Some p-Block Elements : Notes and Study Materials -pdf
- Concepts of Some p-Block Elements
- Some p-Block Elements Master File
- Some p-Block Elements Revision Notes
- Some p-Block Elements MindMap
- NCERT Solution Some p-Block Elements
- NCERT Exemplar Solution Some p-Block Elements
- Some p-Block Elements: Solved Example 1
- Some p-Block Elements: Solved Example 2
- Some p-Block Elements : Practice Paper 1
- Some p-Block Elements : Practice Paper 2
- Some p-Block Elements : Practice Paper 3
Subtopics of Class 11 Chemistry Chapter 11 – The p-Block Elements
- Group 13 Elements: The Boron Family
- Electronic Configuration
- Atomic Radii
- Ionization Enthalpy
- Physical Properties
- Chemical Properties
- Important Trends And Anomalous Properties Of Boron
- Some Important Compounds Of Boron
- Orthoboric Acid
- Diborane, B2H6
- Uses Of Boron And Aluminium And Their Compounds
- Group 14 Elements: The Carbon Family
- Electronic Configuration
- Covalent Radius
- Ionization Enthalpy
- Physical Properties
- Chemical Properties
- Important Trends And Anomalous Behaviour Of Carbon
- Allotropes Of Carbon
- Uses Of Carbon
- Some Important Compounds Of Carbon And Silicon
- Carbon Monoxide
- Carbon Dioxide
- Silicon Dioxide, SiO2
The p-Block Elements Class 11 Notes Chemistry Chapter 11
• p-Block Elements
Elements belonging to groups 13 to 18 of the periodic table are called p-block elements. General electronic configuration: ns2 np1-6 (except for He)
• Group 13 Elements: The Boron Family
Outer Electronic Configuration: ns2np1
Atomic Radii: The atomic and ionic radii of group 13 elements are smaller than the corresponding elements of alkali and alkaline earth metals.
Reason: On moving from left to right in a period the effective nuclear charge increases and the outer electrons are pulled more strongly towards the nucleus. This results in decrease in atomic size.
On moving down the group, both atomic and ionic radii expected to increase due to the addition of a new electron shell with each succeeding element.
Exception: Atomic radius of Ga is less than that of Al due to the presence of poor shedding 10d-electrons in gallium.
Ionisation enthalpies: First ionisation enthalpies of the elements of group-13 are less than those of the elements present in group-2 in the same period.
Reason: The removal of p-electron is much easier than the s-electron and therefore, the first ionisation enthalpies (∆i H1) of the elements of group 13 are lower as compared to the corresponding elements of group 2.
On moving down the group 13 from B to Al the first-ionization enthalpies (∆i H1) decrease due to an increase in atomic size and screening effect which outweigh the effect of increased
There is discontinuity expected in the ionisation enthalpy values between Al and Ga and between In and Tl due to enability of d- and f-electrons which have low screening effect to compensate the increase in nuclear charge.
Electronegativity: Down the group, electronegativity first decreases from B to Al and then increases.
This is due to discrepancies in the atomic size of the elements.
(i) Due to strong crystalline lattice boron has high melting point. Rest of the members of this family have low melting point.
(ii) Boron is extremely hard and black coloured solid and non metallic in nature.
(iii) Other members of this family are soft metals with low melting point and high electrical conductivity.
Oxidation states: The first two elements boron and aluminium show only +3 oxidation state ~ in the compounds but the other elements of this group gallium, indium and thalium also exhibit +1 oxidation state in addition to +3 oxidation state i.e., they show variable oxidation states.
As we move down the group, the stability of +3 oxidation state decreases while that of +1 oxidation state progressively increases.
• Some Compounds of Boron
Physical properties of boric acid:
(i) It is a white crystalline solid.
(ii) It is soft soapy in touch.
(iii) It is sparingly soluble in cold water but fairly soluble in hot water.
(i) In the manufacture of heat resistant borosilicate glazes.
(ii) As a preservative for milk and food stuffs.
(iii) In the manufacture of enamels and glazes in pottery.
Structure of boric acid
(iii) Diborane, (B2H6): The series of compounds of boron with hydrogen is known as boranes.
Diborane is prepared by the reduction of boron trifluoride with LiAlH4 in diethyl ether.
4BF3 + 3LiAlH4 ——–> 2B2H6+ 3LiF + 3AlF3
Laboratory method of preparation. In laboratory diborane is prepared by the oxidation of sodium borohydride with iodine.
2NaBH4 + I2 ——-> B2H6 + 2NaI +H2
Industrial method of preparation. On industrial scale, diborane is prepared by reduction of BF3 with sodium hydride.
(i) Diborane is a colourless, highly toxic gas with a b.p. of 180 K.
(ii) Diborane catches fire spontaneously upon exposure to air.
(iii) Higher boranes are spontaneously flammable in air.
(i) Boranes are readily hydrolysed by water to form boric acid
B2H6(g) + 6H20(Z) ——> 2B(OH)3(aq) + 6H2(g)
(ii) It burns in oxygen evolving an enormous amount of heat
B2H6 + 302 —–> B203 + 3H20
(iii) Reaction with Lewis base:
Diborane on treatment with lewis bases undergo cleavage reactions to form borane which then reacts with Lewis bases to form adducts.
B2H6 + 2NMe3 ——-> 2BH3. NMe3
B2H6 + 2CO ———> 2BH3 .CO
• Group 14 Elements: The Carbon Family Group 14 includes carbon (C), silicon (Si), Germanium (Ge), tin (Sn) and lead (Pb).
General electronic configuration of carbon family is ns2np1.
Carbon: Carbon is the seventeenth most abundant element by weight in the earth’s crust.
(i) It is available as coal, graphite and diamond. In combined state it is present in metal carbonates, hydrocarbons and carbon dioxide gas (0.03%) in air.
(ii) Naturally occurring carbon contains two stable iosotopes 12C and 13C and third isotope 14C. 14C is a radioactive isotope with half life 5770 years and is used for radiocarbon dating.
Covalent radius: Covalent radius expected to increase from C to Si. From Si to Pb small increase is found.
Reason: Due to the addition of a new energy shell in each succeeding element. The increase in covalent radii from Si to Pb is small due to ineffective shielding of the valence electrons by the intervening d- and f orbitals.
Ionization Enthalpy: The first ionization enthalpies of group 14 elements are higher than those of the corresponding group 13 elements.
Reason: Because effective nuclear charge increases and size of the atoms becomes smaller. First ionization enthalpy decreases on moving down the group from carbon to tin.
The decrease is very sharp from carbon to silicon while there is slight increase in the first ionization enthalpy of lead as compared to that of tin.
Electronegativity: Group 14 elements are smaller in size as compared to group 13 elements that’s why this group are slightly more electronegative than group 13. From Si to Pb it is almost same. Small increase in ionization enthalpy from Sn to Pb is due to the effect of increased nuclear charge outweighs the shielding effect due to the presence of additional 4f- and 5d-electrons.
(i) All the elements of group 14 elements are solids. They are less metallic than group 13.
(ii) M.P. and boiling points of group 14 elements are generally high.
Carbon and silicon mostly show +4 oxidation state. Germanium forms stable compounds in +4 state and only few compounds in +2 state.
Tin forms compounds in both oxidation states. Lead forms compounds in +2 state are stable and in +4 state are strong oxidising agents.
• Anomalous Behaviour of Carbon
Carbon, differs from the rest of the member of its family. The main reason for the anomalous behaviour is:
(i) exceptionally small atomic and ionic size
(ii) higher ionization enthalpy
(iii) absence of d-orbitals in the valence shell.
(iv) Higher electronagativity.
It can be explained as follows:
=> Since carbon has only s and p-orbitals it can accommodate only four pairs of electrons ; other member can expand their covalence due to the presence of d-orbitals.
=> Carbon can form Pπ-Pπ multiple bonds with itself and other atoms having small size and high electronegativity.
Heavier elements do not form Pπ-Pπ bonds because their atomic orbitals are too
large and diffuse to have effective overlapping.
=> Carbon atoms have the tendency to link with one another through covalent bonds to form chains and rings. This property is called catenation.
Down the group property to show catenation decreases.
Lead does not show catenation.
• Allotropes of Carbon
The property of an element to exist in two or more forms which have different physical properties but identical chemical properties is called allotropy and different forms are called allotropes. Carbon exists in two allotropic forms:
(i) Crystalline (ii) Amorphous
Crystalline form of carbon: Diamond, Graphite, Fullerenes Diamond: In diamond each carbon atom undergoes sp3 hybridisation. Each carbon is tetrahedrally linked to four other carbon atoms. The C—C bond length is 154 pm.
(i) It is the hardest substance on earth.
(ii) It is used as an abrasive for sharpening hard tools in making dyes and in manufacture of tungsten filaments.
Graphite: In graphite, carbon is sp2-hybridized. Graphite has a two-dimensional sheet like structure consisting of a number of hexagonal rings fused together. Layers are held by van der Waals forces and distance between two layers is 340 pm.
(i) Graphite conducts electricity along the sheet.
(ii) It is very soft and slippery.
(iii) Used as a dry lubricant in machines running at high temperature, where oil cannot be used as a lubricant.
Fullerenes: Fullerenes was discovered collectively by three scientists namely E. Smalley, R.F. Curl and H.W. Kroto.
Fullerenes is prepared by heating of graphite in an electric arc in the presence of inert gas such as helium or argon.
The sooty material formed by the condensation of vapourised Cn small molecules consists of mainly with smaller quantity of C70 and traces of other fullerenes consisting of even number of carbon atoms up to 350 or above.
Fullerenes are cage like molecules. C60 molecule has a shape like soccer ball and called Buckminsterfullerenes. It is the most stable.
It contains 20 six-membered rings and 12 five-membered rings.
Six-membered rings are fused to both the other six-membered rings and five-membered rings but the five-membered rings are connected only to six-membered rings.
All the carbon atoms are equal and they undergo sp2-Kybridization.
(i) Fullerenes being covalent are soluble in organic solvents.
(ii) It also forms platinum complexes.
Amorphous allotropic forms of carbon coke: It is a greyish black hard solid and is obtained by destructive distillation.
Wood charcoal: It is obtained by strong heating of wood in a limited supply of air.
Animal charcoal: It is obtained by the destructive distillation of bones.
Uses of carbon:
(i) Graphite fibre are used for making superior sports goods such as tennis and badminton rackets, fishing rods.
(ii) Being good conductor graphite is used for making electrodes for batteries and industrial electrolysis.
(iii) Being highly porous, activated charcoal is used for absorbing poisonous gases in gas masks. It is used to decolourize sugar.
(iv) Carbon black is used as black pigment in black ink and as filler in automibile tyres.
(v) Coke is extensively used as reducing agent in metallurgy.
(vi) Diamond is a precious stone.
• Some Important Compounds of Carbon and Silicon
Preparation: It is prepared by direct oxidation of C in limited supply of oxygen.
(i) Carbon monoxide is a colourless and odourless gas.
(ii) It is almost insoluble in water.
(iii) It is powerful reducing agent and reduces almost all metal oxides except alkali and alkaline earth metal oxides.
(iv) In CO molecule there are one σ (sigma) and two π bonds between carbon and oxygen.
: C = O :
(v) It is highly porous in nature. It forms a complex with haemoglobin which is about 300 times more stable than the oxygen-haemoglobin complex. This prevents haemoglobin in the red blood corpuscles from carrying oxygen round the body, there by causing suffocation ultimately leading to death.
Preparation: It is prepared by complete combustion of carbon and carbon containing fuels in
(i) It is a colourless and odourless gas.
(ii) It is slightly soluble in water. When C02 dissolves in water only some of the molecules react with water to form carbonic acid.
(iii) It is not poisonous like CO.
But increase in combustion of fossil fuels and decomposition of limestone for cement manufacture increase of C02 in the atomosphere is one of the main reasons of green house effect.
Silicon dioxide (Si02)
Silicon dioxide, commonly known as silica, occurs in various crystallographic forms.
For example, Quartz, Cristobalite and thermite are some of the crystalline forms of silica.
Silicon dioxide is a covalent three dimensional network solid.
Each silicon atom is covalently bonded in a tetrahedral manner to four oxygen atoms.
Each oxygen atom in turn covalently bonded to another silicon atoms as shown below:
(i) In normal form silica is very less reactive.
(ii) At elevated temperature it does not reacts with halogens, dihydrogen and most of the acids and metals. But it reacts with HF and NaOH.
Si02 + 2NaOH —–> Na2Si03 + H2O
Si02+ 4HF ——–> SiF4+ 2H20
(i) Quartz is extensively used as a piezoelectric material.
(ii) Silica gel is used as adsorbent in chromatography.
(iii) An amorphous form of silica, kieselghur is used in filtration plants.
• P-Block elements: Contains, metals, non-metals and metalloids.
• General configuration: ns2np1-6
– Boron is a typical non-metal and the other members are metals.
– Boron halides are considered to behave like Lewis acids.
– Boric acid is a Lewis acid.
– Borax is a white crystalline solide formula is Na2 [B4O5(OH)4] . 8H20
– Aluminium exhibits +3 oxidation state.
– Allotropy: The important allotropes of carbon are diamond, graphite, and fullerenes.
– The members of carbon family exhibit +4 and +2 oxidation state. The tendency to show +2 oxidation state increases among heavier elements.
– Lead in +2 state is stable whereas in +4 oxidation state it is a strong oxidising agent.
– Carbon monoxide is neutral whereas C02 is acidic in nature.
– Carbon monoxide having lone pair of electrons on C forms metal carbonyls.
– Carbon monoxide forms a haemoglobin complex which is deadly poisonous due to its higher stability.
– Zeolites are complex aluminium silicates.
1 Marks Questions
1.How many groups are there in p-block?
Ans. There are six groups of p-block elements in the periodic table numbering from 13 to 18.
2.What is ‘inert pair effect’?
Ans. The occurrence of oxidation states two unit less than the group oxidation states are sometimes attributed to the ‘inert pair effect’.
3.How does metallic and non-metallic character vary in a group?
Ans. The non-metals and the metals exist only in the p-block of the periodic table. The non-metallic character of elements decreases down the group. In fact the heaviest element in each p-black group is the most metallic in nature.
4.Why do third – period elements expand their covalence above four?
Ans. The third – period elements of p-groups included d-orbital, which can be utilized to form bond and expand octet.
5.Why do heaver elements form π – bonds?
Ans. The heavier elements of p-block elements forms π – bonds because of the combined effect of size and availability of d-orbital’s considerably influences the ability of there elements to form π – bonds.
6.Where does metalloids and non – metals exist?
Ans. It is interesting to note that the non-metals and metalloids exist only in the p-block of the periodic labels.
7.Give the chemical formula of inorganic benzene.
8.Give two examples of electron deficient molecules.
Ans. BF3, B2H6.
9.Arrange the following halides of boron in the increasing order of acidic character: BF3, BCl3, BBr3, BI3.
Ans. BF3 < BCl3 < BBr3 <BI3.
10.Why is boric acid considered as a weak acid?
Ans. Boric acid is not able to release H+ ions on its own. It receive OH– ions from water molecules to complete its octet and in turn release H+ ions.
11.Why is boron metalloid?
Ans. Because, boron resembles both with metals and non-metals, therefore boron is metalloid.
12.Why do boron have unusual high melting point?
Ans. Due to very strong crystalline lattice, boron has unusually high melting point.
13.Why does BF3 act as Lewis acids?
Ans. Boron in its halides has only six electrons in its valence shell. Therefore, it can accept a pair of electrons from any electron-rich molecule. Therefore, it acts as an electron – acceptor and called Lewis acid.
14.What is the electronic configuration of Group -14 elements?
Ans. The electronic configuration is ns2np2.
15.Name the metalloid found in Group 14 element?
Ans. Germanium is a metalloid found in group – 14.
16.Which of the following reacts with water and aqueous solution becomes acidic: SiCl4 or CCl4?
Ans. Si Cl4.
17.Why CCl4 behaves as an electron precise molecule?
Ans. Carbon in CCl4, the number of electrons around the central atom in a molecule is eight and thus is electron precise molecule.
18.Why is lead unaffected by water?
Ans. Lead is unaffected by water, probable because of a protective oxide film formation.
19.What is the common name of recently developed allotrope of carbon i.e. C60 molecule?s
20.How are fullerenes obtained?
Ans. Fullerenes are made by the heating of graphite in an electric arc in the presence of inert gases such as helium or argon.
21.Diamond is the hardest substance known. Why?
Ans. Diamond is the hardest substance on the earth because it is very difficult to break extended covalent bonding.
22.What is water gas?
Ans. The mixture of CO and H2 is known as water gas or synthesis gas.
23.Silicon dioxide is treated with hydrogen fluoride. Explain?
24.What are silicones?
Ans. Simple silicones consists of chains in which alkalyl or phenyl groups
occupy the remaining bonding position on each silicon. They are hydrophobic in nature.
25.What is dry ice?
Ans. Solid CO2 is known as dry ice.
26.What are silicates?
Ans. The structural unit of silicates is SiO44- in which silicon atom is bonded to four oxygen atoms in tetrahedron fashion.
27.Write the resonance structures of carbon dioxide.
Resonance structures of carbon dioxide.
28.What is silica-gel used as?
Ans. Silica gel is used as a drying agent and as a support for chromato graphic materials and catalysts.
29. Write the elements of group 15?
Ans. The elements of group 15 are Nitrogen (N), Phosphorous (P) , Arsenic (As) , Antimony (Sb) and Bismuth (Bi).
30. Write chemical name & formulae of
a) Chile saltpetre
b) Indian saltpetre
Ans. (a) Chile saltpetre – Sodium nitrate –
(b) Indian saltpetre – Potasium nitrate –
31. What is special about the valence configuration of Group 15?
Ans. The valence configuration of 15 group is the s-orbital is completely filled and p-orbital is half filled . This half filled orbital gives extra stability to elements of this group.
32. The atomic radii increases considerably from N to P but very little increase is observed from As to Bi. why?
Ans. There is a considerable increase in size from N to P as expected but due to the presence of completely filled d- orbitals which have very poor shielding effects, the increases in size is very little from As to Bi.
33. Give reason for the following- the first ionization enthalpy of 15th group elements is higher than 16th group elements ?
Ans. Due to extra stability of half filled configuration, the first Ionisation enthalpy of 15th group elements is higher than 16th group configuration
34. How does metallic character vary down the 15 group & why?
Ans. The metallic character increases down the group due to decrease in ionization enthalpy and increase in size of atom.
35. What are the common oxidation states of this group?
Ans. The common oxidation states of the group are -3, +3 & +5.
36. What is the maximum covalence shown by N?
Ans. Nitrogen shows a maximum covalence of +4 because only four orbitals, one S and three P- orbitals are available for bonding in Nitrogen.
37. Bi (v) is a stronger oxidizing agent than Bi(III). Why?
Ans. Bi is more stable in +3 oxidation state in comparison to +5 due to inert pair effect therefore Bi (v) has a strong tendency to act as oxidizing agent.
38. Give an example showing disproportionation of oxidation state of nitrogen?
Here Nitrogen is getting oxidized to a higher oxidation state as well as reduced to a lower oxidation state.
39. Complete and balance –
40. What are the optimum conditions for maximum yield of ammonia?
Ans. The optimum conditions for the production of ammonia are Pa or 200 atm pressure, 700K temperature , and presence of catalyst such as iron oxide with and as promoters.
41. Ammonia is a Lewis base. Why?
Ans. Due to the presence of lone pairs on nitrogen atom of ammonia, it can donate electron pair and acts as a lewis base.
42. Ammonia has higher boiling and melting points than expected. Why?
Ans. In solid and liquid states, ammonia molecules are associated by intermolecular hydrogen bonding. There fore ammonia has higher boiling and melting points.
43. Give reasons-
a) Halogens have smallest atomic radii in their periods
b) The negative electron gain enthalpy of fluorine is less than that of chlorine.
c) All halogens are coloured.
d) The only possible oxidation state of fluorine -1.
e) Fluorine forms only one oxoacid.
f) The stability of hydrides follows the order HF>HCl>HBr>HI.
Ans. (a) Due to maximum effective nuclear charge, halogens have smallest atomic radii.
(b) Due to small size of fluorine atom, there are strong inter electronic repulsions in the small 2p orbital of fluorine and thus incoming electron does not experience much attraction and fluorine has less negative electron gain enthalpy than that of chlorine.
(c) Halogens absorb radiation in visible region which results in oxidation of electrons to higher energy level by absorbing different quanta of radiation, they show different colours.
(d) Since fluorine is most electronegative element and is short of only one electron for completing octet, it shows the only oxidation state of -1.
(e) Due to small size and high electro-negativity, Fluorine forms only one hypohalous acid.
(f) As the size of element increases down the group, the bond dissociation enthalpy for HX bond decreases making the bond weaker and weaker therefore the order of thermal stability is HI < HBr < HCl < HF.
44. Complete and balance-
45. Chlorine water on standing loses its yellow colour. Why?
Ans. On standing chlorine water forms HCl and Hypochlorous acid (HOCl) due to which it loses its colour.
46. Write two uses of chlorine?
Ans. Chlorine is used for
a) Bleaching wood pulp, cotton and textiles
b) Manufacturing dyes, drugs, refrigerants etc.
c) Sterilizing drinking water.
47. Give reasons for the following?
(a) Nitrogen does not show catenation.
(b) exists but does not.
(c) The stability of Hydrides follows the order-
(d) PH3 is a weaker base than.
(e) Molecular nitrogen is chemically inert.
Ans. (a) Nitrogen being small in size has high electron density. Due to strong inter electronic repulsions, N-N single bond is weak & nitrogen does not undergo catenation.
(b) Due to absence of d-orbitals, nitrogen cannot expand its oxidation state to +5 and NCl5 does not exist whereas in P due to presence of empty 3d orbital +5 oxidation state is attained.
(c) As we move down the group 15, atomic radii increases making the bond of element with Hydrogen weaker this decreases the stability of hydrides of heavier elements. Therefore the order of stability is.
(d) As Phosphorous atom is larger than N- atom, the lone pair of electrons is distributed over a large surface area of P-atom than N-atom. Therefore the tendency of P to donate the lone pair of electrons is less.
(e) Molecular nitrogen is inert because N N bond energy is very high due to small size of N- atom and presence of multiple bond.
48. Complete and balance-
49. What are the two isotopes of nitrogen? Write two uses of dinitrogen.
Ans. The two isotopes of nitrogen are .
50. Write the characteristics of pure Ozone?
Ans. Pure ozone is pale blue gas, dark blue liquid and violet black solid.
24. At what concentration ozone is harmful?
Ans. If the concentration of ozone increases above 100 ppm, breathing becomes uncomfortable resulting in headache and nausea and it becomes harmful.
51. Draw the resonating structures of ozone?
52. Explain the oxidizing action of ozone?
Ans. Ozone has a very strong tendency to liberate nascent oxygen according to the equation:-
Therefore it acts as a strong oxidizing agent.
53. How is ozone estimated quantitatively?
Ans. When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (PH = 9.2) iodine is liberated. This iodine can be titrated against a standard solution of sodium thiosulphate to estimate the amount of ozone.
54. Complete and balance:-
55. Give a test to detect the presence of gas?
Ans. When (g) is passed through a violet coloured acidified potassium permanganate solution, it gets decolourised.
56. Which aerosols and oxides deplete ozone layer
Ans. Freon used in aerosol sprays deplete the ozone layer. Nitrogen oxides emitted from the exhaust systems of supersonic jet aeroplanes deplete the ozone layer.
57. Write the members of 16 Group.
Ans. The elements of group 16 are: Oxygen(O), Sulphur (S), Selenium (Se), Tellurium (Te) and Polonium (Po)
58. Give the general electronic configuration of 16 Group.
Ans. The general electronic configuration of 16 group is ns2np4.
59. Discuss the geometry of.
Ans. In the hybridisation of sulphur is. The structure is trigonal bi-pyramidal in which one of the equatorial positions is occupied by a one pair of electrons. The geometry is called see-saw geometry.
60. Complete and balance-
61. Enlist some uses of dioxygen.
Ans. Oxygen is used-
(a) In oxyacetylene welding.
(b) In manufacture of steel.
(c) In hospitals and mountaineering as oxygen cylinders.
62. Write different isotopes of oxygen.
Ans. The isotopes of oxygen are .
63. Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe
Ans. Pt is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite reactive. Hence, oxygen does not react with platinum (Pt) directly.
64. Why is ICl more reactive than ?
Ans. ICl is more reactive than because I-Cl bond in ICl is weaker than I-I bond in .
65. Balance the following equation:
Ans. Balanced equation:
66. How is ammonia manufactured industrially?
Ans. Ammonia is prepared on a large-scale by the Haber’s process.
67. Why does exist but does not (R = alkyl group)?
Ans. N(unlike P) lacks the d-orbital. This restricts nitrogen to expand its coordination number beyond four. Hence, does not exist.
68. Explain why fluorine forms only one oxoacid, HOF.
Ans. Fluorine forms only one oxoacid i.e., HOF because of its high electronegativity and small size.
69. Which one of the following does not exist?
Ans. does not exist.
70. Explain the bleaching action of chlorine?
Ans. The bleaching action of chlorine is due to its tendency to give nascent oxygen so that the substance gets oxidized.
Coloured substances + [O] colourless substance.
2 Marks Questions
1.Why the elements of group 13 are called p-block elements?
Ans.Group 13 elements are called p-block elements because the last electron is present in the p-orbital (np1). The valence shell configurations are B (2s2 2p1), Al (3s2, 3p1), Ga (4s2, 4p1), In (5s2 5p1) Tl (6s2 6p1)
2.The elements B, Al, Ca, In and Tl are placed in the same group of the periodic table. Give reason.
Ans. The elements B, Al, Ga, In and Tl are placed in the same group of the periodic table because each one has the same number of electrons (ns2 np1) in its valance shell.
3.Aluminium forms in not formed why?
Ans. Due to presence of vacant d-orbital’s, Al can expand its octet to form bonds with six fluoride ions whereas B cannot. Boron does not have d-orbital’s.
4.The atomic radius of Ca is less than that of Al. Why?
Ans. This is due to the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium.
5.C and S are always tetravalent but Ge, Sn And Pb show divalency. Why?
Ans. Inert pair is more prominent as we move down the group in p – block elements. Ge, Sn and Pb show divalency due to inert pair effect.
6.Some halides of group 14 elements form complexes of the type . Give reason.
Ans. The halides of the elements having vacant d-orbital’s can form complexes like because in such a case the central atom can increase its coordination number from 4 to 6 due to availability of vacant d–orbital’s.
7.is lead known whereas not. Give reason.
Ans. The main reasons are that
(i) Six large chloride ions cannot be accommodated around Si4+ due to limitation of its size.
(ii) Interaction between lone pair of chloride ion and Si4+ is not very strong.
8.PbI4 does not exist. Why?
Ans.PbI4 does not exist because Pb – I bond initially formed during the reaction does not release enough energy to unpair 6s2 electrons and excite one of them to higher orbital to have four unpaired electrons around lead atom.
9.Why is carbon different from other member of the group?
Ans. Carbon differs from rest of the members of its group due to its smaller size, higher electro negativity, higher ionization enthalpy and unavailability of d-orbital’s.
10.Why does the covalence of carbon not expand beyond four?
Ans. In carbon, only s and p orbital’s are available for bonding and therefore it can accommodate only four pairs of electrons around it. This limit the maximum covalence to four whereas other members can expand their covalence due to the presence of d-orbital’s.
11.Why does carbon show different allotropic forms?
Ans. Due to property of catenation and pπ – pπ bond formation Carbon is able to show different allotropic forms.
12.Silicon has no allotropic form analogous to graphite. Why?
Ans. Due to large size. Si has little or no tendency for pπ – pπ bonding. Whereas carbon atom forms easily pπ – pπ bonds due to smaller size in graphite structure. Hence, Si does not exhibit graphite structure.
13.Why does graphite conduct electricity?
Ans. Graphite forms hexagonal ring and undergoes sp2 hybridization. The electrons are delocalized over the whole sheet. Electrons are mobile and therefore graphite conducts electricity over the sheet.
14.Graphite is used as lubricant. Give reason.
Ans. Graphite has sp2 hybridized carbon with a layer structure due to wide separation and weak inter – layer bonds the two adjacent layers can easily slide over each other. This makes graphite act as a lubricant.
15.How are silicones manufactured?
Ans.They are manufactured by hydrolysis of chlorosilanes –
where R is a methyl or phenyl group.
16.Why does CO2 have a linear shape with no dipole moment
Ans. In CO2 molecule carbon atom undergoes sp hybridization. Two sp hybridized orbital of carbon atom overlap with two p-orbital’s of oxygen atoms to make two sigma bonds while other two electrons of carbon atom are involved in pπ – pπ bonding with oxygen atom. This results in its linear shape [with both c-o bond of equal length (115 pm)] with no dipole moment.
17. Complete and Balance-
Ans. Complete and Balance-
18. All five bonds in are not equal. Give an equation in support of this statement.
Ans. When heated, loses a chlorine molecule this shows that two P- CL bonds are weaker and hence longer than others.
19. Draw the structure of following :-
20. Write two uses of dinitrogen.
Ans. Dinitrogen is used
(a) In the manufacture of ammonia.
(b) As a refrigerant to preserve biological material, food items.
(c) In cryosurgery.
21. Explain the chemistry behind brown ring test for detection of nitrate ions.
Ans. The brown ring test for nitrate ions depends on the ability of to reduce nitrates to nitric oxide, which reacts with Fe2+ to form a brown coloured complex.
22. Give the structures of
(i) sulphurous acid and
(ii) Peroxodisulphurous acid?
Ans. (i) Sulphurous acid
(ii) Peroxodisulphuric acid
23. Write the various steps for preparation of sulphuric acid by contact process?
Ans. Contact process for sulphuric acid:-
Step 1: Burning of sulphur in air to give.
Step 2: Conversion of by reacting it with oxygen in presence of.
Step 3: Absorption of in to give of oleum
Step 4: Dilution of oleum with water to get of desired concentration
24. Name different sulphates formed by sulphuric acid?
Ans. The two type of sulphates are –
(i) Normal sulphate eg.
(ii) acid sulphate eg..
25. Why are pentahalides more covalent than trihalides?
Ans. In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides.
26. Why is the strongest reducing agent amongst all the hydrides of Group 15 elements?
Ans. As we move down a group, the atomic size increases and the stability of the hydrides of group 15 elements decreases. Since the stability of hydrides decreases on moving from to , the reducing character of the hydrides increases on moving from to .
27. Why is less reactive at room temperature?
Ans. The two N atoms in are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, is less reactive at room temperature.
28. How does ammonia react with a solution of ?
Ans.acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion.
29. What is the covalence of nitrogen in?
From the structure of , it is evident that the covalence of nitrogen is 4.
30. What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of ?
Ans. White phosphorous dissolves in boiling NaOH solution (in a atmosphere) to give phosphine, .
31. Write a balanced equation for the hydrolytic reaction of in heavy water.
Ans. All the bonds that are present in are not similar. It has three equatorial and two axial bonds. The equatorial bonds are stronger are stronger than the axial ones. Therefore, when is heated strongly, it decomposes to form .
32. What happens when is heated ?
Therefore, the net reaction can be written as
33. What is the basicity of ?
Since there are three OH groups present in , its basicity is three i.e., it is a tribasic acid.
Concept Insight: Basicity is the Number of hydrogen that are replacable.
34. List the important sources of sulphur.
Ans. Sulphur mainly exists in combined form in the earth’s crust primarily as sulphates [gypsum (), Epsom salt , baryte ] and sulphides [(galena (PbS), zinc blends (ZnS), copper pyrites ].
35. Write the order of thermal stability of the hydrides of Group 16 elements.
Ans. The thermal stability of hydrides decreases on moving down the group. This is due to a decrease in the bond dissociation enthalpy (H-E) of hydrides on moving down the group.
36. Why is a liquid and a gas?
Ans. has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in , which is absent in . Molecules of are held together only by weak van der Waal’s forces of attraction.
Hence, exists as a liquid while as a gas.
37. Complete the following reactions:
Therefore, ozone acts as a powerful oxidising agent.
38. Why does act as a powerful oxidizing agent?
Ans. Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free readical, is very reactive.
Therefore, ozone acts as a powerful oxidizing agent.
39. What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?
Ans. acts as a reducing agent when passed through an aqueous solution containing Fe(III) salt. It reduces Fe(III) to Fe(II) i.e., ferric ions to ferrous ions.
40. Give two examples to show the anomalous behaviour of fluorine.
Ans. Anomalous behaviour of fluorine
(i) It forms only one oxoacid as compared to other halogens that form a number of oxoacids.
(ii) Ionisation enthalpy, electronegativity, and electrode potential of fluorine are much higher than expected.
41. Sea is the greatest source of some halogens. Comment.
Ans. Sea water contains chlorides, bromides, and iodides of Na, K, Mg, and Ca. However, it primarily contains NaCl. The deposits of dried up sea beds contain sodium chloride and carnallite, . Marine life also contains iodine in their systems. For example, sea weeds contain upto 0.5% iodine as sodium iodide. Thus, sea is the greatest source of halogens.
42. Give the reason for bleaching action of .
Ans. When chlorine reacts with water, it produces nascent oxygen. This nascent oxygen then combines with the coloured substances present in the organic matter to oxide them into colourless substances.
Coloured substances + [O]’ Oxidized colourless substance
43. Name two poisonous gases which can be prepared from chlorine gas.
Ans. Two poisonous gases that can be prepared from chlorine gas are
(ii) Mustard gas
44. Why has it been difficult to study the chemistry of radon?
Ans. It is difficult to study the chemistry of radon because it is a radioactive substance having a half-life of only 3.82 days. Also, compounds of radon such as have not been isolated. They have only been identified.
45. Why is helium used in diving apparatus?
Ans. Air contains a large amount of nitrogen and the solubility of gases in liquids increases with increase in pressure. When sea divers dive deep into the sea, large amount of nitrogen dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and it separates from the blood and forms small air bubbles. This leads to a dangerous medical condition called bends. Therefore, air in oxygen cylinders used for diving is diluted with helium gas. This is done as He is sparingly less soluble in blood.
46. Why does the reactivity of nitrogen differ from phosphorus?
Ans. Nitrogen is chemically less reactive. This is because of the high stability of its molecule,. In, the two nitrogen atoms form a triple bond. This triple bond has very high bond strength, which is very difficult to break. It is because of nitrogen’s small size that it is able to form bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen.
47. Why does form hydrogen bond but does not?
Ans. Nitrogen is highly electronegative as compared to phosphorus. This causes a greater attraction of electrons towards nitrogen in than towards phosphorus in . Hence, the extent of hydrogen bonding in is very less as compared to .
48. How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.
Ans. An aqueous solution of ammonium chloride is treated with sodium nitrite.
NO and HNO3are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.
49. Why does nitrogen show catenation properties less than phosphorus?
Ans. Catenation is much more common in phosphorous compounds than in nitrogen compounds. This is because of the relative weakness of the N-N single bond as compared to the P-P single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N-N single bond.
50. Give the disproportionation reaction of .
Ans. On heating, orthophosphorus acid disproportionates to give orthophosphoric acid and phosphine . The oxidation states of P in various species involved in the reaction are mentioned below.
51. Which aerosols deplete ozone?
Ans. Freons or chlorofluorocarbons (CFCs) are aerosols that accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorine-free radicals that combine with ozone to form oxygen.
52. Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Ans. Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.
53. Write two uses of .
Ans. Uses of :
(i) It is used for purifying water.
(ii) It is used as a bleaching agent.
54. Why are halogens coloured?
Ans. Almost all halogens are coloured. This is because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour.
55. Write the reactions of and with water.
56. How can you prepare from HCl and HCl from ? Write reactions only.
Ans.(i) can be prepared from HCl by Deacon’s process.
(ii) HCl can be prepared from Cl2on treating it with water.
57. What inspired N. Bartlett for carrying out reaction between Xe and?
Ans. Neil Bartlett initially carried out a reaction between oxygen and . This resulted in the formation of a red compound, .
Later, he realized that the first ionization energy of oxygen (1175 kJ/mol) and Xe (1170 kJ/mol) is almost the same. Thus, he tried to prepare a compound with Xe and . He was successful and a red-coloured compound, was formed.
58. Write balanced equations for the following:
(i) NaCl is heated with sulphuric acid in the presence of .
(ii) Chlorine gas is passed into a solution of NaI in water.
59. With what neutral molecule is isoelectronic? Is that molecule a Lewis base?
Ans. is isoelectronic to ClF. Also, both species contain 26 electrons in all as shown.
Total electrons = 17 + 8 + 1 = 26
In ClF = 17 + 9 = 26
ClF acts like a Lewis base as it accepts electrons from F to form .
60. How are and prepared?
Ans. (i) can be prepared in two ways as shown.
(ii) can be prepared using .
61. Why do noble gases have comparatively large atomic sizes?
Ans. Noble gases do not form molecules. In case of noble gases, the atomic radii corresponds to van der Waal’s radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, van der Waal’s radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period.
3 Marks Questions
1.Why carbon does not form ionic compounds?
Ans.The electronic configuration of carbon atom is 1s2 2s2 2px1 2py1 and has four valence electrons. In order to form ionic compound, it has to either lose four electrons or gain four electrons. Since very high energy are involved in doing so. Carbon does not form ionic compounds. It completes its octet by sharing of electrons and forms covalent compounds.
2.Why does the heavier elements do not form pπ – pπ multiple bond as carbon do?
Ans.Carbon has the unique ability to form pπ – pπ multiple bond with itself and with other atoms of small size and high electro negativity whereas heavier elements do not from pπ – pπ bonds because their atomic orbital’s are too large and diffuse to have effective overlapping.
3.Why is CO considered poisonous ?
Ans. The highly poisonous nature of CO arises because of its ability to form a complex with haemoglobin which is about 300 times more stable than the oxygen – haemoglobin complex. This prevents haemoglobin in the red blood corpuscles from carrying oxygen round the body and ultimately resulting in death.
4. Write three differences between white and red phosphorous.
|White phosphorous||Red phosphorous|
|1. It is translucent waxy solid||1. It has iron grey Lustre.|
|2. It is poisonous and glows in dark||2. It is non poisonous and does not glow in dark.|
|3. It is less stable and more reactive.||3. It is more stable and less reactive.|
5. Why does dimerise?
Ans. contains odd number of valence electrons. It behaves as a typical odd molecule. On demerisation, it is converted to stable molecule with even number of electrons.
6. Mention the conditions required to maximise the yield of ammonia.
Ans. Ammonia is prepared using the Haber’s process. The yield of ammonia can be maximized under the following conditions:
(i) High pressure
(ii) A temperature of
(iii) Use of a catalyst such as iron oxide mixed with small amounts of and
7. What happens when H3PO3is heated?
Ans. ,on heating, undergoes disproportionation reaction to form and. The oxidation numbers of P in ,, and are +3, – 3, and +5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction.
8. How is estimated quantitatively?
Ans. Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate using starch as an indicator. The reactions involved in the process are given below.
9. How is the presence of detected?
Ans. is a colourless and pungent smelling gas.
It can be detected with the help of potassium permanganate solution. When is passed through an acidified potassium permanganate solution, it decolonizes the solution as it reduces ions to ions.
10. Mention three areas in which plays an important role.
Ans. Sulphuric acid is an important industrial chemicaland is used for a lot of purposes. Some important uses of sulphuric acid are given below.
(i) It is used in fertilizer industry. It is used to make various fertilizers such as ammonium sulphate and calcium super phosphate.
(ii) It is used in the manufacture of pigments, paints, and detergents.
(iii) It is used in the manufacture of storage batteries.
11. Write the conditions to maximize the yield of by Contact process.
Ans. Manufacture of sulphuric acid by Contact process involves three steps.
1. Burning of ores to form
2. Conversion of to by the reaction of the former with
(is used in this process as a catalyst.)
3. Absorption of in to give oleum
The key step in this process is the second step. In this step, two moles of gaseous reactants combine to give one mole of gaseous product. Also, this reaction is exothermic. Thus, in accordance with Le Chatelier’s principle, to obtain the maximum amount of gas, temperature should be low and pressure should be high.
12. Why is for in water?
It can be noticed that
This is because a neutral has a much higher tendency to lose a proton than the negatively charged . Thus, the former is a much stronger acid than the latter.
13. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of and .
Ans. Fluorine is a much stronger oxidizing agent than chlorine. The oxidizing power depends on three factors.
1. Bond dissociation energy
2. Electron gain enthalpy
3. Hydration enthalpy
The electron gain enthalpy of chlorine is more negative than that of fluorine. However, the bond dissociation energy of fluorine is much lesser than that of chlorine. Also, because of its small size, the hydration energy of fluorine is much higher than that of chlorine. Therefore, the latter two factors more than compensate for the less negative electron gain enthalpy of fluorine. Thus, fluorine is a much stronger oxidizing agent than chlorine.
14. Illustrate how copper metal can give different products on reaction with .
Ans. Concentrated nitric acid is a strong oxidizing agent. It is used for oxidizing most metals. The products of oxidation depend on the concentration of the acid, temperature, and also on the material undergoing oxidation.
15. Give the resonating structures of and.
16. The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of hybridisation in and only s-p bonding between hydrogen and other elements of the group].
The above trend in the H-M-H bond angle can be explained on the basis of the electronegativity of the central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. We know that electronegativity decreases on moving down a group. Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing the H-M-H bond angle.
17. Explain why is basic while is only feebly basic.
Ans. is distinctly basic while is feebly basic.
Nitrogen has a small size due to which the lone pair of electrons is concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence, the electron donating capacity of group 15 element hydrides decreases on moving down the group.
18. Nitrogen exists as diatomic molecule and phosphorus as . Why?
Ans. Nitrogen owing to its small size has a tendency to form multiple bonds with itself. Nitrogen thus forms a very stable diatomic molecule, . On moving down a group, the tendency to form bonds decreases (because of the large size of heavier elements). Therefore, phosphorus (like other heavier metals) exists in the state.
19. Why is dioxygen a gas but sulphur a solid?
Ans. Oxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively form bonds and form (O==O) molecule. Also, the intermolecular forces in oxygen are weak van der Wall’s, which cause it to exist as gas. On the other hand, sulphur does not form molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid.
20. Knowing the electron gain enthalpy values for and as -141 and respectively, how can you account for the formation of a large number of oxides having species and not ?
(Hint: Consider lattice energy factor in the formation of compounds).
Ans. Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be.
Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving ion is much more than the oxide involving O-ion. Hence, the oxide having ions are more stable than oxides having . Hence, we can say that formation of is energetically more favourable than formation of .
21. Why are halogens strong oxidising agents?
Ans. The general electronic configuration of halogens is , where n = 2-6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron. Hence, they act as strong oxidizing agents.
22. How are xenon fluorides and obtained?
Ans. and are obtained by a direct reaction between Xe and . The condition under which the reaction is carried out determines the product.
23. Arrange the following in the order of property indicated for each set:
(i) increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI – increasing acid strength.
(iii) – increasing base strength.
Ans. (i) Bond dissociation energy usually decreases on moving down a group as the atomic size increases. However, the bond dissociation energy of is lower than that of and . This is due to the small atomic size of fluorine. Thus, the increasing order for bond dissociation energy among halogens is as follows:
(ii) HF < HCl < HBr < HI
The bond dissociation energy of H-X molecules where X = F, Cl, Br, I, decreases with an increase in the atomic size. Since H-I bond is the weakest, HI is the strongest acid.
On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom decreases. Thus, the basic strength decreases.
24. List the uses of Neon and argon gases.
Ans. Uses of neon gas:
(i) It is mixed with helium to protect electrical equipments from high voltage.
(ii) It is filled in discharge tubes with characteristic colours.
(iii) It is used in beacon lights.
Uses of Argon gas:
(i) Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than N.
(ii) It is usually used to provide an inert temperature in a high metallurgical process.
(iii) It is also used in laboratories to handle air-sensitive substances.
5 Marks Questions
1. Give reasons :-
(a) Oxygen molecule is diatomic where as sulphur molecule is polyatomic.
(b) The most common oxidation state of oxygen is -2.
(c) H2O is liquid whereas is gas at room temperature.
(d) The increasing order of acidic character in 16th group hydrides is.
(e) is exceptionally stable, does not exist.
Ans. (a) Oxygen being small in size forms effective and strong P-P bonds with other oxygen atom. Therefore oxygen molecule is diatomic and discrete whereas Sulphur due to its larger size, its orbitals cannot overlap effectively to form P-P bonds & completes valency by forming bonds with many sulphur atom. Therefore sulphur molecule is polyatomic solid.
(b) Since oxygen is highly electronegative, it has little tendency to give electrons. Therefore its most common oxidation state is -2.
(c) is liquid at room temperature due to presence of intermolecular Hydrogen bonding which is absent in.
(d) As we move down the group, the size of atom increases this make the bond of the element with hydrogen weak. Due to weaker bonds, the bond dissociation enthalpy decreases making the molecule more acidic. Therefore the order of acidic strength is.
(e) is exceptionally stable due to steric reasons. Hydrogen being electropositive or less electronegative than fluorine cannot make the s- electrons of sulphur to participate in bonding. Therefore does not exist.
2. Discuss the different types of oxides.
Ans. A binary compound of oxygen with another element is called oxide. Oxides can be simple or mixed. Simple oxides can be classified as acidic, basic Amphoteric or neutral.
An oxide that combines with water to give an acid is termed acidic oxide e.g., etc.
An oxide that combines with water to give a base is called basic oxide e.g., CaO, BaO etc.
An oxide that shows characteristics of both acids and bases is Amphoteric oxide e.g. .
An oxide that shows characteristic of neither acid nor base is called neutral oxide e.g. CO, NO and.
3. Bond angle in is higher than that in . Why?
Ans. In , P is hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with sp3bonding is changed to pyramidal. combines with a proton to form in which the lone pair is absent. Due to the absence of lone pair in, there is no lone pair-bond pair repulsion. Hence, the bond angle in is higher than the bond angle in .
Concept insite: the long pair-bond pair repulsion is more than bond pair-bond pair repulsion.
4. Comment on the nature of two S-O bonds formed in molecule. Are the two S-O bonds in this molecule equal?
Ans. The electronic configuration of S is .
During the formation of
, one electron from 3p orbital goes to the 3d orbital and S undergoes hybridization. Two of these orbitals form sigma bonds with two oxygen atoms and the third contains a lone pair. p-orbital and d-orbital contain an unpaired electron each. One of these electrons formsbond with one oxygen atom and the other forms bond with the other oxygen. This is the reason has a bent structure. Also, it is a resonance hybrid of structures I and II.
Both S-O bonds are equal in length (143 pm) and have a multiple bond character.
5. Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
Ans. General trends in group15 elements
(i) Electronic configuration: All the elements in group 15 have 5 valence electrons. Their general electronic configuration is .
(ii) Oxidation states: All these elements have 5 valence electrons and require three more electrons to complete their octets. However, gaining electrons is very difficult as the nucleus will have to attract three more electrons. This can take place only with nitrogen as it is the smallest in size and the distance between the nucleus and the valence shell is relatively small. The remaining elements of this group show a formal oxidation state of -3 in their covalent compounds. In addition to the -3 state, N and P also show -1 and -2 oxidation states.
All the elements present in this group show +3 and +5 oxidation states. However, the stability of +5 oxidation state decreases down a group, whereas the stability of +3 oxidation state increases. This happens because of the inert pair effect.
(iii) Ionization energy and electronegativity
First ionization decreases on moving down a group. This is because of increasing atomic sizes. As we move down a group, electronegativity decreases, owing to an increase in size.
(iv) Atomic size: On moving down a group, the atomic size increases. This increase in the atomic size is attributed to an increase in the number of shells.
6. Discuss the trends in chemical reactivity of group 15 elements.
Ans. General trends in chemical properties of group – 15
(i) Reactivity towards hydrogen:
The elements of group 15 react with hydrogen to form hydrides of type , where E = N, P, As, Sb, or Bi. The stability of hydrides decreases on moving down from to .
(ii) Reactivity towards oxygen:
The elements of group 15 form two types of oxides: and , where E = N, P, As, Sb, or Bi. The oxide with the element in the higher oxidation state is more acidic than the other. However, the acidic character decreases on moving down a group.
(iii) Reactivity towards halogens: The group 15 elements react with halogens to form two series of salts: and . However, nitrogen does not form as it lacks the d-orbital. All trihalides (except ) are stable.
(iv) Reactivity towards metals: The group 15 elements react with metals to form binary compounds in which metals exhibit -3 oxidation states.
7. Write main differences between the properties of white phosphorus and red phosphorus.
|White phosphorus||Red Phosphorus|
|It is a soft and waxy solid. It possesses a garlic smell.||It is a hard and crystalline solid, without any smell.|
|It is poisonous.||It is non-poisonous.|
|It is insoluble in water but soluble in carbon disulphide.||It is insoluble in both water and carbon disulphide.|
|It undergoes spontaneous combustion in air.||It is relatively less reactive.|
In both solid and vapour states, it exists as a P4 molecule.
It exists as a chain of tetrahedral P4 units.
8. Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.
Ans. The elements of group 16 are collectively called chalcogens.
(i) Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is , where n varies from 2 to 6.
(ii) Oxidation state:
Asthese elements have six valence electrons , they should display an oxidation state of -2. However, only oxygen predominantly shows the oxidation state of -2 owing to its high electronegativity. It also exhibits the oxidation state of -1 , zero , and +2. However, the stability of the -2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of +2, +4, and +6 due to the availability of d-orbitals.
(iii) Formation of hydrides:
These elements form hydrides of formula , where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type . These hydrides are quite volatile in nature.
9. Describe the manufacture of by contact process?
Ans. Sulphuric acid is manufactured by the contact process. It involves the following steps:
Step (i): Sulphur or sulphide ores are burnt in air to form .
Step (ii): By a reaction with oxygen, is converted into in the presence of as a catalyst.
Step (iii): produced is absorbed on to give (oleum).
This oleum is then diluted to obtain of the desired concentration.
In practice, the plant is operated at 2 bar (pressure) and 720 K (temperature). The sulphuric acid thus obtained is 96-98% pure.
10. How is an air pollutant?
Ans. Sulphur dioxide causes harm to the environment in many ways:
1. It combines with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble.
2. Even in very low concentrations, causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.
3. It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.
11. What are the oxidation states of phosphorus in the following:
(i) (ii) (iii)
(iv) (v) ?
Ans. Let the oxidation state of p be x.
3+x+3(–2) = 0
3 + x – 6 = 0
x – 3 = 0
x = +3
x + 3(–1) =
x – 3 = 0
x = +3
3(+2) + 2(x) = 0
6 + 2x = 0
2x = –6
x = –3
3(+1) + x + 4(–2) = 0
x –5 = 0
x = +5
x + (–2) + 3(–1) = 0
x – 5 = 0
x = +5
12. Give the formula and describe the structure of a noble gas species which is isostructural with:
Ans. (i)is isoelectronic with and has square planar geometry.
(ii)is isoelectronic to and has a linear structure.
(iii)is isostructural to and has a pyramidal molecular structure.