RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements

RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements

Rajasthan Board RBSE Class 12 Chemistry Chapter 7 p Block Elements

RBSE Class 12 Chemistry Chapter 7 Text Book Type Questions

RBSE Class 12 Chemistry Chapter 7 Multiple Choice Type Questions

Question 1.
Which is the most abundant element on the earth’s crust among group 15 elements ?
(a) N
(b) As
(c) P
(d) Sb

Question 2.
A brown gas is obtained when metals are reduced by nitric acid ?
(a) N2O
(b) N2O3
(c) NO2
(d) NO

Question 3.
Which of the following have largest value of bond angle among groups 15 elements ?
(a) NH3
(b) PH3
(c) AsH3
(d) BiH3

Question 4.
Which is the weakest hydrohalic acid ?
(a) HI
(b) HBr
(c) HF
(d) HCl

Question 5.
What is the geometry of XeOF2 ?
(a) Pyramidal
(b) T – Shaped
(c) Octahedral
(d) Tetrahedral

Question 6.
Which of the following has highest ionisation enthalpy ?
(a) P
(b) N
(c) As
(d) Sb

Question 7.
Which of the following oxide has highest acidic character ?
(a) P4O10
(b) SO3
(c) Cl2O7
(d) Al2O3

Question 8.
Which of the following oxo – acid has highest acidic character ?
(a) HClO4
(b) HClO3
(c) HClO2
(d) HClO

Question 9.
Which of the following is known as ‘laughing gas’?
(a) Nitrogen oxide
(b) Nitric oxide
(c) Nitrogen trioxide
(d) Nitrogen pentaoxide

Question 10.
Which of the following halogen has highest electron affinity ?
(a) F
(b) Cl
(c) Br
(d) I

Answers:
1. (a), 2. (c), 3. (a), 4. (c), 5. (b), 6. (b), 7. (c), 8. (a), 9. (a), 10. (b)

RBSE Class 12 Chemistry Chapter 7 Very Short Answer Type Questions

Question 1.
Why are penta – halides are more covalent than trihalides ?
Answer:
The elements of group 15 have five electrons (two in the s – orbital and three in p – orbital) in their respective valence shells. It is difficult to lose three electrons to form E3+ ions. It is even more difficult to lose all the five valence electron (two – s and three – p) to from E5+ ions. Therefore higher elements have no tendency to form ionic compounds. They prefer to form covalent compounds by sharing of electrons. Thus, elements in the +5 oxidation state are more covalent than in the +3 oxidation state.

Question 2.
Why is BiH3 most reducing hydride among all the hydrides of group 15 elements ?
Answer:
As we move down the group -15, the size of the element, increases and therefore the length of the E – H bond increases and its strength decreases. In other words, as we move down the group, the bond with hydrogen can break more easily to evolve H2 gas. Which acts as the reducing agent. Thus Bi-H bond is the weakest amongst the hydrides of elements of group -15, and hence, BiH3 is strong reducing agent.

Question 3.
Why nitrogen is less reactive at room temperature ?
Answer:
Due to the presence of triple bond between the two nitrogen atoms the bond dissociation energy of N2 (941.1 kJ mol-1) is very high therefore nitrogen is less reactive at room temperature. However nitrogen reacts with metals at higher temperature to form nitrides.

Question 4.
How does ammonia react with Cu2+ solution ?
Answer:
Cu2+ ions react with excess of ammonia to form a deep blue coloured complex according to the following reaction :
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 1

Question 5.
What is the valency of nitrogen in N2O3 ?
Answer:
According to the structure of N2O3. It is clear that the valeney of nitrogen in N2O5 will be 5.
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 2

Question 6.
What happens when PCl5 is heated?
Answer:
When PCl5 is heated it form PCl3 and chlorine gas.

Question 7.
Write down a balanced equation for the hydrolysis of PCl5 with heavy water.
Answer:
It reacts with heavy water to form phosphorus oxychloride (POCl3) and deuterium chloride (DCI). POCl3 reacts further with heavy water to form (DCl). POCl3
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 3

Question 8.
What is the basicity of H3PO4?
Answer:
The structure of H3PO4 molecule is given below. It has three P – OH and one P = O bond.
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 4
As it has three ionisable P – OH bonds. Therefore H3PO4 is tribasic.

Question 9.
What happens when H3PO3 is heated?
Answer:
H3PO3 on heating gives orthophosphoric acid and phosphine due to disproportionation reaction.
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 5

Question 10.
H2O is liquid whereas H2S is gas. Why?
Answer:
H2O undergoes extensive intermolecular hydrogen bonding due to greater electro negativity of O than S. So H2O exist as an associated molecules. A larger amount of energy is required to break these hydrogen bonds. Therefore H2O is a liquid at room temperature.
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 6
In contrast, H2S does not undergoes hydrogen bonding as the electro negativity difference between H and S is not appreceable. It exist as discrete molecule which are held together by weak vander Waal’s force of attraction. To break these forces of attraction, only a small amount of energy is required. Therefore. H2S is a gas at room temperature.

Question 11.
Why does O2 act as a powerful oxidising agent?
Answer:
O3 is an endothermic compound. On heating it readily decomposes to give dioxygen and nescent oxygen.

Since nascent oxygen is very reactive, therefore, O3 acts as powerful oxidising agent.

Question 12.
Why is Ka2 << Ka1 for H2SO4 in water?
Answer:
H2SO4 is dibasic acid. It ionises in two stages and hence has two dissociation constants as given below:
(i) H2SO4(aq) + H2O(l) → H3O+(aq) + HSO4(aq) ; Ka1 > 10
(ii) HSO4(aq) + H2O(l) → H3O+(aq) + SO2-4(aq) ; Ka2 = 1.2 x 10-2
Ka1 is greater than Kay, i.e., tendancy to move towards the products is greater in (i) than in (ii). This is because the negatively charged HSO4 ion has much less tendancy to donate a proton to H2O as compared to neutral H2SO4.

Question 13.
Name two poisonous gases which can be prepared from chlorine gas.
Answer:
Two poisonous gases which can be prepared from Cl2 are :
(i) Mustard gas
(ii) Phosgene
These are prepared as follows :
(i) Cl2 is passed through boiling S, when S2Cl2 is formed. This on reaction with ethylene gives mustard gas which was used in second world – war.
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 7

Question 14.
Why is ICl more reactive than I2 ?
Answer:
ICl is more reactive than I2 because I – Cl bond is weaker than I – I bond. A bond between two different atoms is always weaker than that between identical atoms. Consequently, ICl breaks easily to form halogen atoms readily bring out the reactions.

Question 15.
Why is helium used in diving apparatus ?
Answer:
Because of its low solubility (as compared to nitrogen) in blood, a mixture of oxygen and helium is used in diving kit used by deep sea divers.

Question 16.
Balance the following equation:
XeF6 + H2O → XeO2F2 + HF
Answer:
Balance reaction is as follow :
XeF6 + 2H2O2 → XeO2F2 + 4HF

Question 17.
Why has it been difficult to study the chemistry of radon?
Answer:
This is because radon is radioactive element with a short half life (t1/2) of 3.82 days. This makes the study of chemistry of radon difficult.

Question 18.
Given the resonanting structure of NO2 and N2O5.
Answer:
Resonating structure of NO2 are:
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 8
Resonating structure of N2O5 are:
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 9

Question 19.
Why does R, P = O exist but Rg N = O does not (R = alkyl group) ?
Answer:
Due to the absence of d-orbitals, nitrogen atom cannot form pπ – dπ multiple bonds. Thus, N cannot expand its covalency beyond four but in R3N = O, N has a covalency of 5. Therefore, the compound R3N = O does not exist. In contrast, due to the presence of d – orbitals P forms pπ – dπ multiple bonds and hence can expand its covalency beyond 4. Therefore, P forms R3P = O in which the covalency of P is 5.

Question 20.
Explain, why NH3 is basic while BiH3 is only feebly basic.
Answer:
Both NH3 and BiH3 have lone pair of electrons on the central atom and hence should behave as Lewis bases. But NH3 is much more basic than BiH3. This can be explained on the basis of electron density on the central atom. Atomic size of N (70 pm) is much smaller than that of Bi (148 pm), Therefore the electron density on N – atom is much higher than that on Bi-atom. Consequently, the tendency of N in NH3 to donote its pair of electrons is much higher than that of Bi in BiH3. Thus, NH3 is much more basic than BiH3.

Question 21.
Give the disproportionation reaction of H3PO3.
Answer:
H3PO3 on heating undergoes self-oxidation reduction i.e., disproportionation to form PH3 in which P is reduced and H3PO4 in which P is oxidised. Oxidation states of P in H3PO3, PH3 and H3PO4 are + 3, – 3 and
+ 5, respectively
SE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 10

Question 22.
Can PCl5 act as oxidising as well as reducing agent ? Justify.
Answer:
Phosphorus has maximum oxidation state of +5 in PCl5. It cannot increase its oxidation state further and therefore, PCl5 cannot act as reducing agent. However, it can decrease its oxidation number from + 5 to + 3 or some lower value, therefore. PCl5 act as an oxidising agent. For example, it oxidises Ag to AgCl, Sn to SnCl4 etc.
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 11

Question 23.
Which aerosols deplete ozone layer ?
Answer:
Aerosols such as chlorofluorocarbons (CFCs) for example freon (CCl2F2) depelets the O3 layer by supplying Cl free radicals which convert O3 to O2 in the following sequence of reactions :
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 12

Question 24.
Discribe the manufacture of H2SO4 by contact process.
Answer:
Manufacture of H2SO4 by contact process involves three steps:

  1. Burning of sulphur in air to generate SO2.
  2. Conversion of SO2 to SO3 by oxidation with air in the presence of V2O5 as catalyst.
  3. Absorption of SO3 in H2SO4 to obtain oleum H2S2O7. A flow digram for the manufacture of H2SO4 is given below:

RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 13

Question 25.
How is SO2 an air pollutent ?
Answer:

  1. SO2 is strongly irritating to the respiratory tract. SO2 at a concentration of 5 ppm causes throat and eyes irritation causing cough, tears and redness in eyes. It causes breathlessness and affects larynx. i.e, voice box.
  2. SO2 dissolves in rain water and produces H2SO4 which demages building materials especially marble (CaCO3).
    CaCO3 + H2SO3 → CaSO3 + H2O + CO

Question 26.
Why are halogens strong oxidising agents ?
Answer:
Due to low bond dissociation enthalpy, high electronegativity and large electron gain enthalpy, halogens have a strong tendency to accept electrons and thus get reduced.
X + e → X
In other words, halogens act as strong oxidising agents, their oxidising power, however, decreases from F2 to I2 as is evident from their electrode potentials :
E° F2/F = + 2.87 V.E°Cl2/Cl = + 1.36V
Br2/Br = 1.09V, E°I2/I = + 0.54V
Oxidizing agent order of halogens:
F2 > Cl2 > Br2 > I2

Question 27.
Write two uses of ClO2.
Answer:

  1. It is a powerful bleaching agent. Its bleaching power is about 30 times powerful than that of Cl2.
  2. It is used for bleaching flour to make white bread.
  3. ClO2 is a powerful oxidizing agent and chlorinating agent. Large quantity of ClO2 are used for bleaching wood pulp and cellulose and purifying drinking water.

Question 28.
Why are halogens coloured ?
Answer:
The colour of halogens is due to the reason that their molecules absorb light in the visible region. F2 absorbs violet light hence appears pale yellow while iodine absorbs yellow and green light and hence transmists deep violet. Similarly chlorine and bromine appear greenish yellow and orange red respectively.

Question 29.
Write the reactions of F2 and Cl2 with water.
Answer:
F2 being a strong oxidising agent oxidises H2O to O2 or O3. The reactions are given as :
2F2(g) + 2H2O(l) → 4H+(aq) + 4F(aq) + O2(g)
3F2(g) + 3H2O(l) → 6H+(aq) + 6F(aq) + O3(g)
Cl2 on the other hand, reacts with H2O to form hydrochloric acid and hypochlorous acid as per the following equation:
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 14

Question 30.
Why do noble gases have comparatively large atomic size?
Answer:
Number of electrons increases as we move from left to right in a period. The last element in each period is the noble gas. The atomic radius of noble gases are measured with the help of vander Waal’s radius as noble gases do not form molecules. On the other hand the size of other elements are measured by covalent radii. As van der Waal’s radii are larger than covalent radii hence the size of noble gases are comparatively large.

RBSE Class 12 Chemistry Chapter 7 Short Answer Type Question

Question 1. Mention the condition required to maximise the yield of ammonia.
Answer:
Ammonia is prepared by the Haber’s process. Optimum conditions for the manufacture of NH3 are as follow :
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 15
According to the Le Chatelier’s principle to maximise the yield, a high pressure of 200 x 105 Pa is used. To increase the rate of reaction, a temperature around 700 K is used and iron oxide mixed with some K2O and Al2O3 is used as a catalyst. Sometimes, Mo is also used as a promotor to increase the efficiency of the Fe catalyst

Question 2.
Bond angle in PH+4 is higher than that in PH3Why?
Answer:
P in PH3 is sp3 – hybridised. It has three bond pairs and one lone pair around P. Due to stronger lone pairbond pair repulsions than bond pair – bond pair repulsions, the tetrahedral angle decreases from 109°28′, to 93.6°. As a result. PH3 is pyramidal. However, when it reacts with proton, it forms PH+4 ion which has four bond pairs and no lone pair. Now, there are no lone pair-bond pair repulsion. Only four identical bond pair-bond pair interactions exist. PH+4 therefore assumes tetrahedral geometry with a bond angle 109°28° which is higher than in PH3
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 16

Question 3.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2 ?
Answer:
White phosphorus reacts with NaOH to form phosphine as per the following equation :
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 17

Question 4.
List the important sources of sulphur.
Answer:
Sulphur occurs in the earth’s crust in the combind state primarily in the form of sulphates and sulphides.

  • As sulphates : Gypsum, (CaSO4.2H2O), Epsom salt (MgSO4 7H2O), Baryte (BaSO4) etc.
  • As Sulphides: Galena (PbS), Zine blende (ZnS), Copper pyrites (CuFeS2), Iron pyrites (FeS2) etc.

Traces of sulphur occur as H2S and in organic matter such as eggs, proteins, onion, mustard, hair and wool etc.

Question 5.
Write the order of thermal stability of the hydrides of group-16 elements.
Answer:
As the size of the elements increases down the group, the E – H (E stands for elements of group – 16) bond dissociation energy decreases and hence E – H bond breaks more easily. Thus, the thermal stability of hydrides of group-16 elements decreases down the group:
H2O > H2S > H2Se > H2Te > H2Po

Question 6.
Which of the following does not react with oxygen directly ? Zn, Ti, Pt, Fe
Answer:
Platinum is a noble metal. Its ionisation enthalpy is very large. Therefore, it does not react with oxygen directly. In contrast, Zn, Ti and Fe are active metals and hence, directly react with oxygen to form their respective oxides :
Zn + O2 → +ZnO2
Ti + O2 → TiO2
4Fe + 3O2 → 2Fe2O3

Question 7.
Complete the following reactions :
(i) C2H4 + O2 →
(ii) 4Al + 3O2 →
Answer:
(i) C2H4 undergoes combustion to form CO2 and H2O.
C2H4 + 3O2 → 2CO2 + 2H2O
(ii) Al combines with 0, to form alumina.
4Al + 3O2 → 2Al2O3

Question 8.
How is O3 estimated quantitatively?
Answer:
When O3 is treated with excess of KI solution buffered with borate buffer (pH = 9.2) I2 is liberated quantitatively, according to the following equation :
2I(aq) + H2O(l) + O3(g) → 2OH(aq) + I2(s) + O2(g)
The I2 thus liberated is titrated against a standard solution of sodium thiosulphate using starch as an indicator
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 18
Quantitative estimation of O3 is thus carried out.

Question 9.
What happens when sulphur dioxide is passed through an aqueous solution of Fe (III) salt?
Answer:
SO2 acts as a reducing agent and reduces an aqueous solution of Fe (III) salt to Fe (II) salt, given by the equation.
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 19
This is indicated by the removal of green colour and formation of a light brown colour.

Question 10.
Comment on the nature of two S – O bonds formed in SO2 molecules. Are the two S – O
bonds in the molecule equal?
Answer:
In the structure of SO2, S is sp2 – hybridised. Two of the three sp2 – orbitals form two o-bonds while the third contains the lone pair of electrons. S is now left with one half – filled porbital and one half – filled
d – orbitals. They form one pπ – pπ and one pπ – dπ double bond with oxygen atom. Thus, SO2 has bent structure with oxygen atom. So, has bent structure with O – S – O bond angle of 119.50. Due to resonance, the two re-bonds are equal (143 pm)

Question 11.
Mention three areas in which H2SO4 plays an important role.
Answer:
Three areas in which H2SO4 plays an important role are given below:

  1. It is used as electrolyte in storage batteries.
  2. It is used in petrolium refining, detergent industry and in the manufacture of paints, pigments, dyes etc.
  3. H2SO4 is used in the manufacture of fertilizers such as ammonium sulphate, calcium super-phosphate.

Question 12.
Write the conditions to maximise the yeild of H2SO4 by contact process. Answer:
The main step in the preparation of H2SO4 is the oxidation of SO2 to SO3.
2SO2(g) + O2(g) ⇌ 2SO3(g); ∆H°f = – 196.6 kJ/mol-1 The reaction is exothermic and reversible. The forward reaction proceeds with decrease in volume. Therefore according to Le Chatelier’s principle, to maximse the yield of SO3, a low temperature (720 K), a high pressure (2 bar) are required . V2O5 is used as a catalyst.

Question 13.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy compare the oxidising power of F2 and Cl2.
Answer:
The electrode potential of F2 (+ 2.87V) is much higher than that of Cl2 (+ 1.36V) therefore, F2 is a much stronger oxidising agent than Cl2.
This can be explained as under :
Electrode potential depands upon three factors :

  1. bond dissociation enthalpy
  2. electron gain enthalpy
  3. hydration enthaply.

Although electron gain enthalpy of fluorine is less negative (- 333 kJ mol-1) than that of chlorine (- 349 kJ mol-1) the bond dissociation enthalpy of F – F bond is much lower (158.8 kJ mol-1) than that of Cl – Cl bond (246.6 kJ mol-1) and hydration enthalpy of F– ion (515kJ mol-1) is much higher than that of Clion (381 kJ mol-1).
The last two factors are more that compensate the less negative electron gain enthalpy of F2. As a result, electronde potential of F2 is higher than that of Cl2 and hence F2 is a much stronger oxidising agent than Cl2.

Question 14.
Give two examples to show the anamolus behaviour of fluorine.
Answer:

  1. Fluorine exhibits only(-1) oxidation state whereas other halogens exhibit + 1, + 3, + 5 and + 7 oxidation states also. The m.p and b.p of hydrogen halides follow the order :
    HF > HCl > HBr > HI
  2. The anamolus behaviour of fluorine is due to its
    (i) small size
    (ii) highest elctronegativity
    (iii) low F – F bond dissociation enthalpy
    (iv) non-availability of d – orbitals in its valences shell.

This is further explained as under :

  1. Due to its small size, the three lone pair of electrons on each F atom in F – F molecule, repeal the bond pair. As a result F-F bond dissociation energy is lower than that of Cl – Cl bond.
  2. Due to non-availability of d-orbitals in its valence shell, fluorine can not expand its octet. Therefore, fluorine shows only an oxidation state of – 1. All other halogens, due to presence of d-orbitals, show positive oxidation state of +1, +3, + 5 and + 7 besides oxidation state of – 1.

Question 15.
Sea water is the greatest source of some halogens. Comment.
Answer:
Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium, calcium but its main constituent is sodium chloride (2.5% by mass) dried up sea weeds contain sodium chloride and carnallite [KCI MgCl2. 6H2O]. Certain seaweeds contain up to 0.5% of iodine as sodium iodide and chili saltpetre (NaNO3) contains up to 0.2.% of sodium iodide. Thus sea is the greatest source of halogens.

Question 16.
Write the balanced chemical equation for the reaction of Cl2 with hot and concentrated NaOH. Is this reaction is disproportionation reaction ?
Answer:
Cl2 reacts with hot and conc. NaOH forms chloride and chlorate :
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 20
Yes the reaction is disproportionation reaction.

Question 17.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:
Nitrogen exists as a diatomic molecule (N = N). Due to the presence of a triple bond between the two nitrogen atoms the bond dissociation energy is quite large (941.1 kJ mol-1). As a result, nitrogen is inert and unreactive in its elemental state.
In contrast phosphorus exists as a tetraatomic molecule (P4). Since the P-P single bond is much weaker
(213 kJ mol-1) than N = N bond. Therefore, phosphorus is much reactive than nitrogen.

Question 18.
Discuss the trends in chemical reactivity of group-15 elements.
Answer:
Formation of Hydrides :
The elements of group – 15 form volatile hydrides having general formula MH3 e.g. NH3, PH3 AsH3 SbH3, BiH3 or ammonia, phosphine, arsine, stibine and bismuthine respectivily.

(A) Thermal stability:
On moving down the group, the thermal stability of hydrides change in the following order.

Hydrides NH3 PH3 AsH3 SbH3 BiH3
Temparature (k) 1573 K 673 K 733 K 423 K Very unstable

In Contrast, P and H both have an electronegativity of 2.1. Therefore, P – H bond is not polar and hence PH3 does not exhibit H – bonding.

(B) Bond angles :
The hydrides of these elements are pyramidal in shape with a lone pair of electrons in one of the orbital. On moving down the group the bond angles gradually decreases due to decrease in bond pair – bond pair repulsion.

Hydrides Bond angle
NH3 107°
PH3 93.5°
AsH3 91.8°
SbH3 91.3°
BiH3 90°

(C) Reducing Character :
Due to decrease in thermal stability, the reducing character of these hydrides gradually increases in the order.
NH3 < PH3 < AsH3 < BiH3.

Question 19.
Why does NHZ form hydrogen bond but PH3 does not?
Answer:
The electronegativity of N (3.0) is appreciable than that of H (2.1). As a result, N – H bond is quite polar and hence NH3 exhibits intermolecular H – bonding as shown below:
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 21

Question 20.
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.
Answer:
Nitrogen is prepared in the laboratory by heating on equimolar aqueous solution of ammonium chloride and sodium nitrite. The ammonium nitrite formed as a result of double decomposition reaction, decomposes to form dinitrogen.
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 22

Question 21.
How is ammonia manufactured industrially?
Answer:
Ammonia is prepared on industrial scale by Haber’s process.
N2(g) + 3H2(g) ⇌ 2NH3(g) ; ∆H° = – 46.1kJ mol-1
There are two molecules in the product side and 4 molecules in the reactant side. Thus, there is a decrease of pressure as the products are formed. Also heat is evolved when the reactants change into products. According to the Le-Chatelier’s principle, high pressure and low temperature about 200 atm and around 700 K respectivily is required. Iron oxide with small amounts of K2O and Al2O3 act as catalysts.

Question 22.
Illustrate how copper metal can give different products on reaction with HNO3.
Answer:
Copper reacts with HNO3 to give different products under different conditions with dil HNO3, NO gas is evolved
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 23
(ii) With conc HNO3 NO2 is evolved.
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 24

Question 23.
The HNH angle is higher than HPH, HASH and HSbH angles. Why?
Answer:
In all the hydrides of Group 15 elements, the central atom is sp3 – hybridised. Three of the four sp3 – orbitals form three E – H (E stands for element of Group-15) σ – bonds while the fourth contains the lone pair of electron as shown below:
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 25
Since the lone pair repulsions are stronger than the bond pair-bond pair repulsions, therefore, the bond angle decreases from 109°28 to 107.8° in NH3. As we move from N to P to As to Sb, the atomic size goes on increasing. As a result, bond pairs of electrons lie always an away form the central atom. In other words, force of repulsion between the adjacent bond pairs goes on decreasing and consequently, the bond angles keep on decreasing from NH3 to SbH3.

Question 24.
Nitrogen exists as diatomic molecule and phosphorus as P4. Why?
Answer:
Because of its small size and high electronegativity nitrogen forms Pπ – pπ multiple bonds. Therefore, it exists as a diatomic molecule having a triple bond between the two N-atoms. Phosphorus, on the other hand, has large size and lower electronegativity and usually does not form pπ – pπ multiple bonds with itself. Instead it prefers to form
P – P single bonds and hence it exists as tetrahedral, P4 molecules.

Question 25.
Write main difference between the properties of white phosphorus and red phosphorus.
Answer:
Difference between properties of White and Red phosphorus

White phosphorus Red phosphorus
1. White phosphorus has structure as given below:
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 26
It consists discrete tetrahedral P4 molecules.
1. Red phosphorus has the structure as given below:
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 27
It is polumeric, consisting chains of P4 tetrahedran linked together.
2. It is poisonous, insoluble in water but soluble in CS2. 2. It is odourless, non – poisonous insoluble in water as well as in carbon disulphide (CS2).
3. It glows in the dark. 3. It does not glow in the dark.
4. It is more reactive. It catches fire in the air to give dense white fumes of P4O10
P4 + 5O2 → P4O10
4. It is much reactive.

Question 26.
Why does nitrogen show catenation properties less than phophorus?
Answer:
The property of catenation depands upon the strength of the element-element bond. Weaker the bond, smaller is possibility of catenation. Since the N – N (159 kJ mol-1) bond energy or strength is much weaker than P – P (213 kJ mol-1) bond strength hence nitrogen show less catenation properties than phosphours.

Question 27.
Justify the placement of 0,S, Se, Te and Po in the same group of the periodic table in terms of
electronic configuration, oxidation state and formation of hydride.
Answer:
(I) Electronic configuration :
These elements contain six electrons in the valence shell, two in the s-orbital and four in the p-obrital. Hence, they are p – block elements. Their valance shell electronic configuration is ns2np4

Elements Electronic configuration
O(Z = 8) [He] 2s2p4
S(Z = 16) [Ne] 3s3p4
Se(Z = 34) [Ar] 3d10 4s4p4
Te(Z = 52) [kr] 4d10 5s5p4
Po(Z = 84) [Xe] 4f14 5d10 6s6p4

(II) Oxidation state :
The elements of group – 16 show two types of oxidation states.

(a) Positive oxidation state :
The elements of group-16 can attain noble gas configuration by sharing electrons. These elements show
+ 2, + 4 and + 6 oxidation states. Due to the absence of d-orbital in its valence shell oxygen show only + 2 oxidation state such as in OF2, hence it show as a covalent nature.
Other elements of this group have vacent d – orbitals in the valence shall, hence they show + 2, + 4 and + 6 oxidation states. On moving down the group, due to the inert pair effect the stability of + 4 oxidation states increases while that of + 6 oxidation state decreases. Generally + 4 oxidation state is more stable for Se, Te and Po while for S, + 6 oxidation state is more stable.

(b) Negative oxidation state :
The elements can also attain stable inert gas configuration by gaining two electrons. Due to high electronegativity and ionisation energy oxygen shows negative oxidation state in most of its compounds, hence all metal oxides are ionic and contain O2- ions.
However in peroxides such as Na2O2, H2O2, oxygen shows (-1) oxidation state. Due to low electronegativity and low ionisation energy, other elements of this group have less tendency to show – 2 oxidation state. On moving down the group, the tendancy to show – 2 oxidation state decreases. Po does not show – 2 oxidation state

(c) Formation of Hydrides :
The elements of this group form stable, volatile and bivalent hydrides which have the general formula M2H such as H2O, H2S, H2Se, H2 Te and H2Po. Due to the presence of two lone pairs electrons on the central atom, they have bent (V) shapes. The central atom in these hydrides is sp3 hybridised. Some characteristics of these hydrides are given below:

I. Melting and boiling points :
Order of M.P : H2O > H2Te > H2Se > H2S
Order of B.P : H2O > H2Te > H2Se > H2S

II. Covalent Character :
H2O < H2S < H2Se < H2Te

III. Volatility and Thermal stability :
From H2O to H2S, the volatility increases suddenly and then decreases from H2S to H2Te. Hence H2O is least volatile and H2S is most volatile of group-16 elements. Order of thermal stability is
H2O > H2S > H2Se > H2Te

IV. Bond angle :
Order of bond angle is
H2O > H2S > H2Se > H2Te

Question 28.
Why is dioxygen is gas but sulphur a solid?
Answer:
Due to small size and high electro negativity, oxygen atom forms pπ – pπ double bond. (O = O) The intermolecular forces in the oxygen are weak van der Waal’s forces and therefore, oxygen exists as a gas. On the other hand sulphur does not form stable pπ – pπ bonds and do not exists as S2. It is linked by single bonds and form polyatomic complex molecules having eight atoms per molecules (S8) and have puckered ring structure. Therefore Satoms are strongly held together and it exists as solid.

Question 29.
Knowing the electron gain enthalpy values for O → O and O → O2- as -141 and 702 kJmol-1 respectively. How can you account for the formation of a large number of oxides having O2- species and not O ?
Answer:
When oxygen reacts with metal then it from following types of oxides.
(i) M2O
(ii) MO
(iii) MO2
The formation of these oxides involves the following steps.
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 28
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 29
We know that the value of ∆iH2 is much higher than the value of ∆iH1 and the value of ∆egH2 is positive as compared to AegH, but when MO and MO2 compounds are formed then the lattice energy released, in case of MO it is very much higher than the lattice energy relased during the formation of MO2. It is due to the presence of higher charge on both ions in case of MO. Hence the formation of MO is thermodynamically favourable. Due to above reason a large number of oxides having O2- species are formed.

Question 30.
Explain why inspite of nearly the some electro negativity, oxygen forms hydrogen bonding while chlorine does not?
Answer:
Oxygen has smaller size than chlorine. Smaller size of oxygen favours hydrogen bonding.

Question 31.
How can you prepare Cl2 from HCl and HCI from Cl2. Write reactions only.
Answer:
(i) Oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl2 catalyst at 723 K (Deacons’s process).
4HCl + O2 → 2Cl2 + 2H2O

Question 32.
What inspired N. Bertlett for carrying out reaction between Xe and PtF6 ?
Answer:
In 1962, N. Bertlett noticed that platínum hexafluoride (PtF6) is a powerful oxidising agent which combines with molecular oxygen to form ionic compound dioxygen hexafluoroplatinate (v) O+2 [PtF6]
O2(g) + PtF6(aq) → O+2[PtF6]
This indicates that PtFo can oxidised O2 to O+2. Oxygen and Xenon, have some similarities:

  1. The first ionisation enthalpy of Xenone gas (1170 kJ mol-1) is fairly close to that of Oxygen(1166 kJ mol-1)
  2. The molecular diameter of Oxygen and atomic radius of Xenon are similar.

Question 33.
What are the oxidation states of phosphorus in the following
(i) H3PO3
(ii) PCl3
(iii) Ca3P2
(iv) Na3PO4
(v) POF3
Answer:
(i) H3PO3 : + 3
(ii) PCl3 : + 3
(iii) Ca3P3 : – 3
(iv) Na3PO4 : + 5
(v) PO F3 : + 5

Question 34.
Write the balanced equations for the following:
(a) NaCl is heated with sulphuric acid in the presence of MnO2.
(b) Chlorine gas is passed into a solution of NaI in water.
Answer:
(a) 4NaCl + MnO2 + 4H2SO4 → MnCl2 +
4NaHSO4 + Cl2 + 2H2O
(b) 2Nal(aq) + Cl2(g) → 2NaCl(aq) + I2(s)

Question 35.
How are xenon fluorides XeF2 ,XeF4, XeF6 obtained?
Answer:
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 30

Question 36.
What neutral molecule is ClO-1 isoelectronic ? Is that molecule a Lewis base?
Answer:
ClO-1 is isoelectronic with CIF. Yes it is a Lewis base.

Question 37.
Which one of the following does not exist?
(i) XeOF4
(ii) NeF2
(iii) XeF2
(iv) XeF6
Ans.
(ii) NeF2

Question 38.
Given the formula and describe the structure of a noble gas species which is isostructural with :
(i) ICI4
(ii) IBr2
(iii) BrO3
Answer:
(I) XeF4 → Square planer
(II) XeF2 → linear structure.
(III) XeO3 → Pyramidal structure

(I) Structure of XeF4 :
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 31
(II). Structure of XeF2 :
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 32
(III). Structure of XeO3 :
RBSE Solutions for Class 12 Chemistry Chapter 7 p Block Elements image 33

Question 39.
List the uses of neon and argon gases.
Answer:
(A) Uses of Neon gas :

  1. It is used in safety divices for protecting electrical istruments like voltmeters, relays, ractifires etc.
  2. Neon is used for filling sodium vapour lamps.
  3. It is used in Beacon light as safety signal for air navigaters, because its light has high penetration power.

(B) Uses of Argon gas :

  1. It is used for filling electric bulbs because of its inert nature
  2. It is used in laboratory for handling substances that are air sensistive.
  3. Pure argon is used in gas chromaography.

RBSE Class 12 Chemistry Chapter 7 Long Answer Type Questions

Question 1.
Discuss the general characteristics of group 15 elements with reference to their electronic configuation, oxidation state, atomic size, ionisation enthalpy and electro negativity.
Answer:
General characteristic of group – 15 elements :
(A) Electronic configurations :
The atoms of group 15 have five electrons in the outermost shell, two in s and three in p subshell. The general electronic configuration of this group may be expressed as ns2 np3

Elements Electronic configuration
N(Z = 7) [He] 2s2 2p3
P(Z = 15) [Ne] 3s2 3p3
As(Z = 33) [Ar] 3d10 4s2 4p3
Sb(Z = 51) [kr] 4d10 5s2 5p3
Bi(Z = 83) [Xe] 4f14 5d10 6s2 6p3

(B) Oxidation state :
The elements of group-15 have five electrons in their valence shell. They exhibit various oxidation states from – 3 to + 5.

  • The tendancy of the elements to exhibit – 3 oxidation state decreases on moving down from P to Bi due to increase in size and metallic character.
  • On moving down the group, the stability of + 5 oxidatioin state decreases while that of + 3 oxidation state increases due to inert pair effect.

(C) Atomic size:
The atomic and ionic radii of group-15 elements are smaller than the atomic radii of the elements of group 14 elements. On going down the group, the atomic radii increase with increase in atomic number.
Order of Atomic size :
N > P> As > Sb > Bi

(E) Electronegativity :
The electrogativity values of elements of group-15 are higher than the corresponding elements of group-14. On going down the group, the electronegativity value decrease.
Order of electronegativity given as :
N   > P >  As > Sb > Bi
(3.0) (2.1) (2.0) (1.9)  (1.9)

Question 2.
Arrange the following in the order of property indicated for each set :
(i) F2, Cl2, Br2, I2 : Increasing bond dissociation enthalpy.
(ii) HF, HC1, HBr, HI : Increasing acid strength
(iii) NH3, PH3, AsH3, SbH3, BiH3 : Increasing base strength.
Answer:
(i) As bond length increases, bond dissociation enthalpy decreases. Hence, bond dissociation enthalpy decreases from F2 to I2. But the bond dissociation enthalpy of F2 is lower than Cl2 and Br2. It is due to the presence of lone pair of e on fluorine atom which creates greater repulsion due to small size of fluorine. Hence the bond dissociation enthalpy of F, decreases.
I2 < F2 < Br2 < Cl2

(ii) As the atomic size increases the bond dissociation enthalpy of H – X bond decreases and the acidic strength also incerases. The incresing order of acidic strength is
HF < HCl < HBr < HI

(iii) On moving down the group from N to Bi, the atomic size increases, consequently the electron density on the central atom decreases and the basic strength decreases. Therefore, the increasing order of basic strength is as follow : BiH3 < SbH3 < AsH3 < PH3 < NH3

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