Boron family Questions in English

(A) $BCl_3$ is the compound formed.
$2B + 3Cl_2 \rightarrow 2BCl_3$
$BCl_3$ is electron deficient but it does not form a dimer like $AlCl_3$,$GaCl_3$,or $InCl_3$ because its electron deficiency is compensated by the formation of a $p\pi - p\pi$ back-bonding between the lone pair of electrons of chlorine and the empty unhybridized $p$-orbital of boron.
Thus,$X$ is Boron $(B)$.

(B) Amphoteric substances are those that can react with both acids and bases.
$Al_2O_3 \cdot xH_2O$ is an amphoteric oxide.
It reacts with a base $(OH^{-})$ to form aluminate:
$Al_2O_3 \cdot xH_2O_{(s)} + 2OH^{-}{(aq)} \longrightarrow 2AlO_2^{-}{(aq)} + (x+1)H_2O_{(l)}$
It reacts with an acid $(H^{+})$ to form aluminum ions:
$Al_2O_3 \cdot xH_2O_{(s)} + 6H^{+}_{(aq)} \longrightarrow 2Al^{3+}_{(aq)} + (x+3)H_2O_{(l)}$
Therefore,Set $1$ (reaction with base) and Set $3$ (reaction with acid) best demonstrate its amphoteric nature.

(B) In the structure of diborane $(B_2H_6)$,there are four terminal $B-H$ bonds,each of which is a $2$-centre-$2$-electron $(2c-2e^-)$ bond.
There are two bridging $B-H-B$ bonds,each of which is a $3$-centre-$2$-electron $(3c-2e^-)$ bond.
Thus,the number of $2c-2e^-$ bonds is $4$ and the number of $3c-2e^-$ bonds is $2$.

(B) The reaction of diborane $(B_2H_6)$ with oxygen $(O_2)$ is highly exothermic and produces boron trioxide $(B_2O_3)$ and water:
$B_2H_6 + 3O_2 \longrightarrow B_2O_3 + 3H_2O$
The reaction of diborane $(B_2H_6)$ with water $(H_2O)$ produces boric acid $(H_3BO_3)$ and hydrogen gas $(H_2)$:
$B_2H_6 + 6H_2O \longrightarrow 2H_3BO_3 + 6H_2$
Therefore,the products are $B_2O_3$ and $H_3BO_3$ respectively.

(A) $B_2O_3$ is acidic in nature.
$Al_2O_3$ and $Ga_2O_3$ are amphoteric in nature.
$In_2O_3$ and $Tl_2O_3$ are basic in nature.
As we move down the group $13$,the metallic character of the elements increases,which leads to a transition from acidic to amphoteric and finally to basic oxides.

(C) Borax is the salt of a strong base $(NaOH)$ and a weak acid $(H_3BO_3)$.
When it hydrolyses,it produces $NaOH$ and $H_3BO_3$.
Since $NaOH$ is a strong base and $H_3BO_3$ is a weak acid,the resulting solution is basic in nature,not acidic.
Therefore,the statement that it gives an acidic solution is incorrect.

(B) The reaction of diborane $(B_2H_6)$ with excess ammonia $(NH_3)$ depends on the temperature conditions.
At low temperatures,it forms an ionic adduct,$[BH_2(NH_3)_2]^+ [BH_4]^-$.
However,at very high temperatures $(> 200 \, ^\circ C)$,the reaction leads to the formation of inorganic graphite or boron nitride,represented as $(BN)_x$.

(A) $H_3BO_3$ acts as a weak monobasic Lewis acid by accepting an $OH^-$ ion from water: $H_3BO_3 + H_2O \longrightarrow [B(OH)_4]^- + H^+$.
Option $A$ is considered incorrect because $H_3BO_3$ does not act as a Brønsted-Lowry acid (proton donor) by simple dissociation; rather,it acts as a Lewis acid by accepting an $OH^-$ ion from water,which subsequently releases $H^+$ ions.

(C) . $PCl_5$ undergoes hydrolysis to form $POCl_3$ and finally $H_3PO_4$ (phosphoric acid),which is a weak tribasic acid. Thus,statement $a$ is incorrect.
$b$. In borazine $(B_3N_3H_6)$,the $B$ atom is electron-deficient and the $N$ atom has a lone pair. Nucleophiles attack the electron-deficient $B$ atom. Thus,statement $b$ is correct.
$c$. $Al_2Cl_6$ is a non-polar,planar molecule (dimeric structure). Thus,statement $c$ is incorrect.
$d$. $AlF_3$ is an ionic compound with a high lattice energy and does not exhibit covalent hybridization in the solid state. Thus,statement $d$ is incorrect.
Therefore,statements $a, c,$ and $d$ are incorrect.

(B) Diborane $(B_2H_6)$ has a bridged structure.
It contains two terminal $BH_2$ groups and two bridging hydrogen atoms.
The two bridging hydrogen atoms are bonded to the two boron atoms through three-centre two-electron $(3c-2e)$ bonds,often referred to as banana bonds.
Therefore,there are $2$ such bonds in a molecule of diborane.

(A) $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$ is commonly known as Borax.
The polyanion present in its structure is $[B_4O_5(OH)_4]^{2-}$.
This structure consists of two boron atoms in triangular coordination ($BO_3$ units) and two boron atoms in tetrahedral coordination ($BO_4$ units).
Therefore,it contains two triangular and two tetrahedral units.

(A) Inorganic benzene (borazine,$B_3N_3H_6$) has a structure similar to benzene where $B$ atoms have a partial negative charge and $N$ atoms have a partial positive charge due to the coordinate bond nature of the $B-N$ bond.
When $HCl$ reacts with borazine,the $H^{+}$ ion (proton) attacks the nucleophilic $B^{-}$ center first.
Subsequently,the $Cl^{-}$ ion attacks the electrophilic $N^{+}$ center.
This process repeats three times to form the addition product $B_3N_3H_9Cl_3$.

(C) The hydrolysis reactions are as follows:
$BCl_3 + 3H_2O \rightarrow H_3BO_3 + 3HCl$
$NCl_3 + 3H_2O \rightarrow NH_3 + 3HOCl$
From the reactions,it is clear that $BCl_3$ produces $HCl$ upon hydrolysis,while $NCl_3$ produces $HOCl$ upon hydrolysis.
Therefore,the correct statement is that $NCl_3$ on hydrolysis gives $HOCl$ but $BCl_3$ gives $HCl$.

(D) Boron can be obtained by $(I)$ thermal decomposition of diborane,$(II)$ pyrolysis of boron triiodide (van Arkel) method,and $(III)$ reduction of boron trichloride with hydrogen.
However,it cannot be obtained by the electrolysis of fused boron trichloride.
This is because electrometallurgy is used for the extraction of highly electropositive metals such as sodium,potassium,magnesium,calcium,and aluminium,whereas boron is a non-metal.

(B) When a mineral acid (such as $HCl$) is added to an aqueous solution of borax $(Na_{2}B_{4}O_{7} \cdot 10H_{2}O)$,it reacts to form orthoboric acid $(H_{3}BO_{3})$.
The chemical equation for the reaction is:
$Na_{2}B_{4}O_{7} + 2HCl + 5H_{2}O \rightarrow 2NaCl + 4H_{3}BO_{3}$
Thus,orthoboric acid is the product formed.
Hence,the correct answer is option $B$.

(B) $AlCl_3$ on hydrolysis gives $Al(OH)_3$.
Hydrolysis is the chemical breakdown of a compound due to reaction with water.
The balanced chemical equation for the hydrolysis of aluminum chloride is:
$AlCl_3 + 3H_2O \rightarrow Al(OH)_3 + 3HCl$
Hence,option $B$ is correct.

(D) White fumes are due to the presence of $HCl$ gas.
Anhydrous aluminium chloride is hydrolysed partially with the moisture in the atmosphere to give $HCl$ gas.
This $HCl$ combines with the moisture in the air and appears white in colour.
$AlCl_{3} + 3H_{2}O \longrightarrow Al(OH)_{3} + 3HCl$ (white fumes)
Hence,the correct answer is option $D$.

(D) Sublimation is the process where a solid directly converts into gas without reaching the liquid state.
$AlCl_3$ is a solid that directly converts into gas upon heating at $180 \ ^\circ C$ under atmospheric pressure.

(C) Boron and silicon exhibit a diagonal relationship due to their similar ionic potential (charge/size ratio).
One of the key characteristics of this relationship is that both elements form acidic oxides ($B_2O_3$ and $SiO_2$) which dissolve in strong alkali solutions to form borates and silicates,respectively.
$B_2O_3 + 2NaOH \rightarrow 2NaBO_2 + H_2O$
$SiO_2 + 2NaOH \rightarrow Na_2SiO_3 + H_2O$

(B) The chemical formula of borax is $Na_{2}[B_{4}O_{5}(OH)_{4}] \cdot 8H_{2}O$.
The polyanion unit is $[B_{4}O_{5}(OH)_{4}]^{2-}$.
In this structure,there are $4$ boron atoms.
Two boron atoms are $sp^{2}$ hybridized,which results in a trigonal planar geometry.
Two boron atoms are $sp^{3}$ hybridized,which results in a tetrahedral geometry.
Therefore,the structure contains $2$ tetrahedral and $2$ planar triangular units.

(A) In diborane $(B_2H_6)$,each boron atom undergoes $sp^3$ hybridization.
In the bridged structure of diborane,the $B-H-B$ angle is $83^{\circ}$,not $180^{\circ}$.
Each boron atom is linked to two terminal hydrogen atoms via $2c-2e$ bonds and two bridged hydrogen atoms via $3c-2e$ bonds. Thus,statement $3$ is correct.
Diborane has $8$ atoms ($2$ Boron and $6$ Hydrogen). Total valence electrons = $2 \times 3 + 6 \times 1 = 12$. These $12$ electrons are involved in bonding ($4$ terminal $B-H$ bonds and $2$ bridged $B-H-B$ bonds). Thus,statement $4$ is correct.
Therefore,statements $1, 3,$ and $4$ are correct.

(D) Anhydrous aluminium chloride exists as a dimer $(Al_2Cl_6)$ in the vapour state at low temperatures,but at high temperatures,it dissociates into $AlCl_3$ monomeric molecules.
It acts as a strong Lewis acid due to the incomplete octet of the $Al$ atom.
It sublimes at $180 \ ^oC$ under vacuum.
Therefore,both statements $(A)$ and $(C)$ are considered correct in the context of its physical properties.

(C) The thermal decomposition of borax $(Na_2B_4O_7 \cdot 10H_2O)$ produces sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$:
$Na_2B_4O_7 \cdot 10H_2O \xrightarrow{\text{Heat}} 2NaBO_2 + B_2O_3 + 10H_2O$.
Thus,$X$ is $B_2O_3$.
When $B_2O_3$ reacts with $Cr_2O_3$ upon heating,it forms chromium$(III)$ metaborate $(Cr(BO_2)_3)$,which is a green-colored bead:
$Cr_2O_3 + 3B_2O_3 \xrightarrow{\text{Heat}} 2Cr(BO_2)_3$.
Thus,$Y$ is $Cr(BO_2)_3$.

(D) The conversion of $Borax$ $(Na_2B_4O_7 \cdot 10H_2O)$ to $H_3BO_3$ is achieved by treatment with $HCl$:
$Na_2B_4O_7 + 2HCl + 5H_2O \rightarrow 2NaCl + 4H_3BO_3$
$H_3BO_3$ on heating gives $B_2O_3$:
$2H_3BO_3 \xrightarrow{\Delta} B_2O_3 + 3H_2O$
Finally,$B_2O_3$ is reduced to crystalline boron using $Al$:
$B_2O_3 + 2Al \xrightarrow{\Delta} 2B + Al_2O_3$
Note: Reduction with $Mg$ yields amorphous boron. Thus,$X = HCl$ and $Y = Al$.

(C) $Al(OH)_3$ is an amphoteric hydroxide that dissolves in a strong base like $NaOH$ to form a soluble complex.
In aqueous solution,the aluminum ion maintains its coordination number of $6$.
Therefore,the reaction is: $Al(OH)_3(s) + OH^-(aq) + 2H_2O(l) \to [Al(OH)_4(H_2O)_2]^-(aq)$.
Thus,the correct species formed is $[Al(H_2O)_2(OH)_4]^-$,which is option $C$.

(B) The correct formula for borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$.
$1$. Borax is a primary standard used in acid-base titrations.
$2$. The structure of the $[B_4O_5(OH)_4]^{2-}$ ion consists of two tetrahedral $BO_4$ units and two triangular $BO_3$ units sharing corners. It contains $5 \, B - O - B$ bonds,not $4$.
$3$. Aqueous solution of borax acts as a buffer because it is a salt of a strong base $(NaOH)$ and a weak acid $(H_3BO_3)$.
Therefore,the statement in option $B$ is incorrect.

(A) $H_3BO_3$ acts as a weak monobasic Lewis acid.
$(i)$ The reaction $B(OH)_3 + NaOH \rightleftharpoons Na[B(OH)_4]$ is an equilibrium process.
$(ii)$ Upon the addition of cis-$1, 2-$diol (such as glycerol or mannitol) to the $H_3BO_3$ solution,a stable chelate complex is formed.
$(iii)$ This chelation removes the $[B(OH)_4]^-$ ions from the solution,shifting the equilibrium to the right according to Le Chatelier's principle,thereby increasing the acidic strength of $H_3BO_3$.

(B) The reactions are as follows:
$1. \ Al + 6HCl(aq.) \rightarrow 2AlCl_3 + 3H_2(g) \uparrow$ (Gas $P$ is $H_2$)
$2. \ 2Al + 2NaOH(aq.) + 6H_2O \rightarrow 2Na[Al(OH)_4] + 3H_2(g) \uparrow$ (Gas $Q$ is $H_2$)
Comparing the products:
- Gas $P$ and $Q$ are both $H_2$,so they are the same.
- $X$ $(AlCl_3)$ and $Y$ $(Na[Al(OH)_4])$ are both water-soluble.
- $Al$ is amphoteric as it reacts with both acids and bases.
- $H_2$ gas is inflammable.
Therefore,the statement 'Gas $P$ and $Q$ are different' is incorrect.

(D) The reaction is $2BF_3 + 6LiAlH_4 \xrightarrow{\text{Ether}} B_2H_6 + 6LiF + 6AlH_3$. Here,$(X)$ is $B_2H_6$ (diborane).
In $B_2H_6$,there are $12$ valence electrons involved in bonding ($6$ from $2B$ and $6$ from $6H$).
It contains $4$ terminal $B-H$ bonds,which are $2$-centre-$2$-electron $(2c-2e)$ bonds.
It contains $2$ bridging $B-H-B$ bonds,which are $3$-centre-$2$-electron $(3c-2e)$ bonds.
$B_2H_6$ reacts with $NH_3$ to form different products depending on conditions,such as $B_2H_6 \cdot 2NH_3$ (at low temperature) or $B_3N_3H_6$ (inorganic benzene) at high temperature. Thus,the statement that $X$ does not react with $NH_3$ is incorrect.

(B) In the $NCl_3$ molecule,the $H_2O$ molecule attacks the less electronegative $Cl$ atom rather than the central $N$ atom.
The hydrolysis reaction is: $NCl_3 + 3H_2O \to NH_3 + 3HOCl$.
Here,$NH_3$ is not an oxyacid of the central atom $N$,whereas in other cases like $BF_3$,$SF_4$,and $PCl_5$,the central atom forms its respective oxyacids (e.g.,$H_3BO_3$,$H_2SO_3$,and $H_3PO_4$).

(C) Marine plants,such as seaweeds and kelp,are a rich and important natural source of $Iodine$.

(A) The strength of oxyacids increases with the increase in the oxidation state of the central atom.
In $ClO_3(OH)$ (which is $HClO_4$),the oxidation state of $Cl$ is $+7$.
In $ClO_2(OH)$ (which is $HClO_3$),the oxidation state of $Cl$ is $+5$.
In $SO(OH)_2$ (which is $H_2SO_3$),the oxidation state of $S$ is $+4$.
In $SO_2(OH)_2$ (which is $H_2SO_4$),the oxidation state of $S$ is $+6$.
Comparing the oxidation states,$Cl$ in $HClO_4$ has the highest oxidation state $(+7)$ and is the most electronegative,making $ClO_3(OH)$ the strongest acid among the given options.

(D) At room temperature $(298 \ K)$,$Br_2$ (Bromine) is a liquid non-metal. $Hg$ (Mercury) is a liquid metal. $Ga$ (Gallium) has a melting point of $302.9 \ K$ $(29.8 \ ^\circ C)$,which is very close to room temperature,and it exists as a liquid in warm conditions. Therefore,all the given elements can exist in the liquid state.

(A) The thermal decomposition of orthoboric acid $(H_3BO_3)$ occurs in steps:
$1$. $H_3BO_3 \xrightarrow{100^{\circ} C} HBO_2 (\text{Metaboric acid}) + H_2O$
$2$. $4HBO_2 \xrightarrow{160^{\circ} C} H_2B_4O_7 (\text{Tetraboric acid}) + H_2O$
$3$. $H_2B_4O_7 \xrightarrow{\text{red hot}} 2B_2O_3 + H_2O$
Given $T_1 < T_2$,$X$ is $HBO_2$ (metaboric acid) and $Y$ is $H_2B_4O_7$ (tetraboric acid).

(A) In the $13^{th}$ group,as we move down the group,the stability of the $+1$ oxidation state increases due to the inert pair effect,while the stability of the $+3$ oxidation state decreases.
For $+3$ ions,the stability order is $Al^{3+} > Ga^{3+} > In^{3+} > Tl^{3+}$. Thus,$Ga^{3+} < In^{3+} < Tl^{3+}$ is incorrect.
For $+1$ ions,the stability order is $Ga^{+} < In^{+} < Tl^{+}$.
For $Tl$,the $+1$ state is more stable than the $+3$ state $(Tl^{+} > Tl^{3+})$.
For $Ga$,the $+3$ state is more stable than the $+1$ state $(Ga^{3+} > Ga^{+})$.
Therefore,the incorrect statement is $Ga^{3+} < In^{3+} < Tl^{3+}$.

(B) When $Borax$ $(Na_2B_4O_7 \cdot 10H_2O)$ is dissolved in water,it undergoes hydrolysis to form orthoboric acid $(H_3BO_3)$ and sodium hydroxide $(NaOH)$.
$Na_2B_4O_7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$.
Since $NaOH$ is a strong base and $H_3BO_3$ is a weak acid,the resulting aqueous solution is alkaline (basic) in nature.
This basic nature is responsible for the cleansing action of $Borax$.

(C) Borax,also known as sodium tetraborate,is a white crystalline mineral.
The correct chemical formula for borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$,which contains two tetrahedral and two trigonal boron atoms.

(A) The reaction $2Na[Al(OH)_4] + CO_2 \longrightarrow 2Al(OH)_3 \downarrow + Na_2CO_3$ involves the formation of aluminum hydroxide,$Al(OH)_3$,which is an insoluble solid (precipitate).
Therefore,this is a precipitate formation reaction.

(A) The hydrolysis of boron trioxide $(B_2O_3)$ with water results in the formation of orthoboric acid $(H_3BO_3)$.
The chemical equation is: $B_2O_3 + 3H_2O \longrightarrow 2H_3BO_3$.
In $H_3BO_3$,the oxidation state of boron is $+3$.
Since the maximum oxidation state of boron is $+3$,the acid is named with the $-ic$ suffix (Boric acid).
Therefore,the product is an oxy acid with an $-ic$ suffix.

(A) The reaction is: $\underline{B}F_3 + 3H_2O \longrightarrow H_3BO_3 + 3HF$.
In the product $H_3BO_3$ (Boric acid),the oxidation state of Boron is $+3$.
The maximum oxidation state of Boron is $+3$.
Since the oxidation state of the central atom in the oxy acid is equal to its group number,the suffix used is $-ic$.
Therefore,$H_3BO_3$ is Boric acid,which ends with the $-ic$ suffix.

(B) The reaction for the hydrolysis of $H_4P_2O_5$ (pyrophosphorous acid) is:
$H_4P_2O_5 + H_2O \longrightarrow 2H_3PO_3$
In $H_3PO_3$ (phosphorous acid),the oxidation state of $P$ is $+3$.
Acids of phosphorus with oxidation state $+3$ are named with the $-ous$ suffix (e.g.,phosphorous acid).
Therefore,the product is an oxy acid with an $-ous$ suffix.

(A) The reaction is: $\underline{B}Cl_3 + 3H_2O \longrightarrow H_3BO_3 + 3HCl$.
In $H_3BO_3$ (Boric acid),the oxidation state of Boron is $+3$.
The name of the acid $H_3BO_3$ is Boric acid,which ends with the $-ic$ suffix.
Therefore,the product is an oxy acid with an $-ic$ suffix.

(A) The reaction given is the hydrolysis of pyrophosphoric acid $(H_4P_2O_7)$:
$H_4P_2O_7 + H_2O \longrightarrow 2H_3PO_4$
In the product $H_3PO_4$ (phosphoric acid),the phosphorus atom is in its $+5$ oxidation state.
Oxy acids of phosphorus with the $+5$ oxidation state are named with the suffix $-ic$.
Therefore,the product $H_3PO_4$ is phosphoric acid,which has an $-ic$ suffix.

(B) The reaction is: $\underline{I}Cl_3 + 2H_2O \longrightarrow HIO_2 + 3HCl$.
In the product $HIO_2$,the oxidation state of iodine is $+3$.
Iodine in $+3$ oxidation state forms iodous acid,which is named with the $-ous$ suffix.
Therefore,the product is an oxy acid with an $-ous$ suffix.

(A) The given reaction is the hydrolysis of pyroboric acid ($H_4B_2O_5$ or $B_2O_3 \cdot 2H_2O$).
The reaction is: $H_4B_2O_5 + H_2O \longrightarrow 2H_3BO_3$.
The product formed is $H_3BO_3$,which is Boric acid.
In Boric acid $(H_3BO_3)$,the oxidation state of Boron is $+3$. Since it is the highest oxidation state for Boron,the acid is named with the suffix $-ic$ (Boric acid).
Therefore,the product is an oxy acid with an $-ic$ suffix.

(A) The reaction is the hydrolysis of perboric acid ($H_4B_2O_6$ or $2HBO_3$).
$H_4B_2O_6 + 2H_2O \longrightarrow 2H_3BO_3 + H_2O_2$.
The product $H_3BO_3$ is boric acid,which is an oxyacid of boron with the suffix $-ic$.
Since the product is an oxyacid with an $-ic$ suffix,the correct assignment is $A$.

(B) The reaction is $ICl + H_2O \longrightarrow HIO + HCl$.
In the product $HIO$ (hypoiodous acid),the oxidation state of iodine is $+1$.
Iodine forms oxyacids like $HIO$ $(+1)$,$HIO_2$ $(+3)$,$HIO_3$ $(+5)$,and $HIO_4$ $(+7)$.
According to $IUPAC$ nomenclature for halogen oxyacids,the acid with the lowest oxidation state $(+1)$ is named with the prefix 'hypo-' and the suffix '-ous'.
Therefore,$HIO$ is hypoiodous acid,which has an $-ous$ suffix.

(D) $PCl_5$ acts as a chlorinating agent and reacts with many compounds to replace oxygen or increase the oxidation state of metals.
$BCl_3$ is already a chloride of boron in its maximum oxidation state $(+3)$.
Since boron cannot expand its oxidation state further and there is no oxygen to replace,$BCl_3$ does not react with $PCl_5$.
In contrast,$Hg_2Cl_2$ can be oxidized to $HgCl_2$,$FeCl_2$ can be oxidized to $FeCl_3$,and $S_2Cl_2$ can react with $PCl_5$ to form $SCl_2$ or other sulfur chlorides.

(A) In $BCl_3$,the central Boron atom has only $6$ valence electrons in its valence shell after forming three covalent bonds with Chlorine atoms.
Since it has fewer than $8$ electrons,it is considered an electron-deficient compound (hypovalent).
In $PCl_3$,$P$ has $8$ electrons (octet complete).
In $PCl_5$,$P$ has $10$ electrons (expanded octet).
In $NH_3$,$N$ has $8$ electrons (octet complete).

(C) The general formula for alum is $M_{2}SO_{4} \cdot M'_{2}(SO_{4})_{3} \cdot 24H_{2}O$,where $M$ is a monovalent cation (e.g.,$K^+$,$NH_4^+$) and $M'$ is a trivalent cation (e.g.,$Al^{3+}$,$Fe^{3+}$).
$A$ pseudo alum has the general formula $MSO_{4} \cdot M'_{2}(SO_{4})_{3} \cdot 24H_{2}O$,where $M$ is a divalent cation (e.g.,$Mn^{2+}$) and $M'$ is a trivalent cation (e.g.,$Al^{3+}$).
Therefore,$MnSO_{4} \cdot Al_{2}(SO_{4})_{3} \cdot 24H_{2}O$ is a pseudo alum.