Boron family Questions in English

(N/A) Certain important trends can be observed in the chemical behaviour of group-$13$ elements. The trichlorides,bromides,and iodides of all these elements are covalent in nature and are hydrolysed in water.
Species like tetrahedral $[M(OH)_{4}]^{-}$ and octahedral $[M(H_{2}O)_{6}]^{3+}$ exist in aqueous medium,except for boron.
The monomeric trihalides,being electron-deficient,are strong Lewis acids. Boron trifluoride $(BF_{3})$ easily reacts with Lewis bases such as $NH_{3}$ to complete the octet around boron.
Due to the absence of $d$-orbitals,the maximum covalence of $B$ is $4$. Since $d$-orbitals are available in $Al$ and other elements,their maximum covalence can exceed $4$.
Most other metal halides (e.g.,$AlCl_{3}$) are dimerised through halogen bridging (e.g.,$Al_{2}Cl_{6}$). The metal species completes its octet by accepting electrons from the halogen in these bridged molecules.

(N/A) Borax is the most important compound of boron.
Formula: The actual structure contains the tetranuclear units $[B_{4}O_{5}(OH)_{4}]^{2-}$,so the correct formula is $Na_{2}[B_{4}O_{5}(OH)_{4}] \cdot 8 H_{2}O$.
Physical Properties: It is a white crystalline solid. It dissolves in water to give an alkaline solution due to hydrolysis:
$Na_{2}B_{4}O_{7} + 7 H_{2}O \rightarrow 2 NaOH + 4 H_{3}BO_{3}$ (Orthoboric acid).
Chemical Properties (Borax Bead Test): On heating,borax loses water molecules and swells. On further heating,it turns into a transparent liquid that solidifies into a glass-like material called a borax bead:
$Na_{2}B_{4}O_{7} \cdot 10 H_{2}O$ $\xrightarrow{\Delta} Na_{2}B_{4}O_{7}$ $\xrightarrow{\Delta} 2 NaBO_{2} + B_{2}O_{3}$ (Sodium metaborate and Boric anhydride).
The metaborates of transition metals have characteristic colors,allowing for identification in the laboratory. For example,heating borax with $CoO$ on a platinum wire loop forms a blue $Co(BO_{2})_{2}$ bead.

(N/A) Physical properties :
$\rightarrow$ Orthoboric acid,$H_3BO_3$,is a white crystalline solid with a soapy touch.
$\rightarrow$ It is sparingly soluble in water but highly soluble in hot water.
$\rightarrow$ It has a layer structure in which planar $BO_3$ units are joined by hydrogen bonds as shown in the figure.
Chemical properties :
$\rightarrow$ It can be prepared by acidifying an aqueous solution of borax :
$Na_2B_4O_7 + 2 HCl + 5 H_2O \rightarrow 2 NaCl + 4 B(OH)_3$
$\rightarrow$ 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 :
$B(OH)_3 + 2 HOH \rightarrow [B(OH)_4]^- + H_3O^+$
$\rightarrow$ On heating,orthoboric acid above $370 \ K$ forms metaboric acid,$HBO_2$,which on further heating yields boric oxide,$B_2O_3$ :
$H_3BO_3$ $\xrightarrow{\Delta} HBO_2$ $\xrightarrow{\Delta} B_2O_3$

(N/A) Boric acid ($H_3BO_3$ or $B(OH)_3$) is not a protic acid.
$A$ protic acid is defined as a substance that donates a proton $(H^+)$ in an aqueous solution.
Boric acid acts as a weak monobasic Lewis acid.
It accepts a lone pair of electrons from the $OH^-$ ion of water to form the $[B(OH)_4]^-$ complex,thereby releasing $H^+$ ions into the solution:
$B(OH)_3 + 2H_2O \longrightarrow [B(OH)_4]^- + H_3O^+$

On heating orthoboric acid $(H_{3}BO_{3})$ at $370 \ K$ or above,it loses a water molecule to form metaboric acid $(HBO_{2})$.
On further heating,metaboric acid loses more water to yield boric oxide $(B_{2}O_{3})$.
The reactions are as follows:
$H_{3}BO_{3} \xrightarrow{370 \ K} HBO_{2} + H_{2}O$
$2HBO_{2} \xrightarrow{\Delta} B_{2}O_{3} + H_{2}O$

(A) $(i)$ The salt $X$ is alkaline to litmus,indicating it is a salt of a strong base and a weak acid. Upon strong heating,it swells to form a glassy material $Y$. This behavior is characteristic of borax $(Na_2B_4O_7 \cdot 10H_2O)$.
$Na_2B_4O_7 \cdot 10H_2O + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$
$(ii)$ On heating,borax loses water and swells to form a glassy bead $Y$ consisting of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
$Na_2B_4O_7 \cdot 10H_2O \xrightarrow{\Delta} Na_2B_4O_7 + 10H_2O$
$Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3 (Y)$
$(iii)$ When concentrated $H_2SO_4$ is added to a hot solution of borax,white crystals of orthoboric acid $(Z = H_3BO_3)$ are formed.
$Na_2B_4O_7 \cdot 10H_2O + H_2SO_4 \rightarrow Na_2SO_4 + 4H_3BO_3 (Z) + 5H_2O$

(N/A) Preparation:
- The simplest boron hydride known is diborane. It is prepared by treating boron trifluoride with $LiAlH_{4}$ in diethyl ether.
$4 BF_{3} + 3 LiAlH_{4} \rightarrow 2 B_{2}H_{6} + 3 LiF + 3 AlF_{3}$
- $A$ convenient laboratory method for the preparation of diborane involves the oxidation of sodium borohydride with iodine.
$I_{2} + 2 NaBH_{4} \rightarrow B_{2}H_{6} + 2 NaI + H_{2}$
- Diborane is produced on an industrial scale by the reaction of $BF_{3}$ with sodium hydride.
$2 BF_{3} + 6 NaH \stackrel{450 \ K}{\longrightarrow} B_{2}H_{6} + 6 NaF$
Physical properties:
- Diborane is a colourless,highly toxic gas with a boiling point of $180 \ K$.
- Diborane catches fire spontaneously upon exposure to air.
- It burns in oxygen,releasing an enormous amount of energy.
Chemical properties:
- Boranes are readily hydrolyzed by water to give boric acid.
$B_{2}H_{6(g)} + 6 H_{2}O_{(l)} \rightarrow 2 B(OH)_{3(aq)} + 6 H_{2(g)}$
- Diborane undergoes cleavage reactions with Lewis bases $(L)$ to give borane adducts,$BH_{3} \cdot L$.
$B_{2}H_{6} + 2 NMe_{3} \rightarrow 2 BH_{3} \cdot NMe_{3}$
$B_{2}H_{6} + 2 CO \rightarrow 2 BH_{3} \cdot CO$
- Reaction of ammonia with diborane initially gives $B_{2}H_{6} \cdot 2 NH_{3}$,which is formulated as $[BH_{2}(NH_{3})_{2}]^{+}[BH_{4}]^{-}$; further heating gives borazine,$B_{3}N_{3}H_{6}$,known as "inorganic benzene" due to its ring structure with alternate $BH$ and $NH$ groups.
$3 B_{2}H_{6} + 6 NH_{3}$ $\rightarrow 2[BH_{2}(NH_{3})_{2}]^{+}[BH_{4}]^{-} \stackrel{\Delta}{}$ ${\longrightarrow} 2 B_{3}N_{3}H_{6} + 12 H_{2}$

(N/A) Diborane $(B_2H_6)$ is an electron-deficient compound with only $12$ valence electrons. The two $B$ atoms and four terminal $H$ atoms lie in the same plane,while the two bridging $H$ atoms lie in a perpendicular plane (one above and one below). The terminal $B-H$ bonds are $2c-2e$ (two-center-two-electron) bonds,and the bridging $B-H-B$ bonds are $3c-2e$ (three-center-two-electron) bonds.
$(b)$ Boric acid $(H_3BO_3)$ has a layered structure. Each planar $BO_3^{3-}$ unit is linked to others through $H$ atoms. The $H$ atoms form covalent bonds with the oxygen atoms of the $BO_3$ unit and hydrogen bonds with the oxygen atoms of adjacent $BO_3$ units.

(N/A) When heated,borax loses water of crystallization and swells. On further heating,it melts to form a transparent liquid,which solidifies into a glass-like material known as borax bead.
$Na_{2}B_{4}O_{7} \cdot 10H_{2}O \xrightarrow{\Delta} Na_{2}B_{4}O_{7} + 10H_{2}O$
$(b)$ Boric acid acts as a weak monobasic Lewis acid in water. It accepts an $OH^{-}$ ion from water to form the tetrahydroxoborate ion.
$B(OH)_{3} + 2H_{2}O \rightleftharpoons [B(OH)_{4}]^{-} + H_{3}O^{+}$
$(c)$ Aluminium reacts with dilute $NaOH$ to form sodium tetrahydroxoaluminate$(III)$ and releases hydrogen gas.
$2Al(s) + 2NaOH(aq) + 6H_{2}O(l) \rightarrow 2Na[Al(OH)_{4}](aq) + 3H_{2}(g)$
$(d)$ $BF_{3}$ acts as a Lewis acid and $NH_{3}$ acts as a Lewis base. They react to form an adduct,completing the octet of boron.
$F_{3}B + :NH_{3} \rightarrow F_{3}B \leftarrow NH_{3}$

(N/A) $(i) \, 2 BF_{3} + 6 LiH \rightarrow B_{2}H_{6} + 6 LiF$
$(ii) \, B_{2}H_{6} + 6 H_{2}O \rightarrow 2 H_{3}BO_{3} + 6 H_{2}$ (Boric acid)
$(iii) \, 2 NaH + B_{2}H_{6} \rightarrow 2 Na[BH_{4}]$ (Sodium borohydride)
$(iv) \, H_{3}BO_{3} \xrightarrow{\Delta } HBO_{2} + H_{2}O$ (Metaboric acid)
$4 HBO_{2} \xrightarrow{\Delta } H_{2}B_{4}O_{7} + H_{2}O$
$H_{2}B_{4}O_{7} \xrightarrow{\Delta } 2 B_{2}O_{3} + H_{2}O$ (Tetraboric acid and Boron trioxide)
$(v) \, 2 Al + 2 NaOH + 6 H_{2}O \rightarrow 2 Na[Al(OH)_{4}] + 3 H_{2}$
$(vi) \, B_{2}H_{6} + 2 NH_{3} \rightarrow 2 BH_{3} \cdot NH_{3}$

(N/A) Boron,being an extremely hard refractory solid with a high melting point,low density,and very low electrical conductivity,finds many applications.
$1$. Boron fibers are used in making bullet-proof vests and light composite materials for aircraft.
$2$. The boron-$10$ $({}^{10}B)$ isotope has a high ability to absorb neutrons; therefore,metal borides are used in the nuclear industry as protective shields and control rods.
$3$. The main industrial application of borax and boric acid is in the manufacture of heat-resistant glasses (e.g.,Pyrex),glass-wool,and fiberglass.
$4$. Borax is also used as a flux for soldering metals,for heat,scratch,and stain-resistant glazed coatings on earthenwares,and as a constituent of medicinal soaps.
$5$. An aqueous solution of orthoboric acid is generally used as a mild antiseptic.

(N/A) $Aluminium$ is a bright silvery-white metal with high tensile strength. It possesses high electrical and thermal conductivity. On a weight-to-weight basis,the electrical conductivity of $Aluminium$ is twice that of $Copper$.
$Aluminium$ is used extensively in industry and everyday life. It forms alloys with $Cu, Mn, Mg, Si,$ and $Zn$. $Aluminium$ and its alloys can be shaped into pipes,tubes,rods,wires,plates,or foils,and therefore,find uses in packaging,utensil making,construction,and the aerospace and transportation industries.
The use of $Aluminium$ and its compounds for domestic purposes has been reduced considerably due to their potential toxic nature.

(N/A) Properties of $B$: $(i)$ $B$ has a high melting point,low density,and very low electrical conductivity.
$(ii)$ It is an extremely hard,high-temperature resistant solid element.
Properties of $Al$: $(i)$ It is a silvery-white lustrous metal.
$(ii)$ It possesses high electrical and thermal conductivity.
$(iii)$ By weight,$Al$ has twice the conductivity of $Cu$.

(N/A) Aluminium is characterized by its high strength-to-weight ratio,meaning it is very light yet strong.
It can be alloyed with metals like $Cu, Mn, Mg, Si,$ and $Zn$ to significantly enhance its mechanical properties,such as tensile strength and corrosion resistance.
Additionally,it is highly malleable and ductile,allowing it to be shaped into complex aerodynamic structures.
Due to these properties,aluminium alloys are ideal for the construction of aircraft bodies.

(A) The group $13$ elements are Boron $(B)$,Aluminum $(Al)$,Gallium $(Ga)$,Indium $(In)$,and Thallium $(Tl)$.
They possess a general valence shell electronic configuration of $ns^{2} np^{1}$.

(B) The atomic radius of group $13$ elements generally increases down the group due to the addition of new shells.
However,$Ga$ has a smaller atomic radius than $Al$ due to the poor shielding effect of $d$-electrons,which increases the effective nuclear charge.
The correct increasing order is: $B < Ga < Al < In < Tl$.

(A) The hydrated forms of group $13$ elements $Al$,$Ga$,and $In$ exist as polymers in nature due to their ability to form coordinate bonds and bridge structures.
Boron,being a non-metal with a small atomic size,does not form such polymeric hydrated structures.

(A) The hydroxides of group $13$ elements show a trend in their acidic and basic character.
$B(OH)_3$ is weakly acidic.
$Al(OH)_3$ and $Ga(OH)_3$ are amphoteric in nature,meaning they react with both acids and bases.
$In(OH)_3$ and $Tl(OH)_3$ are basic in nature.

(N/A) The structure of diborene $(B_2H_4)$ is analogous to ethene $(C_2H_4)$.
It consists of two boron atoms bonded to each other with a $B-B$ bond length of $134 \text{ pm}$.
Each boron atom is further bonded to two terminal hydrogen atoms with a $B-H$ bond length of $119 \text{ pm}$.
The $H-B-H$ bond angle is approximately $120^{\circ}$, and the $B-B-H$ bond angle is approximately $97^{\circ}$.
The structure is planar, similar to ethene, where the boron atoms are $sp^2$ hybridized.

(B) Aluminum reacts with dilute $HCl$ to form the hydrated aluminum chloride complex and releases hydrogen gas.
The balanced chemical equation is:
$2 Al + 6 HCl + 12 H_2O \rightarrow 2[Al(H_2O)_6]Cl_3 + 3 H_2$

(N/A) $Na_{2}B_{4}O_{7} \cdot 10H_{2}O \xrightarrow{\Delta} Na_{2}B_{4}O_{7} + 10H_{2}O$
$Na_{2}B_{4}O_{7} \xrightarrow{\Delta} 2NaBO_{2} + B_{2}O_{3}$
The mixture of $NaBO_{2}$ (sodium metaborate) and $B_{2}O_{3}$ (boric anhydride) forms a transparent,glassy bead known as the borax bead.

(N/A) Structure: $Al_2Cl_6$ exists as a dimer where two $AlCl_3$ units are joined by two chlorine bridges. Each $Al$ atom is $sp^3$ hybridized and bonded to four chlorine atoms in a tetrahedral geometry.
Uses of $AlCl_3$:
$1$. It acts as a strong Lewis acid.
$2$. It is used as a catalyst in Friedel-Crafts alkylation and acylation reactions.
$3$. It is used in the manufacture of dyes,drugs,and perfumes.

(N/A) Boron exists in nature as two stable isotopes:
$(i)$ ${}^{10}B$ $(19 \%)$
(ii) ${}^{11}B$ $(81 \%)$

(C) Group $13$ elements have a valence shell electronic configuration of $ns^2 np^1$.
Due to the inert pair effect,the stability of the $+1$ oxidation state increases down the group $(Al < Ga < In < Tl)$.
While $B$ and $Al$ primarily show $+3$ oxidation state,$Ga, In,$ and $Tl$ exhibit both $+1$ and $+3$ oxidation states due to the increasing reluctance of the $ns^2$ electrons to participate in bonding.

(A) Aluminum reacts with aqueous alkali (like $NaOH$) to form sodium tetrahydroxoaluminate$(III)$ and releases hydrogen gas.
The balanced chemical equation is:
$2 Al(s) + 2 NaOH(aq) + 6 H_{2}O(l) \rightarrow 2 Na[Al(OH)_{4}](aq) + 3 H_{2}(g)$

(B) In group $13$,elements other than $B$ (Boron) can form complexes with different coordination numbers.
$Al$,$Ga$,and $In$ can form tetrahedral species like $\left[M(OH)_{4}\right]^{-}$ and octahedral species like $\left[M(H_{2}O)_{6}\right]^{3+}$ in an aqueous medium.
$Tl$ (Thallium) typically shows $+1$ oxidation state stability and does not readily form these specific octahedral complexes in the same manner as $Al, Ga,$ and $In$.

(B) The most common laboratory preparation of diborane $(B_{2}H_{6})$ involves the oxidation of sodium borohydride $(NaBH_{4})$ with iodine $(I_{2})$ in a solvent like diglyme.
The balanced chemical equation is:
$2NaBH_{4} + I_{2} \rightarrow B_{2}H_{6} + 2NaI + H_{2}$

(N/A) The industrial preparation of diborane $(B_{2}H_{6})$ involves the reaction of boron trifluoride $(BF_{3})$ with sodium hydride $(NaH)$ at $450 \ K$:
$2 BF_{3} + 6 NaH \xrightarrow{450 \ K} B_{2}H_{6} + 6 NaF$

(N/A) The preparation of inorganic benzene (borazine) involves the reaction of diborane with ammonia at high temperatures:
$3 B_2H_6 + 6 NH_3 \rightarrow 2 B_3N_3H_6 + 12 H_2$
Alternatively,it can be represented as a two-step process:
$3 B_2H_6 + 6 NH_3$ $\rightarrow 3 [BH_2(NH_3)_2]^+ [BH_4]^-$ $\xrightarrow{\Delta} 2 B_3N_3H_6 + 12 H_2$

(N/A) Borax $(Na_2B_4O_7 \cdot 10H_2O)$ is used as a flux for soldering metals.
It is also used for making heat,scratch,and stain-resistant glazed coatings on earthenwares and pottery.

(A) The dilute aqueous solution of $H_3BO_3$ (orthoboric acid) is used as a mild antiseptic and antibiotic.

(A) $Al$ (Aluminum) forms various important alloys with metals such as $Cu$ (Copper),$Mn$ (Manganese),$Mg$ (Magnesium),and $Zn$ (Zinc).
$Si$ (Silicon) is also commonly added to improve casting properties,although it is a metalloid.

(N/A) Aluminium:
$1$. Aluminium foils are used as wrappers for food materials.
$2$. The fine dust of the metal is used in paints and lacquers.
$3$. Aluminium,being highly reactive,is used in the extraction of chromium and manganese from their oxides.
$4$. Aluminium wires are used as electricity conductors.
$5$. Alloys containing aluminium are light and very useful.
Copper:
$1$. Copper is used in making wires for the electrical industry and for water and steam pipes.
$2$. It is also used in several alloys that are tougher than the metal itself,e.g.,brass (with zinc),bronze (with tin),and coinage alloy (with nickel).

(A) The freezing point of ammonia $(NH_3)$ is $198.4 \ K$.
The boiling point of ammonia $(NH_3)$ is $239.7 \ K$.
Therefore,the correct values are $198.4 \ K$ and $239.7 \ K$.

(A-Q, B-S, C-R,P, D-Q,R) $A-Q$: $Bi^{+3}$ undergoes hydrolysis to form bismuthyl ion $(BiO)^{+}$.
$B-S$: $[AlO_2]^{-}$ (aluminate ion) on dilution with water precipitates $Al(OH)_3$.
$C-R, P$: $SiO_4^{-4}$ units condense to form $Si_2O_7^{-6}$ upon acidification or heating.
$D-Q, R$: Borate ions $(B_4O_7)^{-2}$ undergo hydrolysis and acidification to form boric acid $[B(OH)_3]$.

(N/A) In $BCl_3$,Boron is attached to three chlorine atoms by covalent bonds,but the octet of Boron is incomplete. It requires two electrons to complete its octet.
Such electron-deficient molecules have a tendency to accept a pair of electrons to achieve a stable electronic configuration and thus behave as Lewis acids.
$BCl_3$ easily accepts a lone pair of electrons from ammonia $(NH_3)$ to form the adduct $BCl_3.NH_3$:
$H_3N: + BCl_3$ $\rightarrow H_3N$ $\rightarrow BCl_3$
$AlCl_3$ achieves stability by forming a dimer,$Al_2Cl_6$. In the dimer,each Aluminum atom is bonded to four chlorine atoms,two of which are bridging chlorine atoms. The structure is shown below:
(See the provided image for the $Al_2Cl_6$ dimer structure.)

(N/A) Boric acid $(H_3BO_3)$ acts as a weak monobasic Lewis acid in water.
It does not release $H^+$ ions directly. Instead,it accepts a lone pair of electrons from an $OH^-$ ion provided by the water molecule to complete its octet.
The reaction is as follows:
$B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^- + H_3O^+$

(N/A) Boric acid $(H_{3}BO_{3})$ has a layered structure in which planar $H_{3}BO_{3}$ units are joined by hydrogen bonds,forming hexagonal rings as shown in the 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.
In water,boric acid is present in the form of the $[B(OH)_{4}]^{-}$ species.
The hybridization of boron in this species is $sp^{3}$.
The reaction is: $B(OH)_{3} + 2H_{2}O \rightarrow [B(OH)_{4}]^{-} + H_{3}O^{+}$

(N/A) $BCl_3$ and $AlCl_3$ are electron-deficient compounds because the central metal atom has an incomplete octet. Therefore,they act as Lewis acids by accepting a lone pair of electrons.

(N/A) Aluminum is amphoteric in nature; it reacts with acids and bases to produce salt and $H_{2}$ gas. The evolution of $H_{2}$ gas is confirmed by the pop sound produced when a burning matchstick is brought near the mouth of the test tube.
$2 Al(s) + 6 HCl(aq) \rightarrow 2 AlCl_{3}(aq) + 3 H_{2}(g)$
$2 Al(s) + 2 NaOH(aq) + 6 H_{2}O(l) \rightarrow 2 Na[Al(OH)_{4}](aq) + 3 H_{2}(g)$
When $Al$ reacts with concentrated $HNO_{3}$,a thin,non-porous,and inert layer of aluminum oxide $(Al_{2}O_{3})$ forms on the surface of the metal. This layer acts as a protective barrier,preventing further reaction between the metal and the acid,a phenomenon known as passivation.

(N/A) $(1)$ The first ionization enthalpy of $Ga$ $(579 \ kJ \ mol^{-1})$ is higher than that of $Al$ $(577 \ kJ \ mol^{-1})$ because the $d$-electrons present in $Ga$ have a poor shielding effect. This leads to a higher effective nuclear charge,which holds the valence electrons more tightly.
$(2)$ Boron has a very small atomic size and a very high sum of the first three ionization enthalpies $(\Delta_{i}H_{1} + \Delta_{i}H_{2} + \Delta_{i}H_{3})$. The energy required to remove three electrons is too high to be compensated by the lattice energy or hydration energy,so boron does not form $B^{3+}$ ions and instead forms covalent compounds.

(N/A) $Al$ has vacant '$d$' orbitals and can expand its coordination number to form $[AlF_6]^{3-}$.
On the other hand,Boron does not have '$d$' orbitals and cannot expand its covalence beyond $4$,thus it forms $[BF_4]^{-}$ instead of $[BF_6]^{3-}$.
Due to the inert pair effect,the $+2$ oxidation state is more stable than the $+4$ oxidation state for lead $(Pb)$.

(N/A) $(1)$ $BF_3$ undergoes partial hydrolysis. The $HF$ produced reacts with $H_3BO_3$ to form fluoroboric acid,preventing complete hydrolysis:
$4BF_3 + 3H_2O \rightarrow H_3BO_3 + 3H^+ + 3[BF_4]^-$
$(2)$ In graphite,carbon atoms are $sp^2$ hybridized and form $p\pi-p\pi$ double bonds due to their small size and high electronegativity. Silicon,due to its larger atomic size and lower electronegativity,cannot form stable $p\pi-p\pi$ multiple bonds,and therefore cannot form a graphite-like structure.

(A) The reaction of borax $(Na_2B_4O_7)$ with $HCl$ and water is given by:
$Na_2B_4O_7 + 2HCl + 5H_2O \to 2NaCl + 4H_3BO_3$
Thus,$A = Na_2B_4O_7$ and $X = H_3BO_3$.
On heating orthoboric acid $(H_3BO_3)$ at $370 \ K$,it forms metaboric acid $(HBO_2)$:
$H_3BO_3 \xrightarrow[370 \ K]{\Delta} HBO_2 + H_2O$
On further heating above $370 \ K$,metaboric acid $(HBO_2)$ dehydrates to form boron trioxide $(B_2O_3)$:
$2HBO_2 \xrightarrow[> 370 \ K]{\Delta} B_2O_3 + H_2O$
Therefore,$Z = B_2O_3$.

The given reactions represent the synthesis and chemical properties of diborane $(B_{2}H_{6})$.
$(i)$ The reaction of boron trifluoride with lithium aluminium hydride produces diborane: $4BF_{3} + 3LiAlH_{4} \to 2B_{2}H_{6} (X) + 3LiF + 3AlF_{3}$.
$(ii)$ Diborane reacts with water to form orthoboric acid: $B_{2}H_{6} + 6H_{2}O \to 2H_{3}BO_{3} (Y) + 6H_{2}$.
$(iii)$ Diborane undergoes combustion in oxygen to form boron trioxide and water: $B_{2}H_{6} + 3O_{2} \xrightarrow{\Delta} B_{2}O_{3} + 3H_{2}O$.
Therefore,$X = B_{2}H_{6}$ and $Y = H_{3}BO_{3}$.

(A-3, B-4, C-1, D-5, E-2) $A-3, B-4, C-1, D-5, E-2$
$(A)$ In $B_2H_6$,each $B$ atom uses $sp^3$ hybrids for bonding. Out of the four $sp^3$ hybrids on each $B$ atom,one is without an electron shown in broken lines. The terminal $B-H$ 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 referred to as banana bonds.
$(B)$ Gallium with an unusually low melting point $(303 \ K)$,could exist in a liquid state during summer. Its high boiling point $(2676 \ K)$ makes it a useful material for measuring high temperatures.
$(C)$ Borax is used as a flux for soldering metals,and for heat,scratch,and stain-resistant glazed coating to earthenwares.
$(D)$ Zeolites are aluminosilicates and are widely used as catalysts in petrochemical industries for cracking of hydrocarbons and isomerization,e.g.,$ZSM-5$ (a type of zeolite) is used to convert alcohols directly into gasoline.
$(E)$ Quartz,cristobalite,and tridymite are some of the crystalline forms of silica.

(N/A) For Group $13$:
$(A)$ Atomic size: Atomic radius generally increases down the group due to the addition of new shells. However,$Ga$ has a smaller atomic radius than $Al$ due to poor shielding by $d$-electrons.
$(B)$ Ionization enthalpy: The trend is irregular $(B > Al < Ga > In < Tl)$ due to the shielding effect of $d$ and $f$ orbitals and the increase in nuclear charge.
$(C)$ Metallic character: Increases down the group as ionization energy decreases.
$(D)$ Oxidation states: Group $13$ shows $+3$ and $+1$ (due to inert pair effect). Group $14$ shows $+4$ and $+2$.
$(E)$ Nature of halides: Group $13$ forms $MX_3$ type halides. Group $14$ forms $MX_2$ and $MX_4$ type halides.

(N/A) An aqueous solution of borax is acidified with $HCl$ to produce boric acid $(H_{3}BO_{3})$,which appears as a white crystalline solid and is soapy to the touch.
The chemical reaction is:
$Na_{2}B_{4}O_{7} + 2HCl + 5H_{2}O \rightarrow 2NaCl + 4H_{3}BO_{3}$
This solid is acidic in nature. Boric acid is a weak monobasic Lewis acid. It does not act as a protonic acid (it does not release $H^{+}$ ions directly); instead,it acts as a Lewis acid by accepting a lone pair of electrons from a hydroxyl ion $(OH^{-})$ from water:
$B(OH)_{3} + 2H_{2}O \rightarrow [B(OH)_{4}]^{-} + H_{3}O^{+}$

(A-D) $TlCl$ is more stable than $TlCl_3$ because,moving down the group,the lower oxidation state $(+1)$ becomes more stable due to the inert pair effect. $TlCl$ is ionic,while $TlCl_3$ is covalent.
$(B)$ $AlCl_3$ is more stable than $AlCl$ because $Al$ does not exhibit the inert pair effect. $AlCl_3$ is a covalent compound that acts as a Lewis acid.
$(C)$ $InCl$ is more stable than $InCl_3$ due to the inert pair effect,which makes the $+1$ oxidation state more stable for heavier elements. $In$ exhibits both $+3$ and $+1$ oxidation states. The reaction is: $3 InCl_{(aq)} \rightarrow 2 In_{(s)} + In_{(aq)}^{3+} + 3 Cl_{(aq)}^{-}$

$BCl_3$ does not dimerize because of the small size of the boron atom. Boron cannot accommodate four large-sized chloride ions around it due to steric hindrance.
$AlCl_3$ exists as a dimer in which $Al$ uses its vacant $3p$-orbital to coordinate with a $Cl$ atom from another $AlCl_3$ unit to complete its octet.
$AlCl_3$ achieves stability by forming a dimer,$Al_2Cl_6$,where two $Al$ atoms are bridged by two $Cl$ atoms. The structure consists of two $AlCl_4$ tetrahedra sharing a common edge.