Boron family Questions in English

(B) Diborane $(B_2H_6)$ reacts with ammonia $(NH_3)$ under different conditions to yield various products:
$1$. At low temperature,it forms an ionic addition product: $B_2H_6 + 2NH_3 \rightarrow [BH_2(NH_3)_2]^+ [BH_4]^-$ (often written as $B_2H_6 \cdot 2NH_3$).
$2$. At high temperatures,it forms borazine $(B_3N_3H_6)$,which is known as inorganic benzene: $3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 12H_2$.
$3$. At very high temperatures,it forms boron nitride $((BN)_n)$,which is isoelectronic with graphite: $B_2H_6 + 2NH_3 \rightarrow 2BN + 6H_2$.
Therefore,$B_{12}H_{12}$ is not formed in the reaction between diborane and ammonia.

(C) According to Lewis,compounds that can accept a lone pair of electrons are called Lewis acids.
Boron halides,such as $BX_3$,have only $6$ electrons in the valence shell of the boron atom.
Due to this electron deficiency,they can accept a lone pair of electrons from a donor to complete their octet,thus behaving as Lewis acids.

(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.

(A) Statement $(I)$ is correct due to the inert pair effect,where the stability of the $+1$ oxidation state increases down the group $(Ga < In < Tl)$.
Statement $(II)$ is incorrect; Boron has a very high melting point due to its giant covalent structure.
Statement $(III)$ is correct; Boron lacks $d$-orbitals and can accommodate a maximum of $4$ electron pairs in its valence shell.
Statement $(IV)$ is incorrect; the correct order of atomic radius is $Al > Ga < In$ (due to poor shielding of $d$-electrons in $Ga$).
Statement $(V)$ is correct; Aluminium forms a protective oxide layer on its surface when treated with concentrated nitric acid,making it passive.
Thus,statements $(I, III, V)$ are correct.

(B) To find the element $X$ for which $T_2 - T_1$ is maximum,we calculate the difference for each element given in the table:
For $B$: $3923 - 2453 = 1470 \ K$
For $Al$: $2740 - 933 = 1807 \ K$
For $Ga$: $2676 - 303 = 2373 \ K$
For $In$: $2353 - 430 = 1923 \ K$
Comparing these values,the maximum difference is $2373 \ K$,which corresponds to $Ga$.

(D) When aluminium reacts with dilute hydrochloric acid,it forms aluminium chloride and releases hydrogen gas:
$2 Al_{(s)} + 6 HCl_{(aq)} \longrightarrow 2 AlCl_{3(aq)} + 3 H_2 \uparrow$
Similarly,aluminium reacts with aqueous alkali to release hydrogen gas:
$2 Al_{(s)} + 2 NaOH_{(aq)} + 6 H_2O_{(l)} \longrightarrow 2 Na[Al(OH)_4]_{(aq)} + 3 H_2 \uparrow$
Thus,both gases $A$ and $B$ are $H_2$.

(D) . When $Al$ reacts with dil. $HCl$,it liberates $H_2$ gas: $2 Al + 6 HCl \text{ (dil.) } \longrightarrow 2 AlCl_3 + 3 H_2$.
$B$. When $Al$ reacts with conc. $HNO_3$,it becomes passive due to the formation of a protective oxide layer and does not liberate $H_2$ gas.
$C$. Crystalline boron is chemically inert and does not react with acids to liberate $H_2$ gas.
$D$. Anhydrous $AlCl_3$ undergoes hydrolysis with moisture to release $HCl$ fumes,not $H_2$ gas: $AlCl_3 + 3 H_2 O \longrightarrow Al(OH)_3 + 3 HCl$.
Therefore,only statement $A$ is correct.

(D) In an acidic medium $(pH < 7)$,$AlCl_3$ undergoes hydrolysis to form the hexaaquaaluminum$(III)$ complex ion.
The reaction is: $AlCl_3 + 6H_2O \longrightarrow [Al(H_2O)_6]^{3+} + 3Cl^{-}$
This complex ion has an octahedral geometry.

(C) Aluminium reacts with aqueous $NaOH$ to form sodium tetrahydroxoaluminate$(III)$.
The reaction is: $2Al(s) + 2NaOH(aq) + 6H_2O(l) \longrightarrow 2Na[Al(OH)_4](aq) + 3H_2(g)$.
In aqueous solution,the species exists as the octahedral complex $[Al(H_2O)_2(OH)_4]^-$,where the coordination number of $Al$ is $6$.

(C) The atomic radii of group $13$ elements are expected to increase down the group due to the addition of new shells.
However, due to the poor shielding effect of $d$ and $f$ orbitals (lanthanoid contraction), the atomic radius of $Tl$ is slightly larger than or nearly equal to $In$.
The actual observed order is $Al < Ga < In < Tl$.
Wait, checking the trend: $Al$ $(143 \text{ pm})$, $Ga$ $(135 \text{ pm})$, $In$ $(167 \text{ pm})$, $Tl$ $(170 \text{ pm})$.
Thus, the correct order is $Al < Ga < In < Tl$, which means $Tl > In > Ga > Al$.

(B) Boron $(B)$ is a non-metal and does not form a stable triiodide $(BI_3)$ under standard conditions due to its small size and high ionization energy,although $BI_3$ can be synthesized under specific conditions. However,in the context of group $13$ elements and the inert pair effect,$Tl$ (Thallium) is the most distinct. $Tl$ prefers the $+1$ oxidation state over the $+3$ state due to the inert pair effect,making $TlI_3$ highly unstable and it readily decomposes into $TlI$ and $I_2$. Thus,$Tl$ does not form a stable triiodide.

(D) Borax $(Na_2B_4O_7 \cdot 10H_2O)$ dissolves in water to form an alkaline solution.
The hydrolysis reaction is: $Na_2B_4O_7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$.
In this reaction,$NaOH$ is a strong base and $H_3BO_3$ (orthoboric acid) is a very weak acid.
Due to the presence of the strong base $NaOH$,the resulting solution is alkaline in nature.

(C) An oxide that reacts with both acids and bases is known as an amphoteric oxide.
$Al_2O_3$ is an amphoteric oxide.
It reacts with hydrochloric acid $(HCl)$ to form aluminum chloride:
$Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O$.
It also reacts with sodium hydroxide $(NaOH)$ to form sodium aluminate:
$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$.

(A) The correct matches are as follows:
$A$. $H_3BO_3$ (Boric acid) exhibits intermolecular hydrogen bonding in its solid state. Thus,$A-3$.
$B$. $AlCl_3$ exists as a dimer $(Al_2Cl_6)$ in the vapor phase or non-polar solvents. Thus,$B-1$.
$C$. $B_3N_3H_6$ (Borazine) is known as inorganic benzene due to its structural similarity to benzene. Thus,$C-4$.
$D$. $BF_3$ involves back bonding between the lone pair of fluorine and the vacant p-orbital of boron. Thus,$D-2$.
Therefore,the correct sequence is $A-3, B-1, C-4, D-2$.

(C) The molecule $B_2H_6$ (Diborane) contains two bridging hydrogen atoms.
In $B_2H_6$,each boron atom is $sp^3$ hybridized.
Two hydrogen atoms act as bridges between the two boron atoms,forming two $3c-2e$ (three-center-two-electron) bonds,which are also known as hydrogen bridge bonds or banana bonds.

(C) The structure of phosphorous acid $(H_{3}PO_{3})$ contains one $P=O$ bond,two $P-OH$ bonds,and one $P-H$ bond.
Protons attached directly to oxygen atoms in $P-OH$ groups are acidic because the oxygen atom is highly electronegative,which facilitates the release of the $H^+$ ion.
The hydrogen atom directly bonded to the phosphorus atom $(P-H)$ is not acidic because the electronegativity difference between $P$ and $H$ is very small.
Therefore,$H_{3}PO_{3}$ has $2$ acidic protons.

(A) In the structure of diborane $(B_2H_6)$,there are two boron atoms.
Each boron atom is bonded to two terminal hydrogen atoms via normal covalent bonds ($2 \times 2 = 4$ terminal hydrogens).
The two boron atoms are linked by two bridging hydrogen atoms,which form $3c-2e$ (three-center two-electron) bonds.
Therefore,there are $4$ terminal hydrogens and $2$ bridging hydrogens.

(A, D) In diborane $(B_2H_6)$,both boron atoms are $sp^3$ hybridised.
It is diamagnetic because all electrons are paired.
It contains two $3C-2e$ (three-center two-electron) bonds,also known as banana bonds.
There are two types of hydrogen atoms: $4$ terminal hydrogens and $2$ bridging hydrogens.

(D) Diborane $(B_2H_6)$ has a bridged structure where two $BH_2$ units are linked by two bridging hydrogen atoms.
Each bridge consists of a $3$-center-$2$-electron $(3c-2e)$ bond.
Since there are two such bridges,the total number of electrons involved in the bridging bonds is $2 \times 2 = 4$ electrons.
These are often referred to as banana bonds.

(A) The chemical formula of the borate anion in borax is $[B_4O_5(OH)_4]^{2-}$.
By examining the structure of the $[B_4O_5(OH)_4]^{2-}$ ion:
$1$. There are $5$ $B-O-B$ linkages forming the cyclic structure.
$2$. There are $4$ $B-OH$ bonds attached to the boron atoms.
Therefore,the number of $B-O-B$ links is $5$ and the number of $B-OH$ bonds is $4$.

(B) The self-ionization of interhalogen compounds like $BrF_3$ occurs through the transfer of a fluoride ion $(F^-)$.
The reaction is represented as: $2 BrF_3 \rightleftharpoons BrF_2^+ + BrF_4^-$.
Here,$BrF_3$ acts as both an acid and a base,forming the $BrF_2^+$ cation and the $BrF_4^-$ anion.

(C) Ionic radii of trivalent cations increase down the group due to the addition of new shells: $B^{3+} < Al^{3+} < Ga^{3+} < In^{3+} < Tl^{3+}$. Thus,statement $(A)$ is incorrect.
$(B)$ Electronegativity does not decrease regularly down the group due to poor shielding of $d$ and $f$ orbitals: $B > Tl > In > Ga > Al$. Thus,statement $(B)$ is incorrect.
$(C)$ Boron has the smallest atomic size in the group,leading to the highest first ionisation enthalpy: $B > Tl > Ga > Al > In$. Thus,statement $(C)$ is correct.
$(D)$ According to Fajan's rule,small cations with high charge density form covalent bonds. Trichlorides and triiodides of group $13$ elements are covalent in nature. Thus,statement $(D)$ is correct.

(B) The reaction of borax on heating is: $Na_2 B_4 O_7 \xrightarrow{\Delta} 2 NaBO_2 (X) + B_2 O_3 (Y)$.
In the non-luminous flame,$CuSO_4$ reacts with $B_2 O_3$ to form copper$(II)$ metaborate: $CuSO_4 + B_2 O_3 \xrightarrow{\Delta} Cu(BO_2)_2 (Z) + SO_3$. Here,the oxidation state of $Cu$ in $Z$ is $+2$.
In the luminous flame,$Cu(BO_2)_2$ is reduced by carbon in the presence of $NaBO_2$ to form copper$(I)$ metaborate: $2 Cu(BO_2)_2 + 2 NaBO_2 + C \xrightarrow{\Delta} 2 CuBO_2 (Q) + Na_2 B_4 O_7 + CO$. Here,the oxidation state of $Cu$ in $Q$ is $+1$.
Thus,the oxidation states of $Cu$ in $Z$ and $Q$ are $+2$ and $+1$ respectively.

(A) Statement $I$: The standard electrode potential $E^\circ_{Al^{3+}/Al}$ is approximately $-1.66 \ V$,while $E^\circ_{Tl^{3+}/Tl}$ is approximately $+1.26 \ V$. $A$ more negative electrode potential indicates a stronger reducing agent and a more electropositive nature. Therefore,Statement $I$ is true.
Statement $II$: The sum of the first three ionization enthalpies of Boron is extremely high,which prevents the formation of $B^{3+}$ ions,leading it to form only covalent compounds. In contrast,Aluminium has a significantly lower sum of ionization enthalpies,which allows it to form $Al^{3+}$ ions under specific conditions. Therefore,Statement $II$ is true.