Boron family Questions in English

(B) In the nuclear industry,materials with high neutron absorption cross-sections are required for control rods and protective shields. $Metal \ borides$ (such as $B_4C$) are widely used for this purpose because boron has a high capacity to absorb neutrons.

(B) Orthoboric acid $(H_3BO_3)$ is a weak monobasic Lewis acid.
It acts as a Lewis acid by accepting a pair of electrons from $OH^-$ ions in water: $B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^- + H_3O^+$.
It does not release three $H^+$ ions in water; it releases only one $H^+$ ion via the hydrolysis of the boron center.
Therefore,the statement that it releases three $H^+$ ions is incorrect.

(B) The reaction of $BCl_3$ with $H_2$ in the presence of a $Cu-Al$ catalyst at $450^{\circ}C$ produces diborane $(B_2H_6)$ as $X$.
$2BCl_3 + 6H_2 \xrightarrow{Cu-Al, 450^{\circ}C} B_2H_6 + 6HCl$
Methylation of diborane $(B_2H_6)$ with methyl chloride $(CH_3Cl)$ leads to the formation of tetramethyldiborane as $Z$.
$B_2H_6 + 4CH_3Cl \rightarrow (CH_3)_4B_2H_2 + 4HCl$
Thus,$Z$ is $(CH_3)_4B_2H_2$.

(A) The melting points of group $13$ elements are influenced by their metallic bonding and crystal structure.
The melting points of group $13$ elements follow the order: $B > Al > Tl > In > Ga$.
Gallium $(Ga)$ has an unusually low melting point of $303 \ K$ because it exists as $Ga_2$ molecules in the solid state,which are held by weak van der Waals forces.
Therefore,the correct order for $Al$,$Ga$,and $In$ is $Ga < In < Al$.

(C) $AlF_3$ is insoluble in anhydrous $HF$ because it lacks free $F^-$ ions to interact with the $Al^{3+}$ center.
In the presence of $KF$,$KF$ dissociates to provide $F^-$ ions,which react with $AlF_3$ to form a soluble complex.
The reaction is:
$AlF_3 + 3KF \rightarrow K_3[AlF_6]$
This complex dissociates into $3K^+$ and $[AlF_6]^{3-}$.
Therefore,the correct option is $[AlF_6]^{3-}$.

(B) $(I)$ $Ga_2O_3$ is an amphoteric oxide. This is a correct statement as it reacts with both acids and bases:
Reaction with an acid: $Ga_2O_3 + 6 HCl \longrightarrow 2 GaCl_3 + 3 H_2O$
Reaction with a base: $Ga_2O_3 + 2 NaOH \longrightarrow 2 NaGaO_2 + H_2O$
$(II)$ The dimer of aluminium chloride $(Al_2Cl_6)$ has only two $Al-Cl-Al$ bridge bonds,not three. The structure consists of two $AlCl_4$ tetrahedra sharing a common edge.
$(III)$ Boron is a very hard refractory solid with a high melting temperature $(2349 \ K)$,which is a correct statement.
Therefore,statements $I$ and $III$ are correct. The correct option is $(b)$.

(A) The chemical formula for $Cryolite$ is $Na_3AlF_6$. It is an anhydrous mineral and does not contain any water of crystallization.
$Bauxite$ is $Al_2O_3 \cdot 2H_2O$,which contains water molecules.
$Kernite$ is $Na_2B_4O_7 \cdot 4H_2O$,which contains water molecules.
$Borax$ is $Na_2B_4O_7 \cdot 10H_2O$,which contains water molecules.
Therefore,$Cryolite$ is the correct answer.

(B) The reaction $I$ is: $Be(OH)_2 + 2NaOH \text{ (excess)} \rightarrow Na_2[Be(OH)_4] \text{ (X)}$. The covalency of $Be$ in $[Be(OH)_4]^{2-}$ is $4$.
The reaction $II$ is: $Al(OH)_3 + NaOH \text{ (excess)} \rightarrow Na[Al(OH)_4] \text{ (Y)}$. The covalency of $Al$ in $[Al(OH)_4]^{-}$ is $4$.
Thus,the covalencies of $Be$ and $Al$ in $X$ and $Y$ are $4$ and $4$ respectively.

(C) Group $13$ elements (boron family) typically exhibit a valency of $+3$.
Group $16$ elements (oxygen family) typically exhibit a valency of $-2$.
To form a neutral compound,the charges must be balanced.
Using the criss-cross method:
$A^{+3}$ and $B^{-2}$
By swapping the charges,we get $A_2B_3$.
Therefore,the general formula is $A_2B_3$.

(B) The basic character of an oxide increases with an increase in the metallic character of the element.
As we move down the group,the metallic character increases.
Among the given elements ($B$,$Al$,$Ga$,$Tl$),$Tl$ $(Thallium)$ is at the bottom of the group and thus possesses the highest metallic character.
Therefore,$Tl$ forms a basic oxide.
$B$ $(Boron)$ forms an acidic oxide,while $Al$ $(Aluminium)$ and $Ga$ $(Gallium)$ form amphoteric oxides.

(B) Kernite is a mineral of Boron $(B)$ with the formula $Na_2B_4O_6(OH)_2 \cdot 3H_2O$.
Cryolite is a mineral of Aluminum $(Al)$ with the formula $Na_3AlF_6$.
Therefore,$X$ is $B$ and $Z$ is $Al$.

(A) $I$. In the borax bead test,cobalt oxide reacts with $B_2O_3$ to form cobalt metaborate,$Co(BO_2)_2$,which is blue in color. This statement is correct.
$II$. Diborane $(B_2H_6)$ is prepared by the reaction of sodium borohydride $(NaBH_4)$ with iodine $(I_2)$: $2NaBH_4 + I_2 \rightarrow B_2H_6 + 2NaI + H_2$. This statement is correct.
$III$. In diborane $(B_2H_6)$,boron is more electropositive than hydrogen,so the oxidation state of hydrogen is $-1$. This statement is incorrect.
$IV$. Boric acid $(H_3BO_3)$ is a weak monobasic Lewis acid,not a tribasic acid,as it accepts $OH^-$ ions from water. This statement is incorrect.
Therefore,statements $I$ and $II$ are correct.

(C) The correct answer is $C$.
$A$: $TlI_3$ does not exist because $Tl^{3+}$ is a strong oxidizing agent and $I^-$ is a strong reducing agent,leading to the formation of $TlI$ and $I_2$.
$B$: Trihalides like $BCl_3$ undergo hydrolysis to form $[B(OH)_4]^-$,which is a tetrahedral species.
$C$: Diborane $(B_2H_6)$ is an electron-deficient hydride,not an electron-precise hydride,as it has fewer electrons than required for standard covalent bonding.
$D$: Hydrolysis of diborane $(B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 6H_2)$ yields boric acid $(H_3BO_3)$.

(B) The hydrolysis of diborane $(B_2H_6)$ is represented by the reaction: $B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 6H_2$.
Thus,the compound $X$ is orthoboric acid $(H_3BO_3)$.
Regarding the properties of $H_3BO_3$:
$I$. It is a weak monobasic acid,not a tribasic acid,because it acts as a Lewis acid by accepting an $OH^-$ ion from water: $B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^- + H_3O^+$.
$II$. This statement is correct.
$III$. It has a layer structure in the solid state where planar $BO_3$ units are linked by hydrogen bonds. This statement is correct.
$IV$. It is sparingly soluble in cold water but its solubility increases with temperature. Thus,it is not considered 'highly' soluble.
Therefore,the correct statements are $II$ and $III$.

(A) The reaction of $BF_3$ with $LiAlH_4$ in ether produces diborane $(B_2H_6)$ as $X$:
$4BF_3 + 3LiAlH_4 \rightarrow 2B_2H_6 + 3LiF + 3AlF_3$.
$X$ $(B_2H_6)$ reacts with $NH_3$ to form an adduct,which on heating gives inorganic benzene (borazine,$B_3N_3H_6$) as $Z$:
$3B_2H_6 + 6NH_3$ $\rightarrow 3[BH_2(NH_3)_2]^+[BH_4]^-$ $\xrightarrow{\Delta} 2B_3N_3H_6 + 12H_2$.
Borazine $(B_3N_3H_6)$ has a cyclic structure similar to benzene.
It contains $12$ $\sigma$-bonds ($6$ $B-N$ and $6$ $B-H/N-H$ bonds) and $3$ $\pi$-bonds.
Thus,$x = 12$ and $y = 3$.
$(x + y) = 12 + 3 = 15$.

(A) On balancing the equation $(1)$:
$2BF_3 + 6NaH \xrightarrow{450 \ K} B_2H_6 + 6NaF$
So,$X$ is $B_2H_6$ (diborane).
Balancing the equation $(2)$:
$B_2H_6 + 6H_2O \longrightarrow 2B(OH)_3 + 6H_2 \uparrow$
So,$Y$ is $B(OH)_3$ (boric acid).
$I$. $X$ $(B_2H_6)$ is an electron-deficient molecule (contains $3c-2e^-$ bonds). This is correct.
$II$. In $X$ $(B_2H_6)$,there is no direct $B-B$ bond; it has two bridging hydrogen atoms $(B-H-B)$. This is incorrect.
$III$. $Y$ $(B(OH)_3)$ is a weak monobasic Lewis acid,not a tribasic acid. This is incorrect.
$IV$. $Y$ $(B(OH)_3)$ acts as a Lewis acid as it accepts a lone pair of electrons from $OH^-$ ions. This is correct.
Therefore,statements $I$ and $IV$ are correct.

(A) The first reaction is: $4BF_3 + 3LiAlH_4 \xrightarrow{(C_2H_5)_2O} 2B_2H_6 + 3LiF + 3AlF_3$. Here,$X$ is $B_2H_6$ (diborane).
The second reaction is: $B_2H_6 + 2NaH \longrightarrow 2NaBH_4$. Here,$Y$ is $NaBH_4$ (sodium borohydride).
$NaBH_4$ contains the $BH_4^-$ ion,where hydrogen is in the $-1$ oxidation state.
$NaBH_4$ is a well-known reducing agent,not an oxidizing agent.
Therefore,the statement 'It is a good oxidizing agent' is incorrect.

(B) The reaction of diborane $(B_2 H_6)$ with carbon monoxide $(CO)$ under pressure yields the borane carbonyl adduct,$BH_3 \cdot CO$ $(X)$.
$B_2 H_6 + 2 CO \rightarrow 2 BH_3 \cdot CO$
The reaction of diborane $(B_2 H_6)$ with sodium hydride $(NaH)$ in the presence of diethyl ether $((C_2 H_5)_2 O)$ yields sodium borohydride,$NaBH_4$ $(Y)$.
$B_2 H_6 + 2 NaH \xrightarrow{(C_2 H_5)_2 O} 2 NaBH_4$
Therefore,$X = BH_3 \cdot CO$ and $Y = NaBH_4$.

(C) The chemical formula of borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$.
Comparing this with the given expression $Na_2[B_4O_5(OH)_x] \cdot yH_2O$,we get:
$x = 4$
$y = 8$
Therefore,the sum $x + y = 4 + 8 = 12$.

(A) Statement $A$ is correct: $AlCl_3$ forms a dimer $(Al_2Cl_6)$ to complete its octet.
Statement $B$ is correct: $BCl_3$ has only $6$ electrons in the valence shell of boron,making it electron-deficient.
Statement $C$ is incorrect: The standard reduction potential $E^0_{Al^{3+}/Al}$ is $-1.66 \ V$,not $+1.26 \ V$.
Statement $D$ is incorrect: Due to the inert pair effect,the $+1$ oxidation state of thallium $(Tl^+)$ is more stable than the $+3$ oxidation state $(Tl^{3+})$.
Therefore,statements $C$ and $D$ are incorrect.

(C) The Lewis acidity of boron trihalides depends on 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 $F$ atom allows for effective $2p-2p$ back-bonding,which reduces the electron deficiency of the boron atom.
As we move from $F$ to $I$,the size of the halogen atom increases,making the $p\pi-p\pi$ back-bonding less effective.
Therefore,the electron deficiency of the boron atom increases,leading to an increase in Lewis acidic character.
The correct order is $BI_3 > BBr_3 > BCl_3 > BF_3$.

(D) The assertion is false because boron has a small atomic size and lacks $d$-orbitals in its valence shell,which prevents it from expanding its coordination number beyond $4$. Therefore,the formation of an octahedral $[B(OH_2)_6]^{3+}$ complex is impossible.
However,the reason is true because boron is electron-deficient ($sextet$ of electrons) and acts as a Lewis acid,reacting with Lewis bases like $OH^{-}$ to form the stable tetrahedral $[B(OH)_4]^{-}$ ion.
Thus,$A$ is false but $R$ is true.

(A) The element with $ns^2 np^1$ configuration is a member of Group $13$.
Boron $(B)$ is a black-coloured non-metal that exists in crystalline and amorphous forms.
Crystalline boron is very hard and unreactive towards air at room temperature.
Amorphous boron reacts with oxygen in the air to form boron trioxide $(B_2O_3)$,which is acidic in nature:
$4B(s) + 3O_2(g) \rightarrow 2B_2O_3(s)$
$B_2O_3$ further reacts with water to form boric acid:
$B_2O_3(s) + 3H_2O(l) \rightarrow 2B(OH)_3(aq)$

(A) Boron has the atomic number $5$ and electronic configuration $1s^2 2s^2 2p^1$.
Due to the absence of $d$-orbitals in the valence shell,boron cannot expand its coordination number beyond $4$.
Therefore,it cannot form a hexacoordinated complex like $[B(H_2O)_6]^{3+}$.
Thus,the complex $[B(H_2O)_6]^{3+}$ does not exist.

(D) Amphoteric oxides are those that exhibit both acidic and basic properties.
$Ga_2O_3$ is a well-known amphoteric oxide.
$As_4O_{10}$ and $Sb_4O_{10}$ are also amphoteric in nature as they show acidic character with strong bases and basic character with strong acids.
$B_2O_3$ is acidic,and $Tl_2O$ is basic.
Therefore,the set of amphoteric oxides is $Ga_2O_3, As_4O_{10}, Sb_4O_{10}$.

(D) $1$. Diborane $(B_2H_6)$ is an electron-deficient species that acts as a Lewis acid. It reacts with carbon monoxide $(CO)$,a Lewis base,to form a coordinate bond adduct:
$B_2H_6 + 2CO \longrightarrow 2BH_3 \cdot CO$ $(A-II)$
$2$. Diborane burns in oxygen to form boron trioxide $(B_2O_3)$:
$B_2H_6 + 3O_2 \longrightarrow B_2O_3 + 3H_2O$ $(B-I)$
$3$. Diborane undergoes hydrolysis with water to produce boric acid $(H_3BO_3)$:
$B_2H_6 + 6H_2O \longrightarrow 2H_3BO_3 + 6H_2$ $(C-III)$
Thus,the correct match is $A-II, B-I, C-III$.

(A) $Al(OH)_3$ acts as a base and reacts with acid $HCl$ to give salt and water:
$Al(OH)_3 + 3HCl \longrightarrow AlCl_3 + 3H_2O$
$Al(OH)_3$ also acts as an acid and reacts with base $NaOH$ to give salt and water:
$Al(OH)_3 + NaOH \longrightarrow Na[Al(OH)_4]$
Since $Al(OH)_3$ reacts with both acids and bases,it is amphoteric in nature.

(A) $H_3BO_3$ (boric acid) is a weak monobasic Lewis acid.
It does not dissociate to give $H^+$ ions directly.
Instead,it accepts an $OH^-$ ion from water to form $[B(OH)_4]^-$ and releases $H^+$ (as $H_3O^+$),as shown in the reaction:
$B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$

(C) In diborane $(B_2H_6)$,each boron atom is bonded to four hydrogen atoms (two terminal and two bridging).
To accommodate these four bonds,the boron atom undergoes $sp^3$ hybridisation.
Three of the $sp^3$ hybrid orbitals contain one electron each,while one is empty.
The two terminal $B-H$ bonds are normal covalent bonds,while the two bridging $B-H-B$ bonds are three-center two-electron $(3c-2e)$ bonds,often referred to as banana bonds.
Therefore,the hybridisation of boron in diborane is $sp^3$.

(C) Borax has the formula $Na_2[B_4O_5(OH)_4] \cdot 8 H_2O$.
When dissolved in water,it undergoes hydrolysis to form orthoboric acid $(H_3BO_3)$ and sodium hydroxide $(NaOH)$.
Since $NaOH$ is a strong base and $H_3BO_3$ is a very weak acid,the resulting aqueous solution is basic in nature.

(C) On strong heating,orthoboric acid $(H_3BO_3)$ first dehydrates to form metaboric acid $(HBO_2)$ and water.
Upon further heating,metaboric acid decomposes into boric oxide $(B_2O_3)$,also known as boric anhydride,and water.
The overall reaction is:
$2 H_3BO_3 \xrightarrow{\Delta} B_2O_3 + 3 H_2O$

(A) The conversion of borax to crystalline boron proceeds as follows:
$1. Na_2B_4O_7 \cdot 10H_2O + 2HCl \rightarrow 4H_3BO_3 + 2NaCl + 5H_2O$
Here,$(X) = H_3BO_3$.
$2. 2H_3BO_3 \xrightarrow{\Delta} B_2O_3 + 3H_2O$
$3. B_2O_3 + 3Mg \xrightarrow{\Delta} 2B + 3MgO$
Here,$(Y) = Mg$.
Thus,$X$ is $H_3BO_3$ and $Y$ is $Mg$.

(A) . Boron-$10$ is used as a neutron absorber $(III)$.
$B$. Borax is used in the manufacture of heat resistance glasses $(IV)$.
$C$. Boron-fibre is used in bullet proof vests $(II)$.
$D$. Orthoboric acid is used as a mild antiseptic $(I)$.
Therefore,the correct match is $A-III, B-IV, C-II, D-I$.

(B) $I$. The structural formula of borax $(Na_2 B_4 O_7 \cdot 10 H_2 O)$ is $Na_2[B_4 O_5(OH)_4] \cdot 8 H_2 O$,which contains the $\left[B_4 O_5(OH)_4\right]^{2-}$ unit. Thus,statement $I$ is correct.
$II$. The aqueous solution of borax is basic in nature because it hydrolyzes to form $H_3 BO_3$ (a weak acid) and $NaOH$ (a strong base). The reaction is: $Na_2 B_4 O_7 + 7 H_2 O \longrightarrow 4 H_3 BO_3 + 2 NaOH$. Thus,statement $II$ is incorrect.
$III$. In the borax bead test,cobalt $(Co)$ forms cobalt metaborate,$Co(BO_2)_2$,which is blue in color. Thus,statement $III$ is correct.
Therefore,the correct statements are $I$ and $III$.

(C) Diborane $(B_2H_6)$ is a colourless,highly toxic,and pyrophoric gas at room temperature.
It has a repulsively sweet odour and is widely used as a chemical reagent in organic synthesis.
Therefore,the correct option is $(C)$.

(B) The reaction of borax $(Na_2B_4O_7)$ with water is:
$Na_2B_4O_7 + 7H_2O \rightarrow 4H_3BO_3 (B) + 2NaOH (A)$
The reaction of borax with dilute $HCl$ is:
$Na_2B_4O_7 + 2HCl + 5H_2O \rightarrow 2NaCl (C) + 4H_3BO_3 (B)$
From the above reactions,we identify $(A) = NaOH$ and $(C) = NaCl$.
Therefore,the correct option is $(B)$.

(A) $1.$ In orthoboric acid $H_3BO_3$,boron is $sp^2$ hybridized and thus has a planar geometry.
$2.$ In $BCl_3 \cdot NH_3$,boron is $sp^3$ hybridized due to the formation of a coordinate bond with $NH_3$,resulting in a tetrahedral geometry.
$3.$ Borax is a salt of a strong base $(NaOH)$ and a weak acid $(H_3BO_3)$. Therefore,its aqueous solution is basic in nature,not acidic.
Thus,statements $(a)$ and $(b)$ are correct.

(C) The chemical formula of borax is $Na_2 B_4 O_7 \cdot 10 H_2 O$,which is more accurately represented as $Na_2 [B_4 O_5 (OH)_4] \cdot 8 H_2 O$.
$Na_2 B_4 O_7 \cdot 5 H_2 O$ is known as tincalconite (borax pentahydrate).
$Na_2 B_4 O_7 \cdot 7 H_2 O$ and $Na_2 B_4 O_7 \cdot 2 H_2 O$ are not standard common forms of borax.

(D) In $BF_4^{-}$,the oxidation state of $B$ is $+3$. In $AlF_6^{3-}$,the oxidation state of $Al$ is $+3$. Thus,statement $(i)$ is incorrect.
In $BF_4^{-}$,the covalency of $B$ is $4$. In $AlF_6^{3-}$,the covalency of $Al$ is $6$. Thus,statement $(ii)$ is correct.
In $BF_4^{-}$,$B$ is surrounded by $4$ bonds,meaning $8$ electrons in its valence shell,so it obeys the octet rule. Thus,statement $(iii)$ is correct.
$B$ and $Al$ do not have a diagonal relationship; $B$ has a diagonal relationship with $Si$. Thus,statement $(iv)$ is incorrect.
Therefore,statements $(ii)$ and $(iii)$ are correct.

(A) Borax is $Na_2B_4O_7 \cdot 10H_2O$.
When dissolved in water,it undergoes hydrolysis:
$Na_2B_4O_7 + 7H_2O \longrightarrow 2NaOH + 4H_3BO_3$.
Thus,the products formed are sodium hydroxide $(NaOH)$ and orthoboric acid $(H_3BO_3)$.

(A) The reaction of diborane $(B_2H_6)$ with ammonia $(NH_3)$ at low temperature results in the formation of an ionic adduct $X$,which is diammoniate of diborane,$[BH_2(NH_3)_2]^+[BH_4]^-$.
The chemical equation is: $B_2H_6 + 2NH_3 \longrightarrow [BH_2(NH_3)_2]^+[BH_4]^-$.
Upon heating,this compound $X$ undergoes decomposition to form borazine $(B_3N_3H_6)$ and hydrogen gas $(H_2)$: $3[BH_2(NH_3)_2]^+[BH_4]^- \xrightarrow{\Delta} 2B_3N_3H_6 + 12H_2$.

(A) The molecular formula of diborane is $B_2H_6$.
In the structure of diborane,there are $4$ terminal $B-H$ bonds which are $2$-centre-$2$-electron $(2C-2e^-)$ bonds.
Additionally,there are $2$ bridging $B-H-B$ bonds,which are $3$-centre-$2$-electron $(3C-2e^-)$ bonds.
Therefore,the bonding consists of $4$ two-centre-two-electron bonds and $2$ three-centre-two-electron bonds.

(A) Diborane $(B_2H_6)$ reacts with $HCl$ in the presence of $AlCl_3$ as a catalyst to undergo hydrochlorination,which results in the liberation of hydrogen gas $(H_2)$.
The chemical reaction is as follows:
$B_2H_6 + HCl \xrightarrow{AlCl_3} B_2H_5Cl + H_2 \uparrow$

(B) The reaction of $BCl_3$ with $H_2$ in the presence of a $Cu-Al$ catalyst at $450^{\circ}C$ produces diborane $(B_2H_6)$ as product $X$:
$2BCl_3 + 6H_2 \xrightarrow{Cu-Al, 450^{\circ}C} B_2H_6 + 6HCl$
Diborane $(B_2H_6)$ undergoes methylation with $CH_3Cl$ to form tetramethyldiborane,which is $Z$:
$B_2H_6 + 4CH_3Cl \rightarrow (CH_3)_4B_2H_2 + 4HCl$
Thus,$Z$ is $(CH_3)_4B_2H_2$.

(B) Boron reacts with concentrated $HNO_3$ to form orthoboric acid $(H_3BO_3)$ and nitrogen dioxide $(NO_2)$.
The balanced chemical equation is:
$B + 3HNO_3 \longrightarrow H_3BO_3 + 3NO_2$
Therefore,the statement that boron reacts with concentrated $HNO_3$ to give $N_2O$ is incorrect.