NCERT Solutions Class 11 Chemistry Chapter 11
NCERT Solutions For Class 11 Chemistry Chapter 11
NCERT Chemistry Class 11 Chapter 11 deals with the topic of p-block elements. Just to give you an essence, p-block elements are elements where the last electron enters or is found in the outermost p-subshell. Class 11 Chemistry Chapter 11 p-block elements is an important topic for the students.
After studying the chapter, students must solve the exercises given at the back to understand the concept in a better way. To help students, Extramarks offers NCERT Solutions For Class 11 Chemistry Chapter 11 with answers to all the questions given in the back exercise of the chapter. Students can access and download the solutions for free.
NCERT Solutions For Class 11 Chemistry Chapter 11 The p-Block Elements
The main purpose of NCERT solutions for Class 11 Chemistry Chapter 11 is to help students understand the elements that are present in the p-block of the modern periodic table. The reason behind placing importance on this section is to understand how f and d electrons tend to make chemistry even more interesting.
Since this chapter covers a lot of topics, students should refer to NCERT Chemistry Solutions for Class 11 to retain the concepts and get accurate answers to the practise questions listed at the end of the chapter.
Access NCERT Solution For Class 11 Chapter 11 – The p-Block Elements
Along with answers to the questions in the textbook, NCERT Solutions Class 11 Chemistry Chapter 11 provide exemplary problems and worksheets that guide you in understanding the topic thoroughly and help you in your examination. Besides, students who read through or practise these NCERT Solutions will definitely be able to score better marks in the examination.
Chemistry in Class 11 brings a variety of newly introduced topics that students have not read about in the earlier classes. All of these topics are essential in building a strong base in the subject for higher education in colleges, especially for students who plan to build a career in fields related to Chemistry.
After reading every chapter, students should practise the exercises given at the end to understand the concepts better. Referring to NCERT Solutions by Extramarks will help students in clarifying their doubts and they will also be able to know whether they have got the accurate answer for every question.
NCERT Solutions For Class 11 Chemistry Chapter 11 – The p-Block Elements
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Class 11 Chemistry p-Block Elements NCERT Solutions
NCERT Solutions Class 11 Chemistry Chapter 11 has answers to all the questions that are listed in NCERT Class 11 Chapter 11. The solutions have been curated by our subject experts as per the latest term-wise syllabus.
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NCERT Solutions For Class 11 Chemistry Chapter 11 Exercises
Q1. Is Boric acid a protic acid?
A1. Boric acid is an acid, but not a protic acid. It is a weak monobasic acid with low pH.
B(OH)3 + 2H2O -> [B(OH)4] + H3O+
Q2. How do you explain the higher stability of BCI3 as compared to TICI3?
Ans: In the periodic table, group 13 includes elements such as boron and thallium. +1 oxidation state gets more stable as you move down the group in this group. Due to the fact that Boron’s +3 oxidation state is more stable than Tlhallium’s +3 oxidation state, BCl3 is more stable than TlCl3. This is because Thallium’s +3 state is highly oxidising and it quickly reverts to its more stable +1 form.
Benefits Of NCERT Solutions For Chapter 11 For Class 11
NCERT Solutions for Chapter 11 are prepared by subject-matter experts after conducting significant research and as per the CBSE’s latest curriculum. The solutions have in-depth explanations of all the questions, and students can rely on them for a high level of accuracy.
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Q.1 Discuss the pattern of variation in the oxidation states of i) B to Tl and ii) C and Pb.
i) The elements present in group 13 are B, Al, Ga, In and Tl. Among them B and Al have no d- or f- electrons, so they cannot exhibit inert pair effect. They show +3 oxidation states due to the presence of 3 electrons in the valence shell, out of them two electrons are present in the s-orbital and one electron is present in the p-orbital. Other elements, Ga to Tl have either only d electrons or both d- and f- electrons, hence they show oxidation state of +1 and +3 due to the inert pair effect, this effect is more prominent in Tl. As we move down the group, due to increase in the number of electrons in the d- and f- block electrons, orbitals the inert pair effect becomes more prominent and pronounced. From Ga to Tl, the stability of +1 oxidation state increases (Ga < In <tl), of=”” that=”” while=””> Tl).</tl),>
Thus for Tl, +1 oxidation state is more stable than +3 oxidation state.
ii)The elements present in group 14 are C, Si, Ge, Sn, Pb. C and Si do not have d- or f- electrons, hence they don’t exhibit inert pair effect. They show +4 oxidation states due to the presence of 4 electrons in the valence shell, two in s- and two in p- orbital. But the other elements from Ge to Pb contain d- or f- electrons; hence show +2 and +4 oxidation states due to inert pair effect. When we move down the group
, the inert pair effect becomes more prominent and stability of +2 oxidation state increases (Ge<sn
Thus for Pb, +2 oxidation state is more stable than +4 oxidation state.
Q.2 How can you explain higher stability of BCl3 as compared to TlCl3?
The shielding or screening effect follows the order s>p>d>f, thus s-orbital has strong shielding effect and f-orbital has poor shielding effect. In Tl, 3d, 4d, 5d and 4f electrons show poor shielding effect on 6s electrons. As a result, only 6p1 electron of Tl participates in bond formation. Thus Tl shows +1 oxidation state rather than +3 oxidation state. TlCl is stable compound but TlCl3 is unstable.
B doesn’t contain d- or f-electrons, thus can’t show inert pair effect. The three valence electrons ((2s2 2p1) participate in bond formation and hence it can show +3 oxidation state. Thus BCl3 is stable compound than TlCl3.
Q.3 Why does boron triflouride behave as a Lewis acid?
In BF3 molecule, B atom has only 6 electrons in its valence shell and hence is an electron-deficient molecule. To complete its octet, B requires two more electrons, so it easily accepts a pair of electrons from nucleophiles such as F–, CN– etc. Thus, BF3 acts as Lewis acid.
Q.4 Consider the compounds, BCl3 and CCl4. How will they behave with water? Justify.
BCl3 is electron-deficient molecule as B has six electrons in the valence shell. To complete its octet, it accepts a pair of electrons donated by water molecule. Thus, BCl3 undergoes hydrolysis to form boric acid (H3BO3) and HCl.
BCl3 + 3H2O → H3BO3 + 3HCl
In comparison to BCl3, in CCl4 molecule, the C atom has 8 electrons in the valence shell. The compound is an electron-precise molecule and hence it can neither accept nor donate a pair of electrons. So CCl4 does not undergo hydrolysis in water.
Q.5 Is boric acid a protic acid? Explain.
No, boric acid is not a protic acid, it is an electron-deficient molecule having incomplete octet. It doesn’t ionize in water to give a proton rather it accepts a lone pair of electrons from the oxygen atom of H2O molecule to form a hydrated species.
Oxygen atom is electronegative, so +ve charge on oxygen atom pulls the s – electrons of O–H bond towards itself. Thus it releases the proton and B(OH)3 acts as a weak monobasic Lewis acid. When it reacts with NaOH solution, sodium metaborate is formed.
Q.6 Explain what happens when boric acid is heated.
Upon heating boric acid loses water in different stages at different temperature. The final product is boron trioxide or boric anhydride.
Q.7 Describe the shapes of BF3 and BH4–. Assign the hybridization of boron in these species.
In BF3 molecule, boron is sp2–hybridized and therefore, it is a planer molecule. On the other hand, in BH4–, boron is sp3–hybridized, so it has tetrahedral structure.
Q.8 Write reactions to justify amphoteric nature of aluminium.
Al reacts with both acid and alkali to form dihydrogen.
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)
2Al(s) + 2NaOH(aq) + 6H2O(l) → 2Na+[Al(OH)4]–(aq) + 3H2(g)
Q.9 What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species? Explain.
The compounds in which the central atom does not have 8 electrons in the valence shell or those which have 8 valence electrons but can expand its covalency beyond 4 due to the presence of vacant d-orbitals, are called electron deficient compounds.
(i) In BCl3, central atom B contains only 6 electrons in valence shell, thus it is an electron deficient compound. NH3 donates its pair of electrons to BCl3 molecule and Cl3B ← NH3 adduct is formed.
(ii) In SiCl4, the central Si atom has 8 electrons but it contains vacant d orbital, so it can expand its covalency beyond 4. Thus SiCl4 is also an electron-deficient compound.
Q.10 What is the state of hybridization of carbon in a) CO32- b) diamond, c) graphite?
a) Hybridization of C in CO32– is sp2,
b) Hybridization of C in diamond is sp3,
c) Hybridization of C in graphite is sp2.
Q.11 Explain the difference in properties of diamond and graphite on the basis of their structures.
|Diamond is hard. It exists as three-dimensional network solid, thus it is very hard.||Graphite is soft. It has layered structure and two successive layers are held together by weak forces of attraction, thus one layer can slip over the other. Therefore, graphite is soft.|
|Since all the electrons are firmly held in C–C, s-bond, thus no free electrons are available for the conduction of electricity. So, diamond is poor conductor of electricity.||Since only 3 electrons of each carbon are involved in making the hexagonal rings in graphite, and the 4th valence electron of each carbon is free to move, it can conduct electricity. So, graphite is good conductor of electricity.|
|Diamond is transparent due to the high refractive index.||Graphite is black in colour and possesses a metallic lustre.|
Q.12 Rationalize the given statements and give chemical reactions:
• Lead (II) chloride reacts with Cl2 to give PbCl4.
• Lead (IV) chloride is highly unstable towards heat.
• Lead is known not to form an iodide, PbI4.
- Lead has d- and f- orbital electrons, it shows inert pair effect. For that reason, +2 oxidation state of Pb is more prominent rather than +4 oxidation state. Thus lead (II) chloride is more stable than lead (IV) chloride. Therefore, PbCl2 does not react with chlorine to form PbCl4.
- Due to greater stability of +2 oxidation state of Pb over +4 oxidation state (due to inert pair effect); PbCl2 is more stable than PbCl4.
- In PbI4, Pb is in +4 oxidation state. But due to inert pair effect, +4 oxidation state is not stable for Pb. Pb–I bond formed initially during the reaction does not release enough energy to unpair the 6s2 electrons and excite them to 6p–orbital to have 4 unpaired electrons around Pb atom needed for the formation of PbI4. Again, due to the oxidizing power of Pb4+ ion and reducing power of I– ion, PbI4 does not exist.
Q.13 Suggest reasons why the B–F bond lengths in BF3 (130 pm) and BF4–(143 pm) differ.
In BF3, boron is sp2 hybridized, it has vacant 2p- orbital and F has three lone pair of electrons in 2p-orbitals. Because of similar sizes of 2p-orbitals in F and B, pp – pp back bonding occurs in which a lone pair of electrons is transferred from F to B. Thus B–F bond acquires some partial double bond character.
But in BF4– ion, B is sp3– hybridized and does not have any vacant p-orbital available to accept the electrons donated by F atom, so pp–pp back bonding does not occur here. The bonds present in BF4– are purely single bonds. Double bonds are shorter in length than single bonds. So, in BF3 bond length is 130 pm and in BF4– ,it is 143pm.
Q.14 If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.
Both B and Cl atoms have electronegativity difference, so B–Cl bond has polar character and has a definite dipole moment. But polarity also depends upon the geometry of the molecule. BCl3 has planer geometry, the bond angle between the three B–Cl bonds is 120°. Thus the resultant dipole moment of two B–Cl bonds is cancelled by equal and opposite dipole moment of the third B–Cl bond. As a result, overall dipole moment of BCl3 becomes zero. A clear view of dipole moment of BCl3 can be depicted by the following picture.
Q.15 Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is bubbled through. Give reasons.
HF is a covalent compound in anhydrous form and the molecules are held together by hydrogen bonds. It does not give F– ion in the medium, so AlF3 does not dissolve in HF. But NaF is ionic solid, it produces F– ions which react with AlF3 to form the soluble complex.
3NaF + AlF3 → Na3[AlF6]
Sod. hexafluoroaluminate (III)
B has much higher tendency to form complexes than Al, due to its smaller size and higher electronegativity. So, when BF3 is added to the above solution, AlF3 gets precipitated.
Na3[AlF6] + 3BF3 → 3Na[BF4] + AlF3(s)
Sod. Tetrafluroborate (IIII)
Q.16 Suggest a reason as to why CO is poisonous.
Haemoglobin present in red blood cells of human body combines with oxygen to form oxyhaemoglobin. It is a reversible reaction; oxyhaemoglobin carries oxygen from lungs to different tissues of body.
When CO and O2 both are available, haemoglobin preferably combines with CO than O2. CO combines with haemoglobin irreversibly to form carboxyhaemoglobin, which is more stable than oxyhaemoglobin. The percentage of oxygen in blood gets reduced. Thus the oxygen carrying capacity of haemoglobin is destroyed and human being dies of suffocation.
Haemoglobin + CO → Carboxyhaemoglobin
For this reason, CO is poisonous.
Q.17 How is excessive content of CO2 responsible for global warming?
The normal amount of CO2 in air is 0.03%. During respiration, CO2 is evolved and it is utilized by plants during photosynthesis. Thus, CO2 concentration is balanced in the earth. If concentration of CO2 increases beyond 0.03% (due to excessive combustion), some of the CO2 will always remain unutilized. This excess CO2 absorbs heat radiated by earth and some of it is dissipated into the atmosphere, the remaining part is radiated back to the earth. Thus the temperature of the earth increases gradually. This phenomenon is called greenhouse effect and CO2 is called green house gas which is a major contribution to the global warming.
Q.18 Explain structures of diborane and boric acid.
Structure of diborane:
In diborane (B2H6), two types of hydrogen atoms are present-
- The four hydrogen atoms (two on the left and two on the right side of the boron atom) are called the terminal hydrogens, and these 4 H atoms and 2 boron atoms lie in the same plane.
- The remaining two hydrogen atoms, one lying above the plane and another lying below the plane form bridges and are called bridge hydrogens.
The four terminal B-H bonds are normal covalent bonds which are formed by sharing a pair of electrons between B and H atoms. So it is also called two centre electron pair bonds or two centre-two electron bond (2c-2e).
Another two bridge bonds i.e. B—-H—-B are quite different from normal covalent bonds. Each bridge hydrogen atom is bonded with two B atoms by a pair of electrons. So it is called three centre-two electron bonds (3c-2e). The 3c-2e bond resembles to a banana, so it is also called banana bond. This kind of bond is weak.
On the basis of hybridization, explanation of diborane structure:
The electronic configuration of boron atom is as follows:
So it undergoes sp3-hybridization. The two half filled hybride orbitals of each B atom overlap with the half filled orbital of hydrogen atoms to form normal covalent bonds. The third half filled hybrid orbital of one boron atom and the vacant hybrid orbital of 2nd boron atom overlap simultaneously with the half filled orbital of H atom, thus electron cloud spreads over three atoms (2 Boron and 1 H atom). For this reason this bond is called 3 centre-2 electron bonds (3c-2e) or banana bond.
Structure of boric acid: In borate ion (BO33-), the boron atom is sp2-hybridized and has trigonal planer structure. In excited state, boron has three half filled atomic orbitals (2s, 2px and 2py), these three atomic orbitals undergo sp2-hybridization to give three sp2-hybrid orbitals. Each of these orbital overlaps with 2p-orbitals of O–, forming B–O– bonds.
In boric acid, the planer BO33– ions are joined by unsymmetrical hydrogen bonds to give layered structure. The adjacent layers in the crystal of boric acid are held together by weak forces of attraction, so one layer can slide over the other. This property makes boric acid soft and soapy to touch.
Q.19 What happens when?
(a) Borax is heated strongly,
(b) Boric acid is added to water,
(c) Aluminium is treated with dilute NaOH,
(d) BF3 is reacted with ammonia?
a) Borax contains 10 molecules of water of crystallization (Na2B4O7.10H2O). When it is heated, the water molecules get vapourised. When borax is strongly heated, a transparent glassy bead which consists of sodium metaborate (NaBO2) and boric anhydride is formed.
b) Boric acid acts as weak Lewis acid. It gives proton into the aqueous solution.
c) When Al is treated with aqueous solution of NaOH, sodium aluminate and dihydrogen are formed.
2Al(s) + 2NaOH(aq) + 6H2O(l) → 2Na+[Al(OH)4]–(aq) + 3H2(g)
d) BF3 is a Lewis acid because it has incomplete octet and NH3 is a strong base having lone pair of electrons. BF3 accepts a pair of electrons from NH3 to form the corresponding adduct.
Q.20 Explain the following reactions
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper;
(b) Silicon dioxide is treated with hydrogen fluoride;
(c) CO is heated with ZnO;
(d) Hydrated alumina is treated with aqueous NaOH solution.
a) When silicon is heated with methyl chloride in presence of Cu power (acting as a catalyst), a mixture of mono-, di- , tri-methylchlorosilanes with small amount of tetra-methylsilane is formed.
b) Silicon dioxide initially dissolves in HF to form silicon tetra-fluoride and it further reacts with remaining HF to form hydrofluorosilicic acid.
SiO2 + 4HF → SiF4 + 2H2O
SiF4 + 2HF → H2SiF6
c) ZnO is reduced by CO to Zn metal.
ZnO + CO → Zn + CO2
d) Hydrated alumina reacts with NaOH to form sodium meta-aluminate.
Al2O3.2H2O(s) + 2NaOH(aq) + H2O(l) → 2Na[Al(OH)4](aq)
Hydrated alumina Sod. tetrahydroxoaluminate (III)
Q.21 Give reasons:
(i) Conc. HNO3 can be transported in aluminium container.
(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
(iii) Graphite is used as lubricant.
(iv)Diamond is used as an abrasive.
(v) Aluminium alloys are used to make aircraft body.
(vi)Aluminum utensils should not be kept in water overnight.
(vii) Aluminium wire is used to make transmission cables.
i) Al reacts with conc. HNO3 to form a very thin layer of aluminium oxide on its surface which protects it from further reaction.
2Al(s) + 6HNO3(l) → Al2O3(s) + 6NO2(g) + 3H2O(l)
Thus, Al becomes passive and the Al container can be used for transportation of Conc. HNO3.
ii) Al reacts with NaOH to form sodium aluminate and dihydrogen gas is evolved. The pressure of the hydrogen gas thus produced can be used to open clogged drains.
2Al(s) + 2NaOH(aq) + 2H2O(l) → 2NaAlO2(aq) + 3H2(g)
iii) Graphite has a layered structure; two successive layers are held together by weak van der Waals forces of attraction. So the layers can slip over one another. For this reason it can act as lubricant.
iv) Diamond exists as three dimensional network structure in which each carbon atom is tetrahedrally bonded with four neighboring carbon atoms. So it is very hard and can be used as abrasive.
v) The following characteristics of aluminium alloys make them useful for making aircraft body:
- Resistance to corrosion.
vi)Al reacts with water and dissolved oxygen to form thin film of aluminium oxide.
2Al(s) + O2(g) + H2O(l) → Al2O3(s) + H2(g)
Very small amount of Al2O3 dissolves in water to give a few ppm of Al3+ ions. Since Al3+ ions affect our health, so drinking water should not be kept in Al utensils for overnight.
Vii) Al is light weight and good conductor of electricity, so it is used for making transmission cables and for winding the moving coils of dynamo or motor.
Q.22 Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?
The decrease in ionization enthalpy from C to Si is phenomenal due to large increase in atomic size and decrease in screening effect (force of attraction of the nucleus for valance electrons) from C to Si.
Q.23 How would you explain the lower atomic radius of Ga as compared to Al?
Ga has electrons in 3d orbital, which shows poor shielding effect on the valence electrons. So the effective nuclear charge of Ga is greater in magnitude as compared to Al. As a result, the valence electrons of Ga experience greater force of attraction by the nucleus than Al, so atomic size of Ga (135pm) is slightly less than Al (143pm).
Q.24 What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on physical properties of two allotropes?
Allotropy is the existence of an element in two or more different forms which have different physical properties and but identical chemical properties. Different forms of an element are called allotropes. For example, diamond and graphite are the allotropes of carbon.
The impact of structure on physical properties of two allotropes is as follows:
1. Diamond exists in a three dimensional network structure and has high density. Therefore, it is hard. Graphite has two-dimensional sheet like structure and any two consecutive layers are held together by weak van der Waals forces of attraction. Thus one layer can slip over the other and it becomes soft and can be used as lubricant.
2. In diamond, all the electrons are held together by strong sigma bond, as a result, no electrons are available for conducting electricity. So diamond is a poor conductor of electricity. But in graphite, the fourth valance shell electron of each carbon is free to move, therefore, it can conduct the electricity. So graphite is a good conductor of electricity.
3. Due to its high refractive index, diamond can reflect and refract light, so it is transparent. Graphite can’t reflect and refract light, so it is opaque.
Q.25 (a) Classify following oxides as neutral, acidic, basic or amphoteric: CO, B2O3, SiO2, CO2, Al2O3, PbO2, and Tl2O3
(b) Write suitable chemical equations to show their nature.
a) Neutral oxide: CO.
Acidic oxide: B2O3, SiO2, CO2
Basic Oxide: Tl2O3
Amphoteric oxide: Al2O3, PbO2
b) Acidic oxides react with alkalis to form salts. The chemical reactions involved are given below:
B2O3 + 2NaOH → 2NaBO2 + H2O
Boric anhydride Sod. Meta borate
CO2 + 2NaOH → Na2CO3 + H2O
Amphoteric oxides react with both acid and base. The chemical reactions involved are as follows:
Al2O3 + 2NaOH → 2NaAlO2 + H2O
Alumina Sod. Meta aluminate
Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O
PbO2 + 2NaOH → Na2PbO3 + H2O
Lead dioxide Sod. plumbate
2PbO2 + 2H2SO4 → 2PbSO4 + 2H2O + O2
Basic oxides dissolve in acids. For example: Tl2O3 dissolves in acids.
Tl2O3 + 6HCl → 2TlCl3 + 3H2O
Q.26 In some of the reactions, thallium resembles aluminium, whereas in others, it resembles with group I metals. Support this statement by giving some evidences.
Aluminium shows +3 oxidation state in all compounds and Tl also shows +3 oxidation state in some compounds like TlCl3, Tl2O3, etc.
Group I metals show oxidation state of +1. Due to inert pair effect, Tl also shows +1 oxidation state in the compounds like Tl2O, TlCl etc. Like group I oxides, Tl2O is strongly basic in nature.
Q.27 When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.
In a schematic diagram, the reaction can be represented as:
X + NaOH → A (white ppt.) + excess NaOH → B (soluble compound)
A + dil HCl → C
X must be Al, ppt. A is Al(OH)3 and complex B must be sodium tetrahydroxoaluminate(III).
Al + 3NaOH → Al(OH)3 ↓ + 3Na+
Al(OH)3 + NaOH → Na+ [Al(OH)4]–
Sod. Tetrahydroxoaluminate (III)
Compound C is AlCl3. Al(OH)3 reacts with dil. HCl to give compound C.
Al(OH)3 + 3HCl → AlCl3 + 3H2O
Upon heating, A decomposes to compound D. D is Al2O3 (alumina) which is used to extract metals.
2Al(OH)2 → Al2O3 + 3H2O
Q.28 What do you understand by (a) inert pair effect (b) allotropy and (c) catenation?
(a) Inert pair effect: As we move down a group, the tendency of s-electrons of the valence shell to participate in bond formation gradually decreases. This reluctance of s- electrons to participate in bond formation is called inert pair effect.
As we move down the group, additional shells are incorporated (having d- or/and f- orbitals) and d- or f- orbitals have poor shielding effect on ns2 electrons. Due to increase in effective nuclear charge, the s-electrons become more tightly held (more penetrating) and tend to remain paired, therefore, become more reluctant to participate in bond formation.
(b) Allotropy: The phenomenon of existence of an element in two or more different forms which have different physical properties but identical chemical properties is called allotropy and the different forms are called allotropes. Diamond and graphite are the two allotropes of carbon. The allotropes of sulphur are monoclinic sulphur and rhombic sulphur.
(c) Catenation: Catenation is the property of an element which may be defined as the ability of like atoms to link with one another by covalent bonds. Due to the smaller size and high electronegativity of carbon, it shows this unique property. Since bond energy of C–C single bond is very large, it can form long straight or branched C–C chains or rings of different sizes and shapes. For this catenation property, carbon can form a large number of hydrocarbons, no other atom can show. Other than carbon Si shows the catenation property.
Q.29 A certain salt X, gives the following results.
(i) Its aqueous solution is alkaline to litmus.
(ii) It swells up to a glassy material Y on strong heating.
(iii) When conc. H2SO4 is added to a hot solution of X, white crystal of an acid Z separates out.
Write equations for all the above reactions and identify X, Y and Z.
i) Since the aqueous solution of the salt (X) turns red litmus paper blue, then it should be salt of strong base and a weak acid.
ii) Salt (X) swells up to glassy material (Y) on strong heating, so (X) is borax and (Y) must be a mixture of sodium meta-borate and boric anhydride.
iii) When conc. H2SO4 is added to salt X, white crystals of an acid Z separate out, so Z must be orthoboric acid.
The involved reactions are as follows:
Q.30 Write balanced equations for:
(i) BF3 + LiH →
(ii) B2H6 + H2O →
(iii) NaH + B2H6 →
(v) Al + NaOH →
(vi) B2H6 + NH3 →
Q.31 Give one method for industrial preparation and one for laboratory preparation of CO and CO2 each.
Industrial preparation of CO:
Laboratory preparation of CO:
Industrial preparation of CO2
Laboratory preparation of CO2
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
Q.32 An aqueous solution of borax is
(a) Neutral (b) Amphoteric
(c) Basic (d) Acidic
c) Borax is a salt of a strong base (NaOH) and a weak acid (H3BO3), thus it is basic in nature.
Therefore, option (c) is correct.
Q.33 Boric acid is polymeric due to
(a) its acidic nature (b) the presence of hydrogen bonds. (c) its monobasic nature (d) its geometry.
Boric acid is polymeric due to the presence of H-bonds. Thus, (b) is the right option.
Q.34 The type of hybridisation of boron in diborane is (a) sp (b) sp2 (c) sp3 (d) dsp2
The hybridisation of B in diborane is sp3.Therefore, option (c) is correct.
Q.35 Thermodynamically the most stable form of carbon is
(a) Diamond (b) Graphite
(c) Fullerenes (d) Coal
Thermodynamically the most stable form of carbon is graphite. Thus, option (b) is the correct one.
Q.36 Elements of group 14
(a) exhibit oxidation state of +4 only
(b) exhibit oxidation states of +2 and +4
(c) form M2– and M4+ ions
(d) form M2+ and M4+ ions
Group 14 elements show +2 and +4 oxidation states. Due to the inert pair effect, the lower oxidation state becomes more and more stable and the higher oxidation state becomes less stable. Thus option (b) is the correct answer.
Q.37 If the starting material for the manufacture of silicones is RSiCl3, write the structure of the product formed.
Silicones are formed by hydrolysis of alkyltri- chlorosilanes.
Q.38 Write the resonance structures of CO32–and HCO3–.
Resonance structures of CO32- :
Resonance structures of bicarbonate HCO3–
FAQs (Frequently Asked Questions)
1. What makes it important for students to refer to NCERT Solutions for Class 11 Chemistry Chapter 11?
If scoring high marks in chemistry is your goal, then referring to NCERT solutions by Extramarks should be the top thing for you when it comes to preparations. The chapters are explained in a simplified manner to keep all your concepts clear and to the point.
2. What is the total number of elements present in the p-block?
There are a total of 35 p-block elements in the p-block which are further classified into different subgroups. These elements fall into the p orbital based on their valence electrons
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