Boron family Questions in English

(D) All three reactions are standard laboratory or industrial methods for the preparation of diborane $(B_2H_6)$:
$I.$ The reaction of sodium borohydride with boron trifluoride in ether is a common laboratory method: $3NaBH_4 + 4BF_3 \to 2B_2H_6 + 3NaBF_4$.
$II.$ The reaction of sodium borohydride with iodine is another method: $2NaBH_4 + I_2 \to B_2H_6 + 2NaI + H_2$.
$III.$ The reaction of boron trifluoride with sodium hydride at $450 \ K$ is an industrial method: $2BF_3 + 6NaH \to B_2H_6 + 6NaF$.
Therefore,all three reactions are correct.

(C) The thermal decomposition of borax $(Na_2B_4O_7 \cdot 10H_2O)$ results in the formation of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
Thus,$(A) = B_2O_3$.
When boric anhydride $(B_2O_3)$ reacts with metal oxides like $MnO$ upon heating,it forms metal metaborates,which are often colored beads.
$B_2O_3 + MnO \xrightarrow{\Delta} Mn(BO_2)_2$.
Thus,$(B) = Mn(BO_2)_2$.
Therefore,$(A)$ is $B_2O_3$ and $(B)$ is $Mn(BO_2)_2$.

(D) $1$. The basicity of phosphorous acid $(H_3PO_3)$ is $2$,not $3$,because it has two $P-OH$ bonds and one $P-H$ bond.
$2$. Perchloric acid $(HClO_4)$ has the structure $Cl(=O)_3(OH)$ and contains no peroxy linkage.
$3$. Solid $SO_3$ exists as a cyclic trimer $(SO_3)_3$ with a ring structure.
$4$. Borazine $(B_3N_3H_6)$,often called inorganic benzene,has a planar hexagonal structure similar to benzene. Due to its high symmetry,its net dipole moment is $0$.

(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.
As the size of the halogen increases from $F$ to $I$,the effectiveness of $p\pi-p\pi$ back-bonding decreases $(F > Cl > Br > I)$.
Since back-bonding reduces the electron deficiency of the boron atom,the Lewis acidity increases as the back-bonding decreases.
Therefore,the correct order of Lewis acidity is $BF_3 < BCl_3 < BBr_3 < BI_3$.

(D) Borax $(Na_2B_4O_7 \cdot 10H_2O)$ undergoes hydrolysis in water to form $NaOH$ (a strong base) and $H_3BO_3$ (a weak acid). The presence of $NaOH$ makes the solution alkaline.
$(i)$ Borax: $Na_2B_4O_7 + 7H_2O \longrightarrow 2NaOH + 4H_3BO_3$ (Alkaline).
$(ii)$ Potash alum $(K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O)$ is acidic in water due to the hydrolysis of $Al^{3+}$ ions.
$(iii)$ $SO_2$ reacts with water to form sulfurous acid $(H_2SO_3)$,which is acidic.
$(iv)$ $SO_2(OH)_2$ is sulfuric acid $(H_2SO_4)$,which is strongly acidic.
Therefore,only $(i)$ produces an alkaline solution.

(B) In Group $13$,the stability of the $+3$ oxidation state decreases down the group due to the inert pair effect.
As we move from $B$ to $Tl$,the $ns^2$ electrons become increasingly reluctant to participate in bonding.
Therefore,the stability of the $+3$ oxidation state follows the order: $B^{3+} > Al^{3+} > Ga^{3+} > In^{3+} > Tl^{3+}$.

(A) The hydrolysis of boron trichloride $(BCl_3)$ with water proceeds as follows:
$BCl_3 + 3H_2O \longrightarrow H_3BO_3 + 3HCl$
Thus,the products formed are boric acid $(H_3BO_3)$ and hydrochloric acid $(HCl)$.

(D) The basicity of phosphorous acid $(H_3PO_3)$ is $2$, as it contains two $P-OH$ bonds.
Perchloric acid $(HClO_4)$ has no peroxy linkage.
Solid $SO_3$ exists as a cyclic trimer $(S_3O_9)$.
Borazine $(B_3N_3H_6)$ is a planar molecule with a structure similar to benzene, resulting in a net dipole moment of zero $(\mu = 0)$.

(B) An alum is a double salt with the general formula $M_2SO_4 \cdot M'_2(SO_4)_3 \cdot 24H_2O$,where $M$ is a monovalent cation (like $K^+$,$Na^+$,$NH_4^+$) and $M'$ is a trivalent cation (like $Al^{3+}$,$Cr^{3+}$,$Fe^{3+}$).
$Li^+$ is too small to form a stable alum structure.
Therefore,$K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$ (Potash Alum) is a true alum,while the lithium compound is not.

(C) Boric acid $(H_3BO_3)$ acts as a Lewis acid because it accepts a lone pair of electrons from the $OH^-$ ion of water to form $[B(OH)_4]^-$.
It does not dissociate to give $H^+$ ions directly in water,so it is not an Arrhenius acid.
It does not donate a proton $(H^+)$ to water,so it is not a Bronsted acid.
Therefore,it is classified as a Lewis acid.

(C) The acidic or basic nature of hydroxides depends on the ionization energy and the electronegativity of the central atom.
$B(OH)_3$ (also written as $H_3BO_3$) is orthoboric acid.
It acts as a weak Lewis acid because it accepts a lone pair of electrons from $OH^-$ ions in water,releasing $H^+$ ions:
$B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^- + H_3O^+$.
In contrast,$Be(OH)_2$,$Mg(OH)_2$,and $Al(OH)_3$ are amphoteric or basic in nature.

(B) Gallium $(Ga)$ has a very high boiling point $(2403 \ K)$ and a low melting point $(303 \ K)$.
Because of its wide liquid range,it is used in high-temperature thermometers to measure temperatures up to $1000 \ ^\circ C$.

(C) Boron has an electronic configuration of $1s^2 2s^2 2p^1$. In its compounds,boron forms three covalent bonds,resulting in only $6$ electrons in its valence shell. Since it does not complete its octet,it is electron-deficient and acts as a Lewis acid by accepting a lone pair of electrons.

(A) Boron compounds,when heated in a Bunsen burner flame,impart a characteristic green color to the flame. Trimethyl borate,$B(OMe)_3$,is a volatile boron compound that burns with a green-edged flame. This is a standard test for the presence of borates.

(C) Borazole,also known as inorganic benzene,has the chemical formula $B_3N_3H_6$.
It is isostructural with benzene $(C_6H_6)$ and is formed by the reaction of diborane with ammonia.

(C) The structure of diborane $(B_2H_6)$ consists of four terminal hydrogen atoms and two bridging hydrogen atoms.
Terminal hydrogen atoms and boron atoms lie in the same plane.
It contains two $3$-center-$2$-electron $(3c-2e)$ bonds,often called banana bonds,and four $2$-center-$2$-electron $(2c-2e)$ bonds.
It does $NOT$ contain a four-center bond.
Therefore,the statement in option $C$ is incorrect.

(C) Elements of group $13$ (Boron family) have a general valence shell electronic configuration of $ns^2 np^1$.
They typically exhibit an oxidation state of $+3$ in their compounds.
When these elements react with oxygen,they form oxides with the general formula ${R_2}{O_3}$ (where $R$ is the group $13$ element).
For example,Boron forms $B_2O_3$ and Aluminum forms $Al_2O_3$.

(B) Potash alum,$KAl(SO_4)_2 \cdot 12H_2O$,is a double salt.
When dissolved in water,it undergoes hydrolysis to form $Al(OH)_3$ (a weak base) and $H_2SO_4$ (a strong acid).
Due to the presence of a strong acid,the resulting aqueous solution is acidic in nature,not basic.
Therefore,the statement that its aqueous solution is basic is incorrect.

(D) The reaction of aluminum with metal oxides is highly exothermic,releasing a large amount of energy. This process is known as the thermite reaction. $2Al + Fe_2O_3 \rightarrow Al_2O_3 + 2Fe + \text{Heat}$. This heat is sufficient to melt the metal,making it useful in thermite welding for joining railway tracks.

(C) $BF_3$ does not exist as a dimer because the fluorine atoms are small and cannot effectively bridge the boron atoms due to steric hindrance and the lack of vacant orbitals in fluorine to form a stable bridge. $BCl_3$,$BBr_3$,and $BI_3$ are stronger Lewis acids than $BF_3$ due to the back-bonding effect in $BF_3$. Ammonal is indeed a mixture of aluminum powder and ammonium nitrate used as an explosive. $BF_3$ is a covalent molecule and does not ionize,hence it does not conduct electricity in the liquid state.

(A) Aluminium chloride $(AlCl_3)$ is an electron-deficient compound with only $6$ electrons in the valence shell of the $Al$ atom.
To achieve a stable octet configuration,$AlCl_3$ undergoes dimerization to form $Al_2Cl_6$.
In this dimer,each $Al$ atom achieves a coordination number of $4$ by forming coordinate bonds with chlorine atoms,thereby completing its octet.

(C) Aluminum is an amphoteric metal. It reacts with both acids and bases to liberate hydrogen gas.
$1$. Reaction with $HCl$: $2Al(s) + 6HCl(aq) \rightarrow 2AlCl_3(aq) + 3H_2(g)$
$2$. Reaction with $NaOH$: $2Al(s) + 2NaOH(aq) + 6H_2O(l) \rightarrow 2Na[Al(OH)_4](aq) + 3H_2(g)$
In both cases,$H_2$ gas is evolved.

(D) $1$. The reaction of $Al$ with aqueous $NaOH$ produces sodium aluminate $(NaAlO_2)$ and releases $H_2$ gas: $2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2$. Thus,$X = Al$ and $Y = NaAlO_2$.
$2$. When $CO_2$ is passed through the aqueous solution of $NaAlO_2$ $(Y)$,it forms aluminum hydroxide $(Al(OH)_3)$ as a precipitate: $2NaAlO_2 + 3H_2O + CO_2 \rightarrow 2Al(OH)_3 \downarrow + Na_2CO_3$. Here,$Z = Al(OH)_3$.
$3$. Heating $Al(OH)_3$ $(Z)$ at $1200 \ ^\circ C$ results in the formation of alumina $(Al_2O_3)$: $2Al(OH)_3 \xrightarrow{1200 \ ^\circ C} Al_2O_3 + 3H_2O$.
$4$. Therefore,the correct sequence is $X = Al, Y = NaAlO_2, Z = Al(OH)_3$.

(A) Diborane $(B_2H_6)$ has a bridged structure where two boron atoms are linked by two hydrogen bridges $(B-H-B)$.
In this structure,there are two types of hydrogen atoms: terminal and bridged.
The terminal $H-B-H$ bond angle is approximately $120^o$.
The bridged $H-B-H$ bond angle is approximately $95^o$.
Therefore,the bond angles are $95^o$ and $120^o$.

(A) Aluminum $(Al)$ reacts with atmospheric oxygen to form a thin,non-porous,and stable layer of aluminum oxide $(Al_2O_3)$ on its surface.
This oxide layer acts as a protective barrier,preventing further oxidation or corrosion of the underlying metal.
This phenomenon is known as passivation.

(B) The hydrolysis of $AlCl_3$ occurs because $Al^{3+}$ is a small,highly charged cation that polarizes the $O-H$ bond in water molecules.
The reaction is given by: $AlCl_3 + 3H_2O \rightarrow Al(OH)_3 + 3HCl$.
Thus,$Al(OH)_3$ (aluminum hydroxide) is produced.

(B) Aluminum hydroxide $(Al(OH)_3)$ is amphoteric in nature. When it reacts with a strong base like caustic soda $(NaOH)$,it forms a soluble complex known as sodium tetrahydroxoaluminate$(III)$.
The chemical equation is: $Al(OH)_3 + NaOH \rightarrow Na[Al(OH)_4]$.

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

(A) Orthoboric acid $(H_3BO_3)$ on heating at $370 \ K$ loses a water molecule to form meta-boric acid $(HBO_2)$.
The reaction is: $H_3BO_3 \xrightarrow{\Delta, 370 \ K} HBO_2 + H_2O$.

(D) Orthoboric acid $(H_3BO_3)$ consists of layers of $BO_3^{3-}$ units.
In these units,the boron atom is $sp^2$ hybridized,which results in a planar triangular geometry.
These units are linked together by hydrogen bonds in a two-dimensional sheet structure.

(D) $1$. Boron carbide $(B_4C)$ is extremely hard and is used as an abrasive.
$2$. Boron is added to steel to increase its hardness and strength.
$3$. Boron sesquioxide $(B_2O_3)$ is a key component in the production of borosilicate glass (e.g.,Pyrex).
$4$. Orthoboric acid $(H_3BO_3)$ consists of layers of $BO_3^{3-}$ units held together by intermolecular hydrogen bonding,not intramolecular. Therefore,the statement claiming it has intramolecular $H$-bonding is incorrect.

(B) Diborane $(B_2H_6)$ is prepared by the reaction of magnesium boride $(Mg_3B_2)$ with phosphoric acid or water. The reaction with water is:
$Mg_3B_2 + 6H_2O \rightarrow 3Mg(OH)_2 + B_2H_6$
Therefore,the correct option is $B$.

(B) Borazine $(B_3N_3H_6)$,also known as inorganic benzene,is prepared by the reaction of diborane $(B_2H_6)$ with ammonia $(NH_3)$ in a $1:2$ molar ratio at high temperatures.
The chemical equation is: $3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 12H_2$.

(A) Borazine $(B_3N_3H_6)$ is known as inorganic benzene because it is isostructural and isoelectronic with benzene $(C_6H_6)$. Both possess a hexagonal ring structure with alternating single and double bonds (in resonance). Therefore,borazine and benzene are structurally similar.

(C) $B(OH)_3$ (Boric acid) is a weak monobasic Lewis acid,not a strong acid.
It acts as a Lewis acid by accepting an $OH^-$ ion from water: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$.
It does not donate a proton $(H^+)$ directly,but it reacts with $NaOH$ to form sodium metaborate or sodium tetrahydroxoborate: $B(OH)_3 + NaOH \rightarrow Na[B(OH)_4]$.
Therefore,the statement that it does not form a salt with $NaOH$ is incorrect.

(D) The reducing power of an element is related to its ability to lose electrons,which is determined by its standard oxidation potential.
For Group $13$ elements,the standard electrode potential $(E^\circ)$ values for the reaction $M^{3+} + 3e^- \rightarrow M$ are:
$Al^{3+}/Al = -1.66 \ V$
$Ga^{3+}/Ga = -0.56 \ V$
$In^{3+}/In = -0.34 \ V$
$Tl^{3+}/Tl = +1.26 \ V$
Since $Al$ has the most negative reduction potential,it is the strongest reducing agent.
As we move down the group,the tendency to form $M^{3+}$ ions decreases due to the inert pair effect,and the stability of the $+1$ oxidation state increases.
Therefore,the reducing power decreases in the order: $Al > Ga > In > Tl$.

(A) When borax $(Na_2B_4O_7 \cdot 10H_2O)$ is heated,it loses water of crystallization and swells to form a transparent glassy bead consisting of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
When this bead is heated with cobalt oxide $(CoO)$ in an oxidizing flame,it forms cobalt metaborate,which is blue in color.
The reaction is as follows:
$CoO + B_2O_3 \rightarrow Co(BO_2)_2$ (Cobalt metaborate,blue bead).

(A) Heating hydrated aluminum chloride $(AlCl_3 \cdot 6H_2O)$ results in hydrolysis rather than dehydration. The reaction is: $2(AlCl_3 \cdot 6H_2O) \xrightarrow{\Delta} Al_2O_3 + 6HCl + 9H_2O$. Thus,it yields aluminum oxide $(Al_2O_3)$ instead of anhydrous $AlCl_3$.

(C) Potassium alum is $KAl(SO_4)_2 \cdot 12H_2O$. When $NaOH$ is added to it,initially $Al(OH)_3$ white precipitate is formed: $Al^{3+} + 3OH^- \rightarrow Al(OH)_3 \downarrow$. On adding an excess of $NaOH$,the $Al(OH)_3$ precipitate dissolves to form a soluble complex,sodium tetrahydroxoaluminate: $Al(OH)_3 + OH^- \rightarrow [Al(OH)_4]^-$. Thus,the final result is a clear solution.

(A) Borax $(Na_2B_4O_7 \cdot 10H_2O)$ reacts with mineral acids like hydrochloric acid $(HCl)$ to produce boric acid $(H_3BO_3)$.
The chemical reaction is:
$Na_2B_4O_7 + 2HCl + 5H_2O \rightarrow 2NaCl + 4H_3BO_3$
Thus,hydrochloric acid is added to borax to obtain boric acid.

(A) Boric acid $(H_3BO_3)$ is a mild antiseptic for eyes and is used in the manufacture of optical glass,enamel,and pottery glazes. It is also used as a flux in soldering. However,it is not used in the manufacture of enamel and pottery glass as a primary component in the way it is used for specialized glass; rather,it is used in the preparation of glazes. Given the standard options,all listed uses are technically correct applications of boric acid. However,if this is a multiple-choice question where one must be excluded,it is often noted that boric acid is not used as a primary structural component in the manufacture of ordinary glass or enamel,but rather as a flux or additive. Based on standard chemistry curricula,all these are recognized uses. If the question implies an incorrect usage,none of these are incorrect. Assuming the question asks for a common industrial application,all are correct.

(B) Borazole $(B_3N_3H_6)$ is isoelectronic with benzene $(C_6H_6)$.
However,in borazole,the $B-N$ bond is polar due to the difference in electronegativity between Boron $(2.04)$ and Nitrogen $(3.04)$.
This polarity makes the molecule susceptible to nucleophilic and electrophilic attacks,unlike the non-polar benzene ring.
Therefore,borazole is more reactive than benzene.

(D) Boron $(B)$ is a non-metal and does not react with non-oxidizing acids like $HCl$ or $H_2SO_4$ to liberate $H_2$ gas.
Metals like $Al$,$In$,and $Ti$ are electropositive and react with acids to displace $H_2$ gas.
Therefore,the correct answer is $B$.

(A) The melting point of Group $13$ elements is influenced by their crystal structure and bonding strength.
Boron $(B)$ has an exceptionally high melting point of approximately $2450 \ K$ due to its strong covalent network structure.
In contrast,other elements in the group like Aluminium $(Al)$,Gallium $(Ga)$,and Thallium $(Tl)$ have significantly lower melting points due to their metallic bonding.
Therefore,Boron has the highest melting point among the given elements.

(B) In group $13$ of the periodic table,metallic character increases as we move down the group from top to bottom.
Boron $(B)$ is at the top of the group and is a non-metal or metalloid.
Aluminium $(Al)$,Gallium $(Ga)$,Indium $(In)$,and Thallium $(Tl)$ show increasing metallic character as their atomic size increases and ionization energy decreases.
Therefore,Boron has the least metallic character among the given elements.

(D) The group valency of group $13$ elements is $+3$.
As we move down the group,the stability of the $+1$ oxidation state increases due to the inert pair effect.
Thallium $(Tl)$ shows both $+1$ and $+3$ oxidation states,but the $+1$ oxidation state is more stable than the $+3$ oxidation state.
Therefore,$Tl$ is the element that often does not exhibit the group valency of $+3$ in its stable compounds.

(D) Aluminum $(Al)$ is a metal that typically forms covalent bonds due to its high polarizing power,making its compounds predominantly covalent rather than ionic.
$Al$ is indeed a light metal with high tensile strength.
$Al$ acts as a strong reducing agent,often used in the thermite process.
$Al$ reacts with steam at high temperatures to form aluminum oxide $(Al_2O_3)$ and hydrogen gas $(H_2)$.
Therefore,the statement that $Al$ does not react with steam even at high temperatures is incorrect.