The P - Block Elements

(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.

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.

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.

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

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.

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)

(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.

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)

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.

(i) HCO₃^-

(ii) CO₃^2-

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) CO₃^2-

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.

• 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

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.

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.

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

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.

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.

(a) Diborane

B₂H₆ is an electron-deficient compound. B₂H₆ has only 12 electrons – 6 e^– to 6 H atoms and 3 e^–each 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.

(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

(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

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.

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

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.

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.

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.

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.

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.

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.

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.

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.

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

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.

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.

(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.

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

(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)

(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)→ CaCl_2(aq) + CO_2(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) + H_2(g)

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

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

Boron in diborane is sp³ hybridised.

Graphite is thermodynamically the most stable form of carbon.

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.

(i)

(ii)