Answer

(i) $ \mathrm{B}$ to $ \mathrm{Tl}$

Boron (B), Aluminum (Al), Gallium (Ga), Indium (In), and Thallium (TI) belong to Group 13 of the periodic table.

Boron (B): Typically exhibits a +3 oxidation state.

Aluminum (Al): Also shows a +3 oxidation state.

Gallium (Ga): Can exhibit +1 and +3 oxidation states.

Indium (In): Similar to Gallium, it shows +1 and +3 oxidation states.

Thallium (Tl): Primarily shows a +1 oxidation state.

As we move down the group from Boron to Thallium, the +3 oxidation state becomes less stable, while the +1 oxidation state becomes more stable.

This trend is attributed to the inert pair effect, where the s-electrons (the two electrons in the outermost sorbital) are less likely to participate in bonding as we move down the group. Thus, the stability of the lower oxidation state $(+1)$ increases.

(ii) $ \mathrm{C}$ to $ \mathrm{Pb}$

Identify the Group: Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), and Lead (Pb) belong to Group 14 of the periodic table.

Carbon (C): Typically exhibits a +4 oxidation state.

Silicon (Si): Also shows a +4 oxidation state.

Germanium (Ge): Can exhibit +2 and +4 oxidation states.

Tin (Sn): Similar to Germanium, it shows +2 and +4 oxidation states.

Lead (Pb): Primarily shows a +2 oxidation state.

As we move down the group from Carbon to Lead, the +4 oxidation state becomes less stable, while the +2 oxidation state becomes more stable.

This trend is also due to the inert pair effect, where the s-electrons become less involved in bonding as we go down the group, leading to a preference for the lower oxidation state ( +2 ).

In both groups, as we move down the periodic table, the lower oxidation state becomes more stable due to the inert pair effect, which explains the observed patterns in oxidation states.

Answer

Boron and thallium belong to group 13 of the periodic table. In this group, the +1 oxidation state becomes more stable on moving down the group. $BCl_3$ is more stable than $TlCl_3$ because the +3 oxidation state of $B$ is more stable than the +3 oxidation state of $Tl$. In $Tl$, the +3 state is highly oxidising and it reverts back to the more stable +1 state.

Answer

The electric configuration of boron is $n s^{2} n p^{1}$. It has three electrons in its valence shell. Thus, it can form only three covalent bonds. This means that there are only six electrons around boron and its octet remains incomplete. When one atom of boron combines with three fluorine atoms, its octet remains incomplete. Hence, boron trifluoride remains electron-deficient and acts as a Lewis acid.

Answer

Being a Lewis acid, $BCl_3$ readily undergoes hydrolysis. Boric acid is formed as a result.

$BCl_3+3 H_2 O \longrightarrow 3 HCl+B(OH)_3$

$CCl_4$ completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When $CCl_4$ and water are mixed, they form separate layers.

$CCl_4+H_2 O \longrightarrow$ No reaction

Answer

Boric acid does not ionize in water to give proton. Hence, it is not a protic acid. It is a Lewis acid as it accepts electrons from hydroxide ions.

$ \mathrm{B}(\mathrm{OH})_3(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{~B}\left(\mathrm{OH}_4\right)\right]^{-}(\mathrm{aq})+\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq}) $

therefore, boric acid is not protic acid.

Answer

On heating orthoboric acid $(H_3 BO_3)$ at $370 K$ or above, it changes to metaboric acid $(HBO_2)$.

On further heating, this yields boric oxide $B_2 O_3$.

$ H_3 BO_3 \xrightarrow[370 K]{\Delta} \underset { \text {Metaboric acid }}{HBO_2} \xrightarrow[\text{ red hot }]{\Delta} \underset { \text {Boric oxide }}{B_2 O_3} $

Answer

$BF_3$

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 $s p^{2}$ hybridised orbitals of boron with the sporbitals of three halogen atoms. Boron is $s p^{2}$ hybridised in $BF_3$.

$BH_4{ }^{-}$

Boron-hydride ion $(BH_4)$ is formed by the $s p^{3}$ hybridisation of boron orbitals. Therefore, it is tetrahedral in structure.

Answer

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

(i) $2 Al {(s)}+6 HCl {(aq)} \longrightarrow 2 Al {(aq)}^{3+}+6 Cl {(aq)}^{-}+3 H {2(g)}$

(ii) $2 Al {(s)}+2 NaOH {(aq)}+6 H_2 O {(l)} \longrightarrow 2 Na^{+}[Al(OH)4] {(aq)}^{-}+3 H _2{(g)}$

Answer

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

(i) $BCl_3$

$BCl_3$ is an appropriate example of an electron-deficient compound. $B$ has 3 valence electrons. After forming three covalent bonds with chlorine, the number of electrons around it increases to 6 . However, it is still short of two electrons to complete its octet.

(ii) $SiCl_4$

The electronic configuration of silicon is $n s^{2} n p^{2}$. This indicates that it has four valence electrons. After it forms four covalent bonds with four chlorine atoms, its electron count increases to eight. Thus, $SiCl_4$ is not an electron-deficient compound.

Answer

The resonance structure of $CO_3^{2-}$ is

The resonance structure of $HCO_3^{-}$ is

There are only two resonating structures for the bicarbonate ion.

Answer

The state of hybridisation of carbon in:

(a) $CO_3^{2-}$

$C$ in $CO_3^{2-}$ is $s p^{2}$ hybridised and is bonded to three oxygen atoms.

(b) Diamond

Each carbon in diamond is $s p^{3}$ hybridised and is bound to four other carbon atoms.

(c) Graphite

Each carbon atom in graphite is $sp^2 $ hybridised and is bound to three other carbon atoms

Answer

Answer

(a)

Lead belongs to group 14 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, $ PbCl_4 $ is much less stable than $ PbCl_2 $. However, the formation of $ PbCl_4 $ takes place when chlorine gas is bubbled through a saturated solution of $ PbCl_2 $ .

$ PbCl_2 (s) + Cl_2 (g) \rightarrow PbCl_4 (l) $

(b) 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 $Pb(II)$.

$ PbCl_4(l) \xrightarrow{\Delta} PbCl _2(s) + Cl _2(g) $

(c) Lead is known not to form $PbI_4$. $Pb(+4)$ is oxidising 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.

$Pb(IV)$ oxidises $I^{-}$ to $I^{2}$ and itself gets reduced to $Pb(II)$.

$ PbI_4 \longrightarrow PbI_2+I_2 $

Answer

In $B F_3$ molecule, $B-F$ bond has partial double bond due to presence of $p \pi-p \pi$ backbonding between $B$ and $F$, which removes the electron deficienty of the molecule.

Due to this, $B-F$ bond length is reduced to 130 pm . When $B F_3$ changes to $B F_4$, the hybridization changes from $s p^2$ to $s p^3$.

The double bond character changes to single bond and the $B-F$ bond length becomes 143 pm .

Answer

As a result of the difference in the electronegativities of $B$ and $Cl$, the $B-Cl$ bond is polar in nature. However, the $BCl_3$ molecule is non-polar. This is because $BCl_3$ 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.

Answer

Anhydrous HF is a covalent compound and forms strong intramolecular H bonds. It does not give fluoride ions and cannot dissolve $ \mathrm{AlF}_3$ in it. KF is an ionic compound and contains fluoride ions. It combines with $ \mathrm{AlF}_3$ to form the soluble complex.

$ \mathrm{AlF}_3+3 \mathrm{NaF} \rightarrow \mathrm{Na}_3\left[\mathrm{AlF}_6\right] $

B has small size and high electronegativity. It has mauch higher tendency for complex formation than Al . Hence, when $ \mathrm{BF}_3$ is added to above solution, $ \mathrm{AlF}_3$ is precipitated.

$ \mathrm{Na}_3\left[\mathrm{AlF}_6\right]+3 \mathrm{BF}_3 \rightarrow 3 \mathrm{NaBF}_4+\mathrm{AlF}_3 $

Answer

Carbon monoxide is a colorless, odorless gas that is produced from incomplete combustion of carboncontaining fuels.Hemoglobin is a protein in red blood cells that is responsible for transporting oxygen from the lungs to the rest of the body.

When carbon monoxide enters the bloodstream, it binds to hemoglobin to form a complex known as carboxyhemoglobin.Carbon monoxide has a much higher affinity for hemoglobin than oxygen does. This means that CO will preferentially bind to hemoglobin over oxygen.

As CO binds to hemoglobin, it prevents oxygen from attaching to hemoglobin. This leads to a decrease in the amount of oxygen that can be transported to tissues and organs.

The lack of oxygen in the body can lead to symptoms of poisoning, including headaches, dizziness, confusion, and in severe cases, can be fatal.

Therefore, the reason why carbon monoxide is poisonous is that it binds with hemoglobin to form carboxyhemoglobin, which inhibits the transport of oxygen in the body.

Answer

Carbon dioxide is a very essential gas for our survival. However, an increased content of $CO_2$ 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 sunrays. 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.

Answer

Diborane

$B_2 H_6$ is an electron-deficient compound. $B_2 H_6$ has only 12 electrons - 6 efrom $6 H$ atoms and 3 eeach from $2 B$ atoms. Thus, after combining with $3 H$ atoms, none of the boron atoms has any electrons left. X-ray diffraction studies have shown the structure of diborane as:

2 boron and 4 terminal hydrogen atoms $(H_t)$ lie in one plane, while the other two bridging hydrogen atoms $(H_b)$ 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-centre two-electron $(2 c-2 e)$ bonds, while the two bridging ( $B-H-B)$ bonds are three-centre two-electron $(3 c-2 e)$ bonds.

Boric acid

Boric acid has a layered structure. Each planar $BO_3$ unit is linked to one another through $H$ atoms. The $H$ atoms form a covalent bond with $BO_3$ unit, while a hydrogen bond is formed with another $BO_3$ unit. In the given figure, the dotted lines represent hydrogen bonds.

Answer

(a) When borax is heated strongly, it loses water and swells into the white mass, which on further heating melts to form a transparent glassy solid called borax glass and borax bead.

$\begin{aligned} & \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 1 \mathrm{OH}_2 \mathrm{O} \xrightarrow{\text { heat }} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+10 \mathrm{H}_2 \mathrm{O} \ & \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \xrightarrow{\text { heat }} \underset{\text { (sodium meta borate) }}{2 \mathrm{NaBO}_2}+\mathrm{B}_2 \mathrm{O}_3\end{aligned}$

(b) When boric acid is added to water, it accepts electrons from - OH ion. Boric acid is sparingly soluble in cold water however fairly soluble in hot water.

$ \mathrm{B}(\mathrm{OH})_3+2 \mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}_3 \mathrm{O}^{+}$

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

$2 \mathrm{Al}+2 \mathrm{NaOH}+6 \mathrm{H}_2 \mathrm{O} \rightarrow \underset{\text { Sodium tetrahydroxoaluminate(III) }}{2 \mathrm{Na}^{+}\left[\mathrm{Al}(\mathrm{OH})_4\right]^{-}}+3 \mathrm{H}_2$

(d) $ \mathrm{BF}_3$ (a Lewis acid) reacts with $ \mathrm{NH}_3$ (a Lewis base) to form an adduct. This results in a complete octet around B in $ \mathrm{BF}_3$.

$ \mathrm{F}_3 \mathrm{~B}+: \mathrm{NH}_3 \rightarrow \mathrm{~F}_3 \mathrm{~B} \leftarrow: \mathrm{NH}_3$

Answer

(a) When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about $570 K$, a class of organosilicon polymers called methyl-substituted chlorosilanes are formed.

$CH_3-Cl+Si \ \ \ \ \underset{Cu}{\xrightarrow {\Delta}} \ \ \ \ (CH_3)SiCl+(CH_3) SiCl_2 +CH_3 Sicl_3 + (CH_3)_4Si$

(b) When silicon dioxide $(SiO_2)$ is heated with hydrogen fluoride $(HF)$, it forms silicon tetrafluoride $(SiF_4)$. 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.

$SiO_2+4 HF \longrightarrow SiF_4+2 H_2 O$

The $SiF_4$ formed in this reaction can further react with $HF$ to form hydrofluorosilicic acid.

$SiF_4+2 HF \longrightarrow H_2 SiF_6$

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

$ ZnO +CO \xrightarrow{\Delta} Zn +CO $

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

$ Al_2 O_3 \cdot 2 H_2 O+2 NaOH \longrightarrow 2 NaAlO_2+3 H_2 O $

Answer

(i) Concentrated $HNO_3$ 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.

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

$ 2 Al+2 NaOH+6 H_2 O \longrightarrow 2 Na^{+}[Al(OH)_4]^{-}+3 H_2 $

(iii) 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.

(iv) In diamond, carbon is $s p^{3}$ hybridised. Each carbon atom is bonded to four 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.

(v) Aluminium has a high tensile strength and is very light in weight. It can also be alloyed with various metals such as $Cu, Mn, Mg, Si$, and $Zn$. It is very malleable and ductile. Therefore, it is used in making aircraft bodies.

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

(vii) Silver, copper, and aluminium 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. Aluminium is a very ductile metal. Thus, aluminium is used in making wires for electrical conduction.

Answer

Ionisation 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.

Answer

Gallium (Ga) has a smaller atomic radius than aluminum (Al) because of the poor shielding effect of the 3d electrons in gallium, which means these electrons don’t effectively block the positive charge of the nucleus, causing a stronger pull on the outermost electrons and resulting in smaller atomic radius compared to aluminium; essentially the added d-electrons in gallium don’t shield the nucleus as well as the electrons in aluminum, leading to a greater effective nuclear charge on the outer electrons.

Answer

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

Diamond:

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.

Graphite:

It has $s p^{2}$ hybridised 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.

Answer

(a)

The neutral oxide is $CO$.

The acidic oxides are $ \mathrm{B}_2 \mathrm{O}_3, \mathrm{SiO}_2$ and $ \mathrm{CO}_2$.

The basic oxide is $ \mathrm{Tl}_2 \mathrm{O}_3$.

The amphoteric oxides are $ \mathrm{Al}_2 \mathrm{O}_3$ and $ \mathrm{PbO}_2$.

(b)

The chemical equation are :

$B_2 O_3=$ Acidic

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

$B_2 O_3+2 NaOH \longrightarrow 2 NaBO_2+H_2 O$

$SiO_2=$ Acidic

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

$SiO_2+2 NaOH \longrightarrow 2 Na_2 SiO_3+H_2 O$

$CO_2=$ Acidic

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

$CO_2+2 NaOH \longrightarrow Na_2 CO_3+H_2 O$

$Al_2 O_3=$ Amphoteric

Amphoteric substances react with both acids and bases. $Al_2 O_3$ reacts with both $NaOH$ and $H_2 SO_4$.

$Al_2 O_3+2 NaOH \longrightarrow NaAlO_2$

$Al_2 O_3+3 H_2 SO_4 \longrightarrow Al_2(SO_4)_3+3 H_2 O$

$PbO_2=$ Amphoteric

Amphoteric substances react with both acids and bases. $PbO_2$ reacts with both $NaOH$ and $H_2 SO_4$.

$PbO_2+2 NaOH \longrightarrow Na_2 PbO_3+H_2 O$

$2 PbO_2+2 H_2 SO_4 \longrightarrow 2 PbSO_4+2 H_2 O+O_2$

$Tl_2 O_3=$ Basic

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

$Tl_2 O_3+6 HCl \longrightarrow 2 TlCl_3+3 H_2 O$

Answer

Thallium is a group 13 element with common oxidation states +1 and +3 . Due to inert pair effect +1 oxidation state in Tl is more stable than +3 oxidation state. Al shows +3 oxidation state and alkali metals shows +1 oxidation state. Thus, thallium resembles both Al , and alkali metals.

Like $ \mathrm{Al}, \mathrm{Tl}$ forms compounds such as $ \mathrm{TlCl}_3$ and $ \mathrm{Tl}_2 \mathrm{O}_3$ and like alkali metals, Tl forms compounds such as TlCl and $ \mathrm{Tl}_2 \mathrm{O}$.

Answer

The given metal $X$ gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, $X$ must be aluminium.

The white precipitate (compound A) obtained is aluminium hydroxide. The compound $B$ formed when an excess of the base is added is sodium tetrahydroxoaluminate(III).

$ \underset{\text{ Aluminium }(X)} { 2 \mathrm Al} + \underset{ \text{ Sodium hydroxide }} { \mathrm 3 NaOH} \longrightarrow \underset{{ White ppt.(A) }} {Al(OH)_3} \downarrow+3 Na^{+} $

$\underset{(A)}{Al(OH)_3}+NaOH \longrightarrow \quad \underset{(\text{Soluble complex B})} {Na^{+}[Al(OH)_4]^{-} }$

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

$ \underset{(A)}{Al(OH)_3}+3 HCl \longrightarrow \underset{(C)} {AlCl_3}+3 H_2 O $

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.

$ \underset{(A)}{2 Al(OH)_3} \xrightarrow{\Delta} \underset{(D)} {Al_2 O_3}+3 H_2 O $

Answer

(a) Inert pair effect

As one moves down the group, the tendency of s-block electrons to participate in chemical bonding decreases. This effect is known as inert pair effect. In case of group 13 elements, the electronic configuration is $n s^{2} n p^{1}$ 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 $n s^{2}$ electrons by the $d$ - and $f$ - electrons. As a result of the poor shielding, the $n s^{2}$ 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 $S$.

Answer

(i) Since the aqueous solution of salt $(X)$ is alkaline to litmus, it must be the salt of a strong base and a weak acid.

(ii) Since the salt $(X)$ swells up to a glassy material ( $Y$ ) on strong heating, therefore $(X)$ must be borax and ( $Y$ ) must be a mixture of sodium metaborate and boric anhydride.

(iii) When conc. $ \mathrm{H}_2 \mathrm{SO}_4$ is added to a hot solution of X , i.e, borax, white crystals of an acid (Z) separate out, therefore $(Z)$ must be orthobasic acid. The equations for the reactions involved in the question are :

(i)

$\underset{\text { Borax (X) }}{\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 .10 \mathrm{H}_2 \mathrm{O}} \xrightarrow{\text { Water }} \underset{\text { (Strong alkali) }}{2 \mathrm{NaOH}}+\underset{\text { (Weask acid) }}{\mathrm{H}_2 \mathrm{~B}_4 \mathrm{O}_7+8 \mathrm{H}_2 \mathrm{O}}$

(ii)

$\begin{aligned} & \underset{(X)} {\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O} }\xrightarrow{\text { Heat }} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+10 \mathrm{H}_2 \mathrm{O}, \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7& \xrightarrow{\text { Heat }}{ } \underbrace{2 \mathrm{NaBO}_2+\mathrm{B}_2 \mathrm{O}3}{\text {Glasyy material }}\end{aligned}$

(iii)

$\begin{aligned} & \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \underset{\text { Boric acid (Z) }}{4 \mathrm{H}_3 \mathrm{BO}_3} +\mathrm{Na}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{O}\end{aligned}$

Answer

(i)

$ \underset{\text{(Boron trifluoride)}}{2 BF_3}+\underset{\text{( Lithium hydride)}} {6 LiH} \longrightarrow \underset{\text{(Diborane)}} {B_2 H_6}+\underset{\text{(Lithium fluoride)}}{6 LiF} $

(ii) $ \underset{\text{(Diborane)}}{B_2 H_6}+\underset{\text{( Water )}}{6 H_2 O} \longrightarrow \underset{\text{(Orthoboric acid)}}{2 H_3 BO_3}+ \underset{\text{( Hydrogen)}}{6 H_2}$

(iii) $\underset{\text{(Diborane )}}{B_2 H_6}+\underset{\text{(Sodium hydride)}}{2 NaH} \xrightarrow{\text{ ether }} \underset{\text{( Sodium borohydride)}} {2 NaBH_4}$

(iv)

$ H_3BO_3 \xrightarrow {\Delta} HBD_2 + H_2O $

$2HBO_2 \xrightarrow {\Delta} B_2O_3 +H_2O$

(v) $ Al+2 NaOH+2 H_2 O \longrightarrow 2NaAlO_3+3 H_2 \uparrow$

(vi) $3 B_2 H_6+6 NH_3 \longrightarrow 3[BH_2(NH_3)_2]^{+}[BH_4]^{-} \xrightarrow{\Delta} \underset{(Borazene)} {2 B_3 N_3 H_6}+12 H_2 \uparrow$

Answer

Laboratory preparation of carbon dioxide:

Calcium carbonate reacts with dil HCl to form carbon dioxide.

$ \mathrm{CaCO}_3+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} $

Industrial preparation of carbon dioxide:

Limestone is heated to produce carbon dioxide.

$ \mathrm{CaCO}_3 \xrightarrow{\text { heat }} \mathrm{CaO}+\mathrm{CO}_2 $

Laboratory preparation of carbon monoxide:

Formic acid is dehydrated with conc sulphuric acid at 373 .

$ \mathrm{HCOOH} \xrightarrow[\text { conc. } \mathrm{H}_2 \mathrm{SO}_4]{373 \mathrm{~K}} \mathrm{H}_2 \mathrm{O}+\mathrm{CO} \uparrow $

Industrial preparation of CO:

Steam is passed over hot coke.

$ \mathrm{C}+\mathrm{H}_2 \mathrm{O} \xrightarrow{473-1273 \mathrm{~K}} \underset{\text { water gas }}{\mathrm{CO}+\mathrm{H}_2} $

Answer:(c) basic

An aqueous solution of borax $ \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$ is basic in nature as it on hydrolysis produces $ \mathrm{H}_3 \mathrm{BO}_3$ (a weak acid) and NaOH (a strong base).

$ \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+7 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+4 \mathrm{H}_3 \mathrm{BO}_3 $

Answer:(b) the presence of hydrogen bonds

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

Answer:(c) $sp^3 $

Diborane, $B_2H_6$ has two electrons each, three centered bonds. Each Boron atom (B) is linked with four hydrogen atoms. This makes tetrahedral geometry. Hence each boron atom is $sp^3$ hybridised.

Hence boron in diborane is $sp^3$ hynridised.

Answer:(b) graphite

Graphite is thermodynamically the most stable form of carbon.

Answer:(b) exhibit oxidation state of +2 and +4

The elements of group 14 have 4 valence electrons. Therefore, the oxidation state of the group is +4 . However, as a result of the 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.