Nitrogen family Questions in English

(B) $1$. Phosphine $(PH_3)$ has a rotten fish-like smell,not a rotten egg-like smell. Thus,statement $(a)$ is incorrect.
$2$. In aqueous solution and in the presence of light,$PH_3$ decomposes to give red phosphorus and hydrogen: $PH_{3(aq)} \xrightarrow{\text{light}} P_{(red)} + H_2$. Thus,statement $(b)$ is correct.
$3$. Phosphine is highly reactive and explodes in contact with strong oxidizing agents like $Br_2$,$Cl_2$,or $HNO_3$. Thus,statement $(c)$ is correct.
$4$. Pure phosphine is non-inflammable,but it becomes inflammable due to the presence of $P_2H_4$ or $P_4$ vapors. Thus,statement $(d)$ is correct.
Therefore,statements $(b)$,$(c)$,and $(d)$ are correct.

(B) The thermal decomposition reaction of ammonium dichromate is given by:
$(NH_4)_2Cr_2O_{7(s)} \rightarrow N_{2(g)} + 4H_2O_{(g)} + Cr_2O_{3(s)}$
Ammonium dichromate decomposes on heating to produce nitrogen gas $(N_2)$,water vapour,and solid chromium$(III)$ oxide $(Cr_2O_3)$.

(D) When ammonia $(NH_3)$ reacts with an excess of chlorine $(Cl_2)$,the reaction produces nitrogen trichloride $(NCl_3)$ and hydrogen chloride $(HCl)$.
The balanced chemical equation is:
$NH_3 + 3Cl_2 \rightarrow NCl_3 + 3HCl$
Since $Cl_2$ is in excess,the product formed is nitrogen trichloride $(NCl_3)$.

(A) The thermal decomposition of ammonium salts depends on the nature of the anion.
$1$. Heating $NH_4NO_3$ produces nitrous oxide and water: $NH_4NO_3 \rightarrow N_2O + 2H_2O$.
$2$. Heating $NH_4Cl$ produces ammonia and hydrogen chloride: $NH_4Cl \rightarrow NH_3 + HCl$.
$3$. Heating $(NH_4)_2CO_3$ produces ammonia,carbon dioxide,and water: $(NH_4)_2CO_3 \rightarrow 2NH_3 + CO_2 + H_2O$.
Since $NH_4NO_3$ does not produce $NH_3$ upon heating,the correct option is $A$.

(A) The given reaction is the hydrolysis of peroxodiphosphoric acid $(H_4P_2O_8)$:
$H_4P_2O_8 + 2H_2O \longrightarrow 2H_3PO_4 + H_2O_2$.
In the product $H_3PO_4$ (phosphoric acid),the phosphorus atom is in the $+5$ oxidation state.
Acids of phosphorus in the $+5$ oxidation state are named with the $-ic$ suffix (e.g.,phosphoric acid).
Therefore,the product is an oxyacid with an $-ic$ suffix.

(D) $NF_3$ is a very stable molecule due to the high electronegativity of fluorine and the small size of the nitrogen atom,which prevents the attack of water molecules on the nitrogen center.
Therefore,$NF_3$ does not undergo hydrolysis at room temperature $(R.T.)$.
Since there is no reaction,the product is not an oxy acid.
Thus,the correct description is that the product is not an oxy acid,neither with $-ic$ suffix nor with $-ous$ suffix.

(D) The reaction is: $NCl_3 + 3H_2O \longrightarrow 3HOCl + NH_3$.
In this reaction,the underlined atom is $N$ (nitrogen).
The products formed are $HOCl$ (hypochlorous acid) and $NH_3$ (ammonia).
$NH_3$ is a base,not an oxyacid.
$HOCl$ is an oxyacid of chlorine,but the question asks about the product related to the underlined atom $(N)$.
Since $NH_3$ is not an oxyacid,it does not have an -ic or -ous suffix.
Therefore,the product containing the underlined atom is not an oxyacid.

(B) The reaction is: $\underline{N}_2O_3 + H_2O \longrightarrow 2HNO_2$.
In $N_2O_3$,the oxidation state of nitrogen is $+3$.
$HNO_2$ is nitrous acid,which contains nitrogen in the $+3$ oxidation state.
Since the suffix for the acid with the lower oxidation state is $-ous$,$HNO_2$ is an oxy acid with an $-ous$ suffix.
Therefore,the correct option is $B$.

(A) The reaction for the hydrolysis of $N_2O_5$ is:
$N_2O_5 + H_2O \longrightarrow 2HNO_3$
In $HNO_3$ (nitric acid),the oxidation state of nitrogen is $+5$.
Since the nitrogen is in its highest oxidation state,the resulting acid is named with the $-ic$ suffix (nitric acid).
Therefore,the product is an oxyacid with an $-ic$ suffix.

(B) The complete hydrolysis of $PCl_3$ at room temperature $(R.T)$ is given by the reaction:
$PCl_3 + 3H_2O \longrightarrow H_3PO_3 + 3HCl$
In the product $H_3PO_3$ (phosphorous acid),the oxidation state of phosphorus is $+3$.
Oxy acids of phosphorus with an oxidation state of $+3$ are named with the suffix $-ous$.
Therefore,$H_3PO_3$ is phosphorous acid,which has an $-ous$ suffix.
Thus,the correct option is $B$.

(A) The complete hydrolysis of $PCl_5$ is given by the reaction:
$PCl_5 + 4H_2O \longrightarrow H_3PO_4 + 5HCl$
In this reaction,the phosphorus atom in $PCl_5$ (oxidation state $+5$) is converted to phosphoric acid $(H_3PO_4)$,where phosphorus is also in the $+5$ oxidation state.
Since the oxidation state of phosphorus remains $+5$,the product is phosphoric acid,which ends with the $-ic$ suffix.
Therefore,the correct option is $A$.

(C) The hydrolysis of phosphorus($III$,$V$) oxide,$\underline{P}_4O_8$,involves the reaction with water to produce a mixture of phosphoric acid and phosphorous acid.
The balanced chemical equation is:
$P_4O_8 + 6H_2O \longrightarrow 2H_3PO_3 + 2H_3PO_4$
In this reaction:
$1$. $H_3PO_4$ is phosphoric acid ($-ic$ suffix).
$2$. $H_3PO_3$ is phosphorous acid ($-ous$ suffix).
Since the products are two oxy acids,one with an $-ic$ suffix and the other with an $-ous$ suffix,the correct option is $C$.

(C) The reaction of $Zn$ with dilute $HNO_3$ is:
$4Zn 10HNO_3 (\text{dil.}) \to 4Zn(NO_3)_2 N_2O 5H_2O$
Thus,the gas $(P)$ is $N_2O$ (nitrous oxide).
$1.$ $N_2O$ is a neutral oxide,so it gives a neutral solution in water.
$2.$ $N_2O$ contains $33.3\%$ oxygen by mass,which is higher than the oxygen content in air $(\approx 21\%)$.
$3.$ The brown ring test is given by $NO$ gas,not by $N_2O$. Therefore,the statement that $N_2O$ forms a brown ring with $FeSO_4$ is incorrect.

(A) $1$. Heating of $NH_4ClO_4$: $2 NH_4ClO_4 \xrightarrow{\Delta} N_2 \uparrow + Cl_2 \uparrow + 2 O_2 \uparrow + 4 H_2O$. This reaction produces $N_2$ gas,not $NH_3$.
$2$. Heating of $NH_4Cl$: $NH_4Cl \xrightarrow{\Delta} NH_3 \uparrow + HCl \uparrow$. This reaction liberates $NH_3$ gas.
$3$. $(NH_4)_2CO_3 + 2 NaOH \longrightarrow 2 NH_3 \uparrow + Na_2CO_3 + 2 H_2O$. This reaction liberates $NH_3$ gas.
$4$. $Li_3N + 3 H_2O \longrightarrow 3 LiOH + NH_3 \uparrow$. This reaction liberates $NH_3$ gas.
Therefore,the heating of $NH_4ClO_4$ does not liberate $NH_3$ gas.

(A) $NCl_3 + 3 H_2O \longrightarrow NH_3 + 3 HOCl$
$PCl_3 + 3 H_2O \longrightarrow H_3PO_3 + 3 HCl$
$AsCl_3 + 3 H_2O \longrightarrow H_3AsO_3 + 3 HCl$
$BiCl_3 + H_2O \rightleftharpoons BiOCl + 2 HCl$
In the hydrolysis of $NCl_3$,the products are ammonia $(NH_3)$ and hypochlorous acid $(HOCl)$,not hydrochloric acid $(HCl)$.

(C) The reaction of lead $(Pb)$ with dilute nitric acid $(HNO_3)$ is given by:
$3Pb + 8HNO_3 (\text{dil.}) \xrightarrow{\Delta} 3Pb(NO_3)_2 + 2NO\uparrow + 4H_2O$
Here,$Q$ is $NO$ (Nitric oxide).
$NO$ is a paramagnetic colourless gas. (Statement $A$ is correct).
$NO$ reacts with $O_2$ in the air to form $NO_2$,which is a paramagnetic brown coloured gas. (Statement $B$ is correct).
$NO$ reacts with $FeSO_4$ to form a brown ring complex $[Fe(H_2O)_5(NO)]SO_4$,but it does not combine with $Fe_2(SO_4)_3$. (Statement $C$ is incorrect).
$HNO_2$ undergoes disproportionation to form $HNO_3$ and $NO$. (Statement $D$ is correct).
Therefore,the incorrect statement is $C$.

(C) The reaction is: $3Pb + 8HNO_3 \to 3Pb(NO_3)_2 + 2NO \uparrow + 4H_2O$.
Here,$Q$ is $NO$ (Nitric oxide).
$NO$ has $15$ electrons,making it a paramagnetic,colourless gas.
$NO$ reacts with $O_2$ in air to form $NO_2$ $(2NO + O_2 \to 2NO_2)$,which is a paramagnetic,brown-coloured gas.
$HNO_2$ undergoes disproportionation: $3HNO_2 \to HNO_3 + 2NO + H_2O$,producing $NO$.
$NO$ combines with $FeSO_4$ to form a brown ring complex $[Fe(H_2O)_5(NO)]SO_4$,but it does not combine with $Fe_2(SO_4)_3$ (where $Fe$ is in $+3$ oxidation state).
Therefore,the statement in option $C$ is incorrect.

(B) The reaction is: $Mg_3N_2(s) + 6H_2O(l) \rightarrow 3Mg(OH)_2(s) \downarrow + 2NH_3(g) \uparrow$.
Here,$P$ is $Mg(OH)_2$ and $Q$ is $NH_3$ gas.
Excess $NH_3$ forms coloured complexes with transition metal ions like $Ni^{2+}$,$Cr^{3+}$,and $Cu^{2+}$ due to $d-d$ transitions.
However,$Zn^{2+}$ has a $3d^{10}$ electronic configuration (no unpaired electrons),so the complex $[Zn(NH_3)_4]^{2+}$ is colourless.
Therefore,the correct option is $B$.

(C) The hydrolysis of group $15$ trihalides varies based on the nature of the central atom.
$PCl_3$ undergoes complete hydrolysis to form $H_3PO_3$ and $HCl$.
$NCl_3$ undergoes hydrolysis to form $NH_3$ and $HOCl$.
$SbCl_3$ and $BiCl_3$ undergo partial hydrolysis to form oxychlorides ($SbOCl$ and $BiOCl$ respectively) and $HCl$.
Therefore,the reaction $SbCl_3 + 3H_2O \to H_2SbO_3 + 3HCl$ is incorrect,as the correct reaction is $SbCl_3 + H_2O \to SbOCl + 2HCl$.

(C) Non-metals like phosphorus $(P_4)$ do not react with water but react with strong alkalis like $NaOH$ to form phosphine $(PH_3)$ and sodium hypophosphite $(NaH_2PO_2)$.
The reaction is: $P_4 + 3NaOH + 3H_2O \to PH_3 + 3NaH_2PO_2$.

(A) The thermal decomposition of ammonium dichromate is given by the reaction: $(NH_4)_2Cr_2O_7 \to N_2 + Cr_2O_3 + 4H_2O$. The gas liberated is nitrogen $(N_2)$.
Similarly,the thermal decomposition of ammonium nitrite is: $NH_4NO_2 \to N_2 + 2H_2O$. This also produces nitrogen gas $(N_2)$.

(A) $2NH_4ClO_4 \xrightarrow{\Delta} N_2 \uparrow + Cl_2 \uparrow + 2O_2 \uparrow + 4H_2O$. Here,$NH_3$ is not produced.
$(b)$ $NH_4Cl \xrightarrow{\Delta} NH_3 \uparrow + HCl \uparrow$. $NH_3$ is produced.
$(c)$ $(NH_4)_2CO_3 + 2NaOH \rightarrow 2NH_3 + 2H_2O + Na_2CO_3$. $NH_3$ is produced.
$(d)$ $Li_3N + 3H_2O \rightarrow 3LiOH + NH_3$. $NH_3$ is produced.

(A) $(A). \ N_2O \text{ is diamagnetic and neutral towards water.}$
$(B). \ NO_2 \text{ is paramagnetic and acidic in nature.}$
$(C). \ NO \text{ is paramagnetic and neutral towards water.}$
$(D). \ N_2O_3 \text{ is diamagnetic and acidic in nature.}$

(B) The reaction between lead$(IV)$ oxide $(PbO_2)$ and concentrated nitric acid $(HNO_3)$ is a redox reaction where $PbO_2$ acts as an oxidizing agent.
The balanced chemical equation is:
$PbO_2 + 2HNO_3 \to Pb(NO_3)_2 + H_2O + \frac{1}{2}O_2 \uparrow$
As shown in the equation,oxygen gas $(O_2)$ is evolved during the reaction.

(A) In the structure of $P_4O_{10}$,each phosphorus atom is bonded to four oxygen atoms.
Three of these oxygen atoms are bridging oxygen atoms ($P$-$O$-$P$),and one is a terminal oxygen atom ($P$=$O$).
Therefore,the total number of oxygen atoms attached to each phosphorus atom is $4$.

(D) $1$. Borax $(Na_2[B_4O_5(OH)_4] \cdot 8H_2O)$ contains the $[B_4O_5(OH)_4]^{2-}$ ion,which has $2$ six-membered rings.
$2$. $P_4O_{10}$ has a cage-like structure containing $4$ six-membered rings in a chair conformation.
$3$. Beryl $(Be_3Al_2Si_6O_{18})$ is a cyclic silicate containing $[Si_6O_{18}]^{12-}$ rings,where $12$ oxygen atoms are monovalent (terminal) and $6$ are shared.
$4$. $(HPO_3)_3$ (trimetaphosphoric acid) is a cyclic compound containing $3$ $P-O-P$ linkages,not $P-P$ linkages. Therefore,option $D$ is the incorrect match.

(D) $I.$ $PCl_3$ reacts with moisture to produce $H_3PO_3$ and $HCl$ fumes,which is a correct statement.
$II.$ In the solid state,$PCl_5$ exists as an ionic solid $[PCl_4]^+[PCl_6]^-$,which is a correct statement.
$III.$ In the gaseous phase,$PCl_5$ has a trigonal bipyramidal geometry where the three equatorial bonds are equivalent and the two axial bonds are longer and weaker than the equatorial bonds. Thus,all five bonds are not equivalent. This statement is incorrect.
Therefore,statements $I$ and $II$ are correct.

(C) $1$. Due to the inert pair effect,the stability of the lower oxidation state increases down the group.
$2$. $Bi^{5+}$ is highly unstable because $Bi(III)$ is the most stable oxidation state for Bismuth,hence $Bi^{5+}$ salts do not exist.
$3$. $Pb^{4+}$ is a strong oxidising agent because $Pb(II)$ is more stable than $Pb(IV)$.
$4$. $Sn^{4+}$ is the most stable oxidation state for Tin,so it is not a good oxidising agent.
$5$. $Tl^{+}$ is the most stable oxidation state for Thallium,so it is not a good oxidising agent.
$6$. Therefore,the statement that $Sn^{4+}$ salts are good oxidising agents is incorrect.

(C) $NO_2$ contains an odd number of valence electrons,specifically one unpaired electron on the nitrogen atom.
Due to this unpaired electron,$NO_2$ is paramagnetic and reactive.
To achieve a more stable electronic configuration,two $NO_2$ molecules combine to form the dimer $N_2O_4$,which is diamagnetic.

(A) The reaction between nitric acid $(HNO_3)$ and phosphorus pentoxide $(P_4O_{10})$ is a dehydration reaction.
The balanced chemical equation is:
$4HNO_3 + P_4O_{10} \to 4HPO_3 + 2N_2O_5$
Comparing this with the given equation $HNO_3 + P_4O_{10} \to 4HPO_3 + X$,we can see that $X$ corresponds to $2N_2O_5$.
Therefore,$X$ is $N_2O_5$ (dinitrogen pentoxide).

(NONE) Let us analyze each reaction:
$1$. $4HNO_3 + P_4O_{10} \to 2N_2O_5 + 4HPO_3$: This reaction is correct as $P_4O_{10}$ acts as a dehydrating agent.
$2$. $P_4 + 10SO_2Cl_2 \to 4PCl_5 + 10SO_2$: This reaction is correct.
$3$. $Ca_3P_2 + 6HCl \to 3CaCl_2 + 2PH_3$: This reaction is correct.
$4$. $NH_4Cl + NaNO_2 \xrightarrow{\Delta} N_2 + 2H_2O + NaCl$: This reaction is correct.
Upon reviewing the provided options,all reactions are chemically balanced and represent standard textbook reactions for the preparation of these compounds. However,if this is a multiple-choice question where one must be incorrect,there may be a typo in the provided options. Given the standard chemistry,all these reactions are correct. If forced to choose,please verify the source material for a potential typo in the stoichiometry or products.

(C) The chemical formula of hypophosphorous acid is $H_{3}PO_{2}$.
In its structure,the phosphorus atom is bonded to one oxygen atom via a double bond $(P=O)$,one hydroxyl group $(-OH)$,and two hydrogen atoms directly attached to the phosphorus atom ($P-H$ bonds).
Therefore,there are $2$ hydrogen atoms directly attached to the $P$ atom.

(D) $N_2O_3$,$NO_2$,and $N_2O_5$ are acidic oxides because they react with water to form acids (e.g.,$N_2O_3 + H_2O \rightarrow 2HNO_2$).
$NO$ (Nitric oxide) is a neutral oxide and does not react with water to form an acidic solution.

(A) $1$. $NH_4NO_2$ on heating produces $N_2$ gas: $NH_4NO_2 \rightarrow N_2 + 2H_2O$.
$2$. $NaN_3$ (Sodium azide) on heating produces $N_2$ gas: $2NaN_3 \rightarrow 2Na + 3N_2$.
$3$. $Ba(N_3)_2$ (Barium azide) on heating produces $N_2$ gas: $Ba(N_3)_2 \rightarrow Ba + 3N_2$.
$4$. $NH_2CONH_2$ (Urea) on heating decomposes to produce ammonia $(NH_3)$ and cyanic acid $(HNCO)$,not $N_2$ gas.

(C) Let us analyze each option:
$(A)$ $Mg_2C_3 + 4H_2O \to 2Mg(OH)_2 + C_3H_4$ (Propyne); $Al_4C_3 + 12H_2O \to 4Al(OH)_3 + 3CH_4$ (Methane). Gases are different.
$(B)$ $NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$; $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$. Gases are different.
$(C)$ $NH_4Cl + NaOH \to NaCl + H_2O + NH_3(X)$; $NaNO_3 + 8Al + 5NaOH + 18H_2O \to 8Na[Al(OH)_4] + 3NH_3(Y)$. Both produce $NH_3$. Thus,$X = Y$.
$(D)$ $3Cu + 8dil. HNO_3 \to 3Cu(NO_3)_2 + 4H_2O + 2NO(X)$; $3Ag + 4dil. HNO_3 \to 3AgNO_3 + 2H_2O + NO(Y)$. Both produce $NO$. Thus,$X = Y$.
Note: In many competitive contexts,both $(C)$ and $(D)$ are considered correct as they produce the same gaseous products.

(A) The reaction cycle for the production and interconversion of nitrogen oxides is as follows:
$1$. $NO$ reacts with $O_2$ in air to form $NO_2$: $2NO + O_2 \rightarrow 2NO_2$.
$2$. $NO_2$ reacts with water to form nitric acid: $3NO_2 + H_2O \rightarrow 2HNO_3 + NO$.
$3$. Dilute nitric acid $(HNO_3)$ reacts with dilute metals like copper $(Cu)$ or silver $(Ag)$ to produce $NO$. Specifically,$3Ag + 4HNO_3 \text{ (dilute)} \rightarrow 3AgNO_3 + NO + 2H_2O$.
Comparing this with the given options,option $A$ correctly represents the cycle where $Ag$ reacts with dilute $HNO_3$ to produce $NO$.

(D) $P_4O_{10}$ is a strong dehydrating agent.
When it reacts with concentrated $HNO_3$,it dehydrates the acid to form its anhydride,$N_2O_5$,and phosphoric acid derivative,$HPO_3$.
The balanced chemical equation is:
$4 HNO_3 + P_4O_{10} \rightarrow 2 N_2O_5 + 4 HPO_3$

(C) Laughing gas $(N_2O)$ is prepared by the thermal decomposition of ammonium nitrate $(NH_4NO_3)$.
The reaction is: $NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$.
Alternatively,it can be prepared by heating a mixture of $NH_4Cl$ and $NaNO_3$ as they react to form $NH_4NO_3$ in situ,which then decomposes to produce $N_2O$.

(C) The acidic character of non-metallic oxides increases with an increase in the oxidation state of the central atom and decreases down the group as metallic character increases.
Comparing the oxidation states: $P$ in $P_4O_6$ is $+3$,$P$ in $P_4O_{10}$ is $+5$,$As$ in $As_4O_6$ is $+3$,and $As$ in $As_4O_{10}$ is $+5$.
Since $As$ is more metallic than $P$,its oxides are less acidic than the corresponding oxides of $P$.
Between $As_4O_6$ and $As_4O_{10}$,$As_4O_6$ has a lower oxidation state $(+3 < +5)$,making it the least acidic among the given options.

(A) The reaction of white phosphorus $(P_4)$ with aqueous sodium hydroxide $(NaOH)$ is a disproportionation reaction.
The balanced chemical equation is:
$P_4 + 3NaOH + 3H_2O \longrightarrow PH_3 + 3NaH_2PO_2$
In this reaction,phosphorus is simultaneously oxidized and reduced.
$PH_3$ is phosphine and $NaH_2PO_2$ is sodium hypophosphite.

(D) An acidic anhydride is an oxide that reacts with water to form an acid.
$N_2O_5$ is the anhydride of nitric acid $(HNO_3)$.
The reaction is: $N_2O_5 + H_2O \longrightarrow 2HNO_3$.
$CO$ and $NO$ are neutral oxides,while $ClO_2$ is a mixed anhydride.

(A) Pure phosphine $(PH_3)$ is non-flammable. However,it becomes inflammable due to the presence of impurities like diphosphine $(P_2H_4)$ or white phosphorus $(P_4)$ vapours.

(A) $X$ $(2NO + O_2)$ consists of colourless gases.
$Y$ $(NO_2)$ is a brown-coloured gas and is paramagnetic due to the presence of an odd electron.
$Z$ $(N_2O_4)$ is a colourless solid or liquid and is diamagnetic because all electrons are paired.

(D) $Au$ (Gold) is a noble metal and is chemically inert towards concentrated $HNO_3$.
$Cu$ and $Ag$ react with concentrated $HNO_3$ to produce $NO_2$ gas.
$I_2$ is oxidized by concentrated $HNO_3$ to $HIO_3$ along with the evolution of $NO_2$ gas.

(A) The reaction of bauxite $(Al_2O_3)$ with coke $(C)$ and nitrogen $(N_2)$ at $2075 \ K$ is:
$Al_2O_3 + 3C + N_2 \xrightarrow{2075 \ K} 2AlN + 3CO$
Here,$x$ is aluminium nitride $(AlN)$.
When $AlN$ reacts with water,it undergoes hydrolysis to produce ammonia gas:
$AlN + 3H_2O \rightarrow Al(OH)_3 + NH_3 \uparrow$
Thus,the gas produced is $NH_3$.

(D) Aluminum becomes passive in concentrated oxidizing acids because it forms a thin,protective,non-porous layer of oxide $(Al_2O_3)$ on its surface. This layer prevents further reaction of the metal with the acid. Concentrated $HNO_3$,$H_2CrO_4$,and $HClO_4$ are all strong oxidizing agents that cause this passivation.

(C) The ionization energy of $B$ is very high. The sum of the first three ionization enthalpies of boron is extremely large,which is not compensated by the hydration energy of the $B^{3+}$ ion. Therefore,the $B^{3+}$ ion does not exist in an aqueous solution.

(A) The inert pair effect is the tendency of the two electrons in the outermost $s$-orbital to remain unshared in compounds due to the poor shielding effect of intervening $d$ and $f$ orbitals.
As we move down the group from $B$ to $Tl$,the inert pair effect increases.
Therefore,the stability of the $+1$ oxidation state increases down the group,while the stability of the $+3$ oxidation state decreases.
The correct order of the inert pair effect for Group $13$ elements is $B < Al < Ga < In < Tl$.

(B) The inert pair effect is the tendency of the two electrons in the outermost $s$-orbital to remain unshared in compounds of post-transition metals. This effect becomes more prominent as we move down the group in the $p$-block elements. $Aluminium$ $(Al)$ belongs to group $13$ and is the first member of the group. It does not exhibit the inert pair effect because it lacks the $d$ or $f$ orbitals that would shield the $s$-electrons from the nucleus,and it is too small to have a stable $+1$ oxidation state. Therefore,$Al$ typically shows a $+3$ oxidation state.

(A) Alums are double salts with the general formula $M^I M^{III} (SO_4)_2 \cdot 12H_2O$,where $M^I$ is a monovalent metal ion (like $K^+$,$Na^+$,$NH_4^+$) and $M^{III}$ is a trivalent metal ion (like $Al^{3+}$,$Cr^{3+}$,$Fe^{3+}$). Therefore,all alums contain one monovalent and one trivalent metal.