Boron family Questions in English

(A) Boron belongs to the second period and has a valence shell configuration of $2s^2 2p^1$.
It lacks $d$-orbitals in its valence shell,which limits its maximum covalency to $4$.
Therefore,boron cannot expand its octet to form $BF_6^{3-}$ because it would require a coordination number of $6$ and the presence of $d$-orbitals to accommodate the extra electrons.

(A) The reaction $2x + B_2H_6 \rightarrow [BH_2x_2]^+ [BH_4]^-$ represents the unsymmetrical cleavage of diborane $(B_2H_6)$.
Small amines like $NH_3$ undergo this type of reaction to form ionic products.
For example,$2 NH_3 + B_2H_6 \rightarrow [BH_2(NH_3)_2]^+ [BH_4]^-$.
Therefore,$x$ is $NH_3$.

(C) $1$. The thermal decomposition of $B(OH)_3$ (orthoboric acid) at high temperature yields $B_2O_3$ and water vapor. Thus,$X = B_2O_3$ and $Z = H_2O$ (though the question implies a gas,$H_2O$ vapor is released).
$2$. $B_2O_3$ can be reduced to elemental $Boron$ using $Mg$ or $Fe$.
$3$. $B_2O_3$ reacts with $PCl_5$ or other chlorinating agents to form $BCl_3$. $BCl_3$ reacts with $H_2O$ to regenerate $B(OH)_3$.
$4$. $BCl_3$ reacts with $NH_3$ at high temperature to form inorganic benzene $(B_3N_3H_6)$. Thus,$Y = BCl_3$.
$5$. Therefore,$X = B_2O_3, Y = BCl_3, Z = H_2O$ (Note: The provided options suggest $Z = HCl$ in option $C$,which is consistent with the reaction of $B_2O_3$ with $PCl_5$ or similar reagents to produce $BCl_3$ and $HCl$ gas).

(C) In the structure of diborane $(B_2H_6)$,there are four terminal hydrogen atoms bonded to boron atoms by normal covalent bonds,which are $2C-2e^-$ (two-center-two-electron) bonds.
There are two bridging hydrogen atoms,each of which is bonded to two boron atoms by a $3C-2e^-$ (three-center-two-electron) bond,also known as a banana bond.

(D) $Zn(OH)_2$,$Pb(OH)_2$,and $Al(OH)_3$ are amphoteric hydroxides.
They react with strong bases like $NaOH$ to form soluble complex salts.
$Zn(OH)_2 + 2NaOH \rightarrow Na_2[Zn(OH)_4]$ (Sodium zincate)
$Pb(OH)_2 + 2NaOH \rightarrow Na_2[Pb(OH)_4]$ (Sodium plumbite)
$Al(OH)_3 + NaOH \rightarrow Na[Al(OH)_4]$ (Sodium aluminate)
Therefore,all of these can dissolve in excess $NaOH$.

(A) When ammonium hydroxide $(NH_4OH)$ is added to an aqueous solution of aluminum sulfate $(Al_2(SO_4)_3)$,a white gelatinous precipitate of aluminum hydroxide $(Al(OH)_3)$ is formed.
The reaction is: $Al^{3+} (aq) + 3NH_4OH (aq) \rightarrow Al(OH)_3 (s) + 3NH_4^+ (aq)$.
This precipitate of $Al(OH)_3$ is insoluble in excess of ammonium hydroxide $(NH_4OH)$.

(A) The Lewis acidity of boron trihalides $(BX_3)$ depends on the extent of $p\pi-p\pi$ back-bonding between the halogen atom and the boron atom.
In $BF_3$,the fluorine atom has a small size and its $2p$ orbitals effectively overlap with the empty $2p$ orbital of boron,leading to strong back-bonding.
This reduces the electron deficiency of boron,making $BF_3$ the weakest Lewis acid.
As the size of the halogen increases from $F$ to $I$,the extent of back-bonding decreases,making the boron atom more electron-deficient and thus increasing the Lewis acidity.
The order of Lewis acidity is: $BF_3 < BCl_3 < BBr_3 < BI_3$.

(D) In diborane $(B_2H_6)$, there are two types of $B-H$ bonds: terminal and bridge bonds.
$1$. There are $4$ terminal $B-H$ bonds which are $2c-2e$ bonds and are equal in length $(119 \text{ pm})$.
$2$. There are $2$ bridge $B-H$ bonds which are $3c-2e$ bonds and are equal in length $(134 \text{ pm})$.
$3$. The bond angles in the $B-H-B$ bridge are approximately $97^{\circ}$, and the $H-B-H$ angle for terminal hydrogens is approximately $120^{\circ}$.
Therefore, all the given statements are correct.

(A) The reduction of boron trioxide $(B_2O_3)$ with metals like magnesium $(Mg)$ or sodium $(Na)$ typically yields amorphous boron.
However,the reduction of $B_2O_3$ with aluminum $(Al)$ at high temperatures is a common method used to obtain crystalline boron.
Therefore,the reaction $B_2O_3 + 2Al \rightarrow 2B + Al_2O_3$ is the one that yields the crystalline variety of boron.

(C) Boron carbide $(B_4C)$ is an extremely hard ceramic material used in tank armor,bulletproof vests,and engine sabotage powders,and it is commonly known as Norbia.
While boron carbide is very hard,it is not the hardest substance known; diamond is significantly harder.
Silicon carbide is often used for cutting and grinding due to its durability and cost-effectiveness,whereas boron carbide is typically used for specialized armor applications.
Therefore,the statement that it is the hardest substance is incorrect.

(D) In diborane $(B_2H_6)$:
$1.$ Each boron atom is $sp^3$ hybridized,which is correct.
$2.$ The $H-B-H$ bridge angle is not $180^o$; it is approximately $97^o$,so statement $2$ is incorrect.
$3.$ Each boron atom is bonded to two terminal hydrogen atoms via two terminal $B-H$ bonds,which is correct.
$4.$ Diborane has $12$ valence electrons $(2 \times 3 + 6 \times 1 = 12)$,forming $6$ covalent bonds (each with $2$ electrons),so there are $12$ bonding electrons available,which is correct.
Therefore,statements $1, 3,$ and $4$ are correct.

(D) When $Na_2B_4O_7 \cdot 10H_2O$ (Borax) is heated,it loses water of crystallization and swells. On further heating,it melts to form a clear,transparent glassy bead consisting of $NaBO_2$ (sodium metaborate) and $B_2O_3$ (boric anhydride or boron trioxide).
The reaction is as follows:
$Na_2B_4O_7 \cdot 10H_2O \xrightarrow{\Delta} Na_2B_4O_7 + 10H_2O$
$Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3$
Thus,the transparent glassy mass is a mixture of sodium metaborate and boric anhydride.
Option $(D)$ is correct.

(A) The structure of $B_2H_6$ consists of two boron atoms and six hydrogen atoms.
Four hydrogen atoms are terminal hydrogen atoms,each connected to a boron atom by a normal covalent bond. These are $2$-center-$2$-electron $(2C-2e)$ bonds.
Two hydrogen atoms are bridging hydrogen atoms,each connected to both boron atoms. These bridge bonds are $3$-center-$2$-electron $(3C-2e)$ bonds.
Therefore,the structure contains four $2C-2e$ bonds and two $3C-2e$ bonds.

(B) Incorrect. The ordinary form of borax is $Na_2B_4O_7 \cdot 10H_2O$.
$B$ Correct. Colemanite is $Ca_2B_6O_{11} \cdot 5H_2O$.
$C$ Incorrect. Boronatrocalcite (ulexite) is $NaCaB_5O_9 \cdot 8H_2O$.
$D$ Incorrect. The octahedral form of borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$.

(A) Boron hydrides are classified into two main series: $B_nH_{n+4}$ and $B_nH_{n+6}$.
Members of the $B_nH_{n+6}$ series are generally less stable than those of the $B_nH_{n+4}$ series.
Diborane $(B_2H_6)$ is a colourless,highly reactive gas that is stable at low temperatures but reacts readily with oxygen in a highly exothermic reaction.

(C) The reactions are as follows:
$B_2H_6 + 2KOH + 2H_2O \rightarrow 2KBO_2 + 6H_2$
$2Al + 2KOH + 2H_2O \rightarrow 2KAlO_2 + 3H_2$
In both reactions,a salt of the type $KMO_2$ (where $M$ is $B$ or $Al$) is obtained.
Therefore,option $(C)$ is correct.

(A) Answer:- $(A)$ Metaborate
On heating,borax loses its water of crystallization and swells up to form a fluffy mass. On further heating,it melts to give a clear liquid which solidifies to a transparent glassy bead consisting of sodium metaborate and boric anhydride.
$Na_{2}B_{4}O_{7} \cdot 10H_{2}O$ $\xrightarrow{\Delta} Na_{2}B_{4}O_{7}$ $\xrightarrow{\Delta} 2NaBO_{2} + B_{2}O_{3}$
In the presence of transition metal oxides,the bead forms colored metal metaborates,e.g.,$CoO + B_{2}O_{3} \rightarrow Co(BO_{2})_{2}$. Thus,metaborate is the key component formed.

(B) Boron Nitride $(BN)$ is known as inorganic graphite.
It is a chemical compound with the formula $BN$,consisting of an equal number of boron and nitrogen atoms.
$BN$ is isoelectronic to carbon and exists in a layered structure similar to graphite,which is why it is referred to as inorganic graphite.

(C) Anhydrous $AlCl_3$ is a well-known Lewis acid used as a catalyst in Friedel-Crafts reactions and cracking of petroleum. However,hydrated $AlCl_3$ (often written as $AlCl_3 \cdot 6H_2O$) is primarily used as a mordant in the dyeing industry. Since the question specifically asks for the use of hydrated $AlCl_3$,the correct answer is mordant.

(A) The thermal decomposition of orthoboric acid $(H_3BO_3)$ occurs in steps:
$1$. At $100^{\circ}C$ $(T_1)$,$H_3BO_3$ loses water to form metaboric acid $(HBO_2)$:
$H_3BO_3 \xrightarrow{100^{\circ}C} HBO_2 + H_2O$
$2$. At $160^{\circ}C$ $(T_2)$,metaboric acid $(HBO_2)$ converts into tetraboric acid $(H_2B_4O_7)$:
$4HBO_2 \xrightarrow{160^{\circ}C} H_2B_4O_7 + H_2O$
$3$. On further strong heating (red hot),tetraboric acid decomposes into boron trioxide $(B_2O_3)$:
$H_2B_4O_7 \xrightarrow{\Delta} 2B_2O_3 + H_2O$
Thus,$X$ is metaboric acid and $Y$ is tetraboric acid.

(B) The Lewis acidity of boron halides depends on the extent of $p\pi-p\pi$ back-bonding between the halogen atom and the boron atom.
The tendency for back-donation from the halogen to the vacant $p$-orbital of boron is highest in $BF_3$ due to the small size of the fluorine atom,which allows for effective overlap.
As the size of the halogen increases from $F$ to $I$,the effectiveness of back-bonding decreases,making the boron atom more electron-deficient.
Therefore,the decreasing order of Lewis acidity is $BI_3 > BBr_3 > BCl_3 > BF_3$.

(D) In diborane $(B_2H_6)$,the structure consists of two $BH_2$ groups linked by two bridging hydrogen atoms.
The terminal $B-H$ bonds are normal $2c-2e$ (two-center,two-electron) covalent bonds,while the bridging $B-H-B$ bonds are $3c-2e$ (three-center,two-electron) bonds.
Due to the difference in the nature of these bonds,the terminal $B-H$ bond length $(119 \text{ pm})$ is shorter than the bridging $B-H$ bond length $(134 \text{ pm})$.
Therefore,the statement that all $B-H$ bond distances are equal is false.

(C) The structure of the borax anion is $[B_4O_5(OH)_4]^{2-}$.
This structure consists of two boron atoms in tetrahedral coordination (each bonded to four oxygen atoms) and two boron atoms in triangular coordination (each bonded to three oxygen atoms).
Therefore,statement $c$ is correct: Two boron atoms have four $B-O$ bonds while the other two have three $B-O$ bonds.
Additionally,there are four $-OH$ groups in the structure,each attached to one of the four boron atoms. Thus,each boron atom has one $-OH$ group attached to it.
Therefore,statement $d$ is also correct.
The correct statements are $c$ and $d$.

(D) Boron carbide $B_4C$ is not used for making diborane.
It is used in the extraction of boron,as an abrasive for polishing,and for making bullet-proof clothing.

(A) $BCl_{3}$ is a Lewis acid and $NH_{3}$ is a Lewis base. They react to form an adduct,$BCl_{3} \cdot NH_{3}$,rather than undergoing a substitution reaction. Therefore,the reaction given in option $A$ is incorrect.
$AlCl_{3}$ is highly hydrated in water to form $[Al(H_{2}O)_{6}]^{3+}$ ions,which are octahedral.
$H_{3}BO_{3}$ acts as a Lewis acid in water by accepting $OH^{-}$ ions to form $[B(OH)_{4}]^{-}$.
Borax in water hydrolyzes to form $H_{3}BO_{3}$ and $[B(OH)_{4}]^{-}$,creating a buffer solution with a basic $pH$.

(B) $TlI_3$ contains the $Tl^{3+}$ ion and $I^-$ ions,but it is actually a complex salt containing the $Tl^+$ ion and the triiodide ion,$I_3^-$. Thus,it is formulated as $Tl^+[I_3]^-$.
$CsI_3$ is also a salt containing the $Cs^+$ ion and the triiodide ion,$I_3^-$.
Since both compounds contain the same $I_3^-$ anion and have similar lattice structures due to the large size of the cations,$TlI_3$ is isomorphous with $CsI_3$.

(B) The given reaction is: $2BF_3 + 6NaH \xrightarrow{450 \ K} B_2H_6 + 6NaF$. Thus,$X$ is diborane $(B_2H_6)$.
$1$. $B_2H_6$ undergoes hydrolysis to form boric acid $(H_3BO_3)$,which is a weak monobasic acid. This statement is correct.
$2$. In $B_2H_6$,there is no back bonding; it is an electron-deficient molecule with $3c-2e$ bonds. This statement is incorrect.
$3$. In the structure of $B_2H_6$,the two boron atoms and the four terminal hydrogen atoms lie in the same plane. Thus,the maximum number of atoms in one plane is $6$. This statement is correct.
$4$. $B_2H_6$ burns in $O_2$ to form $B_2O_3$ (boron sesquioxide). This statement is correct.

(B) The reaction of diborane $(B_2H_6)$ with ammonia $(NH_3)$ depends on the molar ratio and temperature.
At low temperatures,$B_2H_6$ reacts with $NH_3$ to form an adduct,$B_2H_6 \cdot 2NH_3$,which is ionic in nature,i.e.,$[BH_2(NH_3)_2]^+ [BH_4]^-$.
However,using $THF$ (tetrahydrofuran) as a solvent allows for the cleavage of the diborane bridge to form the monomeric borane-$THF$ complex,$BH_3 \cdot THF$.
When $NH_3$ is added to this $BH_3 \cdot THF$ complex,it undergoes a substitution reaction to form the desired $BH_3 \cdot NH_3$ adduct.
Therefore,Student $II$ is expected to get the correct final product.

(D) Diborane $(B_2H_6)$ undergoes symmetrical cleavage with Lewis bases like $T.H.F.$,$Me_3N$,and $C_5H_5N$ to form adducts of the type $BH_3 \cdot L$.
However,with small,highly basic amines like $MeNH_2$ or $NH_3$,it undergoes unsymmetrical cleavage to form ionic species of the type $[BH_2(L)_2]^+ [BH_4]^-$.
Therefore,$MeNH_2$ causes unsymmetrical cleavage.

(C) The thermal decomposition of borax $(Na_2B_4O_7 \cdot 10H_2O)$ yields sodium metaborate $(NaBO_2)$ and boron trioxide $(B_2O_3)$.
Thus,$X = B_2O_3$.
The reaction of $B_2O_3$ with chromium$(III)$ oxide $(Cr_2O_3)$ produces chromium$(III)$ metaborate $(Cr(BO_2)_3)$,which is a green-colored bead.
Thus,$Y = Cr(BO_2)_3$.
The balanced chemical equations are:
$Na_2B_4O_7 \cdot 10H_2O \xrightarrow{\Delta} Na_2B_4O_7 + 10H_2O$
$Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3$
$3B_2O_3 + Cr_2O_3 \xrightarrow{\Delta} 2Cr(BO_2)_3$

(A) Boric acid $(H_3BO_3)$ contains three $OH$ groups,but it acts as a monobasic Lewis acid in aqueous solution.
It does not dissociate to release $H^+$ ions.
Instead,it accepts a lone pair of electrons from $OH^-$ ions present in water to form the $[B(OH)_4]^-$ complex ion.
The reaction is: $H_3BO_3 + H_2O \rightleftharpoons [B(OH)_4]^- + H^+$.
Since it releases one $H^+$ ion per molecule through this mechanism,its basicity is $1$.

(D) The chemical formula of orthoboric acid is $H_3BO_3$ or $B(OH)_3$.
In $H_3BO_3$,the boron atom is $sp^2$ hybridized,which gives it a trigonal planar geometry.
The three oxygen atoms and the boron atom lie in the same plane.
Additionally,the three hydrogen atoms attached to the oxygen atoms also lie in the same plane due to the planar structure of the $B(OH)_3$ unit.
Therefore,all $7$ atoms ($1$ $B$,$3$ $O$,and $3$ $H$) are present in the same plane.

(C) The reaction of boron trifluoride with lithium aluminium hydride in diethyl ether produces diborane $(B_2H_6)$:
$4BF_3 + 3LiAlH_4 \rightarrow 2B_2H_6 + 3LiF + 3AlF_3$
Thus,$X$ is diborane $(B_2H_6)$.
Structure of diborane $(B_2H_6)$:
$1$. It contains four $2$-centre-$2$-electron $(2c-2e)$ bonds (terminal $B-H$ bonds).
$2$. It contains two $3$-centre-$2$-electron $(3c-2e)$ bonds (bridging $B-H-B$ bonds).
$3$. Total electrons involved in bonding: $4 \times 2$ (terminal) $+ 2 \times 2$ (bridging) $= 12$ electrons.
$4$. Diborane reacts with $NH_3$ to form various products such as borazine $(B_3N_3H_6)$ or the ionic compound $[H_2B(NH_3)_2]^+[BH_4]^-$.
Therefore,the statement that $X$ does not react with $NH_3$ is incorrect.

(D) Borax,$Na_2[B_4O_5(OH)_4] \cdot 8H_2O$,exhibits the following properties:
$1$. On heating,it loses water of crystallization and swells up to form a transparent glassy bead,known as the intumescent property.
$2$. The structure of the tetraborate anion $[B_4O_5(OH)_4]^{2-}$ consists of two $BO_3$ triangles and two $BO_4$ tetrahedra sharing corners,forming two identical six-membered rings.
$3$. Upon hydrolysis,$1 \ mole$ of Borax produces $2 \ moles$ of $H_3BO_3$ and $2 \ moles$ of $NaOH$. The $2 \ moles$ of $NaOH$ produced are neutralized by $2 \ moles$ of $HCl$ $(NaOH + HCl \rightarrow NaCl + H_2O)$.
Therefore,all the given statements are correct.

(B) The extraction of borax from colemanite $(2CaO \cdot 3B_2O_3)$ involves the following steps:
$1$. Colemanite is boiled with a solution of sodium carbonate $(Na_2CO_3)$:
$2CaO \cdot 3B_2O_3 + 2Na_2CO_3 \to 2CaCO_3 \downarrow + Na_2B_4O_7 + 2NaBO_2$
Here,$CaCO_3$ is the residue $(Z)$.
$2$. The filtrate contains $Na_2B_4O_7$ and $NaBO_2$. Upon concentration and cooling,borax $(Na_2B_4O_7 \cdot 10H_2O)$ crystallizes out as the residue $(Y)$,while sodium metaborate $(NaBO_2)$ remains in the filtrate $(X)$.
$3$. $CO_2$ is passed through the filtrate $(X = NaBO_2)$ to convert it into borax $(Y = Na_2B_4O_7 \cdot 10H_2O)$:
$4NaBO_2 + CO_2 \to Na_2B_4O_7 + Na_2CO_3$
Thus,$X$ is $NaBO_2$ and $Y$ is $Na_2B_4O_7 \cdot 10H_2O$.

(B) The melting point of gallium $(Ga)$ is $30^{\circ} C$ and its boiling point is $2240^{\circ} C$.
Because of this very high boiling point and low melting point,it exists in the liquid state over a wide range of temperatures.
Therefore,it is used in high-temperature thermometers.

(A) $B_2O_3$ is an acidic oxide.
$Al_2O_3$ and $Ga_2O_3$ are amphoteric oxides.
$In_2O_3$ is a basic oxide.

(A) The atomic number of Boron is $5$ and its electronic configuration is $1s^2 2s^2 2p^1$.
Due to the small size of the Boron atom,the ionization enthalpies required to remove three electrons are extremely high.
The sum of the first,second,and third ionization enthalpies is so large that it cannot be compensated by the lattice energy (in case of ionic solids) or hydration energy (in case of aqueous solutions).
Therefore,Boron does not form $B^{3+}$ cations easily.

(C) The hydrolysis of $BF_3$ occurs in two steps:
$1$. $BF_3$ reacts with water to form boric acid $(H_3BO_3)$ and hydrogen fluoride $(HF)$:
$BF_3 + 3 H_2O \rightarrow B(OH)_3 + 3 HF$
$2$. The $HF$ produced reacts with excess $BF_3$ to form fluoroboric acid $(HBF_4)$:
$3 HF + 3 BF_3 \rightarrow 3 H[BF_4]$
Combining these,the overall reaction is:
$4 BF_3 + 3 H_2O \rightarrow B(OH)_3 + 3 H[BF_4]$
Thus,both $H_3BO_3$ and $HBF_4$ are formed.

(D) In diborane $(B_2H_6)$,the two boron atoms are not directly bonded to each other.
Instead,they are linked through two bridging hydrogen atoms,forming two $3c-2e$ (three-center two-electron) bonds,also known as banana bonds.
Each boron atom is also bonded to two terminal hydrogen atoms via normal $2c-2e$ covalent bonds.
Therefore,the correct statement is that boron atoms are linked through hydrogen bridges.

(B) The Lewis acid strength of boron trihalides $(BX_3)$ is determined by the extent of $p\pi-p\pi$ back-bonding between the halogen atom and the boron atom.
In $BF_3$,the fluorine atom has a small size and its $2p$ orbitals effectively overlap with the empty $2p$ orbital of boron,leading to strong back-bonding.
This reduces the electron deficiency of boron,making $BF_3$ the weakest Lewis acid.
As the size of the halogen increases from $F$ to $I$,the effectiveness of $p\pi-p\pi$ back-bonding decreases $(F > Cl > Br > I)$.
Therefore,the electron deficiency of boron increases,and the Lewis acid strength increases in the order: $BF_3 < BCl_3 < BBr_3 < BI_3$.

(A) $(I)$ The Lewis acidity of boron trihalides is governed by the extent of $p\pi-p\pi$ back-bonding between the halogen atom and the boron atom.
$(II)$ In $BF_3$,the small size of the fluorine atom allows for effective $2p-2p$ back-bonding,which significantly reduces the electron deficiency of the boron atom.
$(III)$ As the size of the halogen atom increases from $F$ to $I$,the effectiveness of back-bonding decreases,making the boron atom more electron-deficient and thus more acidic.
$(IV)$ Therefore,the correct order of Lewis acidic strength is $BF_3 < BCl_3 < BBr_3 < BI_3$.

(C) In $H_2S_2O_8$ (peroxodisulfuric acid),there is a peroxy linkage $(-O-O-)$ between the two sulfur atoms. The structure is $HO-SO_2-O-O-SO_2-OH$. Therefore,it does not contain an $S-S$ linkage.

(A) When $AlCl_3 \cdot 6H_2O$ is heated strongly,it undergoes hydrolysis and dehydration to form aluminum oxide $(Al_2O_3)$.
The reaction is: $2(AlCl_3 \cdot 6H_2O) \xrightarrow{\Delta} Al_2O_3 + 6HCl + 9H_2O$.

(A) $(I)$ The Lewis acidity of boron trihalides $(BX_3)$ is governed by the extent of $p\pi-p\pi$ back-bonding between the lone pair of the halogen atom and the vacant $p$-orbital of the boron atom.
$(II)$ As the size of the halogen atom increases from $F$ to $I$,the effectiveness of the $p\pi-p\pi$ back-bonding decreases because the overlap between the $2p$ orbital of boron and the $np$ orbital of the halogen becomes less efficient.
$(III)$ Consequently,$BF_3$ has the strongest back-bonding,making it the weakest Lewis acid,while $BI_3$ has the weakest back-bonding,making it the strongest Lewis acid.
$(IV)$ Therefore,the correct order of Lewis acidity is $BF_3 < BCl_3 < BBr_3 < BI_3$.

(D) $H_3BO_3$ (orthoboric acid) is sparingly soluble in cold water but becomes more soluble in hot water.
It acts as a weak monobasic Lewis acid because it accepts an $OH^-$ ion from water: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$.
It is formed when borax reacts with a mineral acid: $Na_2B_4O_7 + 2HCl + 5H_2O \rightarrow 2NaCl + 4H_3BO_3$.

(B) Step $1$: $Na_2B_4O_7 + 2HCl + 5H_2O \rightarrow 2NaCl + 4H_3BO_3$. Thus,$X$ is an acid (like $HCl$ or $H_2SO_4$).
Step $2$: $2H_3BO_3 \xrightarrow{\Delta} B_2O_3 + 3H_2O$.
Step $3$: $B_2O_3 + 2Al \xrightarrow{\Delta} 2B + Al_2O_3$. Thus,$Y$ is Aluminium $(Al)$,which acts as a reducing agent to extract Boron from its oxide.

(D) monobasic acid is an acid that releases only one $H^+$ ion per molecule in an aqueous solution.
$H_3PO_2$ (hypophosphorous acid) is a monobasic acid because it has only one $P-OH$ bond.
$H_3BO_3$ (boric acid) acts as a monobasic Lewis acid by accepting $OH^-$ from water $(B(OH)_3 + H_2O \rightleftharpoons [B(OH)_4]^- + H^+)$.
$HCl$ is a strong monobasic acid.
$H_3PO_3$ (phosphorous acid) is a dibasic acid because it contains two $P-OH$ bonds,allowing it to release two $H^+$ ions per molecule.