Answer

(c) In gallium, the crystal structure is different, suggesting that Ga consists of almost discrete ${\mathrm{Ga}}_2$ molecule, that’s why its melting point is lowest. Ga exists as liquid from ${30}^{\circ }\mathrm{C}$ upto ${2000}^{\circ }\mathrm{C}$ and hence, it is used in high temperature thermometry.

• Option (a) B: Boron does not exist in a liquid state over a wide range of temperatures. It has a high melting point of about 2076°C and does not melt easily, making it unsuitable for use in high-temperature thermometry.

• Option (b) Al: Aluminum has a melting point of about 660°C and does not remain in a liquid state over a wide range of temperatures. It solidifies quickly after melting, which limits its use in high-temperature measurements.

• Option (d) In: Indium has a melting point of about 157°C and does not remain in a liquid state over a wide range of temperatures. Its liquid range is much narrower compared to gallium, making it less suitable for high-temperature thermometry.

Thinking Process

Lewis acid represent those substances in which octet remains incomplete act as electron acceptors.

Answer

(a) Alkaline earth metals form ionic chloride whereas aluminium chloride is covalent. Despite of sharing electrons with chlorine, the octet of aluminium is incomplete. To complete the octet, it needs electrons and thus, acts as a Lewis acid.

• (b) ${\mathrm{MgCl}}_2$: Magnesium chloride is an ionic compound. Magnesium, being an alkaline earth metal, forms ionic bonds with chlorine. In this ionic compound, magnesium has a complete octet and does not have a tendency to accept electrons, thus it does not act as a Lewis acid.

• (c) ${\mathrm{CaCl}}_2$: Calcium chloride is also an ionic compound. Calcium, another alkaline earth metal, forms ionic bonds with chlorine. In this ionic compound, calcium has a complete octet and does not have a tendency to accept electrons, thus it does not act as a Lewis acid.

• (d) ${\mathrm{BaCl}}_2$: Barium chloride is an ionic compound as well. Barium, being an alkaline earth metal, forms ionic bonds with chlorine. In this ionic compound, barium has a complete octet and does not have a tendency to accept electrons, thus it does not act as a Lewis acid.

Thinking Process

Hybridisation and geometry of a complex can be found by counting number of bond pairs and lone nairs nresent in a central atom.

Answer

(a) Structure of $\mathrm{B}(\mathrm{OH}{)}_4^{-}$is

Hybridisation $-s{p}^3$

Geometry - Tetrahedral

• Option (b) $s{p}^3$, square planar:

  • The $s{p}^3$ hybridization leads to a tetrahedral geometry, not a square planar geometry. Square planar geometry is typically associated with $ds{p}^2$ hybridization.

• Option (c) $s{p}^3{d}^2$, octahedral:

  • The $s{p}^3{d}^2$ hybridization corresponds to an octahedral geometry, which requires six electron pairs around the central atom. In ${\left[\mathrm{B}\left({\mathrm{OH}}_4\right)\right]}^{-}$, there are only four groups (OH groups) around the boron atom, making $s{p}^3{d}^2$ hybridization and octahedral geometry incorrect.

• Option (d) $ds{p}^2$, square planar:

  • The $ds{p}^2$ hybridization leads to a square planar geometry, which is not applicable for ${\left[\mathrm{B}\left({\mathrm{OH}}_4\right)\right]}^{-}$ as it has four groups around the central atom, which corresponds to $s{p}^3$ hybridization and a tetrahedral geometry.

Answer

(a) The acidic nature of oxides changes from acidic to basic through amphoteric on moving down the group.

$\underset{\begin{array}{c}\text{ More acidic }\\ \text{ less basic }\end{array}}{\underbrace{{\mathrm{B}}_2{\mathrm{O}}_3}},\underset{\text{Amphoteric }}{\underbrace{{\mathrm{Al}}_2{\mathrm{O}}_3\text{ and }{\mathrm{Ga}}_2{\mathrm{O}}_3}},\underset{\text{Basic }}{\underbrace{{\mathrm{In}}_2{\mathrm{O}}_3\text{ and }{\mathrm{TI}}_2{\mathrm{O}}_3}}$

• (b) $A{l}_2{O}_3$: This oxide is amphoteric in nature, meaning it can react both as an acid and a base. It is not purely acidic.

• (c) $G{a}_2{O}_3$: Similar to $A{l}_2{O}_3$, this oxide is also amphoteric, exhibiting both acidic and basic properties.

• (d) ${\ln }_2{O}_3$: This oxide is basic in nature, meaning it reacts with acids to form salts and water, and does not exhibit acidic properties.

Answer

(a) The lowest atomic number among the given is boron. Boron has atomic number 5 and do not have vacant $d$-orbital. The element $M$ in the complex ion $M{F}_6^{3-}$ has a coordination number of six. Boron can show maximum coordination number of 4 . Thus, $\mathrm{B}$ can not form complex of the type ${\mathrm{MF}}_6^{3-}$.

• Aluminum (Al): Aluminum has an atomic number of 13 and has vacant 3d orbitals available. Therefore, it can achieve a coordination number of 6 and can form the complex ${\mathrm{AlF}}_6^{3-}$.

• Gallium (Ga): Gallium has an atomic number of 31 and has vacant 4d orbitals available. Therefore, it can achieve a coordination number of 6 and can form the complex ${\mathrm{GaF}}_6^{3-}$.

• Indium (In): Indium has an atomic number of 49 and has vacant 5d orbitals available. Therefore, it can achieve a coordination number of 6 and can form the complex ${\mathrm{InF}}_6^{3-}$.

Answer

(c) Lewis acids are substances whose octet is incomplete and accepts electrons.

Boric acid is a monobasic weak acid. It does not liberate ${\mathrm{H}}^{+}$ion but accepts electrons from ${\mathrm{OH}}^{-}$ion i.e., behaves as Lewis acid.

$H_{3}B{O}_3+H_{2}O⟶B(OH{)}_4^{-}+H^{+}$

or, $B(OH{)}_3+2H_{2}O\to {[B(OH{)}_4]}^{-}+H_3{O}^{+}$

The structure of $H_{3}B{O}_3$ is as shown below, where the octet of boron in $H_{3}B{O}_3$ is incomplete.

• (a) Boric acid does not contain replaceable ${\mathrm{H}}^{+}$ ions. It does not donate protons directly but rather accepts ${\mathrm{OH}}^{-}$ ions from water.

• (b) Boric acid does not give up a proton. Instead, it acts as a Lewis acid by accepting ${\mathrm{OH}}^{-}$ ions from water.

• (d) Boric acid does not combine with protons from water molecules. It accepts ${\mathrm{OH}}^{-}$ ions from water, leading to the release of protons.

Answer

(b) The tendency to form long open or closed atom chains by combination of some atoms in themselves is known as catenation. The catenation is maximum in carbon and decreases down the group.

$$\mathrm{C}»\mathrm{Si}>\mathrm{Ge}\approx \mathrm{Sn}>\mathrm{Pb}$$

This is due to high bond energy of $\mathrm{C}-\mathrm{C}$ bonds. Down the group, size increases and electronegativity decreases, thereby, tendency to show catenation decreases.

• Option (a) is incorrect because it suggests a gradual decrease in catenation tendency from carbon to tin, but it does not account for the significant drop in catenation ability after silicon. The correct order shows a much larger difference between carbon and silicon compared to the other elements.

• Option (c) is incorrect because it places silicon above carbon in terms of catenation tendency, which is not accurate. Carbon has the highest tendency for catenation due to its small size and high bond energy of C-C bonds.

• Option (d) is incorrect because it completely reverses the trend, placing germanium and tin above silicon and carbon. This contradicts the established understanding that carbon has the highest catenation tendency, followed by silicon, with germanium and tin having much lower tendencies.

Answer

(c) Silicon has a strong tendency to form polymers like silicones. The chain length of silicon polymer can be controlled by adding ${\mathrm{Me}}_3\mathrm{SiCl}$ which block the ends as shown below

• (a) ${\mathrm{MeSiCl}}_3$: This compound, methyltrichlorosilane, typically leads to cross-linking and branching in the polymer structure rather than controlling the chain length. It introduces multiple reactive sites, which can result in a more complex, three-dimensional network rather than a linear polymer.

• (b) $M{e}_{2}SiC{l}_2$: This compound, dimethyldichlorosilane, is used to form linear silicone polymers but does not effectively control the chain length. It provides two reactive sites that allow for the continuation of the polymer chain but does not cap the ends to limit the chain length.

• (d) ${\mathrm{Me}}_4\mathrm{Si}$: This compound, tetramethylsilane, is fully substituted and does not have any reactive sites (chlorine atoms) to participate in polymerization. Therefore, it cannot be used to control the chain length of silicone polymers.

Thinking Process

Ionisation enthalpy $\left({\mathrm{\Delta }}_{i}H\right)$ is the energy required to remove a valence electron. On moving down the group 13, there is decrease in first ionisation enthalpy due to an increase in atomic size and screening effect.

Answer

(d) On moving from B to Al, all the ionisation enthalpies decreases as expected and this decrease is due to an increase in atomic size and shielding effect.

On moving from Al to $\mathrm{Ga}$, the ionisation enthalpy increases slightly, because on moving from $\mathrm{Al}$ to $\mathrm{Ga}$, both nuclear charge and shielding effect increase but due to poor shielding by $d$-electron in $Ga$, effective nuclear charge on valence electron increases resulting in $d$-block contraction, that’s why ionisation enthalpies increase.

On moving from Ga to In, again there is slight decrease in ionisation enthalpies due to increased shielding effect by additional ten $4d$ electrons, which outweighs the effect of increased nuclear charge.

On moving from In to $\mathrm{Tl}$, ionisation enthalpies show the increase again because fourteen $4f$ electrons shield valence electron poorely (order of shielding effect $s>p>d>f$ and so effective nuclear charge increases, consequently ionisation enthalpies increase.

• Option (a) is incorrect because it suggests a consistent decrease in ionisation enthalpy from B to Tl, which does not account for the observed increases in ionisation enthalpy from Al to Ga and from In to Tl due to poor shielding by d and f electrons, respectively.

• Option (b) is incorrect because it suggests a consistent increase in ionisation enthalpy from B to Tl, which contradicts the observed decreases in ionisation enthalpy from B to Al and from Ga to In due to increased atomic size and shielding effects.

• Option (c) is incorrect because it suggests that Al has a higher ionisation enthalpy than Ga, which is not true. The ionisation enthalpy of Ga is slightly higher than that of Al due to poor shielding by d-electrons in Ga, leading to an increase in effective nuclear charge.

Answer

(b) Boron is trivalent, we would expect a simple hydride $B{H}_3$. However $B{H}_3$ is not stable. The boron possess incomplete octet and $B{H}_3$ dimerises to form $B_2{H}_6$ molecule with covalent and three centre 2-electron bond. The simplest boron hydride is diborane $B_2{H}_6$.

As seen from the structure drawn, 6 electrons are required for the formation of conventional covalent bond structure by B-atom, whereas in diborane, there are 12 valence electrons, three from each boron atoms and six from the six hydrogen atoms. The geometry of $B_2{H}_6$ can be represented as

The four terminal hydrogen atoms and two boron atoms lie one plane. Above and below the plane, there are two bridging hydrogen atoms. Each boron atom forms four bonds even though it has only three electrons. The terminal $\mathrm{B}-\mathrm{H}$ bonds are regular bonds but the bridge $\mathrm{B}-\mathrm{H}$ bonds are different.

Each bridge hydrogen is bonded to the two boron atoms only by sharing of two electrons. Such covalent bond is called three centre electron pair bond or a multi centre bond or banana bond.

• Option (a) is incorrect: In diborane, not all hydrogen atoms lie in one plane. The structure consists of four terminal hydrogen atoms and two boron atoms lying in one plane, while the two bridging hydrogen atoms lie in a plane perpendicular to this plane.

• Option (c) is incorrect: Diborane does not have four bridging hydrogen atoms. It has only two bridging hydrogen atoms and four terminal hydrogen atoms. The boron atoms and the four terminal hydrogen atoms lie in one plane, while the two bridging hydrogen atoms lie in a perpendicular plane.

• Option (d) is incorrect: Not all atoms in diborane lie in the same plane. The structure consists of four terminal hydrogen atoms and two boron atoms in one plane, with the two bridging hydrogen atoms lying in a plane perpendicular to this plane.

Answer

(a) (i) Reaction of ammonia with diborane gives initially $B_2{H}_6\cdot 2N{H}_3$ which is fermulated as $B{H}_2(N{H}_3{)}_2^{+}$ + $B{H}_4^{-}$ further heating gives borazine, $B_3{N}_3{H}_6$ also called borazole.

$\underset{\begin{array}{c}\text{Diborane }\\ (X)\end{array}}{3B_2{H}_6}+6N{H}_3\stackrel{473K}{\to }\underset{\begin{array}{c}\text{Borazole}\\ (Y)\end{array}}{2B_3{N}_3{H}_6}+12H_2$

Borazole has cyclic structure and is isoelectronic and isosteric with benzene and thus called inorganic benzene or triborine triammine or borazine.

(ii) Diborane can be prepared by the reduction of ${\mathrm{BF}}_3$ with lithium aluminium hydride in diethyl ether.

$$4B{F}_3+3LiAl{H}_4\to \underset{(X)}{2B_2{H}_6}+3Al{F}_3+3LiF$$

• Option (b): $B_2{O}_3,B_3{N}_3{H}_6$

  • $B_2{O}_3$ (boron trioxide) does not react with $N{H}_3$ to form $B_3{N}_3{H}_6$ (borazine). The reaction described in the question specifically involves diborane ($B_2{H}_6$) and not boron trioxide.

• Option (c): $B{F}_3,B_3{N}_3{H}_6$

  • $B{F}_3$ (boron trifluoride) is not the compound $X$ that reacts with $N{H}_3$ to form $B_3{N}_3{H}_6$. Instead, $B{F}_3$ is used to prepare diborane ($B_2{H}_6$) by reduction with lithium aluminium hydride. The compound $X$ in the reaction with $N{H}_3$ is $B_2{H}_6$.

• Option (d): $B_3{N}_3{H}_6,B_2{H}_6$

  • $B_3{N}_3{H}_6$ (borazine) is not the initial compound $X$ that reacts with $N{H}_3$. The initial compound $X$ is $B_2{H}_6$ (diborane), which reacts with $N{H}_3$ to eventually form $B_3{N}_3{H}_6$. The sequence of reactions described in the question does not support $B_3{N}_3{H}_6$ being the starting compound.

Answer

(b) Quartz, cristobalite and tridymite are some of the crystalline forms of silica and they are interconvertable at suitable temperature. Quartz is extensively used as a piezoelectric material.

• (a) Pb (Lead): Lead is not a component of quartz. Quartz is composed of silicon and oxygen, forming silicon dioxide (SiO₂). Lead is a heavy metal and does not contribute to the piezoelectric properties of quartz.

• (c) Ti (Titanium): Titanium is not a component of quartz. Quartz is made up of silicon and oxygen atoms. Titanium is a different element and is not found in the chemical structure of quartz.

• (d) Sn (Tin): Tin is not a component of quartz. Quartz consists of silicon and oxygen atoms. Tin is a separate element and does not play a role in the composition or piezoelectric properties of quartz.

Answer

(d) Reducing agents are those substances which reduces other substances and it self oxidises.

In ${\mathrm{SnCl}}_2$, Sn exists in +2 oxidation state, thus, acts as a strong reducing agent. i.e.,

$$\begin{array}{ll}SnC{l}_2+2FeC{l}_3& \to 2FeC{l}_2+SnC{l}_4\\ SnC{l}_2+2CuC{l}_2& \to 2CuCl+SnC{l}_4\end{array}$$

• (a) ${\mathrm{AlCl}}_3$: Aluminum chloride (${\mathrm{AlCl}}_3$) is not commonly used as a reducing agent. In fact, it is often used as a Lewis acid catalyst in various chemical reactions. Aluminum in ${\mathrm{AlCl}}_3$ is in the +3 oxidation state, which is relatively stable and does not readily undergo oxidation to act as a reducing agent.

• (b) ${\mathrm{PbCl}}_2$: Lead(II) chloride (${\mathrm{PbCl}}_2$) is not a commonly used reducing agent. Lead in ${\mathrm{PbCl}}_2$ is in the +2 oxidation state, which is relatively stable. Additionally, lead compounds are generally more toxic and less reactive as reducing agents compared to other options.

• (c) ${\mathrm{SnCl}}_4$: Tin(IV) chloride (${\mathrm{SnCl}}_4$) is not a reducing agent. In ${\mathrm{SnCl}}_4$, tin is in the +4 oxidation state, which is already a high oxidation state. Therefore, it is more likely to act as an oxidizing agent rather than a reducing agent.

Answer

(c) Carbon dioxide can be obtained as a solid in the form of dry ice allowing the liquified ${\mathrm{CO}}_2$ to expand rapidly.

Note Dry ice is also called cardice. It is obtained when ${\mathrm{CO}}_2$ is cooled under pressure (50-60 atm). Dry ice is used for making cold baths in the laboratory by mixing it with volatile organic solvents. It is also used as a coolant for preserving perishable articles in food industry, curing local burns and in hospitals for surgical operation of sores.

• (a) solid ${\mathrm{NH}}_3$: Solid ammonia (${\mathrm{NH}}_3$) is not referred to as dry ice. Dry ice specifically refers to solid carbon dioxide (${\mathrm{CO}}_2$).

• (b) solid ${\mathrm{SO}}_2$: Solid sulfur dioxide (${\mathrm{SO}}_2$) is not known as dry ice. Dry ice is the term used exclusively for solid carbon dioxide (${\mathrm{CO}}_2$).

• (d) solid ${\mathrm{N}}_2$: Solid nitrogen (${\mathrm{N}}_2$) is not called dry ice. The term dry ice is specifically used for solid carbon dioxide (${\mathrm{CO}}_2$).

Answer

(b) Cement is a product obtained by combining a material rich in lime. $\mathrm{CaO}$ with other material like clay which contains silica ${\mathrm{SiO}}_2$ alongwith the oxides of aluminium, iron and magnesium. The average composition of Portland cement is

CaO(50-60%)

$Si{O}_2$(20-25 %)

$A{l}_2{O}_3$(5-10 %)

$F{e}_2{O}_3$(1-2 %)

$S{O}_2$(1-2 %)

MgO(2-3 %)

Thus, it contains elements of group 2 (Ca), group 13 (Al) and group 14 (Si).

• Option (a) is incorrect: This option only mentions Group 2 elements. However, cement also contains elements from Group 13 (Al) and Group 14 (Si), which are not accounted for in this option.

• Option (c) is incorrect: This option mentions elements from Groups 2 and 13, but it omits Group 14 elements. Cement contains silica ($Si{O}_2$), which is an oxide of silicon, a Group 14 element.

• Option (d) is incorrect: This option mentions elements from Groups 2 and 14, but it omits Group 13 elements. Cement contains alumina ($A{l}_2{O}_3$), which is an oxide of aluminum, a Group 13 element.

Answer

$(a,b)$

On moving down the group from $\mathrm{Al}$ to $\mathrm{Ga}$, atomic radius decrease (exception) due to poor shielding by $d$-electrons. On moving from Al to Ga, shielding effect in $d$-electrons is unable to compensate increased nuclear charge.

Hence, successive increase of atomic radius as expected is not observed.

• (c) Presence of higher orbitals: This option is incorrect because the presence of higher orbitals would generally lead to an increase in atomic radius, not a decrease. Higher orbitals are larger and would contribute to a larger atomic size.

• (d) Higher atomic number: This option is incorrect because a higher atomic number alone does not necessarily result in a smaller atomic radius. While a higher atomic number means more protons and a greater nuclear charge, it also means more electrons, which can increase the atomic radius unless the increased nuclear charge significantly outweighs the electron shielding effect.

Answer

$(b,c)$

Sigma bond formed by $s-s$ overlap, $s-p$ overlap and $p-p$ overlapping. pi ( $\pi$ ) bond formed by $p-p$ overlapping.

The structure of ${\mathrm{CO}}_2$ is

Hybridisation of ${\mathrm{CO}}_2$ is $sp$ and the shape is linear. Due to pi bond, it has $p\pi -p\pi$ bonding between carbon and oxygen.

• (a) $s{p}^3$ hybridisation of carbon: This option is incorrect because $s{p}^3$ hybridisation results in a tetrahedral geometry, not a linear shape. In $s{p}^3$ hybridisation, the carbon atom would form four sigma bonds arranged in a tetrahedral structure, which is not the case for ${\mathrm{CO}}_2$.

• (d) $s{p}^2$ hybridisation of carbon: This option is incorrect because $s{p}^2$ hybridisation results in a trigonal planar geometry, not a linear shape. In $s{p}^2$ hybridisation, the carbon atom would form three sigma bonds arranged in a trigonal planar structure, which is not the case for ${\mathrm{CO}}_2$.

Answer

$(a,b)$

The chain length of the polymer can be controlled by adding ${\left(C{H}_3\right)}_{3}SiCl$ which blocks the ends as shown below

• (c) ${\mathrm{Me}}_3\mathrm{SiCl}$ does not necessarily improve the quality and yield of the polymer. Its primary function is to control the chain length and block the end terminals, not to enhance the overall quality or yield.

• (d) ${\mathrm{Me}}_3\mathrm{SiCl}$ does not act as a catalyst during polymerisation. Catalysts are substances that increase the rate of a chemical reaction without being consumed in the process, whereas ${\mathrm{Me}}_3\mathrm{SiCl}$ is used to control the polymer chain length and block the ends.

Answer

$(b,c)$

Fullerene an allotrope of carbon with finite number of carbon atoms with closed cage structure has been identified. The ${\mathrm{C}}_{60}$ isotope is more dominants. The structure is nearly spherical, like football.

Sphere is formed by the combination of 20 hexagons and 12 pentagons. Graphite is thermodynamically more stable than all allotropes of carbon.

• (a) Fullerenes have dangling bonds: This statement is incorrect because fullerenes do not have dangling bonds. Fullerenes are closed-cage structures where each carbon atom forms three bonds with neighboring carbon atoms, resulting in no unpaired electrons or dangling bonds.

• (d) Graphite is slippery and hard and therefore used as a dry lubricant in machines: This statement is incorrect because while graphite is indeed slippery and used as a dry lubricant, it is not hard. Graphite is soft and slippery due to the weak van der Waals forces between its layers, which allow them to slide over each other easily.

Answer

$(a,b,d)$

The bonding and structure of the boranes are of great interest. They are different from all other hydrides as they are electron deficient.

In diborane, there are 12 valene electrons, three from each $\mathrm{B}$-atom and six from $\mathrm{H}$-atoms. Electron structure shown in figure.

The two bridging $\mathrm{H}$-atoms are in a plane perpendicular to the rest of the molecule and prevent rotation between the two B-atoms.

The terminal $\mathrm{B}-\mathrm{H}$ distances are the same as the bond lengths measured in non-electron deficient compounds. These are assumed to be normal covalent bonds, with two electrons shared between two atoms. We can describe these bonds are two centre two electron bond $(2c-2e)$.

Obviously, they are abnormal bonds as the two bridges involve only one electron each from one of the boron atoms and hydrogen atoms, making a total of four electrons.

According to molecular orbital theory, each B-atom uses $s{p}^3$-hybrid orbitals for bonding. Out of the four $s{p}^3$-hybrid orbitals on each B-atom, one is with out an electron shown in broken lines.

The terminal $\mathrm{B}-\mathrm{H}$ bonds are normal 2 centre -2 electron bonds but the two bridge bonds are 3 centre -2 electron bonds. The 3 - centre -2 electron bridge bonds are also called banana bonds.

• Option (c) is incorrect because out of the six B-H bonds, only two bonds can be described in terms of 3-center 2-electron bonds. The other four B-H bonds are terminal bonds and are described as 2-center 2-electron bonds.

Thinking Process

Resonance is a way of describing delocalised electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by one single Lewis formula.

Answer

(d)

The resonating structure of ${\mathrm{CO}}_2$ are

$\mathrm{O}=\mathrm{C}=\mathrm{O}{\leftrightarrow }^{+}\mathrm{O}\equiv \mathrm{C}{\mathrm{O}}^{-}{\leftrightarrow }^{-}\mathrm{O}\mathrm{C}\equiv {\mathrm{O}}^{+}$

• (a) $\mathrm{O}-\mathrm{C}\equiv \mathrm{O}$: This structure is incorrect because it does not satisfy the octet rule for the carbon atom. Carbon would have only six electrons around it, which is not sufficient to complete its octet.

• (b) $\mathrm{O}=\mathrm{C}=\mathrm{O}$: This structure is actually a correct resonance structure of carbon dioxide. It is the most stable and commonly represented form of ${\mathrm{CO}}_2$.

• (c) ${}^{-}\mathrm{O}\equiv \mathrm{C}-{\mathrm{O}}^{+}$: This structure is incorrect because it implies a triple bond between oxygen and carbon, which is not a common or stable bonding arrangement for carbon dioxide. Additionally, the formal charges do not align with the typical resonance structures of ${\mathrm{CO}}_2$.

Answer

In the $BC{l}_3$, due to small size of $B^{3+},BC{l}_3$ is covalent. Still, the octet of $B$ remains incomplete. In $N{H}_3,N$ has a lone pair of electrons. Hence, $N$ shares lone pair of electron with $B$ to complete the octet. Hence, $BC{l}_3$ acts as Lewis acid and $N{H}_3$ as Lewis bases.

$$H_{3}N:+BC{l}_3\to H_{3}N\to BC{l}_3\text{ or }BC{l}_3\cdot N{H}_3$$

${\mathrm{AlCl}}_3$ forms dimer by completing octet of $\mathrm{Al}$ involving $p$-orbitals to accept electron pair from chlorine.

Answer

Orthoboric acid is less soluble in cold water but highly soluble in hot water. It is a monobasic acid. It does not liberate ${\mathrm{H}}^{+}$ion but accepts ${\mathrm{OH}}^{-}$from water, behaving as a Lewis acid.

$$H_{3}B{O}_3+H_{2}O\to B(OH{)}_4^{-}+H^{+}$$

Octet of boron remains incomplete. Oxygen atom contains lone pair of electrons in water molecule. Hence, instead of donating proton $\left({\mathrm{H}}^{+}\right)$, boric acid accepts ${\mathrm{OH}}^{-}$from water forming $\mathrm{B}(\mathrm{OH}{)}_4^{-}$to complete octet.

Since, electron acceptor substance behaves as Lewis acid, therefore, boric acid acts as a Lewis acid in water.

Answer

Orthoboric acid $H_{3}B{O}_3$, in solid state possesses a layer structure made up of $B(OH{)}_3$ units forming hexagonal rings of $H$-bonding as given below

Each $\mathrm{H}$-atom acts as a bridge between two oxygen atoms of different ${\mathrm{BO}}_3^{3-}$ units.

Boric acid when dissolved in water, acts as Lewis acid forming $B(OH{)}_4^{-}$

$$H_{3}B{O}_3+H_{2}O\to B(OH{)}_4^{-}+H^{+}$$

The hybridisation of boron in $\mathrm{B}(\mathrm{OH}{)}_4^{-}$is $s{p}^3$.

Answer

In trivalent state, the number of electrons around the central atom in a molecule of compounds $BC{l}_3$ and $AlC{l}_3$ will be only six.

Such electron deficient molecules have tendency to accept a pair of electron to achieve stable electronic configuration and thus, act as Lewis acids. The tendency to behave as Lewis acid decreases with the increase in the size down the group.

Answer

(a) Carbon tetrachloride $\left(CC{l}_4\right)$ is a covalent compound while $H_{2}O$ is a polar compound. $CC{l}_4$ does not form $H$-bond with water molecule. Hence, it is immiscible in water.

Further more, $CC{l}_4$ is not hydrolysed by water because of the absence of $d$-orbitals in carbon while $SiC{l}_4$ is readily hydrolysed by water.

$$SiC{l}_4+4H_{2}O\to \underset{\text{Silicic acid }}{Si(OH{)}_4}+4HCl$$

The hydrolysis of $SiC{l}_4$ occurs due to coordination of $O{H}^{-}$with empty $3d$ orbitals in silicon atom of $SiC{l}_4$ molecule.

(b) Carbon atoms have the tendency to link with one another through covalent bonds to form chains and rings. This property is called catenation. This is because $\mathrm{C}-\mathrm{C}$ bonds are very strong.

Down the group, the size increases and electronegativity decreases and thereby, tendency to show catenation decreases. Thus, carbon has a strong tendency for catenation as compared to silicon.

Answer

(a) $C{O}_2$ has a linear structure. Its dipole moment is zero. It is believed that $C{O}_2$ molecule is a resonance hybrid has the following structures.

$$\mathrm{O}=\mathrm{C}=\mathrm{O}\leftrightarrow \mathrm{O}-\mathrm{C}\equiv {\mathrm{O}}^{+}\leftrightarrow {\mathrm{O}}^{+}\equiv \mathrm{C}{\mathrm{O}}^{-}$$

The $C{O}_2$ molecules are held together by weak van der Waals’ forces and thus, it exists as gas. In $Si{O}_2$, due to large electronegative difference between $Si$ and $O$, the $Si-O$ bonds have considerable ionic nature.

Therefore, silica has three dimensional network like structure in which Si-atom is tetrahedrally bonded to four oxygen atoms and each oxygen atom is bonded to two silicon atoms by covalent bonds.

There is no discrete ${\mathrm{SiO}}_2$ molecule. It is a network solid with octahedral coordination.

(b) In silicon, vacant $3d$ orbitals are available due to which it can accomodate electrons from 6 fluorine atoms, thereby forming $S{i}_6^{2-}$ ion. However, in case of $C$ only $2p^2$ filled orbitals are available thus, it cannot expand their covalency more than 4 . Thus, $CC{l}_6^2$ is not known.

Answer

The term inert pair effect is often used in relation to the increasing stability of oxidation states that are 2 less than the group valency for the heavier elements of groups 13, 14, 15 and 16. In group 13 all the elements show +3 oxidation state whereas $\mathrm{Ga}$, In and $\mathrm{TI}$ show +1 oxidation state also.

Boron, being small in size can lose its valence electrons to form ${\mathrm{B}}^{3+}$ ion and shows +3 oxidation state.

+1 oxidation state of $\mathrm{TI},\mathrm{Ga}$ is due to inert pair effect. The outer shell s-electrons $\left(n{s}^2\right)$ penetrate to $(n-1)d$-electrons and thus becomes closer to nucleus and are more effectively pulled towards the nucleus.

This results in less availability of $n{s}^2$ electrons pair for bonding or $n{s}^2$ electron pair becomes inert. This reluctance in the participation of $n{s}^2$ eletrons in bonding is termed as inert pair effect. The inert pair is more effective after $n\ge 4$ and increases with increasing value of $n$.

For groups 14 , in spite of four valence electrons, they do not form $M^{4+}$ or $M^{4-}$ ionic compounds. They form covalent compounds with four bonds.

$\mathrm{Ge},\mathrm{Sn}$ and $\mathrm{Pb}$ also exhibit +2 oxidation state due to inert pair effect. ${\mathrm{Sn}}^{2+}$ and ${\mathrm{Pb}}^{2+}$ show ionic nature. The tendency to form +2 ionic state increases on moving down the group due to inert pair effect.

Answer

All members of group 14 form dioxides of formula $M{O}_2$ i.e., for carbon $C{O}_2$ and silica $Si{O}_2$. $C{O}_2$ has a linear structure since, its dipole moment is zero. Both the oxygen atoms are linked by double bonds. The C-atom is $sp$ hybridised.

$$\begin{array}{rl}& p-p(\pi )p-p(\pi )\\ & \mathrm{O}=\mathrm{C}=\mathrm{O}\\ & \mathrm{sp}-\mathrm{p}(\sigma )sp-p(\sigma )\end{array}$$

However, $\mathrm{C}-\mathrm{O}$ bond length is $1.15A$, which is less than calculated value of $\mathrm{C}=\mathrm{O}$ bond. Hence, it shows resonance hybrid of following structures.

$$\mathrm{O}=\mathrm{C}=\mathrm{O}\leftrightarrow \to {\mathrm{O}}^{-}\mathrm{C}\equiv {\mathrm{O}}^{+}\leftrightarrow \mathrm{O}\equiv {\mathrm{C}}^{-}{\mathrm{O}}^{-}$$

But, the structure of silica is entirely different from that of ${\mathrm{CO}}_2$. $\mathrm{Si}\mathrm{O}$ bonds have a considerable ionic character due to large electronegative difference between $\mathrm{Si}$ and $\mathrm{O}$. As a result, silica has a three dimensional structure in which silica atom is tetrahedrally bonded to four oxygen atoms and each oxygen atom is bonded to two silicon atoms by covalent bonds.

There is no discrete ${\mathrm{SiO}}_2$ molecule but entire crystal as a giant molecule. $\mathrm{Si}-\mathrm{O}$ bonds are strongly making it solid having high melting point.

Answer

The structure of ${\mathrm{SiO}}_2$ crystal is as

When some Si-atoms are replaced by trivalent impurity, holes are created, which are equivalent to positive charge. These holes make the crystal structure as conductor of electricity.

Since, the crystals on the whole are always electrically neutral, the obtained crystal is also electrically neutral.

An electron from a neighbouring atom can come and fill the electron hole, but in doing so it would leave an electron hole at its original position. If it happens, it would appear as if the electron hole has moved in the direction opposite to that of the electron that filled it.

Under the influence of electric field, electrons would move towards the positively charged plate through electronic holes, but it would appear as if electron holes are positively charged and are moving towards negatively charged plate. This type of semiconductors are called $p$-type semiconductors.

Answer

In trivalent state, most of the compounds being covalent are hydrolysed in water, e.g., $BC{l}_3$ on hydrolysis in water form ${[B(OH{)}_4]}^{-}$species, the hybridisation state of $B$ is $s{p}^3$.

$$\begin{array}{rl}BCl+3H_{2}O& \to B(OH{)}_3+3HCl\\ B(OH{)}_3+H_{2}O& \to {[B(OH{)}_4]}^{-}+H^{+}\end{array}$$

$AlC{l}_3$ in acidified aqueous solution form octahedral ${\left[Al{\left(H_{2}O\right)}_6\right]}^3+$ ion. In this complex, the $3d$ orbitals of $Al$ are involved and the hybridisation state of $Al$ is $s{p}^3{d}^2$

$$AlC{l}_3+H_{2}OHCl\to {\left[Al{\left(H_{2}O\right)}_6\right]}^3++3C{l}^{-}(aq)$$

of ${\mathrm{Al}}^{3+}$

$s{p}^3{d}^2$ hybridisation

Answer

Aluminium being amphoteric in nature dissolves both in acids and alkalies evolving ${\mathrm{H}}_2$ gas which burns with a pop sound.

$$\begin{array}{rl}& 2Al+6HCl\to 2AlC{l}_3+3H_2\\ & 2Al+NaOH+2H_{2}O\to \underset{\text{ Sodium meta aluminate }}{2NaAl{O}_2}+3H_2\end{array}$$

But when $Al$ is treated with conc. $HN{O}_3$, a thin protective layer of $A{l}_2{O}_3$ is formed on its surface which prevents further action.

$$2Al+6HN{O}_3\to A{l}_2{O}_3+6N{O}_2+3H_{2}O$$

Answer

(a) In gallium, due to poor shielding of valence electrons by the intervening $3d$ electrons. The nuclear charge becomes effective, thus, atomic radius decreases and hence, the ionisation enthalpy of gallium is higher than that of aluminium.

(b) Due to small size of boron, the sum of its first three ionisation enthalpies is very high. This prevent it to form +3 ions and force it to form only covalent compound. That’s why boron does not exist as ${\mathrm{B}}^{3+}$ ion.

(c) Aluminium forms ${\left[{\mathrm{AlF}}_6\right]}^{3-}$ ion because of the presence of vacant $d$-orbitals so it can expand its coordination number from 4 to 6 . In this complex, Al undergoes $s{p}^3{d}^2$ hybridisation.

On the other hand, boron does not form ${\left[B{F}_6\right]}^{3-}$ ion, because of the unavailability of $d$-orbitals as it cannot expand its coordination number beyond four. Hence, it can form ${\left[B{F}_4\right]}^{-}$ion ( $s{p}^3$ hybridisation).

(d) Due to inert pair effect, $Pb$ in +2 oxidation state is more stable than in +4 oxidation state hence $Pb{X}_2$ is more stable than $Pb{X}_4$.

(e) Due to inert pair effect, tendency to form +2 ions increases down the group, hence ${\mathrm{Pb}}^{2+}$ is more stable than ${\mathrm{Pb}}^{4+}$. That’s why ${\mathrm{Pb}}^{4+}$ acts as an oxidising agent while ${\mathrm{Sn}}^{2+}$ is less stable than ${\mathrm{Sn}}^{4+}$ and hence ${\mathrm{Sn}}^{2+}$ acts as a reducing agent.

$$\begin{array}{rl}\underset{\text{ Reducing agent }}{{\mathrm{Sn}}^{2+}}& \to {\mathrm{Sn}}^{4+}+2e^{-}\\ {\mathrm{Pb}}^{4+}+2e^{-}& \to {\mathrm{Pb}}^{2+}\end{array}$$

(f) Electron gain enthalpy of $\mathrm{Cl}$ is more negative than electron gain enthalpy of fluorine because when an electron is added to $\mathrm{F}$, the added electron goes to the smaller $n=2$ quantum level and suffers significant repulsion from other electrons present in this level.

For $n=3$ quantum level (in $\mathrm{Cl}$ ) the added electron occupies a larger region of space and the electron-electron repulsion is much less.

(g) Due to inert pair effect, $Tl$ in +1 oxidation state is more stable than that of +3 oxidation state. Therefore, $TI{\left(N{O}_3\right)}_3$ acts as an oxidising agent.

(h) Property of catenation depends upon the atomic size of the element. Down the group, size increases and electronegativity decreases, thus the tendency to show catenation decreases. As the size of $\mathrm{C}$ is much smaller than $\mathrm{Pb}$, therefore, carbon show property of catenation but lead does not show catenation.

(i) Unlike other boron halides, $B{F}_3$ does not hydrolyse completely. However, it form boric acid and fluoroboric acid. This is because the $HF$ first formed reacts with $H_{3}B{O}_3$.

$$\begin{array}{rl}B{F}_3+3H_{2}O& \to H_{3}B{O}_3+3HF]\times 4\\ H_{2}B{O}_3+4HF& \to H^{+}{\left[B{F}_4\right]}^{-}+3H_{2}O]\times 3\\ 4B{F}_3+3H_{2}O& \to H_{3}B{O}_3+3{\left[B{F}_4\right]}^{-}+3H^{+}\end{array}$$

(j) In graphite, $\mathrm{C}$ is $s{p}^2$ hybridised. Carbon due to its smallest size and highest electronegativity among group 14 elements has strong tendency to form $p\pi -p\pi$ multiple bonds while silicon due to its larger size and less electronegativity has poor ability to form $p\pi -p\pi$ multiple bonds. That’s why the element silicon does not form a graphite like structure.

Answer

(i)$\underset{Borax}{N{a}_2{B}_4{O}_7}+2HCl+5H_{2}O\to 2NaCl+\underset{\text{ Orthoboric acid }(X)}{4H_{3}B{O}_3}$

(ii) ${\mathrm{H}}_3{\mathrm{BO}}_3\stackrel{\mathrm{\Delta },370\mathrm{K}}{\to }\underset{\text{Metaboric acid}}{{\mathrm{HBO}}_2}+{\mathrm{H}}_2\mathrm{O}$

(iii) $4HB{O}_2\underset{-H_{2}O}{\stackrel{\mathrm{\Delta }>370K}{\to }}\underset{\begin{array}{c}\text{ Tetraboric }\\ \text{ acid }\end{array}}{\left[H_2{B}_4{O}_7\right]}\underset{\text{ Heat }}{\to Red}\underset{\text{ Boron trioxide (Z) }}{2B_2{O}_3+H_{2}O}$

Answer

(i) $4B{F}_3+3LiAl{H}_4\to \underset{\text{ Diborane (X) }}{2B_2{H}_6}+3LiF+3Al{F}_3$

(ii) $B_2{H}_6+6H_{2}O\to \underset{\text{(X) Orthoboric acid}}{2H_{3}B{O}_3}+6H_2$

(iii) $B_2{H}_6+3O_2\stackrel{\mathrm{\Delta }}{\to }{B}_2{O}_3+3H_{2}O$

Answer

A. $\to$ (5)

B. $\to$ (3)

C. $\to$ (4)

D. $\to (1,2)$

A. ${\mathrm{BF}}_4^{-}$Tetrahedral shape $s{p}^3$ hybridisation regular geometry.

B. ${\mathrm{AlCl}}_{3-}$ Octet not complete of $\mathrm{Al}$, act as Lewis acid.

C. $\mathrm{SnO}$ ${\mathrm{Sn}}^{2+}$ can show +4 oxidation state.

D. $Pb{O}_2$ Oxidation state of $Pb$ in $Pb{O}_2$ is +4 . Due to inert pair effect $P{b}^{4+}$ is less stable than $P{b}^{2+}$, acts as strong oxidising agent.

Answer

A. $\to$ (3)

B. $\to (4)$

C. $\to (1)$

D. $\to$ (5)

E. $\to$ (2)

A. $B{H}_3$ is unstable forms diborane $B_2{H}_6$ by 3 centre -2 electron bond show banana bond.

B. Gallium with low melting point and high boiling point makes it useful to measure high temperatures.

C. Borax is used as a flux for soldering metals for heat, scratch resistant coating in earthernwares.

D. Alumino silicate used as catalyst in petrochemical industries.

E. Quartz, is a crystalline form of silica.

Answer

A. $\to$ (2)

B. $\to (3)$

C. $\to (2)$

D. $\to (1)$

E. $\to (2)$

F. $\to (3)$

A. Boron in ${[\mathrm{B}(\mathrm{OH}{)}_4]}^{-}s{p}^3$ hybridised.

B. Aluminium in ${\left[Al{\left(H_{2}O\right)}_6\right]}^{3+}s{p}^3{d}^2$ hybridised.

C. Boron in $B_2{H}_{6}s{p}^3$ hybridised.

D. Carbon in Buckminsterfullerene $s{p}^2$ hybridised.

E. Silicon in ${\mathrm{SiO}}_4^{4-}s{p}^3$ hybridised.

F. Germanium in ${\left[{\mathrm{GeCl}}_6\right]}^{2-}s{p}^3{d}^2$ hybridised.

Answer

(d) Assertion is not correct but reason is correct. Aluminium is trivalent whereas silicon is tetravalent. If aluminium atom replaces a few silicon atoms in three dimensional network of silicon dioxide, the overall structure acquires neutrality. (Also, refer to Q. 30)

Answer

(b) Silicones are a group of organo silicon polymers which have $\left(R_2\mathrm{SiO}\right)$ as a repeating unit. This suggests that silicones are surrounded by non-polar alkyl groups that are water repelling in nature. They have wide applications. It is used for water proofing of fabrics.

Answer

For Group 13

(a) Atomic Size On moving down the group for each successive member, one extra shell of electrons is added and therefore, atomic radius is expected to increase. However, a deviation can be seen.

Atomic radius of $\mathrm{Ga}$ is less than that of Al due to presence of additional $10d$ - electrons, which offer poor screening effect to the outer electron.

(b) Ionisation Enthalpy The ionisation enthalpy values as expected from general trends do not decrease smoothly down the group. The decrease from B to Al is associated with increase in size.

The observed discontinued between $\mathrm{Al}$ and $\mathrm{Ga}$ and between In and TI due to low screening effect of $d$ and $f$-electrons which compensates increased nuclear charge.

(c) Metallic or Electropositive Character Boron is a semi-metal (metalloid) due to very high ionisation enthalpy. All others are metals and metallic character first increases from $\mathrm{B}$ to $\mathrm{Al}$ as size increases. From $\mathrm{Al}$ to $\mathrm{Tl}$ decrease due to poor shielding of $d$ - and $f$-electrons,

(d) Oxidation States As we move down the group, the stability of +3 oxidation state decreases while that of +1 oxidation state progressively increases. In other words, the order of stability of +1 oxidation state increase in the order. $\mathrm{Al}<\mathrm{Ga}<\mathrm{In}<\mathrm{Tl}$. Infact, in $\mathrm{Ga}$, In and $\mathrm{Tl}$, both +1 and +3 oxidation states are observed.

(e) Nature of Halides These elements react with halogens to form trihalids (except ${\mathrm{TH}}_3$ )

$$2E(s)+3X_2(g)\to 2{\mathrm{EX}}_3(s)[X=\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{I}]$$

Boron in halides are electron deficient molecules and behave as Lewis acids. The Lewis character decreases in the order: $B{I}_3>BB{r}_3>BC{l}_3>B{F}_3$

For Group 14

(a) Atomic Size There is considerable increase in covalent radius from C to Si thereafter from $\mathrm{Si}$ to $\mathrm{Pb}$ as small increase in radius is observed. This is due to the presence of completely filled $d$ and $f$ - orbitals in heavier member.

(b) Ionisation Enthalpy The first ionisation enthalpy of group 14 members is higher than the corresponding members of group 13. The influence of inner core electrons is visible here. In general the ionisation enthalpy decreases down the group. Small decrease in $\mathrm{\Delta }$; $\mathrm{H}$ from $\mathrm{Si}$ to $\mathrm{Ge}$ to $\mathrm{Sn}$ and slight increase in $\mathrm{\Delta };\mathrm{H}$ from $\mathrm{Sn}$ to $\mathrm{Pb}$ is the consequence of poor shielding effect of intervening $d$ and $f$-orbitals and increase in size of the atom.

(c) Metallic Character Metallic character increases down the group $\mathrm{C}$ (non-metal) $\mathrm{Si},\mathrm{Ge}$ (metalloid) $\mathrm{Sn},\mathrm{Pb}$ (metals).

(d) Oxidation States The group 14 elements have four electrons in the outermost shell. The common oxidation states exhibited by these elements are +4 and +2 . Carbon also exhibits negative oxidation states. Since, the sum of the first four ionisation enthalpies is very high, compounds in +4 oxidation states are generally covalent in nature. In heavier members the tendency to show +2 oxidation state increases in the $\mathrm{Ge}<\mathrm{Sn}<\mathrm{Pd}$ due to inert pair effect.

(e) Nature of Halides These elements can form halides of formula $M{X}_2$ and $M{X}_4$ (where, $\mathrm{X}=\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{I})$. Except carbon, all other members react directly with halogen under suitable condition to make halides.

Most of $M{X}_4$ are covalent with $s{p}^3$ hybridisation and tetrahedral in structure. Exceptions are $Sn{F}_4$ and $Pb{F}_4$ which are ionic in nature. Heavier members $Ge$ to $Pb$ are able to make halides of formula $M{X}_2$. Stability of halides increases down the group.

Answer

(a) In ${\mathrm{AlCl}}_3,\mathrm{Al}$ has only six electrons in its valence shell. It is an electron deficient species. Therefore, it acts as a Lewis acid (electron acceptor).

(b) In ${\mathrm{BF}}_3$ boron has a vacant $2p$-orbital and fluorine has one $2p$-completely filled unutilised orbital. Both of these orbitals belong to same energy level therefore, they can overlap effectively and form $p\pi -p\pi$ bond. This type of bond formation is known as back bonding.

While back bonding is not possible in $BC{l}_3$, because there is no effective overlapping between the $2p$-orbital of boron and $3p$-orbital of chlorine. Therefore, electron deficiency of $B$ is higher in $BC{l}_3$ than that of $B{F}_3$. That’s why $B{F}_3$ is a weaker Lewis acid than $BC{l}_3$.

(c) In $Pb{O}_2$ and $Sn{O}_2$, both lead and tin are present in +4 oxidation state. But due to stronger inert pair effect, $P{b}^{2+}$ ion is more stable than $S{n}^{2+}$ ion. In other words, $P{b}^{4+}$ ions i.e., $Pb{O}_4$ is more easily reduced to $P{b}^{2+}$ ions than $S{n}^{4+}$ ions reduced to $S{n}^{2+}$ ions. Thus, $Pb{O}_2$ acts as a stronger oxidising agent than $Sn{O}_2$.

(d) ${\mathrm{Tl}}^{+}$is more stable than ${\mathrm{Tl}}^{3+}$ because of inert pair effect.

Answer

When an aqueous solution of borax is acidified with $\mathrm{HCl}$ boric acid is formed.

$$N{a}_2{B}_4{O}_7+2HCl+5H_{2}O\to 2NaCl+\underset{\text{ Boric acid }}{4H_{3}B{O}_3}$$

Boric acid is a white crystalline solid. It is soapy to touch because of its planar layered structure. Boric acid is a weak monobasic acid. It is not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion.

$$B(OH{)}_3+2HOH\to [B(O{H}_4){]}^{-}+H_3{O}^{+}$$

Answer

(a) $TICl$ more stable than $TIC{l}_3$ due to inert pair effect. $TIC{l}_3$ is less stable and covalent in nature but $TICl$ is more stable and ionic in nature.

(b) Due to absence of $d$-orbitals, Al does not show inert pair effect. Hence, its most stable oxidation state is +3 . Thus, $AlC{l}_3$ is much more stable than AlCl. Further, in the solid or the vapour state, $AlC{l}_3$ covalent in nature but in aqueous solutions, it ionises to form $A{l}^{3+}(aq)$ and $C{l}^{-}(aq)$ ions.

(c) Due to inert pair effect, indium exists in both +1 and +3 oxidation states out of which + 3 oxidation state is more stable than +1 oxidation state. In other words, ${\mathrm{InCl}}_3$ is more stable than $\mathrm{InCl}$. Being unstable, $\mathrm{In}\mathrm{Cl}$ undergoes disproportionation reaction.

$$3\mathrm{InCl}(aq)\to 2\ln (s)+{\ln }^{3+}(aq)+3{\mathrm{Cl}}^{-}(aq)$$

Answer

Boron halides do not exist as dimer due to small size of boron atom which makes it unable to accommodate four large sized halide ions. ${\mathrm{AlCl}}_3$ exists as dimer. Al makes use of vacant $3p$ —orbital by coordinate bond i.e., Al atoms complete their octet by forming dimers.

Answer

Due to $p\pi -p\pi$ back bonding, the lone pair of electrons of $\mathrm{F}$ is donated to the $\mathrm{B}$-atom. This delocalisation reduces the deficiency of electrons on $\mathrm{B}$ thereby increasing the stability of ${\mathrm{BF}}_3$ molecule.

Structure of diborane

Due to absence of lone pair of electrons on $H$-atom, this compensation does not occur in $B{H}_3$. In other words, electron deficiency of $B$ stays and hence to reduce its electron deficiency, $B{H}_3$ dimerises to form $B_2{H}_6$.

In $B_2{H}_6$, four terminal hydrogen atoms and two boron atoms lie in one plane. Above and below this plane there are two bridging $H$-atoms. The four terminal $B-H$ bonds are regular while the two bridge ( $\begin{array}{lll}B& H& B\end{array})$ bonds are three centre- two electron bonds.

Answer

(a) Silicones are a group of organosilicon polymers, which have $\left(R_2\mathrm{SiO}\right)$ as a repeating unit. These may be linear silicones, cyclic silicones and cross-linked silicones.

These are prepared by the hydrolysis of alkyl or aryl derivatives of $SiC{l}_4$ like $RSiC{l}_3,R_{2}SiC{l}_2$, and $R_{3}SiCl$ and polymerisation of alkyl or aryl hydroxy derivatives obtained by hydrolysis.

Uses

These are used as sealant, greases, electrical insulators and for water proofing of fabrics. These are also used in surgical and cosmetic plants.

(b) Boron forms a number of covalent hydrides with general formulae $B_n{H}_{n+4}$ and $B_n{H}_{n+6}$. These are called boranes. $B_2{H}_6$ and $B_4{H}_{10}$ are the representative compounds of the two series respectively.

Preparation of Diborane

It is prepared by treating boron trifluoride with ${\mathrm{LiAlH}}_4$ in diethyl ether.

$$4B{F}_3+3LiAl{H}_4\to 2B_2{H}_6+3LiF+3Al{F}_3$$

On industrial scale it is prepared by the reaction of $B{F}_3$ with sodium hydride.

$$2B{F}_3+6NaH\stackrel{450\text{ K}}{⟶}{B}_2{H}_6+6NaF$$

Answer

Since, compound $(A)$ of boron reacts with $NM{e}_3$ to form an adduct $(B)$ thus, compound $(A)$ is a Lewis acid. Since, adduct $(B)$ on hydrolysis gives an acid $(C)$ and hydrogen gas, therefore, $(A)$ must be $B_2{H}_6$ and $(C)$ must be boric acid.

$$\begin{array}{rl}& \underset{\text{ Diborane (A) }}{{\mathrm{B}}_2{\mathrm{H}}_6}+\underset{\text{ Adduct (B) }}{2\mathrm{NMe}}\to 2{\mathrm{BH}}_3.NM{e}_3\end{array}$$ $$\begin{array}{rl}& \underset{\text{ (B) }}{{\mathrm{BH}}_3.{\mathrm{NMe}}_3}+3{\mathrm{H}}_2\mathrm{O}\to \underset{\text{ Boric acid (C) }}{{\mathrm{H}}_3{\mathrm{BO}}_3}+6{\mathrm{H}}_2+{\mathrm{NMe}}_3\end{array}$$