Boron family Questions in English

(N/A) In $BF_3$,the lone pair of electrons on the fluorine atom is donated to the vacant $p$-orbital of the boron atom through $p\pi-p\pi$ back bonding. This delocalization of electrons reduces the electron deficiency of the boron atom,thereby decreasing its Lewis acidic character and increasing the stability of $BF_3$.
In $BH_3$,there is no lone pair of electrons on the $H$ atom to participate in back bonding. Therefore,$BH_3$ dimerizes to form $B_2H_6$ (diborane) to achieve stability.
Structure of $B_2H_6$: In diborane,the four terminal $H$ atoms and two $B$ atoms lie in the same plane. The two bridging $H$ atoms lie above and below this plane. Each boron atom is $sp^3$ hybridized. The bridging $H$ atoms are involved in $3$-center-$2$-electron $(3c-2e)$ bonds,also known as banana bonds.

(N/A) . Silicones are a group of organosilicon polymers,which have $(R_{2}SiO)$ as a repeating unit. The starting materials for the manufacture of silicones are alkyl or aryl substituted silicon chlorides,$R_{n}SiCl_{(4-n)}$,where $R$ is an alkyl or aryl group.
When methyl chloride reacts with silicon in the presence of copper as a catalyst at a temperature of $570 \ K$,various types of methyl substituted chlorosilanes of formula $MeSiCl_{3}$,$Me_{2}SiCl_{2}$,and $Me_{3}SiCl$ with a small amount of $Me_{4}Si$ are formed.
Hydrolysis of dimethyl dichlorosilane,$(CH_{3})_{2}SiCl_{2}$,followed by condensation polymerization yields straight-chain polymers.
Uses: They are used as sealants,greases,and electrical insulators,and for the waterproofing of fabrics. Being biocompatible,they are also used in surgical and cosmetic implants.
$B$. Boranes are the binary compounds of boron and hydrogen,similar to alkanes. These are covalent hydrides with the formula $B_{2}H_{6}$,called diborane.
Preparation of diborane: It is prepared by treating boron trifluoride with $LiAlH_{4}$ in diethyl ether.
$4BF_{3} + 3LiAlH_{4} \rightarrow 2B_{2}H_{6} + 3LiF + 3AlF_{3}$
$B_{2}H_{6}$ is produced on an industrial scale by the reaction of $BF_{3}$ with $NaH$ (sodium hydride):
$2BF_{3} + 6NaH \xrightarrow{450 \ K} B_{2}H_{6} + 6NaF$

(N/A) Compound $(A)$ of boron reacts with $NMe_3$ to form an adduct $(B)$,indicating that $(A)$ is a Lewis acid. Since $(B)$ on hydrolysis yields an acid $(C)$ and $H_2$ gas,$(A)$ is identified as $B_2H_6$,$(B)$ is the adduct $BH_3 \cdot NMe_3$,and $(C)$ is boric acid $(H_3BO_3)$.
The reactions are as follows:
$B_2H_6 + 2NMe_3 \rightarrow 2BH_3 \cdot NMe_3$
$(A) \text{ (Diborane)} + \text{Reactant} \rightarrow (B) \text{ (Adduct)}$
$BH_3 \cdot NMe_3 + 3H_2O \rightarrow H_3BO_3 + NMe_3 + 3H_2$
$(B) + \text{Hydrolysis} \rightarrow (C) \text{ (Boric acid)} + \text{Byproducts}$

(N/A) The element is Boron $(B)$.
Reaction: $BF_3 + NH_3 \rightarrow F_3B \leftarrow NH_3$
Explanation: In $BF_3$,the boron atom has only $6$ electrons in its valence shell (incomplete octet). Therefore,it is electron-deficient and acts as a Lewis acid by accepting a lone pair of electrons from the Lewis base,ammonia $(NH_3)$.

(D) The chemical formula of pyrophosphoric acid is $H_4P_2O_7$.
By observing its structure:
$1$. There are $4$ $P-OH$ bonds (two on each phosphorus atom).
$2$. There are $2$ $P=O$ bonds (one on each phosphorus atom).
$3$. There is $1$ $P-O-P$ linkage connecting the two phosphorus atoms.
Therefore,the number of $P-OH$,$P=O$,and $P-O-P$ bonds are $4, 2$,and $1$ respectively.

(B) $B_2O_3$ and $SiO_2$ are acidic oxides.
Generally,oxides of non-metals are acidic in nature.
$CaO$ and $BaO$ are basic oxides (metal oxides),and $N_2O$ is a neutral oxide.

(B) $B(OH)_{3}$ is a weak Lewis acid,while $H_{3}PO_{3}$ is a dibasic acid. Both are acidic in nature.
$B(OH)_{3}$ is acidic in nature,whereas $Al(OH)_{3}$ is amphoteric in nature.
$NaOH$ and $Ca(OH)_{2}$ are both basic in nature.
$Be(OH)_{2}$ and $Al(OH)_{3}$ are both amphoteric in nature.
Therefore,the set with different natures is $B(OH)_{3}$ and $Al(OH)_{3}$.

(C) Statement $(a)$ is correct: Diborane is prepared by the reaction of $NaBH_{4}$ with $I_{2}$ as follows: $2 NaBH_{4} + I_{2} \rightarrow B_{2}H_{6} + 2 NaI + H_{2}$.
Statement $(b)$ is incorrect: In diborane,each boron atom is $sp^{3}$ hybridized.
Statement $(c)$ is incorrect: Diborane has two bridged $3$ centre$-2$ electron $(3c-2e^-)$ bonds,not one.
Statement $(d)$ is incorrect: Diborane is a non-planar molecule.
Therefore,only statement $(a)$ is correct.

(C) In diborane $(B_2H_6)$,each boron atom undergoes $sp^3$ hybridization,not $sp^2$.
Diborane has a non-planar structure where the two bridging hydrogen atoms lie in a plane perpendicular to the plane containing the four terminal hydrogen atoms and the two boron atoms.
Therefore,the statement that both boron atoms are $sp^2$ hybridized is incorrect.

(C) Analysis of statements:
$(A)$ Incorrect: $B_2H_6$ has two types of $B-H$ bonds: terminal $B-H$ bonds (covalent) and bridging $B-H-B$ bonds ($3$-centre-$2$-electron).
$(B)$ Incorrect: $B_2H_6$ contains two $3$-centre-$2$-electron bonds,not four.
$(C)$ Correct: $B_2H_6$ is an electron-deficient molecule and acts as a Lewis acid.
$(D)$ Correct: $B_2H_6$ can be synthesized by the reaction: $3NaBH_4 + 4BF_3 \rightarrow 2B_2H_6 + 3NaBF_4$.
$(E)$ Incorrect: $B_2H_6$ is a non-planar molecule.
Therefore,statements $(C)$ and $(D)$ are correct.

(D) Borax $(Na_2B_4O_7 \cdot 10H_2O)$ dissolves in water to form orthoboric acid $(H_3BO_3)$ and sodium hydroxide $(NaOH)$.
$Na_2B_4O_7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$.
Since $NaOH$ is a strong base and $H_3BO_3$ is a very weak acid,the resulting aqueous solution is strongly basic.

(A) $Cs$ (Caesium) is used in photoelectric cells due to its low ionization energy.
$Ga$ (Gallium) has a high boiling point and is used in high temperature thermometers.
$B$ (Boron) fibres are used in making bullet-proof vests.
$Si$ (Silicon) based silicones are used in water repellent sprays due to their non-polar alkyl groups.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(B) Neutral oxides are: $NO$,$N_2O$,and $CO$.
Acidic oxides are: $B_2O_3$,$N_2O_5$,$SO_3$,and $P_4O_{10}$.
Therefore,the total number of acidic oxides is $4$.

(B) The given reactions are:
$2BF_{3} + 6NaH \xrightarrow{450 \ K} B_{2}H_{6} + 6NaF$
$B_{2}H_{6} + 2NMe_{3} \rightarrow 2[BH_{3} \leftarrow NMe_{3}]$
Here,$A$ is $B_{2}H_{6}$ (diborane) and $B$ is the adduct $[BH_{3} \leftarrow NMe_{3}]$.
In the adduct $[BH_{3} \leftarrow NMe_{3}]$,the boron atom is bonded to three hydrogen atoms and one nitrogen atom.
The boron atom is $sp^{3}$ hybridized and possesses a tetrahedral geometry.

(A) Boric acid ($H_3BO_3$ or $B(OH)_3$) is a weak monobasic Lewis acid.
It does not act as a proton donor (Brønsted acid) by releasing $H^{+}$ ions directly.
Instead,it acts as a Lewis acid by accepting a lone pair of electrons from the $OH^{-}$ ion of a water molecule.
The reaction is: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$.
This process releases $H^{+}$ ions (as $H_3O^+$) into the solution,which is why it exhibits acidic properties.
Therefore,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation of $A$.

(B) The preparation of borazine $(B_3N_3H_6)$ involves the reaction of diborane $(B_2H_6)$ with ammonia $(NH_3)$ at high temperatures.
The balanced chemical equation is:
$3 B_2H_6 + 6 NH_3 \xrightarrow{\Delta} 2 B_3N_3H_6 + 12 H_2$
Here,$X$ is $B_2H_6$ and $Y$ is $NH_3$.

(B) Assertion $A$ is true: Boron cannot form $BF_{6}^{3-}$ because it lacks $d$-orbitals in its valence shell,which prevents the expansion of its octet.
Reason $R$ is true: The size of the Boron atom is indeed very small.
Conclusion: While both statements are true,the inability to form $BF_{6}^{3-}$ is primarily due to the absence of $d$-orbitals,not just the small size of the atom. Therefore,$R$ is not the correct explanation for $A$.

(B) The borax bead test involves the following reactions:
$Na_{2}B_{4}O_{7} \cdot 10 H_{2}O \xrightarrow{\Delta} Na_{2}B_{4}O_{7} + 10 H_{2}O$
$Na_{2}B_{4}O_{7} \xrightarrow{\Delta} 2 NaBO_{2} + B_{2}O_{3}$
$B_{2}O_{3} + CoO \rightarrow Co(BO_{2})_{2} \text{ (cobalt (II) metaborate)}$
The blue coloured bead is due to the formation of cobalt $(II)$ metaborate,$Co(BO_{2})_{2}$.

(B) The reaction of sodium borohydride with iodine produces diborane,which is a Lewis acid $(X = B_2H_6)$.
$2 NaBH_4 + I_2 \xrightarrow{\text{Diglyme}} B_2H_6(X) + H_2 + 2 NaI$
Heating diborane with ammonia in a $1:2$ molar ratio produces borazine,which is a cyclic compound $(Y = B_3N_3H_6)$ and hydrogen gas $(Z = H_2)$.
$3 B_2H_6 + 6 NH_3 \xrightarrow{\Delta} 2 B_3N_3H_6(Y) + 12 H_2(Z) \uparrow$

(D) When a mixture of diborane $(B_2H_6)$ and ammonia $(NH_3)$ is heated at high temperatures,the final product is borazine $(B_3N_3H_6)$,which is also known as inorganic benzene.
The reaction is given by:
$3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 12H_2$
Borazine is structurally similar to benzene and exhibits similar chemical properties.

(C) The given boron trihalides are electron-deficient molecules that act as Lewis acids by accepting an electron pair.
The Lewis acid strength is determined by the extent of $ppi-ppi$ back-bonding between the halogen atom and the boron atom.
In $BF_3$,the $2p-2p$ back-bonding is most effective due to the similar size of the orbitals,which reduces the electron deficiency of the boron atom significantly.
As the size of the halogen increases from $F$ to $Cl$ to $Br$,the effectiveness of back-bonding decreases $(2p-2p > 2p-3p > 2p-4p)$.
Therefore,the electron deficiency on boron increases in the order $BF_3 < BCl_3 < BBr_3$,making $BBr_3$ the strongest Lewis acid.

(C) $Al$ is the correct answer.
Amphoteric oxides are those oxides that exhibit both acidic and basic properties.
Among the given elements,$Al$ reacts with oxygen to form $Al_2O_3$,which is amphoteric.
$N_2(g) + O_2(g) \longrightarrow 2NO_2(g)$ (Acidic oxide)
$P_4(s) + 5O_2(g) \longrightarrow P_4O_{10}(s)$ (Acidic oxide)
$4Al(s) + 3O_2(g) \longrightarrow 2Al_2O_3(s)$ (Amphoteric oxide)
$2Na(s) + O_2(g) \longrightarrow Na_2O_2(s)$ (Basic oxide)

(B) The hydrolysis of $BCl_3$ results in the formation of boric acid $(X)$.
$BCl_3 + 3H_2O \longrightarrow H_3BO_3 (X) + 3HCl$
When boric acid $(X)$ reacts with sodium carbonate $(Na_2CO_3)$,it produces sodium tetraborate $(Y)$,carbon dioxide,and water.
$4H_3BO_3 + Na_2CO_3 \longrightarrow Na_2B_4O_7 (Y) + CO_2 + 6H_2O$
Therefore,$X$ is $H_3BO_3$ and $Y$ is $Na_2B_4O_7$.

(A) The chemical formula of borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$.
The anionic part of borax is the tetranuclear unit $[B_4O_5(OH)_4]^{2-}$.
In this structure,there are $4$ boron atoms.
Looking at the structure,there are $5$ oxygen atoms bridging the boron atoms,forming $5$ $B-O-B$ linkages.
Therefore,the number of boron atoms is $4$ and the number of $B-O-B$ units is $5$.

(D)
$Al_2O_3$ is an amphoteric oxide (those oxides which show both the properties of acids and bases),so it can react both with acids and alkalis,e.g.
$Al_2O_3 + 6HCl \longrightarrow 2AlCl_3 + 3H_2O$
$Al_2O_3 + 2NaOH + 3H_2O \longrightarrow 2Na[Al(OH)_4]$

(D) The structures of the given oxoacids of phosphorus are as follows:
$H_3PO_2$ (Hypophosphorous acid): Contains $2$ $P-H$ bonds.
$H_3PO_3$ (Phosphorous acid): Contains $1$ $P-H$ bond.
$H_3PO_4$ (Orthophosphoric acid): Contains $0$ $P-H$ bonds.
Thus,the number of $P-H$ bonds in $H_3PO_2, H_3PO_3$ and $H_3PO_4$ are $2, 1$ and $0$ respectively.
Therefore,the correct option is $D$.

(A) The acidity of oxoacids of phosphorus depends on the number of $OH$ groups directly attached to the phosphorus atom. Protons attached directly to the phosphorus atom are not acidic.
$1$. In $H_3PO_2$ (hypophosphorous acid),there is one $P-OH$ group,so it is monobasic (number of acidic protons = $1$).
$2$. In $H_3PO_3$ (phosphorous acid),there are two $P-OH$ groups,so it is dibasic (number of acidic protons = $2$).
$3$. In $H_3PO_4$ (phosphoric acid),there are three $P-OH$ groups,so it is tribasic (number of acidic protons = $3$).
Therefore,the number of acidic protons are $1$,$2$ and $3$ respectively. The correct option is $A$.

(C) Boric acid $(H_3BO_3)$ molecules are linked together by strong intermolecular hydrogen bonding,which results in a layered solid structure at room temperature.
In contrast,$BF_3$ is a monomeric covalent molecule with weak van der Waals forces of attraction between its molecules,making it a gas at room temperature.

(D) The Lewis acidity of boron trihalides is determined by the extent of $p\pi-p\pi$ back bonding between the halogen atom and the boron atom.
In $BF_3$,the $2p-2p$ overlap is most effective,resulting in the strongest back bonding,which reduces the electron deficiency of the boron atom.
As we move from $BF_3$ to $BI_3$,the size of the halogen atom increases,making the $p\pi-p\pi$ overlap less effective $(2p-2p > 2p-3p > 2p-4p > 2p-5p)$.
Consequently,the extent of back bonding decreases,and the Lewis acidity increases in the order: $BF_3 < BCl_3 < BBr_3 < BI_3$.

(A) The chemical formulas for the given compounds are as follows:
$a$. Borax: $Na_2B_4O_7 \cdot 10H_2O$
$b$. Kernite: $Na_2B_4O_7 \cdot 4H_2O$
$c$. Orthoboric acid: $H_3BO_3$
$d$. Borax bead: $NaBO_2$
Therefore,the correct matching is $a-iv, b-ii, c-iii, d-i$.

(B) When borax $(Na_2B_4O_7 \cdot 10H_2O)$ is heated,it first loses its water of crystallization and then melts to form a transparent liquid which solidifies into a glass-like mass consisting of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
The balanced chemical equation is:
$Na_2B_4O_7 \stackrel{\Delta}{\longrightarrow} 2NaBO_2 + B_2O_3$
Thus,the product $X$ is $B_2O_3$.

(C) The basic character of oxides of Group-$13$ elements increases down the group as the metallic or electropositive character of the elements increases.
The trend in the nature of oxides is as follows:
$B_2O_3$ (acidic) < $Al_2O_3$ (amphoteric) < $Ga_2O_3$ (amphoteric) < $In_2O_3$ (basic) < $Tl_2O_3$ (most basic).
Therefore,$Tl_2O_3$ is the most basic oxide.

(B) Boron is non-metallic in nature. It is an extremely hard,black-colored solid that exists in many allotropic forms.
Due to its very strong crystalline lattice,boron has an unusually high melting point and boiling point compared to other members of Group $13$.

Element	$B$	$Al$	$Ga$	$In$	$Tl$
Melting point $(K)$	$2453$	$933$	$303$	$430$	$576$
Boiling point $(K)$	$3923$	$2740$	$2676$	$2353$	$1730$

Thus,both statements are correct.

(B) Borazine $(B_{3}N_{3}H_{6})$ is an inorganic benzene analog.
It exhibits electronic delocalization due to the overlap of $p$-orbitals of $B$ and $N$ atoms.
It is a cyclic compound.
It reacts with water to form boric acid,ammonia,and hydrogen gas:
$B_{3}N_{3}H_{6} + 9H_{2}O \rightarrow 3NH_{3} + 3H_{3}BO_{3} + 3H_{2}$.
Banana bonds are characteristic of diborane $(B_{2}H_{6})$,not borazine.
Therefore,the statement that it contains banana bonds is incorrect.

(A) In the $[BF_4]^-$ ion,Boron forms $4$ covalent bonds with $4$ fluorine atoms. Therefore,the covalency of Boron is $4$.
To find the oxidation state of Boron $(x)$:
$x + 4 \times (-1) = -1$
$x - 4 = -1$
$x = +3$
Thus,the covalency is $4$ and the oxidation state is $+3$.

(D) The chemical formula of Borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$.
Comparing this with the given formula $Na_2B_4O_x(OH)_y \cdot zH_2O$,we get $x = 5$,$y = 4$,and $z = 8$.
Therefore,$x + y + z = 5 + 4 + 8 = 17$.

(B) Boron exists as a giant covalent polymer in its solid state,which results in a very strong crystalline lattice structure.
Due to this strong lattice,a large amount of energy is required to break the bonds,leading to an unusually high melting point $(2453 \ K)$ compared to other group $13$ elements.
Therefore,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(A) The chemical formula for Fluorspar is $CaF_2$,not $BF_3$.
Therefore,the pair Fluorspar $- BF_3$ is incorrect.

(D) Statement $I$ is true: Group $13$ trivalent halides (except $AlF_3$) are covalent in nature and undergo hydrolysis in water.
Statement $II$ is true: In acidified aqueous solution,$AlCl_3$ exists as the octahedral $[Al(H_2O)_6]^{3+}$ ion,where $Al$ is $sp^3d^2$ hybridized.
Therefore,both statements are correct.

(D) The more abundant isotope of boron is $^{11}B$.
Number of neutrons in $^{11}B = 11 - 5 = 6$.
So,$x = 6$.
Amorphous boron reacts with air (oxygen) upon heating to form boron trioxide: $4B + 3O_2 \rightarrow 2B_2O_3$.
In $B_2O_3$,the oxidation state of boron is $+3$.
So,$y = 3$.
Therefore,$x + y = 6 + 3 = 9$.

(A) Incorrect. The correct order of atomic radii is $Tl > In > Al > Ga > B$ due to the poor shielding effect of $d$-orbitals in $Ga$.
$(B)$ Incorrect. Electronegativity first decreases from $B$ to $Al$ and then increases slightly due to the poor shielding effect of $d$-electrons.
$(C)$ Correct. $Al$ reacts with dil. $HCl$ to release $H_2$ gas,but conc. $HNO_3$ forms a protective oxide layer $(Al_2O_3)$ on the surface,making it passive.
$(D)$ Incorrect. $B$ and $Al$ are most stable in the $+3$ oxidation state. The stability of the $+1$ oxidation state increases down the group due to the inert pair effect.
$(E)$ Correct. In $[Al(H_2O)_6]^{3+}$,$Al$ is $sp^3d^2$ hybridized (octahedral geometry).
Therefore,statements $(C)$ and $(E)$ are correct.

(D) Statement-$I$ is correct because Gallium $(Ga)$ has a very high boiling point $(2676 \ K)$ and a low melting point $(302.9 \ K)$,making it suitable for high-temperature thermometers.
Statement-$II$ is false because a thermometer containing Gallium is used to measure high temperatures,not low temperatures like the freezing point of brine $(256 \ K)$.
Therefore,Statement-$I$ is true but Statement-$II$ is false.

(A) The stability of the $+1$ oxidation state increases down the group $13$ due to the inert pair effect,which is the reluctance of the $ns^2$ electrons to participate in bonding.
As we move from $Ga$ to $Tl$,the energy required to unpair the $ns^2$ electrons increases,making the $+1$ oxidation state more stable.
Therefore,the stability order is $Ga^{+1} < In^{+1} < Tl^{+1}$.
Both Assertion $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(B) Statement $(A)$ is correct: Non-metals have higher electronegativity compared to metals due to their smaller size and higher effective nuclear charge.
Statement $(B)$ is incorrect: Non-metals have higher ionisation enthalpy than metals because they hold their valence electrons more tightly.
Statement $(C)$ is correct: Highly reactive metals (low electronegativity) and highly reactive non-metals (high electronegativity) form ionic compounds due to a large electronegativity difference.
Statement $(D)$ is incorrect: Non-metal oxides are generally acidic in nature.
Statement $(E)$ is incorrect: Metal oxides are generally basic in nature.
Therefore,statements $(A)$ and $(C)$ are correct.

(C) The correct matches are as follows:
$A$. Melting point $[K]$: The order is $B > Al > Tl > In > Ga$,which corresponds to $IV$.
$B$. Ionic Radius $[M^{+3} / pm]$: The order is $Tl > In > Ga > Al > B$,which corresponds to $I$.
$C$. $\Delta_{i} H_1 [kJ \ mol^{-1}]$: The order is $B > Tl > Al \approx Ga > In$,which corresponds to $II$.
$D$. Atomic Radius $[pm]$: The order is $Tl > In > Al > Ga > B$,which corresponds to $III$.
Therefore,the correct sequence is $A-IV, B-I, C-II, D-III$.

(A) The reaction $B(OH)_3 + NaOH \rightleftharpoons Na[B(OH)_4]$ is an equilibrium reaction.
By adding a $cis-1,2-diol$ (like glycerol or mannitol),a stable chelated complex is formed with the borate ion $[B(OH)_4]^-$.
This removal of the product $[B(OH)_4]^-$ from the equilibrium mixture shifts the reaction in the forward direction according to Le Chatelier's principle.

(A) . $Bi^{3+} + H_2O \longrightarrow BiO^+ + 2H^+$. This is a hydrolysis reaction.
$B$. $[AlO_2]^- + H_2O + H^+ \longrightarrow Al(OH)_3$. This is acidification of aluminate.
$C$. $2SiO_4^{4-} + 2H^+ \longrightarrow Si_2O_7^{6-} + H_2O$. This is a condensation reaction involving heat (pyrolysis).
$D$. $B_4O_7^{2-} + 2H^+ + 5H_2O \longrightarrow 4B(OH)_3$. This involves both acidification and hydrolysis.
Therefore,the correct matching is $A-Q, B-R, C-P, D-Q, R$.

(C) $H_3BO_3$ (orthoboric acid) is a weak Lewis acid.
It reacts with water as follows:
$H_3BO_3 + H_2O \rightleftharpoons [B(OH)_4]^- + H^+$
It does not act as a proton donor (Brønsted-Lowry acid); instead,it accepts an $OH^-$ ion from water molecules to release a proton.
Therefore,$Statement-1$ is True,but $Statement-2$ is False.
Hence,the correct option is $(C)$.

(D) The correct formula of borax is $Na_2\left[B_4O_5(OH)_4\right] \cdot 8H_2O$.
$1$. Borax contains a tetranuclear unit $\left[B_4O_5(OH)_4\right]^{2-}$,which consists of two $sp^3$ hybridized boron atoms and two $sp^2$ hybridized boron atoms.
$2$. Since there are two $sp^3$ and two $sp^2$ boron atoms,the number of $sp^2$ and $sp^3$ hybridized boron atoms is equal.
$3$. In the structure,each of the four boron atoms is bonded to one terminal hydroxide $(-OH)$ group.
$4$. The boron atoms are not all in the same plane due to the tetrahedral geometry of the $sp^3$ hybridized boron atoms.
Therefore,statements $A$,$C$,and $D$ are correct.

(A) $H_3BO_3$ does not undergo self-ionization. It acts as a weak acid in water by accepting an $OH^-$ ion from water,releasing $H^+$ ions,thus it is a weak electrolyte.
$H_3BO_3 + H_2O \rightarrow [B(OH)_4]^- + H^+$
On adding cis-diols like ethylene glycol,they form stable chelated complexing species with the borate ion,which shifts the equilibrium to the right,thereby increasing the acidity.
$ [B(OH)_4]^- + 2 \text{ ethylene glycol} \rightarrow [B(\text{glycol})_2]^- + 4H_2O $
Orthoboric acid has a layered,two-dimensional structure due to intermolecular hydrogen bonding between the planar $BO_3^{3-}$ units. Therefore,statement $(C)$ is incorrect.