond is weaker than X – X’ bond in interhalogen compounds.

Question 5.
Nitrogen cannot extend its covalency beyond four but it forms wide variety of oxides and also forms oxoacids.

1. Name the two oxoacids of nitrogen.
2. Action of nitric acid with metals depends on many factors. Justify.

Answer:
1. HNO₂, HNO₃

2. The product of oxidation depends upon the concentration of the acid, temperature and the nature of the material undergoing oxidation, e.g.

a. 3Cu + 8HNO₃(dilute) → 3Cu(NO₃)₂ + 2NO + 4H₂O
Cu + 4HNO₃(conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

b. Zinc reacts with dilute nitric acid to give N₂O and with concentrated acid to give NO₂.
4Zn + 4HNO₃(dilute) → 4Zn(NO₃)₂ + 5H₂O + N₂O
Zn + 10HNO₃(conc.) → Zn(NO₃)₂ + 2H₂0 + 2NO₂

Some metals (e.g., Cr, Al) do not dissolve in concentrated nitric acid due to the formation of a passive film of oxide on the surface.

Question 6.

1. Arrange hydrides of group 16 in the order of increasing acidic strength.
2. Draw the structures of any three oxoacids of phosphorus and find out their basicity.

Answer:
1. H₂O < H₂S < H₂Se < H₂Te
2.

Question 7.

1. Noble gases have very low boiling points. Why?
2. XeF₂ and XeF₄ are important Xe compounds. How can we prepare them and what is the action of water on them?

Answer:
1. Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.

2. XeF₂ and XeF₄ are formed by the direct reaction of elements under the specific conditions.

They are readly hydrolysed even by traces of water,
e.g. 2XeF₂(s) + 2H₂O(l) → 2Xe(g) + 4HF(aq) + O₂(g)

Question 8.
It is greenish yellow gas with an offensive smell used in water purification. It partially dissolves in water to give a solution which turns blue litmus to red. When it is passed through NaBr solution bromine is formed.

• Identify the gas.
• Identify the group to which it belongs.
• Write the electronic configuration.
• Write the equation showing its reaction with water.

Answer:

• Chlorine
• Halogen family (17^th group)
• 1s² 2s² 2p⁶ 3s² 3p⁵
• CI₂ + H₂O → HCI + HOCI

Question 9.
Basic character of hydrides of group 15 is due to the presence of lone pair on the central atom.

1. NH₃ is strongly basic while BiH₃ is weakly basic. Why?
2. Differentiate between allotropic forms of phosphorus.

Answer:
1. Due to the presence of lone pair of electrons on nitrogen atom of NH₃. But in BiH₃, due to the large size of Bi the electron density decreases and hence is less basic.

2. The allotropic forms of P are White, Red and Black phosphorus. White phosphorus consists of tetrahedral P₄ molecules. Red phosphorus is polymeric in structure consisting of chains of P₄ tetrahedra linked together. Black phosphorus has a layer type structure and has α and β forms.

Question 10.

1. Give a method for preparation of XeO₃.
2. Deduce the molecular shape of BrF3 on the basis of VSEPR theory.

Answer:
1. XeO₃ is prepared by hydrolysis of xenon hexa fluride in presence of water.
XeF₆ + 3H₂0 → XeO₃ + 6HF

2. In BrF₃ the central atom Br is sp³d hybridised with 3 bond pairs and 2 lone pairs. According to VSEPR theory the expected geometry is trigonal bipyramidal. But due to strong Ip – Ip and Ip – bp repulsions compared to weak bp – bp repulsion it Assumes a bend T – shape as shown below

Question 11.
Account for the following:

1. BiH₃ is the strongest reducing agent among all the hydrides of group 15 elements.
2. Bleaching action of chlorine.

Answer:

1. It is because BiH₃ is least stable among the hydrides of group 15 elements.
2. Bleaching action of chlorine is due to oxidation. The nacent oxygen ([O]) produced is responsible for the bleaching action.

CI₂ + H₂O → 2HCI + [O]
Coloured substance + [O] → Colourless substance

Plus Two Chemistry The p Block Elements NCERT Questions and Answers

Question 1.

1. Noble gases have very low boiling points. Why?
2. XeF₂ and XeF₄ are important Xe compounds. How can we prepare them and what is the action of water on them?

Answer:
1. Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.

2. XeF₂ and XeF₄ are formed by the direct reaction of elements under specific conditions.

They are readly hydrolysed even by traces of water,
e.g. 2XeF₂(s) + 2H₂O(l) → 2Xe(g) + 4HF(aq) + O₂(g)

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

1. Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.
2. Nitrogen does not contain vacant d-orbitals in its valence shell whereas phosphorus contains vacant d-orbitals in its valence shell.
3. Nitrogen has the ability to form triple bond ( N ≡ N) as a result of which its bond enthalpy (941.4 kJ mol^-1) is very high making it less reactive.

Question 3.
Why does NH₃ form hydrogen bond but PH₃ does not?
Answer:
In NH₃, the nitrogen atom forms hydrogen bond because of the following reasons:

• Small size of nitrogen
• High electronegativity (3.0) of nitrogen

Due to more difference of electronegativity between N and H atom the N-H bond is polar forming hydrogen bond. On the contrary, in PH₃ the P-H bond is almost purely covalent due to larger size and lesser electronegativity (2.11) of phosphorus and hence does not form hydrogen bond.

Question 4.
The HNH angle is higher than HPH, HAsH and HSbH angles. Why?
Answer:
Because in NH₃ is sp³ hybridised. Due to lone pair of electrons the bond angle contracts from 109° 28’ to 106.5°. The decreased bond angle in other hydrides is because of the fact that the sp³ hybridisation becomes less and less distinct with increasing size of the central atom i.e., pure p-orbitals are utilised in M-H bonding or in simple words the s- orbital of H atom overlaps with orbital having almost pure p-character. Thus, the bond angle of the hydrides of group 15 decreases as

Question 5.
Can PCI₃ act as an oxidising as well as a reducing agent? Justify.
Answer:
Yes, This is because in PCI₃ phosphorus is in the intermediate oxidising state of -3.
1. As an reducing agent:
The following reactions support the reducing behaviour of PCI₃.

• PCl₃ + SO₂Cl₂ → PCl₅ + SO₂
• PCl₃ + SO₃ → POCl₃ + SO₂

2. As an oxidising agent:
It oxidises metals to their respective chlorides.

• 12Ag + 4PCl₃ → 12AgCl + P₄
• 6Na + PCl₃ → 3NaCl + Na₃P

Question 6.
Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Answer:
Oxygen atom can form hydrogen bond whereas chlorine does not. The tendency for hydrogen bonding depends upon

1. Small size and
2. High electronegativity values

Although the electronegativities of O and Cl are nearly the same yet chlorine does not form hydrogen bond due to its larger size (99 pm) as compared to oxygen (66 pm).

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