Boron family Questions in English

(C) In diborane $(B_2H_6)$,each boron atom is bonded to four hydrogen atoms (two terminal and two bridging).
This results in a tetrahedral geometry around each boron atom.
Therefore,the hybridization of each boron atom is $sp^3$.

(D) The compound $BH_3$ (borane) is electron-deficient and does not exist in the free monomeric form under standard conditions.
It exists as a stable dimer known as diborane,$B_2H_6$,where boron achieves an octet through three-center two-electron $(3c-2e)$ bonds.

(A) The acidic strength of hydrohalic acids $(HX)$ increases as the bond dissociation energy decreases down the group $(F < Cl < Br < I)$.
Since the $H-F$ bond is the strongest due to the small size of the fluorine atom,it has the highest bond dissociation energy.
Therefore,$HF$ does not release $H^+$ ions as easily as the other hydrohalic acids,making it the weakest acid among the given options.

(A) Orthoboric acid $(H_3BO_3)$ acts as a weak Lewis acid in aqueous medium.
It accepts a lone pair of electrons from $OH^-$ ions of water to form the $[B(OH)_4]^-$ complex and releases a proton $(H^+)$.
The reaction is: $H_3BO_3 + H_2O \rightleftharpoons [B(OH)_4]^- + H^+$.
Since it releases only one $H^+$ ion per molecule,it is a monobasic acid.

(D) $Boron$ halides $(BX_3)$ have only $6$ electrons in the valence shell of the central $Boron$ atom.
Due to this electron deficiency,they have a strong tendency to accept a lone pair of electrons from a donor to complete their octet,which is the definition of a $Lewis$ acid.
For example,in $BF_3$,the $Boron$ atom is electron-deficient.

(D) The reaction of diborane $(B_2H_6)$ with potassium hydroxide $(KOH)$ in the presence of water $(H_2O)$ is a redox reaction.
The balanced chemical equation is:
$B_2H_6 + 2KOH + 2H_2O \to 2KBO_2 + 6H_2$.
Comparing this with the given equation $B_2H_6 + 2KOH + 2X \to 2Y + 6H_2$,we find that $X = H_2O$ and $Y = KBO_2$.

(A) Boron $(B)$ shows a diagonal relationship with silicon $(Si)$.
Boron exhibits anomalous behavior in its group due to its small atomic size and the non-availability of $d$-orbitals.
Due to similarities in charge-to-size ratio,it resembles silicon and shows a diagonal relationship with it.

(C) The element $X$ has an outer electronic configuration of $ns^2np^1$. Since it is in the first short period (Period $2$ or $3$),and considering the valence shell configuration,the element is Boron $(B)$ or Aluminum $(Al)$.
For Boron $(B)$,the oxide is $B_2O_3$,and for Aluminum $(Al)$,the oxide is $Al_2O_3$.
Both $B_2O_3$ and $Al_2O_3$ are known to be amphoteric in nature,meaning they can react with both acids and bases.

(D) Tincal is a naturally occurring mineral form of borax.
Its chemical formula is $Na_2B_4O_7 \cdot 10H_2O$.

(B) Borax is a well-known compound of boron with the chemical formula $Na_2B_4O_7 \cdot 10H_2O$.
It is more accurately represented as $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$ because it contains the tetranuclear unit $[B_4O_5(OH)_4]^{2-}$.
Therefore,the correct composition is $Na_2B_4O_7 \cdot 10H_2O$.

(C) Alum,specifically potassium alum $(KAl(SO_4)_2 \cdot 12H_2O)$,is widely used in water purification and treatment processes. It acts as a coagulant to remove suspended impurities and is also used in the process of water softening. Therefore,the correct option is $(C)$.

(C) Aluminium reacts with caustic soda $(NaOH)$ to form sodium meta-aluminate $(NaAlO_2)$ and hydrogen gas.
The balanced chemical equation is:
$2Al + 2NaOH + 2H_2O \to 2NaAlO_2 + 3H_2 \uparrow$

(C) Potash alum,with the chemical formula $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$,is commonly used in the process of water purification.
It acts as a coagulant,which helps in settling down the suspended impurities in water.

(B) Colemanite is a naturally occurring mineral of boron.
Its chemical formula is $Ca_2B_6O_{11} \cdot 5H_2O$.

(A) $H_3BO_3$ (orthoboric acid) is a very weak monoprotic Lewis acid,not a strong tribasic acid.
It acts as a Lewis acid by accepting an $OH^-$ ion from water molecules,releasing $H^+$ ions in the process: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$.
Therefore,the statement that it is a strong tribasic acid is incorrect.

(C) In diborane $(B_2H_6)$,each boron atom is bonded to four hydrogen atoms (two terminal and two bridging).
According to the valence bond theory,each boron atom undergoes $sp^3$ hybridisation to form four $sp^3$ hybrid orbitals.
Three of these orbitals contain one electron each,while the fourth one is vacant.
This results in a tetrahedral geometry around each boron atom,confirming $sp^3$ hybridisation.

(A) The balanced chemical equation for the reaction is: $B_2O_3 + 3C + 3Cl_2 \to 2BCl_3 + 3CO$.
Boron trichloride $(BCl_3)$ is obtained by passing chlorine gas over a heated mixture of boron trioxide $(B_2O_3)$ and powdered charcoal $(C)$.

(A) The acidity of oxides generally increases with an increase in the non-metallic character of the central element and its oxidation state.
Among the given oxides,$P_2O_3$ is the oxide of phosphorus,which is a non-metal,while $B_2O_3$,$As_2O_3$,and $Sb_2O_3$ are oxides of elements with more metallic or metalloid character compared to phosphorus.
$P_2O_3$ is the most acidic oxide among the choices provided because phosphorus is the most non-metallic element in this group.

(D) The structure of diborane $(B_2H_6)$ consists of two terminal $BH_2$ groups and two bridging hydrogen atoms.
In this structure, there are two types of $B-H$ bonds:
$1$. Terminal $B-H$ bonds: These are normal $2c-2e$ (two-center, two-electron) covalent bonds.
$2$. Bridging $B-H-B$ bonds: These are $3c-2e$ (three-center, two-electron) bonds, also known as banana bonds.
Since the terminal and bridging bonds have different bond lengths ($119 \ pm$ and $134 \ pm$ respectively) and different bonding characteristics, the statement that all $B-H$ bonds are similar is incorrect.
Therefore, the correct option is $D$.

(D) Aluminium $(Al)$ has a high affinity for oxygen.
At high temperatures,it reacts vigorously with oxygen in the air to form aluminium oxide $(Al_2O_3)$.
This reaction is highly exothermic,meaning it releases a significant amount of heat and light.

(C) Aluminium hydroxide is amphoteric in nature and reacts with excess sodium hydroxide to form a soluble complex.
The chemical reaction is:
$Al(OH)_3 + NaOH \rightarrow NaAlO_2 + 2H_2O$
In aqueous solution,$NaAlO_2$ dissociates to form the aluminate ion,$AlO_2^{-}$.

(D) Boron forms covalent bonds primarily due to its small size.
Additionally,the sum of its first three ionization enthalpies is very high.
This high energy requirement prevents the formation of $B^{3+}$ ions and forces boron to form covalent compounds.

(B) In diborane $(B_2H_6)$,there are two types of hydrogen atoms: terminal and bridging.
The terminal $H-B-H$ bond angle is approximately $122^{\circ}$.
The bridging $H-B-H$ bond angle is approximately $97^{\circ}$.
Therefore,the two angles are nearly $97^{\circ}$ and $122^{\circ}$.

(C) In the $p$-block elements of Group $13$,$Gallium$ $(Ga)$,$Indium$ $(In)$,and $Aluminium$ $(Al)$ are metals. $Boron$ $(B)$ is a metalloid,but it is the only element in this group that exhibits non-metallic properties.

(C) When orthoboric acid $(H_3BO_3)$ is heated at high temperatures,it loses water to form boric anhydride $(B_2O_3)$.
The reaction is as follows:
$2H_3BO_3 \xrightarrow{\Delta} B_2O_3 + 3H_2O$
Thus,the residue left is boric anhydride.

(D) The elements of Group $13$ (Boron family) such as $Al$,$Ga$,and $In$ form dimeric halides in the vapor state or in non-polar solvents to complete their octet.
These dimeric structures are represented as $Al_2Cl_6$,$Ga_2Cl_6$,and $In_2Cl_6$.

(A) Aluminium chloride $(Al_2Cl_6)$ is a covalent dimer in non-polar solvents.
When it dissolves in water,it undergoes hydration to form the stable octahedral complex ion,$[Al(H_2O)_6]^{3+}$,and releases chloride ions.
The reaction is: $Al_2Cl_6 + 12H_2O \rightarrow 2[Al(H_2O)_6]^{3+} + 6Cl^-$.
Therefore,the correct species formed are $[Al(H_2O)_6]^{3+}$ and $Cl^-$ ions.

(D) The correct answer is $(D)$.
$B_4C$ (Boron carbide) is known as one of the hardest materials,ranking just after diamond in hardness.

(A) Borazine,$B_3N_3H_6$,is isoelectronic to benzene and hence,is called inorganic benzene.
Some physical properties of benzene and borazine are also similar.

(C) $B(OH)_3$ is the hydroxide of a non-metal and exhibits acidic nature in aqueous solution by accepting $OH^-$ ions from water.
In contrast,$Be(OH)_2$,$Mg(OH)_2$,and $Al(OH)_3$ are metallic hydroxides that exhibit basic or amphoteric character.

(C) Moissan boron is amorphous boron,obtained by the reduction of $B_2O_3$ with $Na$ or $Mg$.
It contains $95-98\%$ boron and is black in colour.

(D) Boron forms various hydrides with the general formulas $B_n H_{n+4}$ and $B_n H_{n+6}$,such as $B_2H_6$ (diborane).
However,the monomeric species $BH_3$ is unstable and does not exist in free form under standard conditions; it dimerizes to form $B_2H_6$ to complete its octet.

(C) Alumina $(Al_2O_3)$ is an amphoteric oxide.
It reacts with both acids and bases to form salts and water.
For example:
$Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O$ (reacts with acid)
$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$ (reacts with base).

(C) Aluminium is called a self-protecting metal because when it comes in contact with the atmosphere,it is covered entirely with a thin,non-porous layer of aluminium oxide $(Al_2O_3)$.
This layer acts as a protective barrier that prevents further oxidation or corrosion of the underlying metal.

(A) Heating hydrated aluminium chloride $(AlCl_3 \cdot 6H_2O)$ results in hydrolysis rather than dehydration.
The reaction is: $AlCl_3 \cdot 6H_2O \xrightarrow{\Delta} Al(OH)_3 + 3HCl + 3H_2O$.
Further heating of $Al(OH)_3$ yields $Al_2O_3$. Therefore,anhydrous $AlCl_3$ cannot be obtained by heating its hexahydrate.

(C) Anhydrous $AlCl_3$ is prepared by passing dry $HCl$ gas over heated aluminium metal.
The chemical reaction is:
$2Al(s) + 6HCl(g) \to 2AlCl_3(s) + 3H_2(g)$

(D) When aluminum $(Al)$ reacts with a hot concentrated solution of potassium hydroxide $(KOH)$,it forms potassium aluminate $(KAlO_2)$ and releases hydrogen gas $(H_2)$.
The balanced chemical equation is:
$2KOH + 2Al + 2H_2O \to 2KAlO_2 + 3H_2$
Therefore,hydrogen gas is evolved.

(C) Anhydrous aluminium chloride exists as a dimer $(Al_2Cl_6)$ in the vapour state and in non-polar solvents.
It is a strong Lewis acid due to the incomplete octet of the $Al$ atom.
Because of its Lewis acidic nature,it undergoes hydrolysis easily in the presence of moisture.
It sublimes at $178\,^oC$ under atmospheric pressure,but it can sublime at lower temperatures under vacuum.

(A) Potassium alum is the common alum of commerce,and its chemical formula is $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$.
Ammonium alum is $(NH_4)_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$.
Chrome alum is $K_2SO_4 \cdot Cr_2(SO_4)_3 \cdot 24H_2O$.

(C) The hydroxide of boron,$B(OH)_3$ or $H_3BO_3$,acts as a weak monobasic Lewis acid in water because it accepts an $OH^-$ ion from water.
Aluminium hydroxide,$Al(OH)_3$,is amphoteric in nature as it reacts with both acids and bases.
Therefore,the correct statement is that the hydroxide of boron is acidic,while that of aluminium is amphoteric.

(A) $AlCl_3$ forms a dimer and exists as $Al_2Cl_6$ to achieve a higher coordination number of $6$ for the aluminium atom.
In $AlCl_3$,the aluminium atom is electron-deficient with an incomplete octet.
By forming a dimer,each aluminium atom accepts a lone pair of electrons from a chlorine atom of another $AlCl_3$ molecule,thereby completing its octet and increasing its coordination number.

(B) An amphoteric oxide is a substance that can react as both an acid and a base.
$Al_2O_3$ (Aluminum oxide) reacts with both acids and bases to form salts and water,making it amphoteric.
$MgO$ is basic,$Cl_2O_7$ is acidic,and $TiO_2$ is generally considered acidic or weakly amphoteric,but $Al_2O_3$ is the classic example of an amphoteric oxide in this context.

(D) Aluminium is a strong reducing agent and is widely used in the thermite process for the reduction of metal oxides. It is not used as an oxidizer in metallurgy.

(A) In diborane $(B_2H_6)$,there are four terminal $B-H$ bonds which are regular two-center two-electron $(2c-2e)$ bonds.
There are also two bridging $B-H-B$ bonds,known as banana bonds,which are three-center two-electron $(3c-2e)$ bonds.

(A) The acidic strength of boron trihalides depends on the extent of back-bonding between the halogen $p$-orbitals and the empty $p$-orbital of boron.
In $BF_3$,the $2p-2p$ back-bonding is most effective due to the small size of the fluorine atom,which reduces the electron deficiency of boron.
As the size of the halogen increases from $F$ to $I$,the effectiveness of back-bonding decreases $(2p-2p > 2p-3p > 2p-4p > 2p-5p)$.
Therefore,the electron deficiency of boron increases from $BF_3$ to $BI_3$,making $BI_3$ the strongest Lewis acid.
The correct order of acidic strength is: $BF_3 < BCl_3 < BBr_3 < BI_3$.

(B) Boron nitride $(BN)$ is known as inorganic graphite.
It is isoelectronic to carbon and possesses a layered structure similar to graphite.
In this structure,boron and nitrogen atoms are arranged in hexagonal rings,making it structurally analogous to the carbon atoms in graphite.

(A) Aluminium forms a protective layer of aluminium oxide $(Al_2O_3)$ on its surface when exposed to concentrated nitric acid.
This passive layer prevents further reaction between the metal and the acid,making it suitable for storage.
The chemical reaction is:
$2 Al + 6 HNO_3 (\text{conc.}) \rightarrow Al_2O_3 + 6 NO_2 + 3 H_2O$

(C) The structure of phosphoric acid $(H_3PO_4)$ contains three $P-OH$ groups.
The basicity of an oxyacid is determined by the number of $P-OH$ groups present in its structure.
Since $H_3PO_4$ has three $P-OH$ groups,it can release three $H^+$ ions in an aqueous solution.
Therefore,$H_3PO_4$ is a tribasic acid.

(B) $H_3PO_3$ is a dibasic acid because it contains two $P-OH$ bonds.
It can replace two hydrogen atoms to form two series of salts: acid salts (e.g.,$NaH_2PO_3$) and normal salts (e.g.,$Na_2HPO_3$).

(D) The basicity 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 overlap effectively with the empty $2p$ orbital of boron,leading to the strongest back-bonding.
This reduces the electron-deficient nature of boron,making $BF_3$ the least basic among the given trihalides.
Therefore,the order of basicity is $BF_3 < BCl_3 < BBr_3 < BI_3$.