GROUP VA & GROUP VIA ELEMENTS
FILL IN THE BLANKS
Q.01: Fill in the blanks:
(i) The elements ________ of group VA are called metalloids. (Arsenic and Antimony)
(ii) In Birkeland and Eyde’s process is prepared from atmospheric oxygen and nitrogen. (nitric oxide, NO)
(iii) The tendency to form long chain of atoms is called ________. (catenation)
(iv) All the elements of group VIA show the property of________. (allotropy)
(v) Selenium shows two allotropic forms which are called ___ forms. (red and grey)
(vi) Specific gravity of H2SO4 at 18°C is _________. (1.834)
(vii) H2 is produced by reacting H2 SO4 with metals, like_________. (Fe, Mg, Zn, Sn)
(viii) The elements of group VIA exhibit maximum oxidation state of _________. (+6)
(ix) The outermost shell of group________ elements contain six electrons.
(x) Oxygen shows ________ behaviour due to the presence of unpaired electrons. (paramagnetic)
(xi) Conc. phosphoric acid acts as a ___________. (deliquescent)
(xii) Nitrogen is a gas while other elements of the same group are _______. (solid)
(xiii) Noble metals like gold and platinum are dissolved in _______. (aqua regia)
(xiv) Sulphur is different from oxygen because it shows _________ oxidation states. (variable)
(xv) HNO3 is used in the manufacture of _________ fertilizers. (nitrogenous)
TRUE/FALSE
Q.02: Indicate True or False:
(i) The metallic character in groups VA and VIA elements increases down the group. (TRUE)
(i) The elements of group VA exhibit maximum oxidation state of +5. (TRUE)
(i) Ionization energy of phosphorus is greater than that of nitrogen. (FALSE)
CORRECTION: Ionization energy of phosphorus is lower than that of nitrogen.
(i) The electronegativity of oxygen is greater than all other elements of groups VA and VIA. (TRUE)
(i) V2 O5 is used as a catalyst for the oxidation of SO2 to SO3. (TRUE)
(i) The oxides of nitrogen are basic in nature. (FALSE)
CORRECTION: The oxides of nitrogen are acidic in nature.
(i) Aqua regia is prepared by mixing one part of conc.
(i) HNO3 with three parts of conc. HCl. (FALSE)
CORRECTION: Aqua regia is prepared by mixing 3 parts of conc. HNO3 with one part of conc. HCl.
(i) TNT is prepared by the reaction of nitric acid with toluene. (TRUE)
(i) P2O3 when reacts with cold water gives phosphorus acid and with hot water it gives phosphoric acid. (TRUE)
(i) Sulphur occurs in many organic compounds of animal and vegetable origins. (TRUE)
MCQs
Q.03: Multiple choice questions.
(i) Out of all the elements of group VA, the highest ionization energy is possessed by:
(a) N
(b) P
(c) Sb
(d) Bi
ANSWER: (a) N
EXPLANATION: Ionization energy is inversely proportional to the size of atom. The smaller the size is, the greater is the ionization energy. Since, nitrogen is the top member of group VA, having the smallest atomic size, its ionization energy will the highest among the whole group.
(ii) Among group VA elements, the most electronegative element:
(a) Sb
(b) N
(c) P
(d) As
ANSWER: (b) N
EXPLANATION: Nitrogen is the most electronegative element in group VA because it has the smallest atomic size so has maximum ability to attract the shared pair of electrons.
(iii) Oxidation of NO in air produces:
(a) N2O
(b) N2O3
(c) N2O4
(d) N2O5
ANSWER: (c) N2O4
EXPLANATION: At 600oC, NO is oxidized to give reddish-brown nitrogen dioxide (NO2) gas. 2NO + O2 ® 2NO2
(iv) The brown gas formed, when metal reduces HNO3:
(a) N2O5
(b) N2O3
(c) NO2
(d) NO
ANSWER: (c) NO2
EXPLANATION: When metals like Cu, Pb, Hg, Sn, Zn etc. react with conc. HNO3, they release a reddish-brown gas NO2 as by product.
(v) Laughing gas is chemically:
(a) NO
(b) N2O
(c) NO2
(d) N2O4
ANSWER: (b) N2O
EXPLANATION: Dinitrogen oxide or nitrous oxide is called laughing gas because when its mixture with oxygen is inhaled for a sufficient long time, it causes hysterical laughter.
(vi) Out of all the elements of group VIA, the highest melting and boiling points is shown by the element:
(a) Te
(b) Se
(c) S
(d) Pb
ANSWER: (a) Te
EXPLANATION: Tellurium has the highest melting point (450oC) and boiling point (990oC) among all the members of group VIA.
(vii) SO3 is not absorbed in water directly to form H2SO4 because:
(a) The reaction does not go to completion
(b) The reaction is quite slow
(c) The reaction is highly exothermic
(d) SO3 is insoluble in water
ANSWER: (c) The reaction is highly exothermic
EXPLANATION: SO3 is not dissolved in water to get H2SO4 because the reaction is highly exothermic and a dense acid fog is formed which does not condense easily. Therefore, SO3 is first dissolved in 98% H2SO4 to form oleum or pyro-sulphuric acid (H2S2O7). Oleum is then easily diluted by water to any required concentration.
(viii) Which catalyst is used in contact process?
(a) SO3
(b) Ag2O
(c) V2O5
(d) Fe2O3
ANSWER: (c) V2O5
EXPLANATION: Pre-heated gases at 400-500oC are passed through vertical iron columns packed with the catalyst V2O5. Here, SO2 is oxidized to SO3.
2SO2(g) + O2(g) ® 2SO3(g)
(ix) Which of the following species has the maximum number of unpaired electrons?
(a) O2
(b) O2+
(c) O2—
(d) O22—
ANSWER: (a) O2
EXPLANATION: Molecular oxygen has two unpaired electrons in its antibonding molecular orbitals p★(2py) and p★(2pz).
Q.04: (i) How does nitrogen differ from other elements of its group?
(ii) Why does aqua regia dissolve gold and platinum?
(iii) Why the elements of group VIA other than oxygen show more than two oxidation states?
(iv) Write down a comparison of the properties of oxygen and sulphur.
(v) Write down the equation for the reaction between conc. H2 SO4 and copper and explain what type of reaction is it.
ANSWERS:
(i) Differences of Nitrogen:
(1) Nitrogen is a gas while other elements of this group are solids.
(2) Nitrogen does not have allotropes while other elements of this group have allotropes.
(3) Nitrogen shows oxidation states of -3, -2,-1, +1, +3 +4 and +5, while other elements can show only +3 and +5 oxidation states.
(4) Other members of the group can make use of d-orbitals in their bonding, while nitrogen does not have d-orbitals to use in bonding. Nitrogen, therefore, cannot make five covalent bonds like others elements of the group.
Ans: (ii) “The mixture of one volume of concentrated HNO3 and three volumes of concentrated HCl is called aqua regia.” It is used to dissolve noble metals like gold and platinum. This is due to the liberation of atomic or nascent chlorine (Cl) during the reaction. This nascent chlorine is very reactive and reacts with Au and Pt to change them into their soluble chlorides.
HNO3(conc.)+3HCl(conc.) ⟶ NOCl(aq)+Cl2(g)+ 2H2O(ℓ)
NOCl ⟶ NO(g) + [Cl] (g)
Au(s) + 3[Cl] (aq) ⟶ AuCl3(aq)
Pt(s) + 4[Cl] (aq) ⟶ PtCl4(aq)
Ans: (iii) The elements of group VIA, except oxygen, show a covalency of +2, +4 and +6, like SCl2, SCl4 and SCl6. +2 oxidation state is shown due to the 2 unpaired electrons in p orbitals. +4 oxidation state is shown when one electron is promoted from p orbital to the vacant d-orbital. While +6 oxidation state is shown when another electron from s-orbital is promoted to the next vacant d-orbital. Oxygen does not have d-orbitals in valence shell, so its oxidation state is limited to only -2 & +2 (with F only).
Ans: (iv) Similarities Between Oxygen and Sulphur:
(1) Both oxygen and sulphur have same outermost electronic configuration of ns2np4.
(2) Both oxygen and sulphur are usually divalent.
(3) Both oxygen and sulphur exhibit allotropic forms.
(4) Both are typical non-metals.
Ans: (v) Reactions Between Conc. H2SO4 and Cu:
Cu(aq) + 2H2SO4(aq) ⟶ CuSO4(aq)+ 2H2O(ℓ) + SO2(aq)
The change in oxidation number of Cu from 0 to +2 in the products shows that it is an oxidation-reduction reaction in which copper reduces sulphuric acid and in return is itself oxidized.
Q.05: (a) Explain the Birkeland and Eyde’s process for the manufacture of nitric acid.
Birkland and Eyde’s Process for the Manufacture of HNO3:
N2(g)+ O2(g) ⟶ 2NO(g)
2NO(g) + O2(g) ⟶ 2NO2(g)
2NO2(g)+ H2O(ℓ) ⟶ HNO3(aq)+ HNO2(aq)
3HNO2(g) ⟶ HNO3(aq)+ 2NO(g) + H2O(ℓ)
(b) Which metals evolve hydrogen upon reaction with nitric acid? Illustrate along with chemical equations.
Magnesium, calcium and manganese react with dilute HNO3 to evolve hydrogen gas.
Mn(s) + 2HNO3(aq) ⟶ Mn(NO3)2(aq)+ H2(g)
Mg(s) + 2HNO3(aq) ⟶ Mg(NO3)2(aq)+ H2(g)
(c) What is meant by fuming nitric acid?
Fuming Nitric Acid: “A fuming liquid consisting of conc. HNO3 saturated with nitrogen dioxide (NO2) is called fuming nitric acid.” It is a red liquid, made by bubbling NO2 gas through concentrated HNO3.
Q.06: (a) Sulphuric acid is said to act as an acid, an oxidizing agent and a dehydrating agent, describe two reactions in each case to illustrate the truth of this statement.
ANSWER:
H2SO4 as Dehydrating Agent: H2SO4 is a strong acid and neutralizes alkalies or carbonates, forming salts and water.
H2SO4(aq)+ 2NaOH(aq) ⟶ Na2SO4(aq)+ 2H2O(ℓ)
H2SO4(aq)+ Na2CO3(aq) ⟶ Na2SO4(aq)+ H2O(ℓ) + CO2(g)
H2SO4 as Dehydrating Agent: H2SO4 has a great affinity for water, so it acts as a dehydrating agent and eliminates water from different compounds. e.g.,
(COOH)2 + H2SO4(conc.) ⟶ CO2(g)+ CO(g) + H2O(ℓ)
Oxalic acid
HCOOH(aq) + H2SO4(conc.) ⟶ CO(g) + H2O(ℓ)
Formic acid
H2SO4 as Oxidizing agent:
H2SO4 acts as a strong oxidizing agent. For example, it oxidizes C, H2S & HBr as:
C(s) + 2H2SO4(aq) ⟶ CO2(g)+ 2SO2(g)+ 2H2O(ℓ)
H2S(g) + H2SO4(aq) ⟶ S(s) + SO2(g)+ 2H2O(ℓ)
2HBr(aq) + H2SO4(aq) ⟶ Br2(g)+ SO2(g)+ 2H2O(ℓ)
(b) Give the advantages of contact process for the manufacture of sulphuric acid.
ANSWER:
Advantages of Contact Process:
(1) In this method, about 100% pure H2SO4 can be produced.
(2) In contact process, solid catalyst (Pt or V2O5) is used, so there is no chance of nitrogen oxide impurities as in lead chamber process.
Q.07: (a) Describe the chemistry of the industrial preparation of sulphuric acid from sulphur by the contact process.
Ans: Consult textbook at page 70—72.
(b) Why is SO3 dissolved in H2SO4 and not in water?
Ans: In contact process, SO3 is not dissolved in water to get H2SO4 because the reaction is highly exothermic and a dense acid fog is formed which does not condense easily. Therefore, SO3 is first dissolved in 98% H2SO4 to form oleum or pyro-sulphuric acid (H2S2O7). Oleum is then easily diluted by water to any required concentration.
H2SO4(aq)+ SO3(g) ⟶ H2S2O7(ℓ)
H2S2O7(ℓ)+ H2O(ℓ) ⟶ 2H2SO4(aq)
(c) Explain the action of sulphuric acid on metals along with chemical equations.
Ans: Reactions of H2SO4 with Metals:
(1) Cold dilute acid reacts with almost all metals to produce hydrogen gas and sulphate salts.
Fe + H2SO4 ⟶ FeSO4 + H2
Zn + H2SO4 ⟶ ZnSO4 + H2
Mg + H2SO4 ⟶ MgSO4 + H2
Sn + H2SO4 ⟶ SnSO4 + H2
(2) Cold concentrated H2SO4 does not react with most of the metals like Cu, Ag, Hg, Pb, Au.
(c) With certain metals hot concentrated sulphuric acid gives metal sulphates, water and SO2.
Cu(aq) + 2H2SO4(aq) ⟶ CuSO4(aq)+ 2H2O(ℓ) + SO2(aq)
2Ag(aq) + 2H2SO4(aq) ⟶ Ag2SO4(aq)+ 2H2O(ℓ) + SO2(aq)
Hg(aq) + 2H2SO4(aq) ⟶ HgSO4(aq)+ 2H2O(ℓ) + SO2(aq)
Q.08: Describe the preparation of NO2 gas. Also give its reactions.
ANSWER:
Preparation of NO2:
1) NO2 can be prepared by heating Pb(NO3)2:
2Pb(NO3)2(s) ⟶ 2PbO(s) + 4NO2(g) + O2(g)
2) NO2 can be prepared by reacting conc. HNO3 with Cu:
Cu(s)+4HNO3(conc.) ⟶ Cu(NO3)2(aq)+ 2H2O(ℓ) + 2NO2(g)
Q.09:How PCl3 and PCl5 can be used for the preparation of other chemical compounds.
Ans: PCl3 reacts with different substances to form different products:
(1) It combines with chlorine to form phosphorus pentachloride:
PCl3(l) + Cl2(g) ⟶ PCl5(s)
(2) It combines with atmospheric oxygen slowly to form phosphorus oxychloride.
2PCl3(l) + O2(g) ⟶ 2POCl3(s)
(3) It is soluble in organic solvents, but readily reacts with water to form phosphorus acid.
PCl3(l) + 3H2O(l) ⟶ H3PO3(aq) + 3HCl(aq)
(4) It reacts with alcohols and carboxylic acids forming the respective chloro derivatives and H3PO3.
3CH3OH(l) + PCl3(l) ⟶ 3CH3Cl(aq)+ H3PO3(aq)
3CH3COOH(l) + PCl3(l) ⟶ 3CH3COCl(aq)+ H3PO3(aq)
Q.10: Answer the following question:
(i) Describe “Ring test” for the confirmation of the presence of nitrate ions in solution.
ANSWER:
(i) Ring test is used to detect the presence of nitrate ions in a salt solution. Concentrated H2SO4 is added slowly to the mixture of salt solution and freshly prepared FeSO4 solution. The formation of a brown coloured ring of addition compound, FeSO4.NO, at the junction of two layers, confirms the presence of nitrate ions in the solution.
FeSO4(aq)+NO(g) ⟶ FeSO4.NO(aq)
(Brown ring)
(ii) NO2 is a strong oxidizing agent. Prove the truth of this statement giving examples.
ANSWER:
NO2 as Oxidizing Agent:
H2S(g) + NO2(g) ⟶ H2O(ℓ) + S(s) + NO(g)
2KI + 2NO2(g) ⟶ 2KNO2(aq)+ I2(s)
(iii) Write down the chemical equations and names of the products formed as a result of the reaction of HNO3 with arsenic and antimony.
As(s) + 5HNO3(aq) ⟶ H3AsO4(aq )+ 5NO2(g) + H2O(ℓ)
(Arsenic acid)
Sb(s) + 5HNO3(aq) ⟶ H3SbO4(aq) + 5NO2(g) + H2O(ℓ)
(Antimonic acid)
(iv) Give the methods of preparation of PCl3.
Preparation of PCl3:
(1) It is usually prepared by melting white phosporus in a retort in an inert atmosphere of CO2 and current of dried chlorine is passed over it. The vapours of PCl3 are collected in a flask kept in an ice-bath.
2P(s) + 3Cl2(g) ⟶ 2PCl3(l)
(2) It may also be prepared by the action of phosphorus with thionyl chloride.
2P(g) + 4SOCl2(l) ⟶ 2PCl3(l) +2SO2(G) + S2Cl2(S)
(v) P2O5 is a powerful dehydrating agent. Prove giving example.
ANSWER:
P2O5 as Dehydrating Agent:
2HNO3(aq)+ P2O5(s) ⟶ N2O5(g)+ 2HPO3(aq)
H2SO4(aq)+ P2O5(s) ⟶ SO3(g)+ 2HPO3(aq)
2CH3COOH(aq) + P2O5(s) ⟶ (CH3CO)2O(ℓ) + 2HPO3(aq)
Acetic anhydride
C2H5OH(ℓ) + P2O5(s) ⟶ C2H4(g) + 2HPO3(aq)
Q.11: Complete and balance the following chemical equation:
(i) P + NO ⟶
(ii) NO + Cl2 ⟶
(iii) H2S + NO ⟶
(iv) Pb(NO3)2 ⟶
(v) NO2 + H2O ⟶
(vi) NO2 +H2SO4 ⟶
(vii) HNO3 + HI ⟶
(viii) HNO2 + NH3 ⟶
(ix) HNO2 + CO(NH2)2 ⟶
(x) KNO3 + H2SO4 ⟶
ANSWER:
(i) 2P(s) + 5NO2(g) ⟶ P2O5(s) + 5NO(g)
(ii) 2NO(g) + Cl2(g) ⟶ 2NOCl(g)
(iii) H2S(g) + 2NO(g) ⟶ H2O(g) + N2O(g) + S(s)
(iv) 2Pb(NO3)2(s) ⟶ 2PbO(s) + 4NO2(g) + O2(g)
(v) 2NO2(g) + H2O(l) ⟶ HNO3(aq) + HNO2(aq)
(vi) H2SO4(aq) + NO2(g) + 2FeSO4(aq) ⟶ Fe(SO4)3(aq) + H2O(ℓ) + NO(g)
(vii) 2HNO3(conc.) + 6HI(aq) ⟶ 4H2O(ℓ) + 2NO(g) + 3I2(s)
(viii) HNO2(aq) + NH3(g) ⟶ NH4NO2(aq)
(ix) 2HNO2(aq) + CO(NH2)2(aq) ⟶ 2N2(g) + CO2(g) + H2O(l)
(x) KNO3(aq) + H2SO4(aq) ® KHSO4(aq) + HNO3(g)
Q.12: Describe the methods of preparation of phosphorus pentoxide and explain its reactions.
ANSWER:
Phosphorus Pentoxide, P2O5 or P4O10:
Preparation:
It is prepared by burning phosphorus in excess of dry air.
P4(s) + 5O2(g) ⟶ 2P2O5(s)
Properties of Phosphorus Pentoxide: It is a white hygroscopic powder having a faint, garlic like odour due to the presence of traces of P2O3. It sublimes at 360°C.
Reactions:
(1) With cold water phosphorus pentoxide forms metaphosphoric acid.
P2O5(s) + H2O(l) ⟶ 2HPO3(aq)
With hot water, it forms orthophosphoric acid.
P2O5(s) + 3H2O(l) ⟶ 2H3PO4(aq)
(2) It is a powerful dehydrating agent, thus, with HNO3, H2SO4, CH3COOH and C2H5OH, it gives N2O5, SO3, (CH3CO)2O and C2H4, respectively.
2HNO3(aq) + P2O5(s) ⟶ N2O5(g) + 2HPO3(aq)
H2SO4(aq) + P2O5(s) ⟶ SO3(g) + 2HPO3(aq)
2CH3COOH(aq) + P2O5(s) ⟶ (CH3CO)2O(l) + 2HPO3(aq)
2C2H5OH(aq) + P2O5(s) ⟶ C2H4(g) + 2HPO3(aq)
Q.13: Discuss the trends in physical properties of group VIA elements.
Consult textbook at page 68—69.