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NCERT Chemistry Class 11 P Block Elements: S Block elements are the elements in periodic table from group 13 to 18. All the non-metals and metalloids exist only in the p-block of the periodic table. In this chapter students will learn the basic physical and chemical properties of the p block elements, uses of p block elements like diamond, graphite carbon in the industries and their properties.
Boron, carbon, nitrogen, oxygen, fluorine and helium are some important elements of p-block groups. Their valence shell electronic configuration is ns2 np1-6 (except for He). The inner core of the electronic configuration may, however, differ and this difference in inner core of elements greatly influences their physical properties such as atomic and ionic radii, ionisation enthalpy, etc. as well as chemical properties. Students can check here NCERT class 11 Chemistry P-block elements solutions prepared by the subject experts.
NCERT Class 11 Chemistry solution P-bock elements provided here will help students to understand the topics thoroughly and score good marks in this section. Students preparing for entrance exams can refer NCERT Solutions Chemistry class 12.
NCERT Chemistry Class 11 P-Block Elements Topics Covered
Candidates can check here the list of all topics that are covered under P-block elements chapter of NCERT Chemistry class 11
- Group 13 Elements: The Boron Family
- Electronic Configuration
- Atomic Radii
- Ionization Enthalpy
- Electronegativity
- Physical Properties
- Chemical Properties
- Important Trends and Anomalous Properties of Boron
- Some Important Compounds of Boron
- Borax
- Orthoboric Acid
- Diborane, B2H6
- Uses of Boron and Aluminium and Their Compounds
- Group 14 Elements: The Carbon Family
- Electronic Configuration
- Covalent Radius
- Ionization Enthalpy
- Electronegativity
- Physical Properties
- Chemical Properties
- Important Trends and Anomalous Behaviour of Carbon
- Allotropes of Carbon
- Diamond
- Graphite
- Fullerenes
- Uses of Carbon
- Some Important Compounds of Carbon and Silicon
- Carbon Monoxide
- Carbon Dioxide
- Silicon Dioxide, SiO2
- Silicones
- Silicates
- Zeolites
THE P -BLOCK ELEMENTS Additional Questions
Assertion and Reason:
Directions:
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
11.1. Assertion: Non-metals readily form anions, in contrast to metals whichreadily form cations.
Reason: In general, non-metals have higher ionisation enthalpies and higherelectronegativities than the metals.
Answer: (a)
11.2. Assertion: The compounds formed by highlyreactive non-metals with highly reactive metalsaregenerally ionic while compounds formed between non-metals themselves are largely covalent in character.
Reason: There are large differencesin the electronegativities of highly reactive non-metals and highly reactive metals whereas small differences in theelectronegativities among the non-metals themselves.
Answer: (a)
11.3. Assertion: PbI4 does not exist.
Reason: Pb—I bond initially formedduring the reaction does not release enoughenergy to unpair6s2 electrons and excite oneof them to higher orbital to have four unpairedelectrons around lead atom.
Answer: (a)
11.4. Assertion: Heavier elements form pπ– pπ bonds.
Reason: Their atomic orbitals are too large and diffuse to have effective overlapping.
Answer: (d) Heavier elementsdo not form pπ– pπ bonds because their atomicorbitals are too large and diffuse to haveeffective overlapping.
11.5. Assertion: Boron halides form addition compounds with NH3.
Reason: Boron halides are Lewis bases.
Answer: (c) Boron halides are Lewis acids and can accept a pair of electrons from amines to form addition product.
MCQs
11.1 Which of the following statement is incorrect?
- Non-metals and metalloids exist only in the p-block of the periodic table.
- The non-metallic character of elements decreases down the group.
- The change of non-metallic to metallic character can be best illustrated by the nature of oxides they form.
- The non-metal oxides are basic whereas metal oxides are acidic or neutral in nature.
Answer: (d) is an incorrect statement because non-metal oxides are acidic or neutral whereas metal oxides are basic in nature.
11.2 Anhydride of carbonic acid is:
(a) H2CO3 (b) CO3 (c) CO- (d) CO2
Answer: (d) CO2 is regarded as anhydride of carbonic acid.
H2CO3 ————> H2O + CO2
11.3 What is producer gas?
(a) mixture of CO and N2 in the ratio of 2 :1
(b) mixture of CO and N2 in the ratio of 1 :1
(a) mixture of CO and N2 in the ratio of 1 :2
(a) mixture of CO and N2 in the ratio of 3 :1
Answer: (a) Producer gas is a mixture of CO and N2 in the ratio of 2 :1
11.4 The state of hybridisation of B in BH4 is
(a) sp (b) sp2 (c)sp3 (d) none
Answer: (b) sp2.
11.5 The basic building unit of all silicates is
(a) SiO24- (b) SiO44- (c) SiO42- (d) SiO-
Answer: (b) SiO44- is the basic unit of all silicates.
Question Answers
Q.11.1 Why Boron differs from aluminium and other members of group 13 in a number of properties?
Answer: Boron shows anomalous behaviour due to small atomic size, high electronegativity, high ionization energy and absence of d-orbital of B.
11.2 What is catenation?
Answer:Carbon atoms have the tendency to linkwith one another through covalent bonds to
form chains and rings. This property is calledcatenation. This is because C—C bonds are
very strong.
Q.11.3 Why does the tendency for catenation decreases in the order C > Si > Ge > Sn?
Answer: The catenation depends on the strength of the element-element bond. As we move down the group 14, the element-element bond energies decrease rapidly, viz. C–C (355 kJ mol–1), Si–Si (222 kJ mol–1), Ge–Ge (167 kJ mol–1) and Sn–Sn (155 kJ mol–1), so the tendency for catenation decreases in the order C > Si > Ge > Sn.
11.4. What is dry ice? Why is it so called?
Answer: Carbon dioxide can be obtained as a solid in the form of dry ice by allowing the liquified CO2 to expand rapidly.
Unlike ordinary ice it does not melt and hence does not wet the surface on which it is kept. Thus, it is called dry ice.
11.5 CO2 is gas while SiO2 is solid at room temperature. Give reason.
Answer: The molecules of CO2 are held together by weak van der Waals forces of attraction which can be easily overcome by collisions of the molecules at room temperature. Consequently, CO2 is a gas.
On the other hand, silicon atom forms four single covalent bonds with O-atom which are tetrahedrally arranged and form a three-dimensional structure. Thus, SiO2 is a high melting solid.
11.6 What are Fullerenes? How are they prepared?
Answer: Fullerenes are the allotropes of carbon. Its structure is like a soccer ball.
They are prepared by heating graphite in electric arc in presence of inert gases such as helium or argon.
11.7 Give the chemical reaction as evidence for each of the following observations.
(i) Tin (II) is a reducing agent whereas lead (II) is not.
(ii) Gallium (I) undergoes disproportionation reaction.
Answer: (i) Due to inert pair effect Pb2+ is more stable than Pb4+. Whereas Sn4+ is more stable than Sn2+.
(ii) 3Ga+ —–> 2Ga + Ga3+
This is because Ga3+ is more stable than Ga+.
The p- Block Elements Solutions and FAQs
Q 11.1. Discuss the pattern of variation in the oxidation states of (i) B to Tl and (ii) C to Pb.
A 11.1
(i) B to Tl
Common oxidation states are +1 and +3. The stability of +3 oxidation state decreases from B to Tl while +1 oxidation state increases from B to Tl.
(ii) C to Pb
The common oxidation states are +4 and +2. Stability of +4 oxidation state decreases from C to Pb.
Q 11.2. How can you explain higher stability of BCl3 as compared to TlCl3?
A 11.2 BCl3 is quite stable. Because there is absence of d- and f-electrons in boron three valence electrons (2s2 2px1) are there for bonding with chlorine atom. In Tl the valence s-electron (6s2) are experiencing maximum inert pair effect. Thus, only 6p1 electron is available for bonding. Therefore, BCl3 is stable but TlCl3 is comparatively unstable.
Q 11.3. Why does borontrifluoride behave as a Lewis acid?
A 11.3 In BF3, central atom has only six electrons after sharing with the electrons of the F atoms. It is an electron-deficient compound and thus behaves as a Lewis acid.
Q 11.4. Consider the compounds, BCl3, and CCl4. How will they behave with water? Justify.
A 11.4 In BCl3, there is only six electrons in the valence shell of B atom. Thus, the octet is incomplete and it can accept a pair of electrons from water and hence BCl3 undergoes hydrolysis. Whereas, in CCl4, C atom has 8 electrons and its octet is complete. That’s why it has no tendency to react with water.
Q 11.5. Is boric acid a protic acid? Explain.
A 11.5 Boric acid is a Lewis acid since it accepts electrons from hydroxyl ion of H2O molecule. It is not a protic acid.
B (OH)3 + 2HOH →[B (OH)4]– + H3O+
Q 11.6. Explain what happens when boric acid is heated.
A 11.6 On heating boric acid above 370 K, it forms metaboric acid, HBO2 which on further heating yields boric oxide B2O3.
H3B2O3 → HBO2 → B2O3
Q 11.7. Describe the shapes of BF3 and BH4–. Assign the hybridisation of boron in these species.
A 11.7 In BF3, boron is sp2 hybridized.
∴ shape of BF3 = planar.
In [BH4]–, boron is sp3 hybridized, thus the shape is tetrahedral.
Q 11.8. Write reactions to justify amphoteric nature of aluminium.
- A 11.8 Aluminium reacts with acid as well as base. This shows amphoteric nature of aluminium.
- 2Al(s) + 6HCl(dil.) →2AlCl3(aq) + 3H2(g)
- 2Al(s) + 2NaOH(aq) + 6H2O(l) →2Na+ [Al(OH)4]–(aq) + 3H2(g)
Q 11.9. What are electron deficient compounds?
Are BCl3 and SiCl4 electron deficient species? Explain.
A 11.9 Electron deficient compounds are those in which the central atom in their molecule has the tendency to accept one or more electron pairs. They are also known as Lewis acid. BCl3 and SiCl4 both are electron deficient species.
Since, in BCl3, B atom has only six electrons. Therefore, it is an electron deficient compound.
In SiCl4 the central atom Si has 8 electrons but it can expand its covalency beyond 4 due to the presence of d-orbitals.
Q 11.10. Write the resonance structures of .
CO32- and HCO3–.
Q 11.11. What is the state of hybridisation of carbon in: (a) CO3^2- (b) diamond (c) graphite?
A 11.11 (a) CO32- (sp2) (b) Diamond (sp3) (c) Graphite (sp2)
Q 11.12. Explain the difference in properties of diamond and graphite on the basis of their structures.
A 11.12 Diamond has a crystalline lattice where each carbon atom undergoes sp3 hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion. The C–C bond length is 154 pm. The structure extends in space and produces a rigid three- dimensional network of carbon atoms. It is very difficult to break extended covalent bonding and, therefore, diamond is a hardest substance on the earth.
Graphite has layered structure in which the layers are held by van der Waals forces and distance between two layers is 340 pm. Each layer is composed of planar hexagonal rings of carbon atoms. C—C bond length within the layer is 141.5 pm. Each carbon atom in hexagonal ring undergoes sp2 hybridisation and makes three sigma bonds with three neighbouring carbon atoms. Fourth electron forms a π bond. The electrons are delocalised over the whole sheet. Electrons are mobile and, therefore, graphite conducts electricity along the sheet. Graphite cleaves easily between the layers and, therefore, it is very soft and slippery. That is why graphite is used as a dry lubricant in machines running at high temperature, where oil cannot be used as a lubricant.
Q 11.13 Rationalise 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.
A 11.13
- PbCl2 + Cl2→ PbCl4.
This is because Pb can show +2 oxidation state more easily than +4 due to inert pair effect. - PbCl4→ PbCl2 + Cl2
Because Pb2+ is more stable than Pb4+ due to inert pair effect. - PbI4 does not exist because I- ion being a powerful reducing agent reduces Pb4+ ion to Pb2+ ion in solution.
- Pb4+ + 2I– → Pb2+ + I2
Q 11.14. Suggest reasons why the B-F bond lengths in BF3 (130 pm) and BF4– (143 pm) differ.
A 11.14 The difference in bond length is due to the difference in the state of hybridisation. In BF3 ‘B’ is sp2 hybridised and in BF4– ‘B’ is sp3 hybridised.
Q 11.15. If B-Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.
A 11.15 B-Cl bond has dipole moment because of polarity. In BCl3,since the molecule is symmetrical (planar), the polarities cancel out and hence the dipole moment is zero
Q 11.16. 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.
A 11.16
- Since, anhydrous HF is a covalent compound and weak acid due to high bond dissociation energy. AlF3 does not dissolve in HF.
- Whereas NaF is ionic compound.
- 3NaF + AlF3→ Na3[AlF6]
- Na3[AlF6] + 3BF3(g) → AlF3 + 3Na+ [BF]–
Q 11.17. Suggest a reason as to why CO is poisonous.
A 11.17 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.
Q 11.18 How is excessive content of CO2 responsible for global warming?
A 11.18 Excess of CO2 absorbs heat radiated by the earth. Some of it is dissipated into the atmosphere while the remaining part is radiated back to the earth. As a result, temperature of the earth increases. This is the cause of global warming.
Q 11.19 Explain structures of diborane and boric acid.
A 11.19 Boric acid contains planar BO33- ions which are linked together through hydrogen bonding shown in the fig.
Q 11.20 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 11.20
- (a) When borax is heated strongly, it loses water and swells into the white mass, which on further heating melts to form a transparent glassy solid called borax glass and borax bead.
- Na2B4O7.10H2O → Na2B4O7 + 10H2O
- Na2B4O7 → 2NaBO2 + B2O3
- (b) When boric acid is added to water, it accepts electrons from –OH ion. Boric acid is sparingly soluble in cold water however fairly soluble in hot water.
- B(OH)3 + 2H2O → [B(OH)4]− + H3O+
- (c) Al reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.
- 2Al + 2NaOH + 6H2O → 2Na+[Al(OH)4]− + 3H2
- (d) BF3(a Lewis acid) reacts with NH3 (a Lewis base) to form an adduct. This results in a complete octet around B in BF3.
- BF3+:NH3→F3B←:NH3
Q 11.21 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 11.21
- (a) Silicon is heated with methyl chloride at high temperature in the presence of copper catalyst at 537 K, methyl substituted chlorosilanes MeSiCl3, Me2SiCl2, Me3SiCl and Me4Si are formed.
- (b) When silicon dioxide is treated with hydrogen fluoride, first SiF4 is formed and then hydro fluorosilicic acid is obtained.
- SiO2 + 4HF → SiF4 + 2H2O
- SiF4+2HF → H2SiF6
- (c) When CO is heated with ZnO, ZnO is reduced to Zn metal.
- CO + ZnO → CO2 + Zn
- (d) When hydrated alumina is treated with aqueous NaOH solution, it dissolves to form sodium meta aluminate.
- Al2O3.2H2O+2NaOH→2NaAlO2+3H2O
Q 11.22. 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) Aluminium utensils should not be kept in water overnight. (vii) Aluminium wire is used to make transmission cables.
A 11.22
- (i) Al reacts with conc. HNO3 to form a very thin film of aluminium oxide on its surface which protects it from further reaction. That is why conc. HNO3 can be transported in aluminium container.
- 2Al(s) + 6HNO3(conc.) → Al2O3(s) + 6NO2(g) + 3H2O(l)
- (ii) NaOH reacts with Al to evolve H2 gas. Thus, the pressure of the gas produced can be used for opening or cleaning clogged drains.
- 2Al(s) + 2NaOH(aq) + 2H2O(l) → 2NaAlO2(aq) + 3H2(g)
- (iii) Graphite has layered structure which are held by weak van der Waals forces. Thus, graphite cleaves easily between the layers, therefore it is very soft and slippery. That is why it is used as lubricant.
- (iv) Diamond is used as anabrasive because it is an extremely hard substance.
- (v) Alloys of aluminiumis used to make aircraft body due to some of its property like toughness, lightness and resistant to corrosion.
- (vi) Generally, aluminium metal does not react with water quickly but, when it is kept overnight, it reacts slowly with water in presence of air.
- 2Al(s) + O2(g) + H2O(l) → Al2O3(S) + H2(g) a very small amount of (in ppm) Al3+ produced in the solution is injurious to health if the water is used for drinking purposes.
- (vii) Aluminium is generally unaffected by air and moisture and it is also good conductor of electricity. That is why it is used in transmission cables.
Q 11.23. Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?
A 11.23 Because there is increase in atomic size on moving from carbon to silicon, the screening effect increases. Thus, the force of attraction of nucleus for the valence electron in silicon decreases as compared to carbon. Therefore, the ionization enthalpy decreases from carbon to silicon.
Q 11.24. How would you explain the lower atomic radius of Ga as compared to Al?
A 11.24 Due to poor shielding effect of d-electrons in Ga, the electrons in gallium experience great force of attraction by nucleus as compared to Al. That is why Ga has lower atomic radius as compared to Al.
Q 11.25. 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?
A 11.25 Allotropes: Allotropes are the different forms of an element which are having same chemical properties but different physical properties due to their structures.
In diamond, carbon is sp3 -hybridized. Since, diamond is three-dimensional network solid, it is hardest substance with high density whereas graphite has a layered structure. The various layers are formed by van der Waals forces of attraction that’s why graphite is soft and slippery.
Q 11.26. (a) Classify following oxides as neutral, acidic, basic or amphoteric: (b) Write suitable equations to show their nature.
CO, B2O3, SiO2, CO2, Al2O3, PbO2, Tl2O3
- (a) Neutral — CO
Acidic — B2O3, SiO2, CO2 - Basic — Tl2O3
- Amphoteric — Al2O3,PbO2
(b)-CO does not react with acid as well as base at room temperature.
Being acidic B2O3, SiO2 and CO2 react with alkalis to form salts. - B2O3 + 2NaOH à 2NaBO2 + H2O
- SiO2 + NaOH à 2Na2SiO3 + H2O
- CO2 + 2NaOH à Na2CO3 + H2O
- Being Amphoteric, Al2O3, PbO2 react with acids and bases.
- Al2O3 + 2NaOH à 2NaAlO2 + H2O
- Al2O3 + 3H2SO4à (Al2SO4)3+ 3H2O
- PbO2 + 2 NaOH à Na2PbO3 + H2O
- 2PbO2 + 2 H2SO4à 2PbSO4 + 2H2O + O2
- Being basic Tl2O3 dissolves in acids
- Tl2O3 + 6HCl à 2TlCl3 + 3H2O
Q 11.27. In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.
A 11.27 Tl belongs to group 13 and shows both the oxidation state +1 and +3 due to inert pair effect. Tl forms basic oxide like group I elements. TlO2 is strongly basic.
Q 11.28. 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.
A 11.28
- The compounds X, A, B, C and D are aluminium, aluminium hydroxide, sodium tetrahydrozoaluminate (III), aluminium chloride and alumina.
- Aluminium reacts with NaOH to form white ppt of Al(OH)3.
- 2Al + 3NaOH → Al(OH)3↓ + 3Na+
- Al(OH)3 reacts with NaOH to form Na+[Al(OH)4]−.
- Al(OH)3 + NaOH → Na+(Al(OH)4)−
- Al(OH)3 reacts with HCL to form AlCl3.
- Al(OH)3 + 3HCl → AlCl3 + 3H2O
- When Al2O3 is heated, Al2O3 is obtained.
- 2Al(OH)3 → Al2O3 + 3H2O
Q 11.29 What do you understand by (a) inert pair effect (b) allotropy and (c) catenation?
A 11.29
(a) Inert pair effect: When the pair of electrons in the valence shell does not take part in bond formation, then this effect is called as inert pair effect.
(b)Allotropy: It is the property of the element by which an element can exist in two or more forms which have same chemical properties but different physical properties due to their structures.
(c)Catenation: The tendency to link with one another through covalent bonds to form chains and rings. This property is called catenation.
For example, carbon forms chains with (C-C) single bonds and also with multiple bonds (C = C or C = C).
Q 11.30. 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. (Advance)
A 11.30 The compounds X, Y and Z are borax, sodium metaborate + boric anhydride and boric acid respectively.
- When borax is heated, it first swells and then forms a transparent glass like bead of sodium meta borate and boric anhydride.
Na2B4O7à 2NaBO2 + B2O3 + 10H2O
(Borax) (sodium metaborate) (Boric anhydride)
- Aqueous solution of borax is alkaline due to formation of strong base NaOH.
- Hence, it turns red litmus blue.
- Na2B4O7 +7 H2O → 4H3BO3 + 2NaOH
- Borax reacts with sulphuric acid to form boric acid and sodium sulphate.
- Na2B4O7 + H2SO4 + 5 H2O → 4H3BO3 + Na2SO4
11.31. Write balanced equations for: Write balanced equations for: (i) BF3 + LiH → (ii) B2H6 + H2O → (iii) NaH + B2H6 → (iv) H3BO3(heat) → (v) Al + NaOH → (vi) B2H6 + NH3 →
A 11.31
The balanced equations are given below:
(i) 2BF3 + 6LiH → B2H6 + 6LiF
(ii) B2H6 + 6H2O → 2H3BO3(orthoboric acid)
(iii) 2NaH + B2H6 →2Na[BH4](sodium borohydride)
(iv) H3BO3 →HBO2(metaboric acid) + H2O
4HBO2 →H2B4O7 → 2B2O3(boron trioxide) + H2O
(v) Al+3NaOH→Al(OH)3 + 3Na
3 B2H6 + 6NH3 → 2B3N3H6 + 12H2
Q 11.32. Give one method for industrial preparation and one for laboratory preparation of CO and CO2 each.
A 11.32 Laboratory preparation of carbon monoxide:
Formic acid is dehydrated with concentrated sulphuric acid at 373 K.
HCOOH → H2O + CO↑
Commercial preparation of CO:
Steam is passed over hot coke.
C + H2O → CO + H2
Laboratory preparation of carbon dioxide:
Calcium carbonate reacts with dilute HCl to form carbon dioxide.
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
Industrial preparation of carbon dioxide:
Limestone is heated to produce carbon dioxide.
CaCO3 → CaO + CO2
Q 11.33. An aqueous solution of borax is (a) neutral (b) amphoteric (c) basic (d) acidic
A 11.33 Borax is a salt of a strong base (NaOH) and a weak acid (H3BO3), therefore, it is basic in nature, i.e., option (c) is correct.
Q 11.34. Boric acid is polymeric due to (a) its acidic nature (b) the presence of hydrogen bonds (c) its monobasic nature (d) its geometry
A 11.34 Boric acid is polymeric due to the presence of H-bonds. Therefore, option (b) is correct.
Q 11.35. The type of hybridisation of boron in diborane is (a) sp (b) sp2(c) sp3(d) dsp2
A 11.35 In B2H6, B is sp3-hybridized. Therefore, option (c) is correct.
Q 11.36 Thermodynamically the most stable form of carbon is (a)diamond (b) graphite (c) fullerenes (d) coal
A 11.36 Thermodynamically the most stable form of carbon is graphite, i.e., option (b) is correct.
Q 11.37. Elements of group 14 (a) exhibit oxidation state of +4 only (b) exhibit oxidation state of +2 and +4 (c) form M2-and M4+ ions (d) form M2+ and M4+ ions.
A 11.37 Due to inert pair effect, elements of group 14 exhibit oxidation states of +2 and +4. Thus, option (b) is correct.
Q 11.38. If the starting material for the manufacture of silicones is RSiCl3, write the structure of the product formed. (Advance)
A 11.38 Hydrolysis of aikyltrichlorosilanes gives cross-linked silicones.
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