Boron family Questions in English

(B) $B_2H_6$ and $B_2O_3$ react with $NH_3$ to form boron nitride $(BN)$.
$1.$ $B_2H_6$ reacts with $NH_3$ at high temperatures to form borazine $(B_3N_3H_6)$,which upon further heating yields boron nitride $(BN)$:
$3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 12H_2$
$B_3N_3H_6 \xrightarrow{\Delta} (BN)_n$
$2.$ $B_2O_3$ reacts with $NH_3$ at high temperatures $(1200^{\circ}C)$ to produce boron nitride:
$B_2O_3 + 2NH_3 \xrightarrow{1200^{\circ}C} 2BN + 3H_2O$
$B$ is an element,not a compound. $HBF_4$ reacts with $NH_3$ to form ammonium tetrafluoroborate $(NH_4BF_4)$.

(A)

Property	$B$	$Al$	$Ga$	$In$	$Tl$
Atomic radius $(pm)$	$88$	$143$	$135$	$167$	$170$
Electronegativity	$2.0$	$1.5$	$1.6$	$1.7$	$1.8$
Density $(g/cm^3)$	$2.35$	$2.7$	$5.9$	$7.31$	$11.85$
$IE_1$ $(kJ/mol)$	$801$	$577$	$579$	$558$	$589$

$A$. Atomic radius order: $B < Ga < Al < In < Tl$. Thus,statement $A$ is incorrect.
$B$. Electronegativity order: $Al < Ga < In < Tl < B$. Thus,statement $B$ is correct.
$C$. Density order: $B < Al < Ga < In < Tl$. Thus,statement $C$ is incorrect.
$D$. $1^{st}$ Ionisation Energy order: $In < Al < Ga < Tl < B$. Thus,statement $D$ is correct.
Therefore,the correct statements are $B$ and $D$.

(B) $(i)$ Statement $(A)$ is true: Due to the inert pair effect,$Tl^{+}$ is more stable than $Tl^{3+}$. Therefore,$Tl^{3+}$ acts as a powerful oxidising agent to gain electrons and form the more stable $Tl^{+}$ state.
$(ii)$ Statement $(B)$ is true: The standard reduction potential $E^0_{Al^{3+}/Al} = -1.66 \ V$ is very low,meaning $Al^{3+}$ is not easily reduced.
$(iii)$ Statement $(C)$ is false: $Tl^{3+}$ is unstable in solution due to the inert pair effect.
$(iv)$ Statement $(D)$ is true: $Tl^{+}$ is more stable than $Tl^{3+}$ due to the inert pair effect.
$(v)$ Statement $(E)$ is true: $Al^{3+}$ is the most stable oxidation state for Aluminum,and $Tl^{+}$ is the most stable oxidation state for Thallium.
Thus,statements $(A), (B), (D), \text{ and } (E)$ are correct.

(D) In the $p$-block elements of group $13$,the metallic character increases as we move down the group from $B$ to $Tl$.
As metallic character increases,the basic nature of the oxides increases.
$B_2O_3$ is acidic,$Al_2O_3$ and $Ga_2O_3$ are amphoteric,while $In_2O_3$ and $Tl_2O_3$ are basic.
Therefore,among the given options,$In_2O_3$ is the most basic oxide.

(A) $BCl_3$ is a monomer because the boron atom is small and cannot accommodate four chlorine atoms around it due to steric hindrance,and it is electron-deficient with an incomplete octet.
$AlCl_3$ exists as a dimer $(Al_2Cl_6)$ in the vapor state or in non-polar solvents to complete the octet of the aluminum atom through coordinate bonding,where chlorine atoms act as bridges.

(D) $Al_{2}O_{3}$ is an amphoteric metallic oxide because it reacts with both acids and bases.
$Al_{2}O_{3} + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}O$
$Al_{2}O_{3} + 2NaOH \rightarrow 2NaAlO_{2} + H_{2}O$

(B) When a mineral acid (like $HCl$) is added to an aqueous solution of Borax $(Na_{2}B_{4}O_{7} \cdot 10H_{2}O)$,it reacts to form orthoboric acid $(H_{3}BO_{3})$.
The chemical reaction is:
$Na_{2}B_{4}O_{7} + 2HCl + 5H_{2}O \rightarrow 4H_{3}BO_{3} + 2NaCl$

(B) . Boric acid $(B(OH)_{3})$ is a weak monobasic acid. It does not dissociate to give $H^{+}$ ions directly; instead,it acts as a Lewis acid by accepting a lone pair of electrons from the oxygen atom of a water molecule to form the $[B(OH)_{4}]^{-}$ complex.

(B) The reaction of a Lewis acid like $BCl_{3}$ with $LiAlH_{4}$ in an ether medium produces diborane $(B_{2}H_{6})$,which is a highly toxic gas.
$4BCl_{3} + 3LiAlH_{4} \rightarrow 2B_{2}H_{6} + 3AlCl_{3} + 3LiCl$
When diborane $(B_{2}H_{6})$ is heated with ammonia $(NH_{3})$,it forms borazine $(B_{3}N_{3}H_{6})$,which is commonly known as inorganic benzene.
$3B_{2}H_{6} + 6NH_{3} \rightarrow 2B_{3}N_{3}H_{6} + 12H_{2}$
Thus,the toxic gas is $B_{2}H_{6}$.

(C) $AlCl_3$ forms a dimer and exists as $Al_2Cl_6$ because it is an electron-deficient compound with an incomplete octet.
To complete its octet,the aluminium atom accepts a lone pair of electrons from the chlorine atom of another $AlCl_3$ molecule,forming a coordinate bond.
This process allows the aluminium atom to increase its coordination number from $3$ to $4$.

(A) Due to the inert pair effect,the stability of the $+1$ oxidation state increases down the group $13$ elements because the $ns^2$ electrons become more reluctant to participate in bonding.
Thus,the stability order for the $+1$ state is $Tl^{+} > In^{+} > Ga^{+}$.
Conversely,the stability of the $+3$ oxidation state decreases down the group.
Thus,the stability order for the $+3$ state is $Ga^{3+} > In^{3+} > Tl^{3+}$.
Therefore,the correct order is $Tl^{+} > In^{+} > Ga^{+}$ and $Ga^{3+} > In^{3+} > Tl^{3+}$.

(A) Inorganic benzene,also known as borazine $(B_3N_3H_6)$,is isoelectronic with benzene $(C_6H_6)$.
It consists of a hexagonal ring with alternating boron and nitrogen atoms,where each boron and nitrogen atom is bonded to one hydrogen atom.
The structure is stabilized by resonance,similar to benzene,and is often represented with partial charges on the atoms $(B^{\delta-}-N^{\delta+})$ due to the difference in electronegativity between boron and nitrogen.
Option $A$ correctly depicts this alternating $B-N$ hexagonal structure.

(A) The melting point order for Group $13$ elements is $Ga < In < Al < Tl < B$. Thus,option $A$ is incorrect.
The electronegativity order is $Tl < In < Ga < Al < B$. Thus,option $B$ is incorrect.
The density order is $B < Al < Ga < In < Tl$. Thus,option $C$ is correct.
The atomic radius order is $B < Al > Ga < In < Tl$. Thus,option $D$ is correct.
Note: In many competitive contexts,this question is considered to have multiple incorrect statements ($A$ and $B$). However,based on standard trends,$A$ is the most frequently cited incorrect order.

(B) The general electronic configuration of group $13$ elements is $ns^2 np^1$.
Due to the presence of these valence electrons,the elements exhibit oxidation states of $+3$ and $+1$.
As we move down the group,the stability of the $+1$ oxidation state increases due to the inert pair effect,which is the reluctance of the $ns^2$ electrons to participate in bonding.
Therefore,the possible oxidation states are $+1$ and $+3$.

(B) Statement $(i)$ is incorrect: The atomic radius of $Al$ $(143 \ pm)$ is actually larger than that of $Ga$ $(135 \ pm)$ due to the poor shielding effect of $d$-electrons in $Ga$.
Statement $(ii)$ is correct: Boron exists in several allotropic forms such as $\alpha$-rhombohedral, $\beta$-rhombohedral, and $\beta$-tetragonal.
Statement $(iii)$ is correct: The melting point of $Ga$ $(303 \ K)$ is the lowest among group $13$ elements due to its unique crystal structure.
Therefore, statements $(ii)$ and $(iii)$ are correct.

(B) In the periodic table of elements,gallium $(Ga)$ belongs to the boron family (group $13$).
This group includes the elements boron $(B)$,aluminium $(Al)$,gallium $(Ga)$,indium $(In)$,and thallium $(Tl)$.
Therefore,aluminium $(Al)$ belongs to the same family as gallium $(Ga)$.
The electronic configuration of these elements shows they have $ns^2 np^1$ valence shell configuration.
$Ga (Z=31) = [Ar] 3d^{10} 4s^2 4p^1$
$Al (Z=13) = [Ne] 3s^2 3p^1$

(A) Aluminium oxide $(Al_2O_3)$ is amphoteric in nature and reacts with an aqueous solution of sodium hydroxide $(NaOH)$ to form a soluble complex,sodium hexahydroxyaluminate$(III)$.
The balanced chemical equation is:
$Al_2O_3 + 2NaOH + 3H_2O \longrightarrow 2Na[Al(OH)_4]$ or more commonly represented as $Al_2O_3 + 6NaOH + 3H_2O \longrightarrow 2Na_3[Al(OH)_6]$ depending on the concentration. In the context of the Bayer process,the product is $Na[Al(OH)_4]$. However,given the options provided,$Na_3[Al(OH)_6]$ is the standard complex representing the hexahydroxyaluminate ion.

(B) The structure of diborane $(B_2H_6)$ consists of two boron atoms and six hydrogen atoms.
There are $4$ terminal hydrogen atoms,each bonded to a boron atom by a $2$-centre-$2$-electron $(2c-2e)$ bond. Thus,$X = 4$.
There are $2$ bridging hydrogen atoms,each involved in a $3$-centre-$2$-electron $(3c-2e)$ bond with the two boron atoms. Thus,$Y = 2$.
The value of $(X+Y) = 4 + 2 = 6$.

(A) . Aluminium becomes passive in concentrated $HNO_3$ due to the formation of a protective oxide layer on its surface,so it does not liberate $H_2$ gas. This statement is incorrect.
$B$. Borazole $(B_3N_3H_6)$ has a structure similar to benzene. It contains $12$ $\sigma$ bonds ($6$ $B-H$ and $N-H$ bonds,and $6$ $B-N$ bonds) and $3$ $\pi$ bonds. This statement is correct.
$C$. Gallium oxide $(Ga_2O_3)$ is amphoteric in nature as it reacts with both acids and bases. This statement is correct.
$D$. $BF_3$ has an incomplete octet on the Boron atom,making it an electron-deficient species and thus a Lewis acid. This statement is correct.

(C) On an industrial scale,diborane $(B_2H_6)$ is prepared by the reaction of boron trifluoride $(BF_3)$ with sodium hydride $(NaH)$ at $450 \ K$.
The chemical equation for this reaction is:
$2BF_3 + 6NaH \xrightarrow{450 \ K} B_2H_6 + 6NaF$
Therefore,the correct option is $C$.

(B) $1$. Borax $(Na_2B_4O_7 \cdot 10H_2O)$ in water hydrolyzes to give $NaOH$ and $H_3BO_3$. Since $NaOH$ is a strong base and $H_3BO_3$ is a weak acid,the solution is alkaline. This statement is correct.
$2$. Orthoboric acid $(H_3BO_3)$ is a weak monobasic Lewis acid. It acts as an acid by accepting a hydroxyl ion $(OH^-)$ from water,not by donating protons. Therefore,calling it a tribasic acid is incorrect.
$3$. Metaboric acid $(HBO_2)$ on heating gives boron trioxide $(B_2O_3)$,which is an acidic oxide. This statement is correct.
$4$. $LiBH_4$ is a well-known reducing agent used in organic synthesis. This statement is correct.
Thus,the incorrect statement is $B$.

(C) ) Incorrect: Aluminium is amphoteric and reacts with both $HCl$ and $NaOH$ to liberate $H_2$ gas.
$B$) Incorrect: The formula of sodium metaborate is $NaBO_2$. $Na_3BO_3$ is sodium orthoborate.
$C$) Correct: Boric acid $(H_3BO_3)$ acts as a weak monobasic Lewis acid by accepting $OH^-$ from water: $B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^- + H_3O^+$.
$D$) Correct: Due to the inert pair effect,the stability of the $+1$ oxidation state increases down the group $13$ $(Al < Ga < In < Tl)$,making $Tl^+$ more stable than $Tl^{3+}$.

(B) $(i)$ $2NaBH_4 + I_2 \rightarrow B_2H_6 + 2NaI + H_2$. This statement is correct.
$(ii)$ $B_2H_6 + 3O_2 \rightarrow B_2O_3 + 3H_2O + \text{Energy}$. This statement is correct as diborane is highly flammable.
$(iii)$ $B_2H_6 + 6H_2O \rightarrow 2B(OH)_3 + 6H_2$. $B(OH)_3$ (orthoboric acid) acts as a weak monobasic Lewis acid by accepting $OH^-$ from water: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$. Thus,statement $iii$ is incorrect.

(B) The reaction of the Lewis acid $BF_3$ $(X)$ with $LiAlH_4$ in ether produces diborane,$B_2H_6$ $(Y)$,which is a highly toxic gas.
$4BF_3 + 3LiAlH_4 \rightarrow 2B_2H_6 + 3LiF + 3AlF_3$
When $B_2H_6$ $(Y)$ is heated with $NH_3$,it forms borazine $(B_3N_3H_6)$,which is known as inorganic benzene.
$3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 12H_2$
When $B_2H_6$ $(Y)$ burns in oxygen,it produces $B_2O_3$ $(Z)$ and $H_2O$.
$B_2H_6 + 3O_2 \rightarrow B_2O_3 + 3H_2O$
$B_2O_3$ is a non-metallic oxide,and non-metallic oxides are acidic in nature. Therefore,'$Z$' is an acidic oxide.

(C) In the structure of diborane $(B_2H_6)$,there are two types of hydrogen atoms: terminal and bridging.
$1$. The terminal $B-H$ bonds are $2$-center-$2$-electron $(2C-2e)$ bonds.
$2$. The bridging $B-H-B$ bonds are $3$-center-$2$-electron $(3C-2e)$ bonds.
$3$. The angle $\theta_1$ represents the $H-B-H$ bond angle between the two bridging hydrogen atoms,which is approximately $97^{\circ}$.
$4$. The angle $\theta_2$ represents the $H-B-H$ bond angle between the two terminal hydrogen atoms,which is approximately $120^{\circ}$.
Therefore,$\theta_1 = 97^{\circ}$ and $\theta_2 = 120^{\circ}$.

(C) $(i)$ $2NaBH_4 + I_2 \longrightarrow B_2H_6 + 2NaI + H_2$
$(ii)$ $2BF_3 + 6NaH \xrightarrow{450 \ K} B_2H_6 + 6NaF$
$(iii)$ $4BF_3 + 3LiAlH_4 \longrightarrow 2B_2H_6 + 3LiF + 3AlF_3$
$(iv)$ $3B_2H_6 + 6NH_3 \xrightarrow{\text{heat}} 2B_3N_3H_6 + 12H_2$
In reaction $(i)$,$H_2$ is produced as a byproduct.
In reaction $(iv)$,$H_2$ is produced as a byproduct.
Therefore,in reactions $(i)$ and $(iv)$,hydrogen is one of the products.

(B) Boron fibres are used for making bullet-proof vests.
Metal borides are used as protective shields because they have the capacity to absorb neutrons.
Borax is used in the manufacture of glass and glass wool to increase the mechanical strength and durability of the glass.
Therefore,all three statements are correct.

(B) The metallic radii of Group $13$ elements are as follows:
$Al = 143 \ pm$
$Ga = 135 \ pm$
$In = 167 \ pm$
$Tl = 170 \ pm$
The metallic radius generally increases down the group, but $Ga$ has a smaller radius than $Al$ due to the poor shielding effect of $d$-electrons, which increases the effective nuclear charge.
Therefore, the correct matching is $A-II, B-I, C-IV, D-III$.

(A) The order of atomic radii of group $13$ elements is $B < Ga < Al < In < Tl$. This is due to the poor shielding effect of $3d$-electrons in $Ga$,which makes its atomic radius slightly smaller than that of $Al$. Thus,statement $(i)$ is correct.
The correct order of ionization enthalpy for group $13$ elements is $B > Tl > Ga > Al > In$. This anomaly is due to the poor shielding effect of $d$ and $f$ orbitals. Thus,statement $(ii)$ is incorrect.
Boron trioxide $(B_2O_3)$ is acidic in nature because it is a non-metal oxide and reacts with water to form boric acid $(H_3BO_3)$. Thus,statement $(iii)$ is incorrect.
Therefore,statements $(ii)$ and $(iii)$ are not correct.

(A) The chemical formulas for the given compounds are as follows:
Borax: $Na_2B_4O_7 \cdot 10H_2O$ (contains $10$ water molecules).
Kernite: $Na_2B_4O_7 \cdot 4H_2O$ (contains $4$ water molecules).
Bauxite: $Al_2O_3 \cdot 2H_2O$ (contains $2$ water molecules).
Therefore,the number of water molecules are $10, 4, 2$ respectively.

(D) Boron exhibits allotropy and exists in several crystalline forms such as $\alpha-$rhombohedral,$\beta-$rhombohedral,and $\beta-$tetragonal,as well as an amorphous form. Therefore,Statement $I$ is incorrect.
Boron is an extremely hard,black-coloured solid with a very high melting point due to its strong covalent crystal lattice structure. Therefore,Statement $II$ is correct.

(C) The electronegativity of $Thallium$ $(Tl)$ is $1.62$ and $Aluminium$ $(Al)$ is $1.61$ on the $Pauling$ scale,making $Al$ slightly more electropositive than $Tl$.
Boron is a poor conductor of electricity at low temperatures.
$^{10}B$ isotope has a high ability to absorb neutrons,not $^{11}B$.
An aqueous solution of orthoboric acid $(H_3BO_3)$ is a weak acid and is commonly used as a mild antiseptic for eyes and skin.

(D) Boron exists as a very hard,black,non-metallic solid with a very strong crystalline lattice,which gives it an unusually high melting point.
Boron is a non-metal and is a poor conductor of electricity,whereas other members of the group are metals and conduct electricity.
The density of boron is relatively low compared to other elements in the group.
The $^{10}B$ isotope has a high ability to absorb neutrons,making it useful in nuclear reactors as control rods and protective shields.

(C) Due to its smaller size and unavailability of $d$-orbitals,boron exhibits properties that contrast with other elements of the boron family.
$I$. Boron trihalides do not form dimeric structures; they complete their octet through back bonding. Other members of this group can form dimeric structures.
$II$. Boron does not show $+1$ as a stable oxidation state. As we move down the group,the stability of the $+1$ oxidation state increases due to the inert pair effect. The $+1$ oxidation state of Boron is highly unstable.
$III$. The maximum covalency of Boron is four,as it lacks $d$-orbitals and cannot accommodate more than eight electrons in its valence shell.
$IV$. Boron does not form $BF_{6}^{3-}$ ion because it cannot expand its octet beyond eight electrons.

(B) The given reaction is the hydrolysis of boric acid in water:
$H_3BO_3 + 2H_2O \longrightarrow [B(OH)_4]^{-} + H_3O^{+}$
Comparing this with the given equation $P + Q \longrightarrow [B(OH)_4]^{-} + H_3O^{+}$,we get:
$P = H_3BO_3$
$Q = 2H_2O$
In this reaction,$H_3BO_3$ acts as a Lewis acid by accepting an $OH^{-}$ ion from water.

(D) Borax $(Na_2B_4O_7 \cdot 10H_2O)$ dissolves in water to form an alkaline solution due to hydrolysis.
The reaction is: $Na_2B_4O_7 \cdot 10H_2O + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$.
Here,$H_3BO_3$ is orthoboric acid,which is the product of boron $(X)$.

(A) When borax $(Na_2B_4O_7 \cdot 10H_2O)$ is dissolved in water,it undergoes hydrolysis to form orthoboric acid and sodium hydroxide.
The reaction is as follows:
$Na_2B_4O_7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$
Since $NaOH$ is a strong base,the resulting solution is alkaline in nature.

(A) Diborane $(B_2H_6)$ is an electron-deficient molecule.
In its structure,the four terminal hydrogen atoms and two boron atoms lie in the same plane.
The two bridging hydrogen atoms lie above and below this plane.
The terminal $B-H$ bonds are normal $2-$centered$-2-$electron $(2c-2e)$ bonds with a bond angle of $120^{\circ}$ $(x)$.
The bridging $B-H-B$ bonds are $3-$centered$-2-$electron $(3c-2e)$ bonds,which are commonly known as banana-bonds or tau-bonds.
The bridging $H-B-H$ bond angle $(y)$ is $97^{\circ}$.

(B) $H_3BO_3$ is a weak monobasic Lewis acid. It does not dissociate to give $H^{+}$ ions directly in water.
Instead,it acts as a Lewis acid by accepting an $OH^{-}$ ion from the water molecule.
This process releases a proton $(H^{+})$ from the water molecule,thereby increasing the acidity of the solution.
The reaction is: $B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^{-} + H_3O^{+}$.
Therefore,$H_3BO_3$ is considered an acid because its molecule accepts $OH^{-}$ from water,releasing a proton.

(B) Boron compounds are often employed as Lewis acids due to their strong electrophilic nature,which is caused by a vacant $p$-orbital that can accept electrons.
Boron behaves as a Lewis acid because it has only $3$ valence electrons,leaving its octet incomplete.
For example,$B_2H_6$ is an electron-deficient compound.
The electronic configuration of Boron is $B (Z=5) = [He] 2s^2 2p^1$.
This results in a total of $6$ electrons in the valence shell of the central atom in compounds like $BF_3$,which is less than the $8$ required for a complete octet.

(A) $BF_3 \xrightarrow{LiAlH_4} B_2H_6 (X)$
$3B_2H_6 + 6NH_3 \xrightarrow{\Delta} 2B_3N_3H_6 (Y) + 12H_2$
Compound $X$ is $B_2H_6$ (Diborane).
Compound $Y$ is $B_3N_3H_6$ (Borazine),which is known as inorganic benzene.

(A) When boron trichloride $(BCl_3)$ reacts with water,it undergoes hydrolysis to form boric acid $(H_3BO_3)$ and hydrochloric acid $(HCl)$.
The balanced chemical equation is:
$BCl_3 + 3H_2O \longrightarrow H_3BO_3 + 3HCl$

(A) Boron halides $(BX_3)$ have a central boron atom bonded to three halogen atoms.
In these molecules,boron has only $6$ electrons in its valence shell (an incomplete octet).
Due to this electron deficiency,they act as Lewis acids by accepting a lone pair of electrons from a Lewis base to complete their octet.

(B) From the given chemical reactions:
$1$. The reaction of diborane with excess ammonia at high temperature produces inorganic benzene $(B_3N_3H_6)$,so $X = B_3N_3H_6$.
$2$. The reaction of sodium hydride with boron trifluoride produces diborane $(B_2H_6)$,so $Y = B_2H_6$.
$3$. The hydrolysis of diborane produces orthoboric acid $(H_3BO_3)$,so $Z = H_3BO_3$.
Therefore,the correct sequence is $X = B_3N_3H_6, Y = B_2H_6, Z = H_3BO_3$.
Hence,option $(b)$ is correct.

(A) Diborane $(B_2H_6)$ can be prepared by the following reactions:
$I. 4BF_3 + 3LiAlH_4 \xrightarrow{\text{ether}} 2B_2H_6 + 3LiF + 3AlF_3$
$II. 2BF_3 + 6NaH \xrightarrow{450 \ K} B_2H_6 + 6NaF$
$IV. 2NaBH_4 + I_2 \longrightarrow B_2H_6 + 2NaI + H_2$
Reaction $III$ $(Na_2B_4O_7 + H_2O)$ is a hydrolysis reaction of borax which does not produce diborane.
Thus,reactions $I, II,$ and $IV$ are used to prepare diborane.

(A) The explanations of the given statements are as follows:
$(i)$ $H_3BO_3$ (orthoboric acid) is a monobasic Lewis acid. In aqueous solution,it accepts an $OH^-$ ion from water to form $[B(OH)_4]^-$ and releases $H^+$,acting as a Lewis acid.
$B(OH)_3 + H_2O \rightleftharpoons [B(OH)_4]^- + H^+$
$(ii)$ The correct formula for borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$. This statement is correct.
$(iii)$ $NaBH_4$ (sodium borohydride) is a well-known reducing agent used in organic chemistry to reduce aldehydes and ketones to alcohols. This statement is correct.
Since all three statements are correct,the correct option is $A$.

(D) Borax $(Na_2B_4O_7 \cdot 10H_2O)$ dissolves in water to form an alkaline solution due to hydrolysis.
The reaction is as follows:
$Na_2B_4O_7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$
Here,$NaOH$ is a strong base and $H_3BO_3$ (orthoboric acid) is a very weak acid.
Since the solution contains a strong base,the resultant solution is basic in nature.
Therefore,the $pH$ of the solution is greater than $7$,which falls in the range of $7 - 14$.

(A) The structure of $B_2H_6$ (diborane) consists of two $BH_2$ groups in the same plane.
There are $4$ terminal $B-H$ bonds,which are $2$-centered-$2$-electron $(2c-2e)$ bonds.
There are $2$ bridge $B-H-B$ bonds,which are $3$-centered-$2$-electron $(3c-2e)$ bonds.
There is no direct $B-B$ bond in $B_2H_6$.
Thus,the number of $BH_2$ groups in one plane is $2$,terminal $B-H$ bonds is $4$,$B-B$ bonds is $0$,and $B-H-B$ bridge bonds is $2$.

(C) Statement $a$ is correct: Borax bead test involves heating borax with a metal salt like $Co^{2+}$,forming a characteristic blue-coloured cobalt metaborate,$Co(BO_2)_2$.
Statement $b$ is correct: The structural formula of borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$.
Statement $c$ is incorrect: Boron trihalides $(BX_3)$ are electron-deficient compounds with an incomplete octet on the boron atom,making them Lewis acids,not Lewis bases.
Therefore,statements $a$ and $b$ are correct.

(C) The structure of diborane $(B_2H_6)$ consists of two boron atoms and six hydrogen atoms.
There are four terminal $B-H$ bonds,which are $2$-centre-$2$-electron $(2c-2e)$ bonds.
There are two bridging $B-H-B$ bonds,which are $3$-centre-$2$-electron $(3c-2e)$ bonds.
Therefore,the number of $2c-2e$ bonds is $4$ and the number of $3c-2e$ bonds is $2$.