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The P - Block Elements

Class 11th Chemistry Part Ii Bihar Board Solution
Exercise
  1. Discuss the pattern of variation in the oxidation states of (i) B to Tl (ii) C to Pb.…
  2. How can you explain higher stability of BCl3 as compared to TlCl3 ?…
  3. Why does boron triflouride behave as a Lewis acid?
  4. Consider the compounds, BCl3 and CCl4. How will they behave with water? Justify.…
  5. Is boric acid a protic acid? Explain.
  6. Explain what happens when boric acid is heated.
  7. Describe the shapes of BF3 and BH4-. Assign the hybridization of boron in these species.…
  8. Write reactions to justify amphoteric nature of aluminium.
  9. What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species?…
  10. Write the resonance structures of CO32- and HCO3-.
  11. What is the state of hybridization of carbon in (a) CO32- (b) diamond (c) graphite?…
  12. Explain the difference in properties of diamond and graphite on the basis of their…
  13. Rationalise the given statements and give chemical reactions: • Lead(II) chloride reacts…
  14. Suggest reasons why the B-F bond lengths in BF3 (130 pm) and BG4- (143 pm) differ.…
  15. If B-Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.…
  16. Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF.…
  17. Suggest a reason as to why CO is poisonous.
  18. How is excessive content of CO2 responsible for global warming?
  19. Explain structures of diborane and boric acid.
  20. What happens when (a) Borax is heated strongly, (b) Boric acid is added to water, (c)…
  21. Explain the following reactions (a) Silicon is heated with methyl chloride at high…
  22. Conc. HNO3 can be transported in aluminium container. Give reasons:…
  23. A mixture of dilute NaOH and aluminium pieces is used to open drain. Give reasons:…
  24. Graphite is used as lubricant. Give reasons:
  25. Diamond is used as an abrasive. Give reasons:
  26. Aluminium alloys are used to make aircraft body. Give reasons:
  27. Aluminium utensils should not be kept in water overnight. Give reasons:…
  28. Aluminium wire is used to make transmission cables. Give reasons:…
  29. Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?…
  30. How would you explain the lower atomic radius of Ga as compared to Al?…
  31. What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and…
  32. (a) Classify following oxides as neutral, acidic, basic or amphoteric: CO, B2O3, SiO2,…
  33. In some of the reactions thallium resembles aluminium, whereas in others it resembles with…
  34. When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which…
  35. What do you understand by (a) inert pair effect (b) allotropy and (c) catenation?…
  36. A certain salt X, gives the following results. (i) Its aqueous solution is alkaline to…
  37. Write balanced equations for: (i) BF3 + LiH → (ii) B2H6 + H2O → (iii) NaH + B2H6→ (iv) (v)…
  38. Give one method for industrial preparation and one for laboratory preparation of CO and…
  39. An aqueous solution of borax isA. neutral B. amphoteric C. basic D. acidic…
  40. Boric acid is polymeric due toA. its acidic nature B. the presence of hydrogen bonds C.…
  41. The type of hybridisation of boron in diborane isA. sp B. sp^2 C. sp^3 D. dsp^2…
  42. Thermodynamically the most stable form of carbon isA. diamond B. graphite C. fullerenes D.…
  43. Elements of group 14A. exhibit oxidation state of + 4 only B. exhibit oxidation state of +…
  44. If the starting material for the manufacture of silicones is RSiCl3, write the structure…

Exercise
Question 1.

Discuss the pattern of variation in the oxidation states of

(i) B to Tl

(ii) C to Pb.


Answer:

(i) Group 13 elements have their electronic configuration of ns2 np1 and the oxidation state exhibited by these elements should be 3. Apart from these two electrons boron and aluminum, other elements of this group exhibit both + 1 and + 3 oxidation states. Boron and aluminum show oxidation state of + 3. This is because of the inert pair effect. The two electrons, which are present in the S-shell do not participate in bonding as they are strongly attracted by the nucleus. As we move down the group, the inert pair effect become more prominent. Therefore Ga ( + 1) is unstable and TI ( + 1) is very stable


On moving down the group, the stability of the + 3 oxidation state gets decreased.



(ii) The electronic configuration of group 14 elements is ns2 np2. Hence, the most common oxidation state exhibited by them should be + 4. On moving down the group, the + 2 oxidation state becomes more and more common and the higher oxidation state becomes less stable because of the inert pair effect. Si and C mostly show the + 4 state. Although Sn, Ge, and Pb show both the + 4 and + 2 states, the stability of higher oxidation states decreases than lower oxidation state on moving down the group.




Question 2.

How can you explain higher stability of BCl3 as compared to TlCl3 ?


Answer:

Thallium and boron belong to group 13 of the periodic table and + 1 oxidation state becomes more stable on moving down the group. Boron is more stable than thallium because + 3 state of thallium is highly oxidizing and it reverts back to more stable + 1 state.



Question 3.

Why does boron triflouride behave as a Lewis acid?


Answer:

The electronic configuration of boron is ns2 np1. It contains 3 electrons in its valence shell. Thus, it can form only 3 covalent bonds which means that there are only 6 electrons around boron and its octet remains incomplete. When 1of the boron’s atom combines with 3 fluorine atoms, its octet (8) remains incomplete. Therefore, boron trifluoride remains electron-deficient and acts as Lewis acid.


Question 4.

Consider the compounds, BCl3 and CCl4. How will they behave with water? Justify.


Answer:

Being a Lewis acid, BCl3 readily undergoes hydrolysis. Boric acid is formed as a result.

BCl3 + 3H2O → 3HCl + B(OH)3


CCl4 completely resists hydrolysis. Carbon does not have any vacant orbital. Therefore, it cannot accept electrons from water to form an intermediate. Separate layers are formed when CCl4 and water are mixed.


CCl4 + H2O → NoReaction



Question 5.

Is boric acid a protic acid? Explain.


Answer:

Boric acid is a weak monobasic acid which behaves as a Lewis acid. So, it is not a protic acid.

B(OH)3 + 2HOH → [B(OH)4]- + H3O +


It behaves as an acid by accepting a pair of electrons from OH ion.



Question 6.

Explain what happens when boric acid is heated.


Answer:

On heating orthoboric acid at 370 K or above, it changes to metaboric acid and On further heating, this yields boric oxide B2O3.

H3BO3→ HBO2(metaboric acid) → B2O3 (boric acid)



Question 7.

Describe the shapes of BF3 and BH4. Assign the hybridization of boron in these species.


Answer:

(i) BH4-

Boron-hydride ion (BH4 ) is formed by the sp3 hybridization of boron orbitals. Therefore, it is a tetrahedral structure.



(ii) BF3


As a result of its small size and high electronegativity, boron tends to form monomeric covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by the overlap of three sp2 hybridized orbitals of boron with the sp orbitals of 3 halogen atoms. Boron is sp2 hybridized in BF3.




Question 8.

Write reactions to justify amphoteric nature of aluminium.


Answer:

A substance is called amphoteric if it displays both characteristics of acids and bases. Aluminium dissolves in both acids and bases, showing amphoteric behaviour.

(i) 2Al(s) + 6HCl(aq) → 2Al3 + (aq) + 6Cl(aq) + 3H2(g)


(ii) 2Al(s) + 2NaOH(aq) + 6H2O(l) → 2Na + [Al(OH)4](aq) + 3H2(g)



Question 9.

What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species? Explain.


Answer:

In an electron-deficient compound, the octet of electrons is not complete, i.e., the central metal atom has an incomplete octet. Hence, it needs electrons to complete its octet.

(i) SiCl4


The electronic configuration of silicon is ns2 np2 . This indicates that it has 4 valence electrons. After it forms 4 covalent bonds with 4 chlorine atoms, its electron count increases to 8. Thus, SiCl4 is not an electron-deficient compound.


(ii) BCl3


It is an appropriate example of an electron-deficient compound. B has three valence electrons. After forming 3 covalent bonds with chlorine, the number of electrons around it increases to six. However, it is still short of 2 electrons to complete its octet.



Question 10.

Write the resonance structures of CO32- and HCO3-.


Answer:

(i) HCO3-



(ii) CO32-




Question 11.

What is the state of hybridization of carbon in (a) CO32– (b) diamond (c) graphite?


Answer:

The state of hybridization of carbon in:


(a) Graphite


1. Each carbon atom in graphite is sp2 hybridized and is bound to 3 other carbon atoms.


2. Graphite exists as sheets of hexagonal arrays of carbon. Each carbon atom in graphite thus has a trigonal planar geometry, which implies sp2 hybridization. Moreover, the p orbitals axial to the hexagonal sheets help to bond the layering sheets together.


(b) CO32-


1. C in CO32 is sp2 hybridized and is bonded to 3 oxygen atoms.


2. No. of single bonds between carbon and other atoms:3


3. No. of lone pairs on carbon atom:0


4. Now sum up the bond pairs and lone pairs = 3 + 0=3. sp2 that is, one s and two p orbitals.


(c) Diamond


1. Each carbon in diamond is sp3 hybridized and is bound to 4 other carbon atoms.


2. The electronic configuration of carbon is 1s2 2s2 2p2, i.e. with four valence electrons spread in the s and p orbitals.


3. In order to create covalent bonds in diamond, the s orbital mixes with the three p orbitals to form sp3 hybridization.



Question 12.

Explain the difference in properties of diamond and graphite on the basis of their structures.


Answer:




Question 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.


Answer:

• Lead belongs to group fourteen of the periodic table. The two oxidation states displayed by this group is + 2 and + 4. On moving down the group, the + 2 oxidation state becomes more stable and the + 4 oxidation state becomes less stable. This is because of the inert pair effect. Hence, PbCl4 is much less stable than PbCl2. However, the formation of PbCl4 takes place when chlorine gas is bubbled through a saturated solution of PlCl2.

• On moving down group IV, the higher oxidation state becomes unstable because of the inert pair effect. Pb(IV) is highly unstable and when heated, it reduces to Pb(II).


• Lead is known not to form PbI4. Pb ( + 4) is oxidizing in nature and I is reducing in nature. A combination of Pb(IV) and iodide ion is not stable. Iodide ion is strongly reducing in nature. Pb(IV) oxidizes I to I2 and itself gets reduced to Pb(II).


PbI4 → PbI2 + I2



Question 14.

Suggest reasons why the B–F bond lengths in BF3 (130 pm) and BG4 (143 pm) differ.


Answer:

The B–F bond length in BF3 is shorter than the B–F bond length in 𝐵𝐹4-.

BF3is an electron deficient species. With a vacant p-orbital on boron, the fluorine and boron atoms undergo pn–pn back-bonding to remove this deficiency. This imparts a double bond character to the B–F bond.



This double-bond character causes the bond length to shorten in BF3 (130 pm). However, when BF3 coordinates with the fluoride ion, a change in hybridization from sp2 (in BF3) to sp3 (in 𝐵𝐹4) occurs. Boron now forms 4σ bonds and the double-bond character is lost. This accounts for a B–F bond length of 143 pm in 𝐵𝐹4 ion.




Question 15.

If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.


Answer:

As a result of the difference in the electronegativities of Cl and B, the B–Cl bond is naturally polar. However, the BCl3 molecule is non-polar. This is because BCl3 is trigonal planar in shape. It is a symmetrical molecule. Hence, the respective dipole moments of the B–Cl bond cancel each other, thereby causing a zero dipole moment.



Question 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.


Answer:

Hydrogen fluoride is a covalent compound and has a very strong intermolecular hydrogen-bonding. Thus, it does not provide ions and aluminum fluoride does not dissolve in it. Sodium fluoride is an ionic compound and when it is added to the mixture, Alf dissolves. This is because of the availability of free F. The reaction involved in the process is:

AlF3 + 3NaF → Na3[AlF6]


Aluminum fluoride gets precipitated out of the solution when boron trifluoride is added to the solution. This happens because the tendency of boron to form complexes is much more than that of aluminum. Therefore, when boron trifluoride is added to the solution, B replaces Al from the complexes according to the following reaction:


Na3[AlF6] + 3BF3→ 3Na[BF4] + AlF3



Question 17.

Suggest a reason as to why CO is poisonous.


Answer:

Carbon monoxide is highly poisonous due to its ability to form a complex with hemoglobin. The former prevents Hb from binding with oxygen. Thus, a person dies because of suffocation on not receiving oxygen The CO–Hb complexly is more stable than the O2–Hb complex. It is found that the CO–Hb complex is about 300 times more stable than the O2–Hb complex.



Question 18.

How is excessive content of CO2 responsible for global warming?


Answer:

Carbon dioxide is an essential gas for our survival. However, an increased content of CO2 in the atmosphere poses a serious threat. An increment in the combustion of fossil fuels, decomposition of limestone, and a decrease in the number of trees has led to greater levels of carbon dioxide. Carbon dioxide has the property of trapping the heat provided by sun rays. Higher the level of carbon dioxide, higher is the amount of heat trapped. This results in an increase in the atmospheric temperature, thereby causing global warming.



Question 19.

Explain structures of diborane and boric acid.


Answer:

(a) Diborane

B2H6 is an electron-deficient compound. B2H6 has only 12 electrons – 6 e to 6 H atoms and 3 eeach from 2 B atoms. Thus, after combining with 3 H atoms, none of the boron atoms has any electrons left. X-ray diffraction studies have shown the structure of diborane as:



2 boron and 4 terminal hydrogen atoms (Ht) lie in one plane, while the other two bridging hydrogen atoms (Hb) lie in a plane perpendicular to the plane of boron atoms. Again, of the two bridging hydrogen atoms, one H atom lies above the plane and the other lies below the plane. The terminal bonds are regular two-center two-electron (2c – 2e ) bonds, while the two bridging (B–H–B) bonds are three-center two-electron (3c – 2e ) bonds.


(b) Boric acid has a layered structure. Each planar BO3 unit is linked to one another through H atoms. The H atoms form a covalent bond with a BO3 unit, while a hydrogen bond is formed with another BO3 unit. In the given figure, the dotted lines represent hydrogen bonds.




Question 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?


Answer:

(a) When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.


(b) When boric acid is added to water, it accepts electrons from OH ion.


B(OH)3 + 2HOH → [B(OH)4] + H3O +


(c) Aluminium reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.


2Al(s) + 2NaOH(aq) + 6H2O(l) → 2Na + [Al(OH)4](aq) + 3H2(g)


(d) BF3 (a Lewis acid) reacts with NH3 (a Lewis base) to form a product. This results in a complete octet around B in BF3.


F3B + :NH3 → F3B ←:NH3



Question 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.


Answer:

(a) When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about 537 K, a class of organosilicon polymers called methyl substituted chlorosilane MeSiCl3, Me2SiCl2, Me3SiCl, and Me4Si) are formed.



(b) When silicon dioxide (SiO2) is heated with hydrogen fluoride (HF), it forms silicon tetrafluoride (SiF4). Usually, the Si–O bond is a strong bond and it resists any attack by halogens and most acids, even at a high temperature. However, it is attacked by HF.


SiO2 + 4HF → SiF4 + 2H2O


The SiF4 formed in this reaction can further react with HF to form hydro-fluorosilicic acid.


SiF4 + 2HF → H2SiF6


(c) When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.


ZnO(s) + CO(g) → Zn(s) + CO2(g)


(d) When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.


Al2O3.2H2O + 2NaOH → 2NaAlO2 + 3H2O



Question 22.

Give reasons:

Conc. HNO3 can be transported in aluminium container.


Answer:

Concentrated HNO3 can be stored and transported in aluminium containers as it reacts with aluminium to form a thin protective oxide layer on the aluminium surface. This oxide layer renders aluminium passive.



Question 23.

Give reasons:

A mixture of dilute NaOH and aluminium pieces is used to open drain.


Answer:

Sodium hydroxide and aluminium react to form sodium tetrahydroxyaluminate(III) and hydrogen gas. The pressure of the produced hydrogen gas is used to open blocked drains.

2Al + 2NaOH + 6H2O → 2Na + [Al(OH)4] + 3H




Question 24.

Give reasons:

Graphite is used as lubricant.


Answer:

Graphite has a layered structure and different layers of graphite are bonded to each other by weak van der Waals’ forces. These layers can slide over each other. Graphite is soft and slippery. Therefore, graphite can be used as a lubricant.



Question 25.

Give reasons:

Diamond is used as an abrasive.


Answer:

In diamond, carbon is sp3 hybridized. Each carbon atom is bonded to 4 other carbon atoms with the help of strong covalent bonds. These covalent bonds are present throughout the surface, giving it a very rigid 3-D structure. It is very difficult to break this extended covalent bonding and for this reason, diamond is the hardest substance known. Thus, it is used as an abrasive and for cutting tools.



Question 26.

Give reasons:

Aluminium alloys are used to make aircraft body.


Answer:

Aluminum has the high tensile strength and it is light weight. It can also be alloyed with various metals such as Si, Mg, Cu, Mn, and Zn. It is very malleable and ductile. Therefore, it is used in making of aircraft bodies.



Question 27.

Give reasons:

Aluminium utensils should not be kept in water overnight.


Answer:

The oxygen present in water reacts with aluminum to form a thin layer of aluminum oxide. This layer prevents aluminum from further reaction. However, when water is kept in an aluminum vessel for long periods of time, some amount of aluminum oxide may dissolve in water. As aluminum ions are harmful, water should not be stored in aluminum vessels overnight.



Question 28.

Give reasons:

Aluminium wire is used to make transmission cables.


Answer:

Silver, copper, and aluminum are among the best conductors of electricity. Silver is an expensive metal and silver wires are very expensive. Copper is quite expensive and is also very heavy. Aluminum is a very ductile metal. Thus, aluminum is used in making wires for electrical conduction.



Question 29.

Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?


Answer:

Ionization enthalpy of carbon (the first element of group 14) is very high (1086 kJ/mol). This is expected owing to its small size. However, on moving down the group to silicon, there is a sharp decrease in the enthalpy (786 kJ). This is because of an appreciable increase in the atomic sizes of elements on moving down the group.



Question 30.

How would you explain the lower atomic radius of Ga as compared to Al?


Answer:

Atomic radius in pm



Although Ga has one shell more than Al, its size is lesser than Al. This is because of the poor shielding effect of the 3d-electrons. The shielding effect of d-electrons is very poor and the effective nuclear charge experienced by the valence electrons in gallium is much more than it is in the case of Al.



Question 31.

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?


Answer:

Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes.


The rigid 3-D structure of diamond makes it a very hard substance. In fact, diamond is one of the hardest naturally-occurring substances. It is used as an abrasive and for cutting tools.


It has sp2 hybridized carbon, arranged in the form of layers. These layers are held together by weak van der Walls’ forces. These layers can slide over each other, making graphite soft and slippery. Therefore, it is used as a lubricant.



Question 32.

(a) Classify following oxides as neutral, acidic, basic or amphoteric:

CO, B2O3, SiO2, CO2, Al2O3, PbO2, Tl2O3

(b) Write suitable chemical equations to show their nature.


Answer:

SiO2= Acidic

Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium silicate.


SiO2 + 2NaOH → 2Na2SiO3 + H2O


Tl2O3 = Basic


Being basic, it reacts with acids to form salts. It reacts with HCl to form thallium chloride.


Tl2O3 + 6HCl → 2TlCl3 + 3H2O


CO = Neutral


CO2 = Acidic


Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium carbonate.


CO2 + 2NaOH → Na2CO3 + H2O


B2O3 = Acidic


Being acidic, it reacts with bases to form salts. It reacts with NaOH to form sodium metaborate.


B2O3 + 2NaOH → 2NaBO2 + H2O


PbO2 = Amphoteric


Amphoteric substances react with both acids and bases. PbO2 reacts with both NaOH and H2SO4.


PbO2 + 2NaOH → Na2PbO3 + H2O


2PbO2 + 2H2SO4→ 2PbSO4 + 2H2O + O2


Al2O3= Amphoteric


Amphoteric substances react with both acids and bases. Al2O3 reacts with both NaOH and H2SO4.


Al2O3 + 2NaOH → NaAlO2


Al2O3 + 3H2SO4→ Al2(SO4)3 + 3H2O



Question 33.

In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.


Answer:

Thallium belongs to group 13 of the periodic table. The most common oxidation state for this group is + 3. However, heavier members of this group also display the + 1 oxidation state. This happens because of the inert pair effect. Aluminum displays the + 3 oxidation state and alkali metals display the + 1 oxidation state. Thallium displays both the oxidation states. Therefore, it resembles both aluminum and alkali metals. Thallium, like aluminum, forms compounds such as TlCl3 and Tl2O3. It resembles alkali metals in compounds Tl2O and TlCl.



Question 34.

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.


Answer:

The given metal X gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, X must be aluminum. The white precipitate (compound A) obtained is aluminum hydroxide. The compound B formed when an excess of the base is added is sodium tetrahydroxy aluminate(III)

Now, when dilute hydrochloric acid is added to aluminum hydroxide, aluminum chloride (compound C) is obtained.


Also, when compound A is heated strongly, it gives compound D. This compound is used to extract metal X. Aluminium metal is extracted from alumina. Hence, compound D must be alumina.



Question 35.

What do you understand by (a) inert pair effect (b) allotropy and (c) catenation?


Answer:

(a) Inert-pair effect

As one moves down the group, the tendency of s-block electrons to participate in chemical bonding decreases. This effect is known as inert pair effect. In case of group of group 13 elements, the electronic configuration is ns2np1 and their group valency is + 3. However, on moving down the group, the + 1 oxidation state becomes more stable. This happens because of the poor shielding of the ns2 electrons by the d- and f- electrons. As a result of the poor shielding, the ns2 electrons are held tightly by the nucleus and so, they cannot participate in chemical bonding.


(b) Allotropy


Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes. For example: carbon exists in three allotropic forms: diamond, graphite and fullerenes.


(c) Catenation


The atoms of some elements such as carbon can link with one another through strong covalent bonds to form long chains or branches. This property is known as catenation. It is most common in carbon and quite significant in Si and S.



Question 36.

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.


Answer:

The given salt is alkaline to litmus. Therefore, X is a salt of a strong base and a weak acid. Also, when X is strongly heated, it swells to form substance Y. therefore, X must be borax. When borax is heated, it loses water and swells to form sodium metaborate. When heating is continued, it solidifies to form a glassy material Y. hence Y must be a mixture of sodium metaborate and boric anhydride.

Na2B4O7 + 7H2O → 2NaOH + 4H3BO3



When concentrated acid is added to borax, white crystals of orthoboric acid(Z) are formed.



Z=Orthoboric acid



Question 37.

Write balanced equations for:

(i) BF3 + LiH →

(ii) B2H6 + H2O →

(iii) NaH + B2H6

(iv)

(v) Al + NaOH →

(vi) B2H6 + NH3


Answer:

(i) 2BF3 + 6LiH → B2H6 + 6LiF


(ii) B2H6 + 6H2O → 2H3BO3 + 6H2


(iii) 2NaH + B2H6→ 2NaBH4


(iv) 4H3BO3 → 4HBO2→ H2B4O7→ 2B2O3


(v) 2Al + 2NaOH → 2Na + [Al(OH)4]-(aq) + 3H2


(vi)



Question 38.

Give one method for industrial preparation and one for laboratory preparation of CO and CO2 each.


Answer:

(a) Carbon dioxide

In the laboratory, CO2 can be prepared by the action of dilute hydrochloric acid on calcium carbonate. The reaction involved is as follows:-


CaCO3 + 2HCl(aq)→ CaCl2(aq) + CO2(g) + H2o


CO2 is commercially prepared by heating limestone. The reaction involved is:-


CaCO3→ CaO + CO2


(b) Carbon monoxide


In the laboratory, CO is prepared by the dehydration of formic acid with conc. H2SO4 at 373k. the reaction involved is:-


HCOOH H2O + CO↑


CO is commercially prepared by [assing steam over hot coke. The temperature should be 473-1273K. the reaction is:-


C(s) + H2O → CO(g) + H2(g)



Question 39.

An aqueous solution of borax is:
A. neutral

B. amphoteric

C. basic

D. acidic


Answer:

Borax is a salt of a strong base(NaOH) and a weak acid (H3BO3). It is therefore basic in nature.


Question 40.

Boric acid is polymeric due to
A. its acidic nature

B. the presence of hydrogen bonds

C. its monobasic nature

D. its geometry


Answer:

Boric acid is polymeric because of the presence of hydrogen bonds. In the given figure, the dotted lines represent hydrogen bonds.



Question 41.

The type of hybridisation of boron in diborane is
A. sp

B. sp2

C. sp3

D. dsp2


Answer:

Boron in diborane is sp3 hybridised.


Question 42.

Thermodynamically the most stable form of carbon is
A. diamond

B. graphite

C. fullerenes

D. coal


Answer:

Graphite is thermodynamically the most stable form of carbon.


Question 43.

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


Answer:

The elements of group 14 have 4 valence electrons. Therefore, the oxidation state of the group is + 4. However, as a result of inert pair effect, the lower oxidation state becomes more and more stable and the higher oxidation state becomes less stable.


Therefore, this group exhibits + 4 and + 2 oxidation states.



Question 44.

If the starting material for the manufacture of silicones is RSiCl3, write the structure of the product formed.


Answer:

(i)


(ii)