WBCHSE Class 11 Chemistry Some P Block Elements Notes

Class 11 Chemistry Some P Block Elements Group-13 Elements (Boron Family) Introduction

The valence shell electronic configuration of the elements of group-13 is ns²np¹ where n = 2-6.  It becomes clear from the electronic configurations that boron (B) and aluminium (Al) have noble gas cores, gallium (Ga) and indium (In) have noble gas cores plus 10 d-electrons and thallium (Tl) have noble gas cores plus 14 F -electrons and 10 d-electrons. Thus electronic configuration of the elements of group 13 is more complex compared to those of groups and 2.

This difference in electronic configuration affects the chemistry of the elements of this group.

Electronic configuration of group 13 – elements:

Occurrence Of Group-13 Elements

1. The elements present in group 13 of the periodic table are boron (B), aluminium (Al), gallium (Ga), indium (In) and thallium (Tl). Except for horon, which is a non-metal, all other elements of this group are metals. The non-metallic character of boron is due to its small atomic size, high ionization enthalpy and comparatively high electronegativity.

2. Boron is a fairly rare element which occurs to a very small extent (0.0001% by mass) in the earth’s crust Natural boron consists of two isotopes: ¹⁰B (19%) and ¹¹B (81%). Boron does not occur in free state as it is highly reactive.It occurs mainly as orthoboric acid & as minerals like

Borax –  Na₂(B₄O₅(OH₄).8H₂O -8H₂O

Kernite – Na₂(B₄O₅(OH₄).2H₂O

Colemanite – Na₂(B₃O₄(OH₃)₂.2H₂O

3. Aluminium is the most abundant metal, and the third most abundant element (8.3%by mass) in the earth’s crust after oxygen (45.5%) and silicon (27.7%). The important minerals of aluminium are:

Bauxite – (Al₂O₃ – 2H₂O)

Cryolite -(Na₃AlF₆)

Orthoclase (feldspar) – KAlSi₃O₈, 

Mica (Muscovite)-  KAl ₂(AlSi₃O₁₀)(F, OH)₂ etc.

4. Gallium, indium and thallium are quite less abundant and occur in traces in sulphide minerals.

5. The highest concentration of Ga (0.1-1%) is found in a rare mineral known as germanite (a sulphide complex of Zn, Cu, Ge and As).

6. Traces of and Tl are available in sulphide ores of zinc and lead respectively

General Trends In Atomic And Physical Properties Of Group-13 Elements

Some important atomic and physical properties of group-13 most elements are given in the following table

Some atomic and physical properties of group-13 elements:

Trends in different atomic and physical properties of group-13 elements with explanations:

Atomic and ionic radii

1. Atomic and ionic radii of group-13 elements are smaller as compared to the corresponding elements of group-2.

Explanation:

On moving from left to right in the periodic table, i.e., on moving from group-2 to group-13in a given period, the magnitude of nuclear charge increases but the new electron is added to the same shell. Since the electrons in the same shell do not screen each other and the effective nuclear charge increases, the outermost electrons experience greater nuclear charge and are pulled more strongly towards the nucleus. As a result, atomic size decreases. The same is true in the case of ionic radius.

2. On moving down the group, both atomic and ionic radii are expected to increase due to the addition of new electronic shells. However, the observed atomic radius of Ga = 135pm is slightly lesser than that of Al = 143pm.

Explanation:

On moving from Al (Z = 13) to Ga (Z = 31), the d-orbitals are filled by electrons. Since the d-orbitals are larger, these intervening electrons in d-orbitals do not screen the nucleus effectively. As a result, the effective nuclear charge experienced by the electrons in Ga is greater than that experienced by the electrons in Al. Hence, the atomic radius of Ga is slightly less than that of Al. The Ionic radii, however, increase regularly on moving down the group

Ionisation enthalpy

1. First ionisation enthalpies (Δ_iH₁) of group-13 elements are lower than the corresponding elements ofgroup-2.

Explanation:

• The first electron, in the case of group-13 elements (ns²np¹), is to be removed from a p-orbital, while in the case of group-2 elements, it is to be removed from an s-orbital.
• Due to greater penetration of the s-orbital, the s-electron is nearer to the nucleus and is more tightly held by the nucleus than a p-electron of the same principal shell.
• The removal of the s-electron requires a greater amount of energy compared to p- the electron and because of this, the values of first ionisation enthalpies (Δ_iH_i) of the elements of group-13 are low as compared to the corresponding elements of group-2.
• The second and third ionisation enthalpies of these elements are, however, quite high because the second and third electrons are to be removed from ns-orbital.

2. On moving down the group from B to Al, the first ionisation enthalpy, (Aÿ) decreases sharply. However, the value of (Δ_iH₁) of Ga is slightly higher than that ofAl, while that ofTl is much higher than that of.

Explanation:

1. The sharp decrease in (Δ_iH₁) value from B to Al is expected because an increase in atomic size and screening effect (caused by to addition of a new shell) outweighs the effect of ‘increased nuclear charge.

2. The element Ga has ten 3d-electrons which do not screen as  N, much as s- and p-electrons. Therefore, due to poor shielding of 3d-electrons, the effective nuclear charge acting on Ga is slightly higher than that on Al.  Due to this, the (Δ_iH₁) value of Ga is slightly higher than that of Al, even though a new shell has been added on going from Al to Ga.

3. The same explanation can be offered on going from Into Tl. Tl has fourteen 4 f-electrons having a very poor screening effect and because of this, there occurs an unexpected increase in tyre effective nuclear charge, for which (Δ_iH₁) of Tl becomes much higher than that of In.

4. The order of (Δ_iH₁) values of group-13 elements is B > Al < Ga > In < Tl. However, this trend is not observed in the (Δ_iH₂) and (Δ_iH₃) values of these elements and this is because once the outermost p-electron is removed, it is not easy to remove the second and third electrons due to a large increase in effective nuclear charge. As expected, the first three ionisation enthalpies of these elements follow the order: Δ_iH₁< Δ_iH₂ < Δ_iH₃

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Oxidation states:

The atoms of group-13 elements have three valence electrons, two in the s -s-subshell and one in the p -subshell. Therefore, it becomes clear from their electronic configurations that +3 is expected to be the most common oxidation state of these elements. Therefore, the group oxidation state of group-13 element is +3.

Due to the small size of the boron, the sum of its first three ionisation enthalpies is very high. Therefore, it cannot lose its valence electrons to form B³⁺ ion rather it forms covalent bonds with other atoms, for example, BH₃ or B₂H₆.

The sum of the first three ionisation enthalpies of Al is much lower than that of B so, it can form an Al³⁺ ion, for example, AlCl₃. Al is a highly electropositive metal.

B and Al exhibit only a +3 oxidation state but Ga, In and Tl show a +3 as well as a +1 oxidation state and on moving down the group the stability of the +3 oxidation state decreases while that of +1 oxidation state progressively increases.

The stability of +1 oxidation state follows the order:

Al < Ga <In < Tl. In the case of Tl, the +1 oxidation state is very much more stable than the +3 oxidation state.

Explanation:

The stability of the +1 oxidation state that increases down the group can be explained in terms of the inert pair effect. On moving down the group, the tendency of electrons of the valence shell to participate in bond formation decreases due to poor shielding of these electrons from the attraction of the nucleus by the intervening d- and f- electrons.

This reluctance or inertness ofthe s-electrons to participate in bond formation is called the inert pair effect. Since the magnitude of this effect increases down the group, the +1 oxidation state becomes more and more stable down the group as compared to the +3 oxidation state. The inert pair effect is maximum in the case of Tl and therefore, it shows mainly +1 oxidation state. Due to lesser stability, Tl³⁺salts act as strong oxidising agents. This is evident from its electrode potential data:

Tl³⁺(aq) + 2e → Tl+(aq);E° = +1.25V

The inert pair effect may also be explained by the fact that as the size of the atom increases from Al to Tl, the energy required to unpair the ns² -electrons is not compensated by the energy released due to the formation of two additional bonds.

Electropositive and metallic character

The elements of group- 13 are less electropositive or metallic as compared to the elements of group 2. On moving down the group, the electropositive character of the elements first increases from B to Al and then decreases from Al to Tl.

Electropositive and metallic Explanation:

1. Elements of group-13 are smallerin size and the sum of the three ionisation enthalpies Δ_iH₁+ Δ_iH₂ + Δ_iH₃ needed to form M³⁺ ions is much higher than the sum of two ionisation enthalpies, Δ_iH₁+ Δ_iH₂ < Δ_iH₃ for the corresponding bigger-sized elements belonging to alkaline earth metals needed to form M²⁺ ions. For this reason, the elements of group 13 are less electropositive than the elements of group 2.

2. Boron has the highest sum of the first three ionisation enthalpies among the elements of group 13. Because of this, it has very little tendency to lose electrons and hence it is the least electropositive among group-13 elements. It is a non-metal and a poor conductor of electricity.

3. On moving from B to Al, the sum of the first three ionisation enthalpies decreases considerably (6887 to 5137kJ-mol^-1 ) due to an increase in atomic size and hence, Al has a much higher tendency to lose electrons, i.e., Al is sufficiently electropositive. All is a metal and a good conductor of electricity.

4. Because of the increasingly poor shielding effect of 3d -electrons in Ga, 4d -electrons in and 4 f-electrons in Tl, the effective nuclear charge gradually increases and as a consequence, they exhibit lesser electropositive and metallic character.

Density

Because of smaller atomic and ionic radii, the elements of group 13 have a higher density as compared to the elements of group 2. On moving down the group, density increases.

Density Explanation:

On moving down the group, the density of these elements increases because the extent of the increase in atomic mass is greater than the extent of the increase in atomic size.0 On moving from B to Tl, both atomic mass and no. of electrons in the inner d- and f- subshell increases. Due to the lower shielding effect of d-and f- electrons, the effective nuclear charge increases from B to Tl. As a result, from B to Tl, the atomic size does not increases much.

Melting and boiling points

Elements of group 13 do not show a regular trend in their melting points. The melting points decrease from B to Ga and then increase from Ga to Tl.

Melting and boiling points Explanation:

This irregular trend is probably due to unusual crystal structures of B and Ga. The much higher melting point of B is due to its giant covalent polymeric crystal structure consisting of icosahedral units with B-atoms at all 12 corners and each B-atom is bonded to five equidistant neighbours resulting in much stronger attractive forces. In contrast, Ga consists of discrete Ga₂ molecules so its melting point is exceptionally low (303K). However, the boiling points of these elements decrease regularly on moving down the group.

Gallium remains liquid over a vast range of temperatures and no other low-melting metal can compare with it. Molten Ga begins to boil only when heated to a temperature of 2276K. Due to this unusual property, gallium is used in thermometers required for measuring very high temperatures (>1000°C).

Electronegativity

Elements of group-13 are more electronegative than the elements of group-1 (alkali metals) and group-2 (alkaline earth metals). On moving down the group, the electronegativity first decreases from B to Al and then increases marginally.

Electronegativity Explanation:

•  Because of smaller atomic size and higher nuclear charge, the electronegativities of group-13 elements are higher than the corresponding elements of group-1 and 2.0 On moving down the group from B to Al, the atomic size increases considerably and as a result, the attraction of the nucleus for the electrons decreases and hence the electronegativity decreases.
• On moving from Al to Tl, the atomic size increases but at the same time effective nuclear charge increases due to poor shielding of the inner d and f-electrons. As a result, the attractive force of the nucleus for the electrons increases and hence the electronegativity increases

Boron

• Boron is the first member of group 13 of the periodic table. There are three electrons in its valence shell (ls²2s²2p¹).
• It exhibits anomalous behaviour and differs from the other members of its family. The reasons behind its exceptional behaviour can be attributed to the
• Exceptionally small atomic size as compared to other elements of its group,
• Much higher ionisation enthalpy and absence of d orbitals in its outermost or valence shell. Boron forms electron-deficient compounds which act as Lewis acids

Anomalous behaviour of boron:

Some important points of distinction between boron and the other members of its family (especially the next member Al) are discussed in the given table

Anomalous properties of boron:

1. Occurrence

Boron does not exist in a free state in nature. It is always found in the combined state as boric acid and borates. Boron occurs in two isotopic forms, ¹⁰B (19%) and ¹¹B (81%). Its abundance in the earth’s crust is very low (about 0.001%)

Important minerals of boron:

Boron may be obtained from the jets of steam which erupt from the volcano as boric acid and also from the water of the hot spring of Tuscany in small amounts as boric acid.

2. Properties of boron

Boron Physical properties:

1. Boron is an extremely hard solid (next to diamond) having a much higher melting point (2450K) and this is because of its three-dimensional network structure. Its boiling point is 3923 K.

2. Boron exists in two allotropic forms namely:

1. Amorphous,
2. Crystalline.

Crystalline boron is of three types:

1. α – Rhombohedral
2. β – Rhombohedral and
3. ϒ – Tetragonal.

The building units of all these forms are B₁₂ icosahedral units with 20 faces and boron atoms at all the 12 comers or vertices.

The melting and boiling point of boron is 2450K and 3923K respectively. The reason behind such high melting and boiling points is attributed to very strong attractive forces among the B₁₂ units as well as its closely packed stable crystal structure.

Chemical properties of crystalline boron

• Crystalline boron is chemically very inert. It is not oxidised even when heated with oxygen Crystalline boron is not attacked by HCl or HF. It is not affected by various oxidising acids such as shot and cone. HNO₃, H₂SO₄ etc.
• When it is fused with Na₂O₂ or Na₂CO₃ and KNO₃ at high temperatures, sodium borate (Na₃BO₃) is obtained.

Chemical properties of amorphous boron

1. Reaction with air:

When amorphous boron is heated in air at 700°C, it bums with a red flame and undergoes oxidation to form boron trioxide. Boron nitride is also formed by its reaction with N₂ gas of air

4B + 3O₂ → 2B₂O₃; 2B + N₂ → 2BN

2. Reaction with strong alkali:

When amorphous boron is fused with NaOH or KOH at a temperature greater than 773K,it forms borate salts and liberates H₂ gas

3. Reaction with oxidising adds:

Boron is not affected by non-oxidising acids such as hydrochloric add. However, it reacts with oxidising adds like cone. H₂SO₄ and HNO₃ form boric add.

3H₂SO₄ + 2B → 2H₂BO₂ + 3SO₂

6HNO₃ + 2B→2H₃BO₃ + 6NO₂

4. Reaction with halogens:

Boron bums in fluorine gas to form boron trifluoride. Boron reacts with chlorine at high temperatures to form boron trichloride

2B + 3F₂→ 2BF₃ ;2B + 3Cl₂→2BCl₃

An aqueous solution of BC13 is addicting because it undergoes hydrolysis to form a mixture of HCl and boric acid.

BCl₃+ 3H₂O →  3HCl + H₃BO₃

5. Reaction with metals (oxidising property):

The binary compounds of boron with elements having electronegativity lower than boron itself (e.g., metal) are called borides. When boron is heated with a metal at high temperature in an electric arc furnace, borides are obtained (B acts as an oxidant). Borides are hard, inert and have special properties

3Mg + 2B→ Mg₃B₂; 3Ca + 2B→ Ca₃ B₂

6. Reaction with carbon:

When boron is heated with carbon having comparable size and electronegativity at high temperatures in an electric arc furnace, very hard covalent boron carbide (B₄C) is obtained. It is even harder than diamonds.It is used as an abrasive. 4B + C→ B₄C It can also be prepared by reducing B₂O₃with coke at high temperature (2500°C) in an electric furnace

2B₃O₃ → B4C + 6CO↑

Reducing property: When boron is heated strongly with SiO₂ and CO₂, it reduces these oxides to give Si and C respectively.

2B₂O₃ + 7C →  B₄C + 6CO↑

7. Reducing property:

When boron is heated strongly with SiO₂ and CO₂, it reduces these oxides to give Si and C respectively.

8. Reaction with water: Red hot boron reduces steam to yield B₂O₃ and dihydrogen

Uses of boron

• Boron, an extremely hard refractory solid with a high melting point, low density and very low electrical conductivity, finds many applications which are as follows:
• Boron fibres having enormous tensile strength are used in making bullet-proof vests and as reinforcement materials in space shuttles and aircraft.
• Because of the high tendency of isotopes to absorb neutrons, metal borides are used in nuclear reactors as protective shields and control rods.
• It is used in the steel industry (instead of using expensive metals like Mo, Cr and W) for manufacturing special types of hard steel
• Its compounds such as borax and boric add are used for making heat-resistant glass (i.e., p a mild antiseptic.
• Boron compounds are used as rocket fuels because of their high energy/mass ratio.
• Boron carbide (B₄C) is used as an abrasive for polishing or grinding.
• Boron is used as a semiconductor for making electronic devices glass), glass-wool and gÿre glass
• An aqueous solution of orthoboric acid is used as

Some Important Compounds Of Boron

1. Borax, Na₂B₄O₇.10H₂O or Na₂[B₄O₅(OH)₄] 8H₂O

Borax or sodium tetraborate decahydrate which occurs naturally as tincal (suhaga) in certain dried-up lakes is the most important compound of boron. Borax contains the tetranuclear units

Therefore, its correct formula is Na₂[B₄O5(OH)₄]-8H₂O.

Borax Preparation

1. From tincal:

Naturally occurring borax or tincal, which contains about 50% borax is boiled with water, concentrated and then filtered to remove insoluble impurities. The filtrate is then concentrated and cooled when crystals of borax separate out.

2. From colemanite:

Finely powdered mineral, colemanite (Ca₂ B₆O₁₁) is boiled with sodium carbonate solution and CaCO₃, Na₂B₄O₇ and NaBO₂ are obtained

Precipitate of CaCO₃ is filtered off and the filtrate is then concentrated and cooled to get the crystals of borax. A current of CO₂ is passed through the mother liquor when sodium metaborate presentient gets converted into borax

4NaBO₂ + CO₂→ Na₂B₄O₇(Borax) + Na₂CO₃

3. From boric acid:

Borax may also be obtained by neutralising boric acid with sodium carbonate. Crystals of Na₂B₄O₇.10H₂O separate on cooling.

4H₃BO₃ + Na₂CO₃ →  Na₂B₄O₇ + 6H₂O + CO₂↑

Borax Physical properties:

1. It is a white crystalline solid.
2. It is less soluble in cold water but more soluble in hot water.
3. When ordinary borax is recrystallised from water at a higher temperature (≈60°C), crystals of sodium tetraborate pentahydrate (Na₂B₄O₇.5H₂O) separate. This is called ‘goldsmith’s sugar.

Borax Chemical properties

1. Nature of aqueous solution:

The aqueous solution of borax is alkaline in nature and this is because borax undergoes hydrolysis to form the strong alkali, NaOH and the weak acid, boric acid. It act as a buffer.

As the aqueous solution of borax is alkaline, it can be titrated against an acid using an orange indicator

Na₂B₄O₇ + 2HCl + 5H₂O→ 4H₃BO₃ + 2NaCl

When phenolphthalein is added to an aqueous solution of borax, the solution becomes pink in colour. However, when glycerol (a polyhydroxy compound) is added to the solution, it becomes colourless again.

Aqueous solution Explanation:

Since the aqueous solution of borax is alkaline in nature, it becomes pink when phenolphthalein is added to it. When glycerol is added to that solution, it combines with B(OH)₄ and removes it from the equilibrium by forming a stable chelate complex. As a consequence, the equilibrium shifts to the right making boric acid a strong acid. Because of the increased concentration of H⁺ ions, complete neutralisation of OH^– ions occurs and the solution becomes colourless again.

H₃BO₃ + H₂O  ⇌  H⁺ B(OH)₄^–; K_a = 6 × 10^-10

2. Reaction with caustic soda:

When a calculated amount of NaOH is added to borax, sodium metaborate is obtained.

Na₂B₄O₇ + 2NaOH→4NaBO₂ + H₂O

Reaction with sulphuric acid: When a calculated quantity of concentrated sulphuric acid is added to a hot concentrated solution of borax, boric acid is produced

3.Reaction with ethanol and sulphuric acid:

When borax is heated with ethanol and concentrated H₂SO₄, vapours of triethyl borate are formed which on ignition bum with a green-edged flame.

Na₂B₄O₇+ H₂SO₄ + 5H₂O →  Na₂SO₄ + 4H₃BO₃

H₃BO₃ + 3C₂H₅OH→B(OC₂H₅)₃ (Triethyl borate)+  3H₂O

This reaction is used as a test for the detection of borate ion (BO^3-₃) in qualitative analysis.

4. Action of heat:

When borax is heated strongly in the flame of a Bunsen burner, it loses its water of crystallisation and swells up to form a puffy mass. On further heating, the

mass turns into a transparent liquid which Nolldllics to form a bead that consists of sodium metaborate (NaBO₂) and boric anhydride (B₂O₃)

Preparation til boron from borax

A hot and concentrated solution of borax reacts with concentrated H₂SO₄ to form I boric acid (H₃BO₃). Boric added when heated tit high temperature, successively dissociates to form boron d ioxide, (B₂O₃). Boron trioxide when heated with Mg-powdor produces boron.

Na₂B₄O₇ + H₂SO₄+ 5H₂O→Na₂SO₄ + 4HaBO₃

Borax bead test

The borax bead test is very Important In a qualitative analysis for the detection of coloured metal ions like Cu²⁺, Ni²⁺, Co²⁺, Cr²⁺ etc.

• At first, a hot platinum loop is touched with borax and then heated in a Bunsen burner’s flame.
• Borax at first swells up and finally melts to form a colourless bead in the loop.
• The hot loop is touched with the salt under investigation and heated at first in the oxidising flame and then in the reducing flame.
• The metal ion is identified from the colour of the bead. This test is called the borax bead test.

Borax Bead Test:

Reactions:

1. Metallic compounds undergo decomposition on heating to form metallic oxides.

2M(NO₃)₂ → 2MO + 4NO₂ + O₂

2MSO₄ → 2MO + 2SO₂ + O₂

(M = Cu, Fe, Co, Ni, Mn, Cr)

2. The basic metallic oxides dissolve in the acidic diborane trioxide(BO) of the borax bead and form coloured metal metaborate salts

MO + BoO₃ → M(BO₂)₂

3. Copper Iron and other metallic salts form -ic metaborate in oxidising flame and -metaborates in reducing flame

Examples:

1. Reactions with copper salt:

In oxidising flame, cupric metaborate (blue) is formed

In reducing flame cuprous metaborate is formed

2. Reactions with iron salt:

In oxidising flame, ferric metaborate (yellow) is formed

In reducing flame ferrous metaborate (green) is formed

3. Reactions with cobalt salt: Both in oxidising and reducing flame, cobalt metaborate (blue) is formed

As the oxidation states of Co (+2) and Cr (+3) remain unchanged, the colour of the bead obtained from them is the same for both reducing and oxidising flame.

Uses of borax

Borax is used:

• In the manufacture of heat-resistant borosilicate (pyrex) glass,
• For preparing medicinal soaps,
• As a flux in soldering metals.
• In the candle industry as a stiffening agent,
• In softening water
• For the borax bead test,
• In the manufacture of perborate
• Na₂(OH)₂B(O — O)2B(OH)₂ 6H₂O, is an important cleansing and bleaching agent used in washing powders.

Orthoboric acid or boric acid, H₂BO₃o r B(OH)₃

The trivial name of orthoboric acid is boric acid

Orthoboric acid Preparation:

1. From colemanite:

Sulphur dioxide is passed through a hot concentrated solution of the mineral cole, Win’ll the resulting solution is concentrated and cooled, and crystals of boric add separate out. Calcium bisulphite being highly soluble In water remains dissolved In the mother liquor.

Ca₂ B₆O₁₁  (Colemanite)+ 4SO₂ + 11H₂ O →  2Ca(HSO₃)₂ (Calcium bisulphite)+ 6H₃BO₃ ( Boric acid)

2. From borax:

When a hot concentrated solution of borax is treated with hydrochloric acid or sulphuric acid, boric acid Is obtained. The resulting solution is concentrated and then cooled when crystals of boric acid separate out

Na₂B₄O₇ + 2HCl + 5H₂O →4H₃BO₃ + 2NaCI

Na₂B₄O₇ + H₂SO₄ + 5H₂O → 4H₃BO₃ + Na₂SO₄

3. From boron compounds by hydrolysis:

Certain boron compounds such as halides, hydrides and nitrides on boiling with water (hydrolysis) produce boric add.

BCI₂ + 3H₂O → H₃BO₃+ 3HCl

B₂H₆(dlborane) + 6H₂O→2H₃BO₃ + 6H₂↑

BN(boron nitride) + 3H₂O→ H₃BO₃ + NH₃

Orthoboric acid’s Physical properties

• It Is a white needle-like crystalline solid with a soft soapy touch.
• It Is sparingly soluble in cold water but highly soluble in hot water.
• It Is steam volatile.

Orthoboric acid Chemical properties

1. Addle nature:

It Is a very weak monobasic acid (K_a = 6 × 10^-10). It does not donate protons like most protonic acids. In fact, due to the small size of B and the presence of only six electrons Its valence shell B(OH)₃ behaves as a Lewis acid and accepts a pair of electrons from OH” Ion water thereby releasing a proton.

H(OH)₃ + 2H₂O ⇌  [B(OH)₄]^–+ H₃O⁺

B(OH)₃ behaves as a very weak acid (pK_a  = 9.2) because it only partially reacts with water to form [B(OH)₄]^–and H₃O⁺ Ions. So, B(OH)₃ or H₃BO₃ cannot be titrated satisfactorily with NaOH solution because no sharp end point Is obtained.

If some polyhydroxy compound such as glycerol, mannitol or catechol is added to the titration mixture, then boric acid behaves as a strong monobasic add.  This occurs due to the Removal of [B(OH)₄]^– Ion from the equilibrium mixture by the formation of a stable complex with the polyhydroxy compound.

It can then be titrated with NaOH solution and the endpoint can be detected using phenolphthalein as an indicator. ‘

2. Action of heats:

When orthoboric acid is heated, it loses molecules of water in three stages at different temperatures thus forming different products

3. Reaction with ethyl alcohol:

Boric acid reacts with ethyl alcohol In the presence of concentrated sulphuric acid to form triethyl borate

Vapours of triethyl borate burn with a green-edged flame. This test is used for detecting boric acid in qualitative analysis.

It is to be noted that this test can also be performed without using H₂SO₄ However, for detecting borate ions, the presence of H₂SO₄ is required. Therefore, boric acid and borate ions can be distinguished by this test.

4. Reaction with fluoride salt:

Boric acid reacts with fluoride salt in the presence of concentrated H₂SO₄ to form volatile boron trifluoride (BF₃). This compound burns with a green-edged flame.

2H₃BO₃ + 3CaF₂ + 3 H₂SO₄ → 3CaSO₄ + 2BF₃ + 6H₂O

5. Reaction with ammonium bi fluoride:

When boric acid is heated with ammonium bi fluoride, no residue is obtained because all the resulting compounds are gaseous

B₂O₃3(s) + 6NH₄BF₄(S)→  8BF₃(G) + 6NH₃(g) + 3H₂O(g)

6. Reaction with potassium bi fluoride:

When the aqueous solutions of two acidic compounds, boric acid and potassium bifluoride (KHF2) are mixed, an alkaline solution is obtained due to the formation of potassium tetrafluoroborate (KBF₄) and potassium hydroxide (KOH).

B(OH)₃+ 2KHF₂→  KBF₄ + KOH + 2H₂O

Being a Lewis acid, B(OH)₃ has a strong tendency to combine with relatively smaller (F^–) ions to form fluoroborate ion (BF^–) and for this reason, this unbelievable reaction takes place.

Uses of boric acid

Boric acid is used

• As a mild antiseptic for washing eyes under the name Boric lotion
• In the manufacture of heat-resistant borosilicate glass,
• As a preservative for milk and foodstuffs,
• In the manufacture of enamels and glazes for pottery.

Structure Of boric acid:

The shining white crystals ofboric acid contain B(OH)₃ units linked by H -bonds in infinite layers of nearly hexagonal symmetry. Since the adjacent layers in the boric acid crystal are held together with weak attractive forces, one layer can easily slide over the other and hence, boric acid is soft and slippery touch.

Diborane, B₂H₆

Boron hydrides are binary compounds of B and H. Although boron does not combine directly with hydrogen, several boron hydrides collectively called boranes, (in analogy with alkanes) are known. Depending upon their general formulae, these hydrides

Can be divided into several categories of which the following two are the most important:

1. Nido-boranes:

General formula: \(\mathrm{B}_n \mathrm{H}_{n+4} \text {, example} \mathrm{B}_2 \mathrm{H}_6\) (diborane), B₅H₉ (pentaborane-9), B₆H₁₀ (hexaborane- 10), B₈H₁₂(octaborane-12),B₁₀H₁₄ (decaborane-14) etc. 

2. Arachno-boranes:

General formula: \(\mathrm{B}_n \mathrm{H}_{n+6} \text {, example }\) B₅H₁₄ (pentaborane-11), BgH₁₂ (hexa-borane-12), BgH₁₄(octaborane-14) etc. The mostimportant hydride ofboron is diborane (B₂Hg).

Preparation of diborane

1. Laboratory preparation:

Diborane is prepared by the oxidation of sodium borohydride (NaBH₄) with I₂ in a diglyme solution.

Diglyme is a polyether whose formula is CH₃OCH₂CH₂OCH₂CH₂OCH₃

2. From boron trifluoride etherate:

It may be prepared by the reduction of boron trifluoride etherate with lithium aluminium hydride (LiAlH₄) in diethyl ether or sodium borohydride (NaBH4) in diglyme

3. Industrial preparation:

On an industrial scale, diborane is prepared by reducing BF₃ with LiH or NaH. 450K

Diborane Physical properties

Diborane is a colourless, foul-smelling, highly toxic gas having a boiling point of 180K.

Diborane Chemical properties

1.  Thermal stability:

It is stable only at low temperatures. When it is heated at 373-523K in a sealed tube, several higher boranes are obtained

However, by controlling the temperature, pressure and reaction time, various individual boranes can be prepared.

2. Combustibility:

When it is exposed to air, it spontaneously catches fire because of the strong affinity of boron towards oxygen. This reaction forming boric anhydride and water is highly exothermic.

B₂H₆ + 3O₂ →B₂O₃ + 3H₂O; ΔH = – 1976 kJ -mol^-1

The higher boranes also spontaneously in the air.

3. Hydrolysis:

It undergoes ready hydrolysis to produce boric acid.

B₂H₆(g) + 6H₂O(aq)→2H₃BO₃(aq) + 6H₂(g)↑

It reacts with methanol to form trimethyl borate.

B₂H₆ + 6CH₃OH → 2B(OCH₃)₃ + 6H₂↑

4. Reaction with Lewis bases:

When diborane is treated with Lewis base, it undergoes cleavage to form monoborane which then reacts with Lewis base to form an adduct.

B₂H₆ + 2NMe₃→ 2BH₃-NMe₃; B₂H₆ + 2CO → 2BH₃-CO

5. Reaction with ammonia:

When diborane is treated with ammonia, an additional compound is formed. The compound on further heating at about 473 K decomposes to give a volatile compound called borazine (or borazole).

Borazine is isosteric (i.e., the same number of atoms) and isoelectronic (i.e., the same number of electrons) with benzene and its structure is similar to that of benzene. Like benzene, all the atoms in borazine are sp² -hybridised. The n -n-bonding of borazine is dative and it arises due to sideways overlapping of filled p-orbitals of N and empty p-orbitals of B. Because of its similarity with benzene, borazine is also called inorganic benzene.

6. Formation of complex borohydrides:

Diborane reacts with several metal hydrides to form borohydrides containing tetrahedral [BH₄]_ ion.

Both sodium and lithium borohydrides are used as very good reducing agents in the synthesis of organic compounds in the laboratory. These two compounds may also be used as starting material for the preparation of other borohydride compounds.

7. Reaction with alkalis:

Diborane dissolves in strong alkalies such as NaOH or KOH solution to form metaborates and H₂ gas.

B₂H₆ + 2KOH + 2H₂O→ 2KBO₂ + 6H₂(Potassium Metaborate)↑

8. Reaction with halogen acids:

Diborane reacts with halogen acids to form halodiborancs and hydrogen gas. The order of reactivity of halogen acids is: HI > HBr > HCl

9. Reaction with halogens:

Diborane reacts with halogens to form corresponding halodiboranes. The order of reactivity of halogens is Cl₂ > Br₂ > I₂. Thus, chlorine reacts with diborane explosively at room temperature, bromine reacts rapidly at 373 K but iodine reacts slowly at higher temperatures

Uses of diborane:

1. Diborane is used in the preparation of several borohydrides such as LiBH₄ NaBH₄, etc.
2. It is used as a reducing agent in organic synthesis.
3. It is also used as a fuel for supersonic rockets.

Structure Of diborane, B₂H₆:

The structure and bonding of diborane seem to be very interesting. In the excited state, the B atom has the electronic configuration 2s¹2p_x¹ 2p_y¹ and therefore, it has only three electrons available for sharing. Now, 14 electrons (for six B—H and one B—B bond) are required if boron forms all conventional covalent bonds in ethane (C₂H₆)

But there are only 12 electrons (six from two B atoms and six from six H-atoms). Thus, the molecule is short of two electrons and its structure cannotbe similar to that of ethane (C₂H₆)

Based on electron diffraction study:

• Diborane has a bridged structure as given in There are two types of hydrogen atoms in this bridged structure. The two boron atoms and four terminal hydrogen atoms (shown by thick lines) lie in the same plane, while the remaining two hydrogen atoms (shown by dotted lines) lying above and below the plane form bridges and these are called bridge hydrogen atoms.
• The two B-H-B bridges lie in a plane which is nearly perpendicular to the plane containing the terminal B—H bonds.
• There are two bonds in the diborane molecule:
  1. The four terminal B—H bonds are normal covalent bonds, each Being formed by sharing a pair ofelectrons between boron and hydrogen atoms. These are quite strong bonds and called two-centre electron pair bonds or two-centre two-electron bonds (2c-2e bonds),
  2. The two bridge bonds B …. H……B are quite different from the normal electron pair bonds. Each bridge H-atom is bonded to two boron atoms by sharing only one pair of electrons.
• Such bridge bonds are called three centre electron pair bonds or three centres two-electron bonds (3c-2e bonds). Three-centre electron pair bonds or three-centre two-electron bonds are very weak bonds and are often called banana bonds as they resemble bananas in shape.
• Molecules like diborane (B₂H₆) which do not have a sufficient number of electrons to form normal covalent bonds are called electron-deficient molecules.

Based on hybridisation:

Boron atoms (excited state electronic configuration: 2s¹2p_x¹ 2p_y¹ in diborane undergo sp3 -hybridisation.

1. The two half-filled sp³ -hybrid orbitals of each boron atom overlap with the half-filled orbitals of hydrogen atoms to form normal covalent bonds.

2. The third half-filled hybrid orbital of one of the two boron atoms and the vacant orbital of the remaining boron atom overlap simultaneously with the half-filled Is -orbital of a hydrogen atom to form a B……H….B bridge bond

3. This bond involves three atoms (two boron atoms and one hydrogen atom) and contains only two electrons because one overlapping orbital of boron is empty. Hence, this B–‘H-‘-B bond is called three centre electron pair (3c-2e) bonds. Because of its typical shape resembling a banana, it is also called a banana bond

Aluminium

Aluminium, the second member of the boron family (group-13), is the most abundant metallic element in the earth’s crust. It is found in a variety of aluminosilicate compounds such as clay, mica and feldspar. The only ore of aluminium from which the metallic aluminium can be extracted profitably (in industry) is bauxite. Bauxite is hydrated aluminium oxide whose molecular formula is Al₂O₃-2H₂O

Aluminium Physical properties:

• Aluminium is a bluish-white metal with a brilliant lustre. But aluminium easily gets tarnished by the formation of a thin layer of oxide on the surface.
• It is a light metal whose density is 2.73g-cm^-1. Aluminium possesses high tensile strength, yet it is malleable and ductile.

Aluminium is a very good conductor of heat and electricity.

Aluminium Chemical properties:

It is not as reactive as its high negative electrode potential (E° = -1.66V) would imply and this is because there is a very thin layer of oxide on its surface.

1. Action of air:

Al remains unaffected in dry air but in the presence of moist air, a thin film of oxide is formed over its surface. Hence, the metallic of disappears. When burnt in oxygen it produces brilliant light.

4Al + 3O₂ → 2Al₂ O₃+ 772 kcal

The reaction is highly exothermic and the heat evolved is used for the reduction of oxides of Cr, Fe, Mn etc. (known as the thermite process).

2. Action of water: Aluminium decomposes boiling water thereby evolving hydrogen gas.

2Al + 6H₂O→2Al(OH)₃ + 3H₂↑

3. The action of non-metals:

Besides oxygen, aluminium reacts with other non-metals such as nitrogen, sulphur and halogens to form nitride, sulphide and halides respectively

4. Action of acids:

It dissolves both in dilute and concentrated hydrochloric acid and dilute sulphuric acid along with the evolution of hydrogen gas.

2Al + 6HCl→ 2AlCl₃ + 3H₂↑

2Al + 3H₂SO₄→Al₂(SO₄)₃ + 3H₃↑

The reaction with dilute sulphuric acid is very slow probably due to the insolubility of tyre oxide film over the metal in the acid. Hot and concentrated sulphuric acid dissolves aluminium with the evolution of sulphur dioxide (SO₂) gas.

2Al + 6H₂SO₄→Al₂(SO₄)₃ + 3SO₂ + 6H₂O

Dilute and concentrated nitric acid have no action on aluminium and this is due to the formation of an impenetrable oxide layer on its surface. Nitric acid may, therefore, be kept in the aluminium vessel.

5. Action of alkalis:

Aluminium dissolves in hot and cone. NaOH or KOH solutions form sodium aluminate with the evolution of hydrogen gas.

2Al + 2NaOH +2H₂O → 2NaAlO₂  (Sodium aluminate) (Soluble) +  3H₂ ↑

Aluminium reacts with hot and concentrated sodium carbonate (Na2C03) solution to form sodium aluminate, carbon dioxide and hydrogen.

2Al + Na₂CO₂ + 3H₂O  → 2NaAlO₂ (Sodium aluminate) (Soluble) + CO₂ + 3H₂↑

Uses of aluminium:

• Aluminium alloys (duralumin: Al, Mg, Cu and magnalium: Al, Mg) are light and strong and thus, are used in the construction of aircraft, ships and cars.
• It is a better conductor than copper and is used for making electric power cables.
• It is used for making doors, windows, building panels, mobile homes and household utensils.
• Finely divided Al powder is used in preparing aluminium paint and as an ingredient in solid fuels in rockets.
• Aluminium foils are used In wrapping soaps, cigarettes and confectioneries.
• Al is used to extract metals such as Cr, Mn etc., from their ores (thermite process).
• A mixture of ammonium nitrate and Al dust (commonly called ammonal) is used to make bombs and crackers.

Compounds Of Silicon

1. Silicon tetrachloride (SiCI₄ )

Silicon tetrachloride Preparation:

Silicon tetrachloride is prepared by heating either silicon or silicon carbide with chlorine

Silicon tetrachloride Properties and uses:

1. Physical state:  It is a volatile liquid (boiling point: 330.5 K).

2. Hydrolysis:

SiCl₄ undergoes ready hydrolysis to produce silicic acid, Si(OH)₄ which on further heating undergoes partial dehydration to yield silica gel (SiO₂  xH₂O).

Silica gel is an amorphous and very porous solid which contains about 4% of water. It is used as an adsorbent in column chromatography and as a catalyst in the petroleum industry. When the hydrolysis of SiCl₄ is carried out at a much higher temperature, finely powdered silica Is obtained instead of silicic acid

The finely powdered silica thus obtained is used as a thixotropic agent (which reduces viscosity temporarily) in polyester, epoxy paints and resins and as an inert filler in silicon rubber.

3.  Reduction: Reduction of SiCl₄ with H₂ gas gives silicon

Ultrapure silicon used for making transistors, computer chips and solar cells is prepared by this method.

4. Reaction with silicon:

When a mixture of SiCl₄ and Si is pyrolysed, a series of perhalosilanes of the general formula, Si_n Cl_{2n+ 2} where n = 2-6, are obtained.

Chains of perhaloslianes are longer than those of silanes and this is due to the formation of pπ-dπ bonding between a lone pair of electrons present on Cl and the empty d-orbitals of Si

2. Silicones

Silicones e Definition:

The synthetic organosilicon polymers containing repeating R₂SiO units held by Si — O — Si linkages are known as silicones The general formula of these compounds is (R₂SiO)n where R = methyl or aryl group. Commercial silicones are generally methyl derivatives and in some cases phenyl derivatives.

Silicones Preparation

Hydrolysis of dichlorodimethylsilane (obtained by the reaction between methyl chloride and silicon in the presence of Cu as catalyst) followed by polymerisation involving intramolecular dehydration yields straight chain polymers, i.e., silicones.

The length of the polymer can be controlled by the reaction of dimethylsiianol with chlorotrimethylsilane. This blocks the terminal end ofthe polymer as follows—

Silicones Properties:

• Silicones containing short chains are oily liquids; those with medium chains are viscous oils, greases and jellies and those with long chains are rubber-like solids.
• They are stable to heat and are also resistant to oxidation, i.e., they are very inert in nature.
• They are water repellents (hydrophobic) & good electrical insulators.

Silicones Uses:

• Silicones are used for making water-proof cloth and paper.
• These are used as electrical insulators.
• Silicon oils are used in high-temperature baths and vacuum pumps.
• Silicon rubbers are very useful as they can retain their elasticity over a wide range of temperatures.
• These are mixed with paints and enamels to make them resistant to the effects of sunlight, high temperatures and chemicals.
• These are used for preparing vaseline-like greases which are used as lubricants in aeroplanes

3. Silicates

Silicates Definition:

Silicates are compounds in which the anions present are either discrete SiO₄^4- tetrahedral units or several such units joined together by corners, i.e., by sharing one oxygen atom but never by sharing edges The negative charge on the silicate structure is neutralised by positively charged metal ions.

Classification Of silicates:

Depending upon the number of comers (0, 1, 2, 3 or 4) of SiO₄^4- tetrahedral unit shared with another tetrahedral unit through oxygen atoms, silicates are following six types:

1. Orthosllicates:

These are simple silicates which contain discrete SiO₄^{4- tetrahedrons}. Some examples are—zircon: Zr₂[SiO₄] , forsterite: Mg₂[SiO₄] , willemite: Zn₂[SiO₄] and phenacite: Be₂[SiO₄]

2. Pyroslllcates:

When two SiO₄^4- – tetrahedra share one corner (i.e., one oxygen atom), Si₂O₇^6-anion is formed. Silicates containing discrete Si₂O units are called pyrosilicates.

The common examples of phyllosilicates are:

1. Thortveitite: Sc₂[Si₂O₇] and
2. Hemimorphite: Zn₄(OH)₂[Si₂O₇]-H₂O

3. Ring or cyclic silicates:

When two O-atoms per tetrahedron are shared to form closed rings, structures with the general formula, (SiO₃)_n^2n-– are obtained. The silicates containing these anions are called cyclic silicates.

Some common examples are :

1. Wollastonite: Ca₃[Si₃O₉] (containing the cyclic ion,  Si₃O₉^6-  and
2. Beryl: Be₃Al₂[Si₆O₁₈] (containing the cycle ion, [Si₆O₁₈]^-12

4. Chain silicates:

If two oxygen atoms of each tetrahedral unit are so shared that a linear single-strand chain of the general formula, (SiO₃)₂ is formed, then the silicates containing these anions are called chain silicates.

Minerals of this type are called pyroxene and these include:

1. Enstatite: Mg₂[(SiO₃)₂],
2. Diopside: CaMg[(SiO₃)₂] and s
3. Spodumene: LiAl[(SiO₃)₂].

On the other hand, those chain silicates, containing double chain are called amphiboles. Here, two chains are attached through the O-atom. These silicates contain (Si₄O₁₁)_n ^6- ions. Minerals of asbestos are most commonly known as amphiboles.

For example: Crocidolite or blue asbestos [Na₂Fe₅(OH)₂(Si₄O₁₁)₂], amosite or brown asbestos [(Mg, Fe)(OH)₂(Si₄O₁₁)₂

Asbestos is heat and fire-resistant and thus is used as a shed for houses. Fine asbestos fibres, on entering the lungs cause asbestosis which can even result in lung cancer.

5. Sheet silicates:

Sharing of three comers i.e., three O-atoms of each tetrahedron results in the formation of an infinite two-dimensional sheet structure of the formula (Si₄O₅)_n ^2n- . Silicates containing these anions are called sheet silicates.

Some of the common examples are:

1. Kaolinite: [Al₂(OH)₄Si₂O₅] and
2. Alc: [Mg₃(OH)₂Si₄O₁₀] .

Clay also belongs to this class containing (Si₂O₆) ^2-– anions

Three-dimensional silicates:

If all the four comers i.e., all the four O-atoms of each tetrahedron are shared with other tetrahedra, a three-dimensional network structure is obtained. These have a general formula, (SiO₂)_n. Some common examples are quartz, tridymite and cristobalite

When a few silicon atoms in a three-dimensional network of SiO₂ are replaced by Al³⁺ions, the overall structure thus obtained carries a negative charge and is called aluminosilicate. Cations such as Na⁺, K⁺ or Ca²⁺ balance the negative charge. Such three-dimensional aluminosilicates are called zeolites.

A common example is natrolite:

Na₂[Al₂Si₃O₁₈]-2H₂O. Feldspars and ultramarines are two other types of three-dimensional aluminosilicates.

Many open channels of molecular levels are present in the structure of the zeolites. Depending on the shape and size of these open channels, ions or molecules of different shapes and sizes are adsorbed by the zeolites. Thus, zeolites are used as molecular sieves for separating molecules of different sizes.

Other two types of three-dimensional aluminosilicates are:

1. Feldspar example: Orthoclase, KAlSi₃O₈) and
2. Ultramarine example: Ultramarine blue, Na₈(AlSiO₄)₆S₂ )

Class 11 Chemistry Some P Block Elements Short Question And Answers

Question 1. Boron shows anomalous behaviour and differs from j the rest of the members of its family —why?
Answer:

Boron shows anomalous behaviour because of

1. Exceptionally small atomic size,
2. High ionisation enthalpy and
3. Absence of d -d-orbitals in its valence shell

Question 2. Give reasons for which carbon differs from the rest of the members of its family
Answer:

• Exceptionally small atomic size,
• Higher electro¬ negativity,
• Higher ionisation enthalpy and
• Absence of f-orbitals in the valence shell.

Question 3. Diamond is a non-conductor of electricity but a good conductor of heat—why?
Answer:

Due to the absence of free electrons, it is a non-conductor of electricity. It has the highest known thermal conductivity because thermal motion is distributed in its 3D -structure very effectively

Question 4. Explain why the melting and boiling points of boron are much higher.
Answer:

1. Boron exists as a giant covalent polymer having a three-dimensional network structure both in the solid and the liquid states.
2. For this reason, its melting and boiling points are very high;

Question 5. pπ-pπ back bonding occurs in the case of boron halides but not in the case of aluminium halides —why?
Answer:

• The tendency of pπ-pπ back bonding decreases with an increase in the size of the central atom.
• Since Al is larger than B, pn-pn back bonding does not take place in the case of Al

Question 6. (SiH₃)₃N is weaker base than (CH₃)₃N —why?
Answer:

• pπ-dπ Back bonding occurs in (SiH₃)₃N but not in (CH₃)₃N.
• Therefore, the unshared pair ofelectrons is more available in (CH₃)₃N and for this reason, it is more basic

Question 7. N(CH₃)₃ is pyramidal but N(SiH₃)₃ is planar—explain.
Answer:

• Because of d -orbital resonance, the N-atom in N(SiH₃)₃ molecule is sp² – hybridised and therefore, the molecule is planar
• No such d -d-orbital resonance occurs in N(CH₃)₃ and the central N-atom is sp³ -hybridised. For this reason, this molecule is pyramidal.

Question 8. CO gets readily absorbed in ammoniacal silver nitrate solution but CO₂ does not—explain.
Answer:

Due to the presence of an unshared electron pair on carbon, CO acting as a Lewis base, combines with ammoniacal cuprous chloride to form a stable and soluble complex. No such reaction takes place in the case of CO₂ because in it carbon has no unshared electron pair,

Question 9. Which out of anhydrous and hydrous AlCl₃ is more soluble in ether and why
Answer:

Anhydrous AlCl₃ is electron deficient but hydrous AlCl₃ is not Because of this, anhydrous AlCl₃ combines with ether through the formation of a coordinate bond and gets dissolved in

Question 10. Explain why the B—X bond distance in BX3 is shorter than the theoretically expected value.
Answer:

This is due to the pπ-pπ backbonding of the filled p orbital of halogen (X) into the empty p -orbital of boron.

As a result, the B— X bond possesses some double-bond character and hence B —X bond is shorter than expected

Question 11. Although aluminium lies above hydrogen in the electrochemical series, it is quite stable in water and air. Why?
Answer:

Aluminium is a highly reactive electropositive metal. In the presence of water and air, a thin layer of aluminium oxide (Al₂O₃) forms over the metal surface. Consequently, this protective layer prevents further reaction of aluminium with water or air.

So, aluminium is quite stable in water and air, despite being situated above hydrogen in the electrochemical series.

Question 12. Using chemical reactions show that amphoteric. aluminium is
Answer:

Aluminium reacts with acids as well as with bases. So, it is amphoteric.

For example:

Question 13. Graphite acts as a better lubricant on the moon compared to that on earth” Justify the validity ofthe statement.
Answer:

The statement is not correct. Different substances like air, water vapour and other gaseous materials enter into the layers of graphite when it is on earth, thus enhancing its lubricating property. However, the moon is devoid of atmosphere. Thus, due to the absence of water vapour and gaseous substances, the lubricating property of graphite is quite less on the moon

Question 14. Explain why PbCl₄ is a good oxidising agent.
Answer:

Due to the inert pair effect, the Pb²⁺ ion is relatively more stable than the Pb⁴⁺ ion and so the Pb⁴⁺ ion readily gets reduced to Pb²⁺ ion by accepting two electrons. That is why PbCl₄ is a good oxidising agent.

Question 15. Why do nitrogen and carbon monoxide show similarities in their physical properties?
Answer:

Nitrogen (N₂) and carbon monoxide (CO) exhibit structural similarities because both molecules contain the same number of valence electrons (10). Because of structural similarities (similar distribution of electrons in the bonding orbital), they show striking resemblance in their physical properties like vapour density, solubility in water, boiling point, melting point, etc

Question 16. Explain why graphite is used as a solid lubricant for heavy machinery.
Answer:

• Since any two successive layers in graphite are held together by weak forces of attraction, one layer can slip over the other.
• This makes graphite soft and a good lubricating agent for heavy machinery

Question 17. Diamond is a bad conductor of electricity but a very good conductor of heat—explain.
Answer:

• There is no free electron left in the structure of the diamond made up of sp³ hybridised C-atoms and so, the diamond is unable to conduct electricity.
• On the other hand, it is a very good conductor of heat because its structure distributes thermal motion in three dimensions very effectively.

Question 18. Despite being a covalent substance, the melting point of diamond is very high—why?
Answer:

• In diamonds, there is a three-dimensional network of strong covalent bonds.
• Since a large amount of thermal energy is required for the cleavage of these bonds, the melting point of the diamond is very high.

Question 19. CO is an inflammable gas while CO₂ is not—why?
Answer:

The oxidation states of carbon in CO, and CO are +4 and +2 respectively. In any compound, the maximum oxidation state of carbon is +4. Consequently, the carbon atom in CO tends to increase its oxidation number, i.e., CO has the inherent urge to be oxidised. For this reason, it reacts readily with oxygen, i.e.,

It is a combustible gas. 2CO +O₂ →2CO₂ On the other hand, the oxidation state of carbon in CO₂ is maximum (+4) so it does not tend to be oxidised and unlike CO, it is not combustible.

Question 20. [SiF₆]^2- is known to exist whereas [CF₆]^2- does not exist. Explain.
Answer:

• Silicon can extend its coordination number beyond four because it possesses vacant d-orbitals. Hence, [SiF₆]^2-– exists.
• On the other hand, C has no vacant d-orbitals in its valence shell and thus it cannot extend its coordination number beyond four.
• Hence, [CF₆]^2-does not exist

Question 21. An aqueous solution of sodium hydroxide is added dropwise to the solution of gallium chloride in water. A precipitate is initially formed. The precipitate dissolves on further addition of NaOH solution. Explain the observation using suitable chemical reactions
Answer:

On the addition of NaOH solution to a solution of GaCl₃, a gelatinous white precipitate of Ga(OH)₃ is formed which dissolves on adding excess NaOH by forming a soluble complex.

Question 22. Define buckyball. How is it made?
Answer:

C₆₀ -fullerene is called buckyball. It contains 20 six-membered rings and 12 five-membered rings.  It is prepared by heating graphite in an electric arc in the presence of an inert gas like argon or helium.  The C₆₀ and C₇₀ fullerenes hence formed can be separated readily by extraction with benzene or toluene followed by chromatography in the presence of alumina.

Question 23. CO is readily absorbed by ammoniacal cuprous chloride solution but CO₂ is not. Explain.
Answer:

CO has a lone pair ofelectrons on C-atom. Thus, it acts as a Lewis base and forms a soluble complex with an ammoniacal CuCl solution.

CuCl + NH₃ + CO → [Cu(CO)NH₃]⁺Cl^– (Soluble)

CO₂, on the other hand, does not possess a lone pair of electrons on the C-atom. Hence, it does not act as a Lewis base. Thus, it does not dissolve in ammoniacal CuCl solution

Question 24. What is the chemical composition of the borax bead?
Answer:

Borax forms a glassy mass on heating called a borax bead

Thus, the borax bead is a mixture of sodium metaborate (NaBO₂) and boric anhydride (B₂O₃)

Question 25. Silicon in elemental form does not form a graphite-like structure. Explain.
Answer:

Silicon is larger than carbon. Thus, pn-pn bonding is not as effective in the case of Si as in the case of C-atom.  In graphite, pn-pn bonding is effective due to the smaller size of carbon.  Thus, Si does not resemble graphite. Rather, it resembles a diamond and is a poor conductor of electricity

Question 26. Anhydrous aluminium chloride cannot be prepared by heating hydrated aluminium chloride. Why?
Answer:

When hydrated aluminium chloride (AlCl₃– 6H₂O) is heated, it gets hydrolysed by the water of crystallisation and aluminium oxide is formed. Therefore, anhydrous aluminium chloride cannot be prepared by heating hydrated aluminium chloride

Question 27. Why are the dihalides of carbon unstable but the dihalides of tin and lead stable?
Answer:

All the elements of group 14 (except C and Si) form stable dihalides. Due to the inert pair effect, these elements are highly stable in the +2 oxidation state. Hence, tin and lead form stable dihalides. On the other hand, carbon is stable in the +4 state. Therefore, dihalides of carbon are unstable

Question 28.

1. Why PbO₂ is oxidising?
2. Which of the following is the thermodynamically most stable form of carbon? Coke, diamond, graphite, fullerenes.

Answer:

1. Among the group-13 elements B and Al exhibit +3 oxidation state only. On the other hand, Tl shows +3 as well as +1 oxidation states but due to the inert pair effect, it is more stable in +1 oxidation state. So, TlCl exists but AlCl does not

2. Graphite.

Question 29. PbCl₂ is less stable than SnCl4 while PbCl₂ is more stable than SnCl₂. Justify or contradict
Answer:

In the case of group-14 elements the number of d- or f- electrons increases down the group from Ge to Pb. Hence, the inert pair effect becomes gradually more prominent. As a result, the stability of the +4 oxidation state decreases and the +2 oxidation state increases down the group. Consequently, Pb is more stable in the +2 state whereas Sn is more stable in the +4 state. Therefore, PbCl₄ is Jess stable than SnCl₄ while PbCl₂ is more stable than “SnCl₂

Question 30. What happens when at first lesser amount and then an excess amount of NaOH solution is added to the Al₂(SO₄) solution?
Answer:

When a lesser amount of NaOH is added to the solution of Al₂(SO)₃, a white precipitate of Al(OH)₃ forms. In the presence of excess NaOH, the solution becomes clear due to the formation of soluble sodium aluminate (NaAlO₂)

Question 31. Explain with reason: SnCl₂ is a solid ionic compound whereas SnCl₄ is a covalent liquid.
Answer:

SnCl₂, Sn(II): [Xe]₄d¹⁰5s¹⁰

SnCl₄, Sn(IV): [Xe]₄d10

Here with an increase in oxidation state from Sn(II) to Sn(IV) the ionisation potential of the central atom (here Sn) increases which makes the Sn—Cl bonds more covalent in SnCl₄ compared to SnCl₂ — Fajan’s rules.

Hence the SnCl₂ molecules are closely packed due to greater ionic character whereas in SnCl₄ the molecules show weaker London forces of interaction due to their covalent nature. This explains the given observation

Question 32. How can you explain the higher stability of BCl₃ as compared to TlCl₃?
Answer:

Because of the poor shielding effect on s-electrons of the valence shell by the inner d -and f-electrons (i.e., 3d, 4d, 5d, 4f-electrons), the inert pair effect is maximum in Tl. As a result, for Tl only the 6p¹ -electron becomes involved in bond formation. Hence the most stable oxidation state of TI is +1 and not +3. Therefore, TlCl is stable but TlCl₃ is unstable.

On the other hand, due to the absence of d -and f-electrons, B does not exhibit an inert pair effect and all three valence electrons become involved in bond formation. Hence, B exhibits an oxidation state of +3 and thus forms BCl₃. So, BCl₃ is more stable than TlCl₃

Question 33. Why does boron trifluoride behave as a Lewis acid?
Answer:

The B-atom in the BF₃ molecule has only she electrons in its valence shell and thus two more electrons are required to complete its octet. Therefore, BF₃ can easily accept a pair of electrons from basic substances such as NH₃, (C₂H₅)₂O etc. and thus behaves as a Lewis acid.

Question 34. BF₃ is reacted with ammonia?
Answer:

Being a Lewis acid, BF₃ accepts a pair of electrons from NH₃(a Lewis base) to form a complex

Question 35. Aluminium chloride exists as a dimer, but boron trichloride does not. Explain
Answer:

The boron atom is so small that it cannot accommodate four large-sized Cl-atoms around it so it cannot complete its octet by forming a dimer. However, the Al-atom being larger can accommodate four Cl-atoms around it. For this reason, AlCl₃ exists as a dimer in which each Al-atom accepts an unshared pair of electrons from the Cl-atom of another molecule to complete its octet.

Question 36. Sn(II) is a reducing agent but Pb(II) is not—why?
Answer:

Because of inert pair effect, both tin and lead show two oxidation states of +2 and +4. But this effect is more prominent in the case of Pb than in Sn and consequently, +2 oxidation state of Sn is less stable than its +4 oxidation state. Therefore, Sn(II) acts as a reducing agent and gets converted to the more stable Sn(IV) by losing two electrons.

In contrast, the +2 oxidation state of Pb is more stable than its +4 oxidation state due to the prominent inert pair effect. Therefore, Pb(II) does not lose electrons easily and does not act as a reducing agent

Question 37. CO is stable but SiO is not—why?
Answer:

Since electronegativity, has a strong tendency to form pn-pn multiple bonds, it combines with oxygen to form CO which is stabilised by resonance as follows:

⇒ \(: \mathrm{C}=\ddot{\mathrm{O}}: \leftrightarrow: \overline{\mathrm{C}} \equiv \stackrel{+}{\mathrm{O}}:\)

Silicon, on the other hand, due to its bigger size and lower electronegativity, does not tend to form pn-pn multiple bonds. Thus, it does not combine with oxygen to form stable SiO

Question 38. [SiF₆]^2-  is known but [SiF₆]^2- is not. Why?
Answer:

Possible reasons for the non-existence of [SiF₆]^2- are:

1. Six fluorine atoms can be easily accommodated around silicon atoms due to smaller size while six larger chlorine atoms cannot be accommodated around silicon atoms.
2. The unshared pair of electrons present in a relatively small 2p -orbital of F interacts with the d -orbitals of Si better than the unshared pair ofelectrons present in a relatively large 3p -orbital of Cl.

Question 39. Explain why CCI₄ is resistant to hydrolysis but SiCl4 undergo ready hydrolysis.
Answer:

Carbon does not undergo hydrolysis because carbon cannot extend its coordination number beyond four due to the absence of vacant d-orbital in its valence shell. On the other hand, SiCl₄ can undergo ready hydrolysis because Si has a vacant d-orbital in its valence shell and can extend its coordination number beyond four

Question 40. No visible reaction is observed when Al metal is left in contact with concentrated HNQ3. Explain.
Answer:

Al is a reactive metal and hence it initially reacts with a cone. HNO₃ to form Al₂O₃ . The oxide forms a protective layer No the surface of the metal and it becomes positive Thus no visible reaction is observed

Question 41. Thermite reaction cannot be stopped by pouring water. Explain.
Answer:

In a thermite reaction, the oxygen needed for the reaction is supplied by the metal oxide. Thus, stopping the oxygen supply (by pouring water) has no effect.

Further at high temperatures (1270-1300K), Al reacts with H₂O to form H₂ gas which spreads the fire rather than extinguishing it

Question 42. Why were lead sheets used on the floors in the Hanging Gardens of Babylon
Answer:

To prevent water from escaping, the lead sheet was extensively used on the floors in the Hanging Gardens of Babylon (one of the wonders ofthe ancient world that was built during the Egyptian civilisation).

Question 43. Explain why HF is not stored in glass containers the visible reaction is observed
Answer:

SiO₂ present in glass reacts readily with hydrofluoric acid (HF) to form H₂SiF₆ which is soluble.

Hence, HF is not stored in glass containers.

Question 44. What is the state of hybridisation of carbon in

1. CO₃^2-
2. Diamond
3. Graphite?

Answer:

The hybridisation state of C

1. In CO₃^2- is sp²
2. In diamond is sp³
3. In graphite is sp²

Class 11 Chemistry Some P Block Elements Very Short Question And Answers

Question 1. Name one ore of boron and give its formula.
Answer: Colemanite (Ca₂B₆O₁₁-5H₂O).

Question 2. What are the two isotopes present in natural boron?
Answer: (19.6%) and 1gB‘(80.4%).

Question 3. Which element of group 13 has the most stable +1 oxidation state?
Answer: Thallium (Tl) has the most stable +1 oxidation state because of its prominent inert pair effect.

Question 4. Which elements of Gr-13 form amphoteric hydroxide?
Answer: Al and Ga form amphoteric hydroxide

Question 5. Which element of group-13 forms only covalent compounds and why?
Answer:

Because of its small atomic size and high value of the sum of the first three ionisation enthalpies (Δ_iH₁ + Δ_iH₂ + Δ_iH₃), boron forms only covalent compounds.

Question 6. Give the general valence shell electronic configuration of group-13 elements. What is their common oxidation state
Answer: ns²np¹; + 3 (where n = 2 to 6).

Question 7. Which one among group-13 elements has the highest value of ionisation enthalpy?
Answer: Boron.

Question 8. Which element of Gr-13 is the most abundant one?
Answer: Aluminium.

Question 9. Write one physical characteristic of boron in which it differs from the other members of group 13.
Answer: Boron is a non-metal, while other elements of group 13 are metals.

Question 10. Why boron compounds such as BF₃ are called electron-deficient compounds?
Answer: Because the valence shell of B in BF₃ has only six elections. Two more electrons are required to complete the octet.

Question 11. Which of the Gr-13 elements forms acidic oxide?
Answer: Boron forms acidic oxide.

Question 12. Arrange the following compounds in order of decreasing strength as Lewis acid: BCl₃, BBr₃, BF₃
Answer:

BBr₃ > BCl₃ > BF₃ .

Question 13. Which compound is responsible for the green-edged Oame in a test for borate ion?
Answer: Triethyl borate [B(OC₂H₅)₃] .

Question 14. Name the compound which on warming produces pure BF3
Answer: Benzenediazonium fluoroborate \(\left(\mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{N}}_2 \mathrm{BF}_4^{-}\right)\)

Question 15. Explain why BF| has no existence.
Answer: This is because boron cannot extend its valency to six due to the absence of vacant d -d-orbitals

Question 16. What type of cations are identified by the borax bead test
Answer: Cations that show colour are identified by the borax bead test

Example: CO^-2+, Ni²⁺, etc

Question 17. What happens when borax solution is acidified? sp³ -hybridised.
Answer: Boric acid is obtained when the borax solution is acidified (Na₂B₄O₇ + 2HCl + 5H₂O→2NaCl + 4H₃BO₃).

Question 18. How are the BO^3-₃ units in boric acid linked to give 12 icosahedral units, it is an extremely hard solid. layered structure?
Answer:

BO^3-₃ units are linked through hydrogen bonding

Question 19. What is the shape of BO^3-₃ion?
Answer: Trigonal planar because the central B -atom is sp² – hybridised

Question 20. Which compounds are formed on heating boric acid?
Answer:  HBO₂, H₂B₄O₇ and B₂O₃ Using a balanced chemical equation show how B(OH)₃ behaves as a monobasic acid in water.

Question 21. Using a balanced chemical equation show how B(OH)₃ behaves as a monobasic acid in water.
Answer: B(OH)₃ + H₂O->[B(OH)₄] +H⁺. As it produces one free H+ ion in solution, it behaves as a monobasic acid

Question 22. What are the forces involved between the layers of two-dimensional sheets of H₃BO₃?
Answer: Weak van der Waals forces

Question 23. What is the composition of the transparent glassy bead obtained on heating borax
Answer: (NaBO₂+ B₂O₃)

Question 24. What is the structural unit present in all allotropic forms of boron?
Answer: B₁₂ icosahedral units

Question 25. What type of bonds are
Answer: Purely covalent B — H bonds and three-centred two-electron (3c-2e) B…H…..B bridge bonds. present in B2Hg molecule?

Question 26. Explain why boron cannot form B3+ ions
Answer:

This is because of its very small atomic size and large sum ofthe first three ionisation enthalpies.

Question 27. Mention the states of hybridisation of boron in B₂H₆ and BF₃
Answer:

In B₂H₆ , boron is sp³ -hybridised while in BF₃ , boron is sp² -hybridised

Question 28. Crystalline boron is an extremely hard solid why?
Answer:

Because the dimensional network structure involves B₁₂ icosahedral units, it is an extremely hard solid

Question 29. What are boranes?
Answer:

Stable covalent boron hydrides such as B₂H₆, B₄H₁₀, B₅H₉ etc. in analogy with alkanes are called boranes

Question 30. Which two out of five members of the carbon family are distinctly metals
Answer: Sn and Pb are distinctly metals.

Question 31. Wlilch one out of catechol, resorcinol and quinol can be used to titrate boric acid against sodium hydroxide using methyl orange as the indicator?
Answer: Two  OH groups are present at adjacent ring carbons due to which it can form a stable complex with B(OH)^–₄.

Question 32. Carbon forms covalent compounds but lead forms ionic compounds—Why?
Answer:

The ionisation enthalpy of carbon is much higher (1086 kj-mol^-1 )whereas that of lead it is much lower (715 kJ-mol^-1). Because of this, carbon forms covalent compounds but lead forms ionic compounds.

Question 33. Which element of the carbon family has no d-orbital in Its valence shell?
Answer:

Carbon has no d -d-orbital in its valence shell.

Question 34. Among the group-14 elements which is the most electronegative one?
Answer: Carbon is the most electronegative one.

Question 35. Which member of the carbon family has the lowest melting point?
Answer: Tin has the lowest melting point.

Question 36. Out of diamond & graphite which is a good conductor of electricity and which is a good conductor of heat
Answer:

Graphite is a good conductor of electricity, while diamond is a good conductor of heat.

Question 37. Which member of the carbon family has the highest value of first ionisation enthalpy?
Answer: The first ionisation enthalpy value of carbon is the highest.

Question 38. Which member of the carbon family has the maximum tendency to exhibit catenation property?
Answer: Carbon has a maximum tendency to exhibit catenation.

Question 39. What are the structural units of ice and dry ice?
Answer: H₂O and CO₂ respectively

Question 40. Among the group-14 elements which one exhibits pπ-pπ multiple bond
Answer:  Carbon exhibits pπ-pπ multiple bonds.

Question 41. What is the basic building unit of all silicates?
Answer:  SiO₄^4- is the basic building unit of all silicates.

Question 42. What happens when cone. H₂SO₄ is dropped on sugar?
Answer: Sugar charcoal is formed.’

Question 43. What is buckminsterfullerene?
Answer: The C₆O fullerene is known as buckminsterfullerene.

Question 44. What is the state of hybridisation of carbon in
Answer:

1. sp²
2. sp²
3. sp.

Question 45. Which allotrope of C is used as a moderator in atomic reactors and as a solid lubricant for heavy machinery?
Answer: Graphite

Question 46. Mention the oxides of C which are the anhydrides of carbonic acid and formic acid respectively.’
Answer: CO₂ is the anhydride of carbonic acid, while CO is the anhydride of formic acid.

Question 47. Name the gases which are present in producer gas
Answer: Carbon monoxide and nitrogen

Question 48. Out of CO and CO₂ which acts as a ligand and can form a coordinate bond with certain metals why?
Answer:

Due to the presence of a lone pair of electrons on carbon, CO acts as a ligand and forms a coordinate bond with certain metals

Question 49. What is the state of hybridisation of carbon in each of the following:

1. Diamond
2. Graphite
3. Fullerene

Answer:

1. sp³
2. sp²
3. sp

Question 50. What is carborundum?
Answer: Silicon carbide (SiC) is called carborundum.

Question 51. Give an example of a reaction where CO₂  acts as an oxidising agent
Answer:

Question 52. What are zeolites?
Answer: Zeolites are microporous 3D aluminosilicates

Question 53. Write the name of the compound used as a fire extinguisher under the name pyrene.
Answer: Carbon tetrachloride (CCl₄)

Question 54. Name the hardest compound of boron
Answer: Boron carbide or morbid.

Question 55. What is alane?
Answer: Alane is a polymeric hydride form of aluminium with the formula (AlH₃)_n

Question 56. Which two elements of group 13 form amphoteric hydroxides?
Answer: Al and Ga

Question 57.  What are the two stable natural isotopes of boron?
Answer: ¹¹B and ¹⁰B

Question 58. Which of the group-13 elements has the most stable +1 oxidation state?
Answer: TI

Question  59. Which of the Gr-13 elements forms only covalent compounds?
Answer:  Boron

Question 60.  The melting point of boron is very high, even though it is a non-metal—why
Answer: Because B exists as a giant covalent molecule,

Question 61. Which acid is obtained when an aqueous solution of borax is acidified?
Answer: Boric acid (H₃BO₃)

Question 62. Which are called boranes?
Answer:  Covalent boron hydrides,

Question 63. What is the correct structural formula of borax?
Answer: Na₂[B₄O₅(OH)₄].8H₂O

Question 64. What happens when orthoboric acid is heated till red hot?
Answer: At first HBO₂ then , H₂B₄O₇ and finally B₂O₃ forms,

Question 65. What is inorganic benzene? Why is it called so?
Answer: Borazine (B₃N₃H₆), because of structural similarity with benzene

Question 66. What is the common oxidation state of group-13 elements?
Answer: +3

Question 67. Write down the chemical composition of the coloured compound obtained finally in the borax bead test.
Answer: Metal metaborate, MBO₂ or M'(BO₂)₂ or M”(BO₂)₃ [M = monovalent, M’ = divalent, M”= trivalent ),

Question 68. Arrange boron halides in decreasing strength as Lewis acid.
Answer: BI₃ > BBr₃ > BCl₃ > BF₃ ,

Question 69. How can boric acid form polymer?
Answer: By formation of hydrogen bonding

Question 70. Explain why BF^-6₃ does not exist.
Answer: Because boron has no vacant d -d-orbital,

Question 71. What is duralumin? Mention its uses.
Answer: It is an alloy of aluminium; it is used for making aeroplanes, automobile parts, pressure cookers etc.

Question 72. Borazine is more reactive than benzene—why?
Answer:  The C =C bonds in benzene are non-polar but the B — N bonds in borazine are polar

Question 73. Some metals are extracted from their oxides by reducing with aluminium instead of carbon—why?
Answer: Because they form metal carbides,

Question 74. Which out of CCl₄ and SiCl₄ reacts with water and why?
Answer: SiCl₄, because there are vacant d orbitals in the valence shell of silicon

Question 75. What is water gas?
Answer: Equimolar mixture of CO and H

Question 76. Carbon compounds are relatively less reactive—why?
Answer: Because of the much higher bond dissociation enthalpy of the C— C bond

Question 77. What is the value of the dipole moment of carbon suboxide and why?
Answer:

Question 78. Mention the hybridisation state of carbon in CO^-2₃ and CO₂.
Answer: sp² and sp respectively

Question 79. Write the name of a neutral oxide of carbon.
Answer: Carbon Monoxide (CO)

Question 80. What is dry ice?
Answer: Solid carbon dioxide,

Question 81. What is the basic structural unit of silicates?
Answer: SiO

Question 82. Write the general formula of silicones.
Answer:

Question 83. Explain why the compounds of boron are called electron-deficient compounds.
Answer: Boron contains six valence electrons, i.e., its octet is incomplete

Question 84. Explain why BF₃ forms addition compound with NH₃ .
Answer:

For completing its octet the boron atom in BF3 accepts a pair of electrons from the nitrogen atom of ammonia and as a result, an addition compound is obtained

Question 85. Out of InCl₃ and In Cl, which one is more stable and why?
Answer:

Due to the weak inert pair effect, the +3 oxidation state of In is relatively more stable

Question 86. Explain why boron does not form BF₆^3- ion
Answer:

The boron atom does not have a vacant d -d-orbital. Hence, it cannot expand its coordination up to six

Question 87. Metallic aluminium is frequently used as a reducing agent in the extraction of Cr, Mn, Fe etc —why?
Answer:

Because of the much higher affinity of Al for oxygen, Al eliminates oxygen from the oxides of moderately electropositive metals,

Question 88. What is called the mixture containing 95% O₂ and 5% CO₂
Answer: Carbogen,

Question 89. What is the purest allotropic form of amorphous carbon?
Answer: Lamp black,

Question 90. What is the molecular mass of the most available natural fullerene?
Answer: 2HO, (C₆₀fullerene) ,

Question 91. Write the names of two greenhouse gases.
Answer: Carbon dioxide and methane,

Question 92. What is ivory black?
Answer: The black substance obtained on dissolution of Ca-salts present in bone charcoal with HCl,

Question 93. Which out of carbon and silicon forms multiple bonds and why?
Answer: Carbon having a small atomic size and high electronegativity can form multiple bonds by pπ-pπ overlapping,

Question 94. What is the anion present in phyllosilicates
Answer: Si₂O^6-₇

Question 95. Why is orthoboric acid used in talcum powders?
Answer:  Orthoboric acid is a fine white powder that easily mixes with talcum powders and imparts antiseptic properties.

Question 96. Why molten AlBr₃ is a poor conductor of electricity?
Answer:  AlBr₃ is a covalent molecule. As it ionises to a very small extent even in a molten state, it is a poor conductor of electricity.

Question 97. Which glass has the highest percentage of lead? Mention its use.
Answer: Flint glass contains the highest percentage of lead as lead silicate. It is used for optical purposes

Question 98. Carbon exhibits catenation property but lead does not— why?
Answer: The C— C bond dissociation enthalpy is much higher than Pb —Pb bond dissociation enthalpy

Question 99. Which are called methanides?
Answer: The carbides containing C₄– ions are called methanides;

Question 100. (CH₃)₃SiOH is more acidic than (CH₃)₃COH, even though carbon is more electronegative than silicon— explain
Answer: Because of d-orbital resonance, the conjugate base of (CH₃)₃SiOH is relatively more stable

Question 101. Silicon is unable to form structures like graphite—why?
Answer: Due to the larger size and lower electronegativity of Si as compared to C, silicon can’t form double bonds through sp² -hybridisation

Question 102. Mention one property of fullerene which differs from that of diamond and graphite.
Answer: Fullerenes dissolve in organic solvents while diamond or graphite does not

Question 103. Write the formula of the following ore: bauxite
Answer: Al₂O₃-2H₂O

Question 104. A white precipitate is formed when a small amount of a gas is passed through lime water. The precipitate dissolves when excess ofthe gas is passed. What can be the possible gases? How would you identify the gases?
Answer:  The possible gases may be either carbon dioxide or sulphur dioxide

Question 105. Which one is the hardest allotrope of carbon? Answer with reason
Answer: Diamond

Question 106. Write balanced equations for Al + NaOH →
Answer:

2Al + 2NaOH + 6H₂O → 2Na+[Al(OH)₄]^–+ 3H₂

Question 107. What are aquadag and oildag? Mention their uses.
Answer:

A colloidal solution of graphite in water is known as aquadag and a colloidal solution of graphite in oil is known as oildag. These are used as lubricants

Class 11 Chemistry Some P Block Elements Fill In The Blanks

Question 1. Boric acid is a ________________ acid and not an acid
Answer: Lewis, protonic

Question 2. Due to _____________ Tl⁺ ion is more stable than Tl³⁺ ion.
Answer: Inert pair effect,

Question 3. Two types of bonds in diborane are covalent and ___________ bond
Answer: 3c-2e bridge bond,

Question 4. Tl³⁺ ion acts as __________________ agent.
Answer: Oxidising,

Question 5. AlCl₃ is a __________________ Lweis acid than BCl
Answer: Stronger

Question 6. BCl₃ is a__________________ Lweis acid than BF
Answer: Stronger

Question 7. AlF₃ is an __________________compound, but AlCl is a __________________compound
Answer: Ionic, covalent

Question 8. The hydrides of boron are called__________________
Answer: Boranes,

Question 9. Inorganic benzene is chemically known as __________________
Answer: Borazine

Question 10. Anhydrous aluminium chloride exists as a __________________
Answer: Dimer

Question 11. The B—F bond present in BF₃ is __________________ compared to the B — F bond present in BF₄.
Answer: Shorter

Question 12. When H3B03 is strongly heated __________________ obtained
Answer: BO (Boron Trioxide)

Question 13. BN is a crystalline solid having a structure similar to __________________
Answer: Graphite

Question 14. On moving down the group, the stability of the +1 oxidation state of the members of the boron family +3 oxidation state
Answer: Increases And Decreases

Question 15. Except boron all members of boron family are_______________
Answer: Metals

Question 16. Except ______________ all members of the carbon family exhibit allotropy.
Answer: Lead

Question 17. Potassium ferrocyanide reacts with concentrated does not. is used as a fuel. sulphuric acid to form _____________ gas
Answer: CO

Question 18. SnCl₂ acts as a __________________
Answer: Reducing

Question 19. Out of CO and CO₂ __________________ combines with haemoglobin
Answer: CO

Question 20. Due to agent ________________ the +2 oxidation state of group 1 elements gradually becomes stable down the group.
Answer: Inert pair effect

Question 21. Only ________________ of the carbon family does not react with water.
Answer: Lead

Question 22. Carbides which on hydrolysis product CH₄ are called ________________
Answer: Methanides

Question 23. The hydrides of silicon are called ______________ is called ‘sugar of lead
Answer: Silanes(Si_nH_2n+2+ 2

Question 24. ________________ is called ‘sugar of lead
Answer: Pb(CH₃COO)₂

Question 25. Formic acid in dehydration produces
Answer: CO

Question 26. Due to the absence of a complex.
Answer: D orbital

Question 27. Mica is an example of
Answer: Sheet Silicate

Question 28. PbCl₄ exists bu ____________
Answer: PbI₄

Question 29. Out of CO and CO₂ ____________ is used as a fuel
Answer: CO

Question 30. Zircon (ZrSiO₄) is an example of _____________
Answer: Orthosilicate

Question 31. In silicones _________________ units are held by Si —O —Si
Answer: R₂SiO

Question 32. Asbestos ______________ is a silicate mineral existing in
Answer: Mg₃(Si₂O₅)(OH)₄

Class 11 Chemistry Some P Block Elements Warm Up Exercise Question And Answers

Question 1. Give reactions to justify the amphoteric nature of Ga.
Answer:

Question 2. Why does BF₃ form an adduct with ammonia?
Answer:

In the NH molecule NH₃ -atom has a lone pair ofelectrons and in BF₃ molecule 2 electrons are required to complete the octet of the B-atom. Thus, NH₃ (a Lewis base) reacts with BF₃ (a Lewis acid) to form an adduct

H₃ N: + BF₃→ [H₃ N→ BF₃ ]

Question 3. Boron is distinctly non-metallic—why?
Answer:

Boron is distinctly non-metallic because of its small atonfWsize, high ionisation enthalpy high electronegativity.

Question 4. Using chemical reactions shows that boron acts as an oxidising agent as well as a reducing agent.
Answer:

B Is heated with Mg in an electric arc furnace magnesium boride Is formed. I lore, B acts as an oxidising agent.

Question 5. Metal borides having 10B are used in nuclear reactors why?
Answer: 10_B has a greater tendency to absorb a high-energy neutron

Question 6. BO₃^3- has a trigonal planar structure—why?
Answer:

The three half-filled orbitals (2s, 2p_x and 2py) of boron in its excited state undergo sp² -hybridisation. The resulting three sp² -hybridised orbitals overlap with 2p – orbitals of three O^– forming three B – O^– bonds. Thus, BO₃^3- ion has a trigonal planar structure

Question 7. Although boric acid [B(OH)₃] contains three -OH groups, it is sparingly soluble in water—why?
Answer:

Boric acid molecules form cyclic two-dimensional associated giant molecules through intermolecular hydrogen bonds. So it finds little or no opportunity to form hydrogen bonds with water and is hence, sparingly soluble in water

Question 8. Among group-14 elements which one exhibits pπ-pπ multiple bonding?
Answer:

Among all the group-14 elements carbon exhibits pn-pn multiple bonding.

Question 9. Account for the anomalous behaviour of carbon from other group-14 elements.
Answer:

Due to small size, high ionisation enthalpy, high electro- * negativity and unavailability of d-orbitals, the behaviour of carbon is different from other elements of group-14

Question 10. The shape of (SiH₃)₃P is pyramidal. Comment.
Answer:

A larger 3p -orbital of P is unable to participate in efficient pπ-dπ bonding. Thus, the central P-atom undergoes sp³-hybridisation and forms pyramidal (SiH)₃P

Question 11. Which element among the group-14 elements is a metalloid?
Answer: Germanium (Ge)

Question 12. Which compound of lead is used as “Sindoor”
Answer: Red lead (Pb₃O₄)

Question 13. Among the dioxides of group-14 elements. PbO₂ is the strongest oxidising agent— explain.
Answer:

Due to the inert pair effect, the +2 oxidation state of Pb is more stable. Hence, PbO₂ is the strongest oxidising agent among the dioxides of group-14 elements.

Question 14. Give the Lewis acidity order: SiI₄, SiCl₄, SiBr₄ > SiF₄
Answer:

Lewis acidity order: Sil₄ > SiBr₄ > SiCl₄ > SiF₄

Question 15. What is the state of hybridisation of carbon in fullerene?
Answer: In fullerene, carbon is sp² -hybridised.

Question 16. How can you decolourise a sample of slightly brown-coloured impure sugar?
Answer:

When an aqueous solution of brown-coloured impure sugar is heated with activated charcoal, a mixture is obtained, which on filtration gives a colourless solution. The solution is concentrated by heating and then cooled. As a result, colourless crystals of pure sugar are obtained.

Question 17. Give two differences between diamond and graphite.
Answer:

Two characteristic differences between diamond and graphite are:

1. Diamond is hard but graphite is soft
2. Diamond is an insulator while graphite is a good conductor of electricity.

Question 18. CO forms an additional compound but CO₂ does not—why?
Answer:

In the CO molecule, the C-atom has a lone pair of electrons. By donating this lone pair of electrons CO forms an addition compound with metals or non-metals. Also in the CO molecule, the C-atom exhibits an oxidation no. of +2.

So, it can increase its oxidation no. from +2 to +4 by forming additional compounds. On the other and C-atom in the CO₂ molecule does not have any lone pair of electrons. Besides, the C-atom exhibits +4 oxidation state in CO₂. So, it has no opportunity to increase its oxidation number. Thus, CO forms an additional compound but CO₂ does not

Question 19. Explain why blue flame is seen in a coal oven.
Answer:

At the bottom section of the coal oven, carbon burns in the presence of excess oxygen producing CO₂ This CO₂ while moving upwards, is reduced by red hot carbon (coke or coal) to CO in the middle section ofthe oven.

The CO bums in the open air at the top of the oven with a blue flame to form CO₂.

1. At the bottom : C + O₂ →C02
2. At the middle: CO₂ + C→2CO
3. At the top: 2CO +O₂ →2CO₂

Question 20. How will you separate CO and CO₂ from a mixture?
Answer:

When a mixture of CO₂ and CO is passed through Cu₂Cl₂ solution acidified with HCl, CO is absorbed in the die solution but CO₂ escapes without participating in a chemical reaction. The solution thus obtained liberates CO on heating

2CO + Cu₂Cl₂ + 4H₂O→2[CuCl-CO. 2H₂O]

Question 21. How will you confirm that a gas is CO₂ but not SO₂?
Answer:

Both the gases are passed through the K₂Cr₂O₇ solution, SO₂ turns the orange colour of the K₂Cr₂O₇ solution green but CO₂ does not.

Question 22. Write the formula of white asbestos. What type of silicate is it?
Answer:

Formula of white asbestos is Mg₃(OH)₄[Si₂O₅] . It is a type of sheet silicate

Question 23. How can ultrapure silicon be prepared from impure silicon?
Answer:

At first, impure Si is treated with chlorine gas to produce impure SiCl₄, which on distillation forms pure SiCl₄. Pure SiCl₄ thus produced is reduced by H₄ gas to ultrapure silicon

Question 24. Explain why silicones are water-repelling in nature.
Answer:

Silicon chains are surrounded by non-polar organic groups. Thus, they are water-repelling in nature (water is a polar solvent)

Question 25. What are zeolites? Give two important uses of zeolites
Answer:

Zeolites are used

1. For softening of hard water
2. As a molecular sieve to separate molecules of different size