Boron family Questions in English

(A) $Na_2B_4O_7 \cdot 10H_2O \xrightarrow[-10H_2O]{\Delta} Na_2B_4O_7$
$Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3$
$CuO + B_2O_3 \to \underset{(\text{Copper meta borate, blue})}{Cu(BO_2)_2}$
In the borax bead test,the metal oxide reacts with $B_2O_3$ to form a metal meta borate,which gives a characteristic color to the bead.

(D) When borate is heated with ethanol and conc. $H_2SO_4$,triethyl borate is produced,and these vapours burn with a green-edged flame on ignition.
$Na_2B_4O_7 + H_2SO_4 + 5 H_2O \rightarrow Na_2SO_4 + 4 H_3BO_3$
$H_3BO_3 + 3 C_2H_5OH \rightarrow B(OC_2H_5)_3 + 3 H_2O$
Thus,the green-edged flame is due to the formation of Triethyl Borate,which is $(C_2H_5)_3BO_3$.

(C) The reaction between ammonia $(NH_3)$ and hydrogen chloride $(HCl)$ gas produces ammonium chloride $(NH_4Cl)$,which appears as dense white fumes.
$NH_3(g) + HCl(g) \to NH_4Cl(s)$ (dense white fumes)

(A) The vapours of trialkyl borates,such as $B(OMe)_3$,burn with a green-edged flame. This reaction is a characteristic qualitative test for the presence of borates.
The chemical reactions involved are:
$2BO_3^{3-} + 3H_2SO_4 \rightarrow 2H_3BO_3 + 3SO_4^{2-}$
$H_3BO_3 + 3CH_3OH \rightarrow B(OCH_3)_3 + 3H_2O$
Since $B(OMe)_3$ is a trialkyl borate,it imparts a green colour to the flame.

(A) $HgCl_2$ exists as a dimer in the solid state. The mercury atoms are linked by chlorine bridges. While $B_2H_6$ is a dimer of $BH_3$,in the context of typical inorganic chemistry questions regarding metal halides,$HgCl_2$ is frequently cited for its dimeric nature in solid form.

(D) The chemical formula of borax is $Na_2B_4O_7 \cdot 10H_2O$.
Its chemical name is $Sodium \ tetraborate \ decahydrate$.

(C) $AlCl_3$ is a covalent compound that undergoes hydrolysis in the presence of moisture (water vapor) in the air.
The reaction is: $AlCl_3 + 3H_2O \rightarrow Al(OH)_3 + 3HCl$.
The $HCl$ gas produced forms white fumes in the air,which is why $AlCl_3$ appears to fume in moist air.

(C) Boron compounds like $BF_3$ have only $6$ electrons in the valence shell of the central boron atom. Due to this incomplete octet,they are electron-deficient and can accept a pair of electrons,which is the definition of a Lewis acid.

(C) Borazole,also known as inorganic benzene,has the chemical formula $B_3N_3H_6$.
It is isoelectronic with benzene $(C_6H_6)$ and possesses a cyclic structure consisting of alternating boron and nitrogen atoms.

(C) In diborane $(B_2H_6)$,the boron atom undergoes $sp^3$ hybridization.
Each boron atom has four $sp^3$ hybrid orbitals.
Three of these orbitals contain one electron each,while one orbital is empty.
The four terminal hydrogen atoms form four $2c-2e$ (two-center-two-electron) bonds with the boron atoms.
The two bridged hydrogen atoms form two $3c-2e$ (three-center-two-electron) bonds,also known as banana bonds.
Therefore,the statement that it contains four one-center bonds is incorrect.

(A) Aluminium chloride $(AlCl_3)$ exists as a dimer $(Al_2Cl_6)$ in the vapour state or in non-polar solvents.
This occurs because the $Al$ atom in $AlCl_3$ is electron-deficient,having only $6$ electrons in its valence shell.
To complete its octet,$Al$ accepts a lone pair of electrons from the chlorine atom of another $AlCl_3$ molecule,forming a coordinate bond.
This results in the formation of a dimer where $Al$ achieves a coordination number of $4$.

(D) $1$. Aluminum $(Al)$ reacts with $NaOH$ to form sodium aluminate $(NaAlO_2)$ and releases $H_2$ gas: $2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2$. Here,$X = Al$ and $Y = NaAlO_2$.
$2$. When $CO_2$ is passed through the aqueous solution of $NaAlO_2$,aluminum hydroxide $(Al(OH)_3)$ is precipitated: $2NaAlO_2 + 3H_2O + CO_2 \rightarrow 2Al(OH)_3 + Na_2CO_3$. Here,$Z = Al(OH)_3$.
$3$. Heating $Al(OH)_3$ at $1200^{\circ}C$ yields alumina $(Al_2O_3)$: $2Al(OH)_3 \xrightarrow{\Delta} Al_2O_3 + 3H_2O$. Thus,the correct sequence is $X = Al, Y = NaAlO_2, Z = Al(OH)_3$.

(B) In diborane $(B_2H_6)$,there are two types of hydrogen atoms: terminal and bridging.
The terminal $B-H$ bonds are normal covalent bonds,and the angle between them is approximately $120^{\circ}$.
The bridging $B-H-B$ bonds involve $3c-2e$ (three-center two-electron) bonds,and the $H-B-H$ angle in the bridge is approximately $95^{\circ}$.
Therefore,the bond angles are $95^{\circ}$ and $120^{\circ}$.

(C) Aluminium hydroxide $(Al(OH)_{3})$ is amphoteric in nature. When it reacts with caustic soda $(NaOH)$,it forms sodium aluminate $(NaAlO_{2})$ and water. The balanced chemical equation is:
$Al(OH)_{3} + NaOH \rightarrow NaAlO_{2} + 2H_{2}O$.

(A) The hydrolysis of boron trichloride $(BCl_3)$ with water proceeds as follows:
$BCl_3 + 3H_2O \rightarrow H_3BO_3 + 3HCl$.
Here,$BCl_3$ reacts with water to form boric acid $(H_3BO_3)$ and hydrogen chloride $(HCl)$.

(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: $H_3BO_3 \xrightarrow{\Delta} HBO_2 + H_2O$.

(A) The hydrolysis of diborane $(B_2H_6)$ is a vigorous reaction that produces boric acid $(H_3BO_3)$ and hydrogen gas $(H_2)$.
The balanced chemical equation is:
$B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 6H_2$.

(C) Borazine,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 reaction is: $3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 12H_2$.
Therefore,the correct condition is a $2NH_3 : 1B_2H_6$ ratio at high temperature.

(D) The structure shown in option $D$ is incorrect for diborane $(B_2H_6)$.
In diborane,there are two bridging hydrogen atoms $(H_b)$ and four terminal hydrogen atoms $(H_t)$. The boron atoms are $sp^3$ hybridized and the structure involves $3c-2e^-$ bonds (banana bonds) for the bridging hydrogens,not the structure shown in option $D$ which incorrectly depicts a nitrogen atom and incorrect bonding.

(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 produces aluminum oxide $(Al_2O_3)$ instead of anhydrous $AlCl_3$. The other methods (passing dry $HCl$ or $Cl_2$ over hot $Al$) are standard industrial methods to produce anhydrous $AlCl_3$.

(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.

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

(B) In the $p$-block elements of group $13$ (Boron family),the metallic character increases as we move down the group from $B$ to $Tl$.
$B$ (Boron) is a non-metal,while $Al$,$Ga$,$In$,and $Tl$ exhibit metallic properties.
Therefore,$B$ (Boron) has the least metallic character among the given options.

(C) The melting points of Group $13$ elements do not show a regular trend due to the differences in their crystal structures and atomic sizes.
$B$ (Boron) has a very high melting point due to its strong covalent network structure.
$Al$ (Aluminum) has a lower melting point than $B$.
$Ga$ (Gallium) has an unusually low melting point $(303 \ K)$ because it exists as $Ga_2$ molecules in the solid state.
$In$ (Indium) and $Tl$ (Thallium) follow,with $Tl$ having a higher melting point than $In$ due to the inert pair effect and relativistic effects.
The correct order is $B > Al > Tl > In > Ga$.

(D) $BCl_3$ is a covalent molecule with a planar triangular geometry.
It is electron-deficient,having only $6$ electrons in the valence shell of Boron,hence it acts as a strong Lewis acid.
Due to the back-bonding ($p\pi-p\pi$ back-donation) from $Cl$ to $B$,the $B-Cl$ bond acquires some double-bond character.
This makes the $B-Cl$ bond length shorter than a pure single bond.
Therefore,the statement that all $B-Cl$ bonds are longer than a single bond is incorrect.

(A) The Lewis acid strength of boron halides is determined by the extent of $p\pi-p\pi$ back-bonding between the halogen atom and the boron atom.
In $BF_3$,the small size of the fluorine atom allows for effective $2p-2p$ back-bonding with boron,which significantly reduces the electron deficiency of the boron atom.
As the size of the halogen increases from $F$ to $Cl$ to $Br$,the effectiveness of back-bonding decreases ($2p-3p$ for $BCl_3$ and $2p-4p$ for $BBr_3$).
Therefore,the electron deficiency on boron increases as we move from $BF_3$ to $BBr_3$,making $BBr_3$ the strongest Lewis acid and $BF_3$ the weakest.
The correct decreasing order is $BBr_3 > BCl_3 > BF_3$.

(A) The Lewis acid strength of group $13$ trihalides depends on the ability of the central metal atom to accept an electron pair.
In $BCl_3$,the boron atom is small and undergoes $p\pi-p\pi$ back-bonding with chlorine,which reduces its electron deficiency.
As we move down the group from $B$ to $In$,the size of the central atom increases,and the extent of back-bonding decreases significantly.
Therefore,the electron deficiency increases as we move from $BCl_3$ to $InCl_3$.
However,the Lewis acidity is also influenced by the electronegativity and the size of the central atom. For group $13$ trihalides,the order of Lewis acid strength is $BCl_3 > AlCl_3 > GaCl_3 > InCl_3$ due to the effective back-bonding in $BCl_3$ and the increasing metallic character down the group.
Thus,the correct decreasing order is $(1) > (2) > (3) > (4)$.

(A) Highly electropositive metals (like $Mg$) form binary compounds (borides and silicides) with $B$ and $Si$.
Hydrolysis of magnesium boride $(Mg_3B_2)$ yields borane $(B_2H_6)$:
$Mg_3B_2 + 6H_2O \rightarrow 3Mg(OH)_2 + B_2H_6$
Hydrolysis of magnesium silicide $(Mg_2Si)$ yields silane $(SiH_4)$:
$Mg_2Si + 4H_2O \rightarrow 2Mg(OH)_2 + SiH_4$
Since the question specifies that $X$ yields borane and $Y$ yields silane,$X$ must be $B$ and $Y$ must be $Si$.

(B) Boric acid $(H_3BO_3)$ exists as a layered structure in the solid state.
In this structure,the planar $BO_3^{3-}$ units are linked together by hydrogen bonds.
This extensive hydrogen bonding leads to the formation of a two-dimensional polymeric sheet-like structure.

(B) In diborane $(B_2H_6)$,each boron atom is bonded to four hydrogen atoms.
Each boron atom undergoes $sp^3$ hybridization to form four $sp^3$ hybrid orbitals.
Three of these orbitals contain one electron each,while the fourth one is empty.
These orbitals participate in the formation of two $3c-2e$ (three-center two-electron) bonds and four terminal $B-H$ covalent bonds.

(C) Boron carbide $(B_4C)$ is known as one of the hardest materials known,ranking just below diamond and cubic boron nitride in hardness. It is widely used in industrial applications such as tank armor and bulletproof vests due to its extreme hardness and low density.

(C) The nature of hydroxides depends on the ionization energy and electronegativity of the central atom.
$B(OH)_3$ (or $H_3BO_3$) is a weak monobasic Lewis acid because it accepts $OH^-$ ions from water,releasing $H^+$ ions.
$Al(OH)_3$ is amphoteric in nature as it reacts with both acids and bases.
Therefore,boron hydroxide is acidic and aluminum hydroxide is amphoteric.

(C) When an aqueous solution of aluminum chloride $(AlCl_3)$ is heated to dryness,it undergoes hydrolysis due to the high charge density of the $Al^{3+}$ ion. The reaction is: $AlCl_3 + 3H_2O \rightarrow Al(OH)_3 + 3HCl$. Upon further heating,the aluminum hydroxide loses water to form aluminum oxide: $2Al(OH)_3 \rightarrow Al_2O_3 + 3H_2O$. Therefore,the final residue obtained is aluminum oxide $(Al_2O_3)$.

(D) Graphite and $BN$ (Boron Nitride) both possess a similar layered structure. Due to this similarity in structure,$BN$ is often referred to as 'inorganic graphite'.

(C) The borax bead test is used to identify transition metal ions in a sample.
When borax $(Na_2B_4O_7 \cdot 10H_2O)$ is heated,it forms a transparent glassy bead consisting of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
Transition metals react with this bead to form colored metaborates.
Since $Na$ is an alkali metal and not a transition metal,it does not form colored metaborates and thus does not give a positive borax bead test.
$Cr$,$Ni$,and $Mn$ are transition metals and will give characteristic colors in the borax bead test.

(D) Boron halides (e.g.,$BF_3$,$BCl_3$) have a central boron atom with only $6$ electrons in its valence shell.
According to the octet rule,boron requires $8$ electrons to complete its octet.
Since it has an incomplete octet,it acts as an electron pair acceptor,which is the definition of a Lewis acid.

(C) Orthoboric acid,$H_3BO_3$ (or $B(OH)_3$),acts as a weak monobasic Lewis acid in water.
It does not donate a proton $(H^+)$ directly; instead,it accepts a hydroxyl ion $(OH^-)$ from water to release a proton.
The reaction is: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$.
Since it acts as a Lewis acid (electron pair acceptor) rather than a Brønsted-Lowry acid (proton donor),Statement-$1$ is true and Statement-$2$ is false.

(A) The Lewis acidity of boron trihalides $(BX_3)$ depends on the extent of back-bonding between the lone pair of the halogen and the empty $p$-orbital of boron.
In $BF_3$,the fluorine atom has a small size and its $2p$-orbitals overlap effectively with the $2p$-orbital of boron ($2p\pi-2p\pi$ back-bonding),making it the weakest Lewis acid.
As the size of the halogen increases from $F$ to $I$,the effectiveness of back-bonding decreases ($2p\pi-3p\pi$ for $BCl_3$,$2p\pi-4p\pi$ for $BBr_3$,and $2p\pi-5p\pi$ for $BI_3$).
Therefore,$BI_3$ has the least back-bonding and the most available empty $p$-orbital,making it the strongest Lewis acid.
The order of Lewis acidity is $BF_3 < BCl_3 < BBr_3 < BI_3$.

(B) In the solid state,$BeCl_2$ and $AlCl_3$ exist as polymeric structures with chlorine bridges. The structure shown in the image represents the bridged dimer/polymer structure where $M = Be$ or $Al$.
$BeCl_2$ forms a chain structure with $Cl$ bridges,and $AlCl_3$ forms a layer structure with $Cl$ bridges.
Option $A$ is incorrect as the coordination number of $Be$ in $BeCl_2$ is $4$.
Option $C$ is incorrect as inorganic benzene is $B_3N_3H_6$ (borazine).
Option $D$ is incorrect as boric acid $(H_3BO_3)$ is a Lewis acid,not a protonic acid,as it accepts $OH^-$ ions from water.

(C) The electronic configuration of the element is given by filling the orbitals in the order: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^{1-6}$.
Summing the electrons: $2+2+6+2+6+2+10 = 30$ electrons are filled before the $4p$ orbital.
For a $p$-block element,the differentiating electron enters the $4p$ orbital.
If the $4p$ orbital has $1$ electron,the atomic number is $30 + 1 = 31$ (Gallium,$Ga$).
If the $4p$ orbital has $6$ electrons,the atomic number is $30 + 6 = 36$ (Krypton,$Kr$).
Thus,the range of atomic numbers for this configuration is $31 - 36$.

(C) The chemical formula for inorganic benzene,also known as borazine,is $B_3H_6N_3$.
It is isostructural with benzene $(C_6H_6)$ because it has the same number of electrons and a similar ring structure.

(A) Boron is an element of group $13$ and contains $3$ electrons in its valence shell.
When its compound $BH_3$ dimerises to form $(BH_3)_2$ (diborane),each boron atom is surrounded by only $6$ electrons,meaning its octet is incomplete.
Therefore,$(BH_3)_2$ is an electron-deficient compound.
In all other given molecules,the octet of the central atoms is complete.

(C) Boric acid is a weak monobasic Lewis acid. It acts as an acid not by donating a proton directly,but by accepting an $OH^{-}$ ion from water,which releases a proton $(H^{+})$ into the solution.
$B(OH)_3 + H_2O \longrightarrow [B(OH)_4]^{-}_{(aq)} + H^{+}_{(aq)}$

(A) The stability of the $+1$ oxidation state increases down the group $13$ elements as $Al < Ga < In < Tl$.
This is due to the inert pair effect,where the $ns^2$ electrons become increasingly reluctant to participate in bond formation due to poor shielding by $d$ and $f$ orbitals.
As we move down the group,the stability of the $+1$ oxidation state increases,making $Tl^+$ the most stable among the given ions.

(C) Boron nitride $(BN)_x$ is known as inorganic graphite because its structure is similar to that of graphite.
In graphite,carbon atoms are arranged in hexagonal layers.
In boron nitride,boron and nitrogen atoms are arranged in a similar hexagonal layered structure,where each boron atom is surrounded by three nitrogen atoms and vice versa.

(C) Electron-deficient molecules act as Lewis acids.
Among the given molecules,only diborane $(B_2H_6)$ is electron-deficient,meaning it does not have a complete octet.
Thus,it acts as a Lewis acid.
$NH_3$ and $H_2O$ are electron-rich molecules and behave as Lewis bases.

(B) The Lewis acid strength of boron trihalides depends on the extent of $p\pi - p\pi$ back bonding from the halogen atom to the boron atom.
In $BF_3$,the $2p - 2p$ back bonding is most effective due to the small size of the $F$ atom,which significantly reduces the electron deficiency of boron.
As the size of the halogen increases from $F$ to $Cl$ to $Br$,the effectiveness of $p\pi - p\pi$ back bonding decreases $(F > Cl > Br)$.
Consequently,the electron deficiency on the boron atom increases in the order $BF_3 < BCl_3 < BBr_3$.
Therefore,the Lewis acid strength follows the sequence $BBr_3 > BCl_3 > BF_3$.

(C) Boric acid is a Lewis acid,not a protonic acid.
Beryllium typically exhibits a coordination number of $4$.
$B_2H_6 \cdot 2NH_3$ is known as 'inorganic benzene' is incorrect; borazine $(B_3N_3H_6)$ is known as inorganic benzene.
Both $BeCl_2$ and $AlCl_3$ exist as polymeric structures with bridged chloride atoms in the solid state.