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 :

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.

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

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.

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.

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.

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) H₂SO_4(aq) + H₂O_(l) → H₃O⁺_(aq) + HSO^–_4(aq) ; Ka₁ > 10
(ii) HSO^–_4(aq) + H₂O_(l) → H₃O⁺_(aq) + SO^2-_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 HSO^–₄ 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.

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 (t_1/2) 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:

Resonating structure of N₂O₅ are:

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

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.

Question 23.
Which aerosols deplete ozone layer ?
Answer:
Aerosols such as chlorofluorocarbons (CFC_s) for example freon (CCl₂F₂) depelets the O₃ layer by supplying Cl free radicals which convert O₃ to O₂ in the following sequence of reactions :

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:

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 + CO₂ ↓

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
E°_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 :
2F_2(g) + 2H₂O_(l) → 4H⁺_(aq) + 4F^–_(aq) + O_2(g)
3F_2(g) + 3H₂O_(l) → 6H⁺_(aq) + 6F^–_(aq) + O_3(g)
Cl₂ on the other hand, reacts with H₂O to form hydrochloric acid and hypochlorous acid as per the following equation:

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 :

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

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 :

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 > H₂P_o

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) + O_3(g) → 2OH^–_(aq) + I_2(s) + O_2(g)
The I₂ thus liberated is titrated against a standard solution of sodium thiosulphate using starch as an indicator

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.

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.
2SO_2(g) + O_2(g) ⇌ 2SO_3(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 Cl^–ion (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 :

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:

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.

Question 21.
How is ammonia manufactured industrially?
Answer:
Ammonia is prepared on industrial scale by Haber’s process.
N_2(g) + 3H_2(g) ⇌ 2NH_3(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

(ii) With conc HNO₃ NO₂ is evolved.

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:

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: It consists discrete tetrahedral P4 molecules.	1. Red phosphorus has the structure as given below: 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] 2s2 2p4
S(Z = 16)	[Ne] 3s2 3p4
Se(Z = 34)	[Ar] 3d10 4s2 4p4
Te(Z = 52)	[kr] 4d10 5s2 5p4
Po(Z = 84)	[Xe] 4f14 5d10 6s2 6p4

(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 O^2- 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. P_o 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 H₂P_o. 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 → O^2- as -141 and 702 kJmol^-1 respectively. How can you account for the formation of a large number of oxides having O^2- 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.


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 O^2- 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]^–
O_2(g) + PtF_6(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) + Cl_2(g) → 2NaCl_(aq) + I_2(s)

Question 35.
How are xenon fluorides XeF₂ ,XeF₄, XeF₆ obtained?
Answer:

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) ICI^–₄
(ii) IBr^–₂
(iii) BrO^–₃
Answer:
(I) XeF₄ → Square planer
(II) XeF₂ → linear structure.
(III) XeO₃ → Pyramidal structure

(I) Structure of XeF₄ :

(II). Structure of XeF₂ :

(III). Structure of XeO₃ :

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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