Alkaline earth metals Questions in English

(C) Calcium hydroxide,$Ca(OH)_2$,is prepared by adding water to quick lime,$CaO$.
It is a white amorphous powder that is sparingly soluble in water.
The clear aqueous solution of $Ca(OH)_2$ is known as lime water.
$A$ suspension of slaked lime in water is known as milk of lime.

(C) Beryllium has a high charge density and small size,which leads to a high tendency for hydrolysis of its salts in aqueous solution.
Therefore,the statement that its salts rarely hydrolyze is incorrect.
For example,$BeCl_2$ undergoes hydrolysis as follows:
$BeCl_2 + 2 H_2O \longrightarrow Be(OH)_2 + 2 HCl$

(A) $Ca^{2+}$ ions are essential for maintaining the regular beating of the heart as they are involved in the excitation-contraction coupling process in cardiac muscle cells. Therefore,the statement that they are not important is false. $Mg^{2+}$ ions are central to the chlorophyll molecule in plants. $Mg^{2+}$ ions also form complexes with $ATP$ to facilitate energy transfer. $Ca^{2+}$ ions play a crucial role in the blood clotting cascade.

(B) The thermal stability of alkaline earth metal carbonates increases down the group,while alkali metal carbonates are generally more stable than alkaline earth metal carbonates.
The order of thermal stability is: $K_2CO_3 > Na_2CO_3 > CaCO_3 > MgCO_3$.
Since $MgCO_3$ has the lowest thermal stability,it decomposes most easily upon heating to release $CO_2$ gas.
The reaction is: $MgCO_3 \stackrel{\Delta}{\longrightarrow} MgO + CO_2$.

(C) The solubility of alkaline earth metal sulphates decreases down the group. The order is $Mg > Ca > Sr > Ba$.
The magnitude of the lattice energy remains almost constant because the size of the sulphate ion is very large compared to the cations.
However,the hydration energy decreases significantly from $Be^{2+}$ to $Ba^{2+}$ as the size of the cation increases down the group.
Since hydration energy decreases faster than lattice energy,the overall solubility decreases down the group.

(A) . Plaster of Paris is $CaSO_4 \cdot \frac{1}{2} H_2O$ $(i)$.
$B$. Epsomite is $MgSO_4 \cdot 7H_2O$ $(ii)$.
$C$. Kieserite is $MgSO_4 \cdot H_2O$ $(iii)$.
$D$. Gypsum is $CaSO_4 \cdot 2H_2O$ $(iv)$.
Therefore,the correct matching is $A-i, B-ii, C-iii, D-iv$.

(B) The solubility of alkaline earth metal sulphates depends on the balance between hydration enthalpy and lattice enthalpy.
Hydration enthalpy decreases as the size of the cation increases,while lattice enthalpy remains relatively constant for these sulphates due to the large size of the sulphate anion.
The ionic size order is $Be^{2+} < Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+}$.
Since $Be^{2+}$ is the smallest ion,it has the highest hydration enthalpy.
$BeSO_4$ is the only sulphate among the options where the hydration enthalpy is sufficiently high to overcome the lattice enthalpy,making it highly soluble in water.

(A) As we move down the group in alkaline earth metals,the lattice energy decreases more rapidly than the hydration energy for hydroxides,leading to an increase in the solubility of their hydroxides in water.
Conversely,the solubility of sulphates decreases down the group.
Ionization energy and electronegativity also decrease down the group due to the increase in atomic size.

(B) In a peroxide,the oxidation number of $O$ is $-1$,and they contain a peroxide linkage $(-O-O-)$. They react with dilute acids to produce $H_2O_2$.
For $KO_2$: Let the oxidation state of $O$ be $x$. Then $1 + 2x = 0$,so $x = -\frac{1}{2}$. This is a superoxide.
For $BaO_2$: Let the oxidation state of $O$ be $x$. Then $2 + 2x = 0$,so $x = -1$. This is a peroxide.
For $MnO_2$ and $NO_2$: These are dioxides where the oxidation state of $O$ is $-2$. Therefore,they are not peroxides.
Thus,$BaO_2$ is the correct answer.

(A) The reactions are as follows:
$A \xrightarrow{\Delta} \text{colourless gas} (CO_2) + \text{residue} (CaO)$
$\text{Residue} (CaO) + H_2O \rightarrow Ca(OH)_2 (B)$
$Ca(OH)_2 + \text{excess } CO_2 \rightarrow Ca(HCO_3)_2 (C)$
$Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 (A) + H_2O + CO_2$
Thus,the compound $A$ is $CaCO_3$.

(D) Equimolar solutions of the given alkaline earth metal chlorides in water undergo hydrolysis to form their respective hydroxides.
The basic strength of alkaline earth metal hydroxides increases down the group as the metallic character increases and the ionization energy decreases.
The order of basic strength is $Mg(OH)_2 < Ca(OH)_2 < Sr(OH)_2 < Ba(OH)_2$.
Since $Ba(OH)_2$ is the strongest base among the given options,its solution will have the highest concentration of $OH^-$ ions,resulting in the highest $pH$.

(A) The hydration energy of sulphates decreases as we move down the group $II$ in the periodic table.
$Mg^{2+}$ is the smallest ion among the group $II$ alkaline earth metal ions,which results in a very high charge density.
Due to its small size,the hydration energy of $Mg^{2+}$ is significantly high,making it greater than its lattice energy.
As a result,$MgSO_4$ is highly soluble in water,whereas the solubility of other sulphates like $SrSO_4$ and $BaSO_4$ decreases due to their lattice energy being higher than their hydration energy.

(D) The solubility of alkaline earth metal sulphates depends on the balance between lattice enthalpy and hydration enthalpy.
For a compound to be highly soluble,its hydration enthalpy must be greater than its lattice enthalpy.
As we move down the group,the size of the metal cation increases,which causes the hydration enthalpy to decrease significantly more than the lattice enthalpy.
$Be^{2+}$ is the smallest cation among the alkaline earth metals,resulting in a very high hydration enthalpy that outweighs its lattice enthalpy.
Therefore,$BeSO_4$ is highly soluble in water,while the solubility decreases down the group from $BeSO_4$ to $BaSO_4$.

(D) The setting of Plaster of Paris $(CaSO_4 \cdot \frac{1}{2}H_2O)$ involves its reaction with water to form Gypsum $(CaSO_4 \cdot 2H_2O)$. This process is known as hydration,where the hemihydrate converts into a dihydrate crystal structure.

(D) The thermal stability of alkaline earth metal carbonates increases as the size of the metal cation increases.
This is because the larger cation polarizes the carbonate ion less effectively,making the $C-O$ bond stronger and the compound more stable.
The ionic radii of the cations follow the order: $Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+}$.
Therefore,the thermal stability follows the order: $MgCO_3 < CaCO_3 < SrCO_3 < BaCO_3$.
Thus,the decreasing order of thermal stability is $BaCO_3 > SrCO_3 > CaCO_3 > MgCO_3$.

(A) The solubility of alkaline earth metal sulphates decreases down the group.
This is because the hydration enthalpy of the metal cation decreases more rapidly than the lattice enthalpy as the size of the cation increases.
Therefore,the solubility order is $BeSO_4 > MgSO_4 > CaSO_4 > SrSO_4 > BaSO_4$.

(B) Gypsum,$CaSO_4 \cdot 2H_2O$,on heating to about $120\ ^\circ C$ forms a compound known as Plaster of Paris.
The chemical composition of Plaster of Paris is represented by $2CaSO_4 \cdot H_2O$ or $CaSO_4 \cdot \frac{1}{2} H_2O$.
The balanced chemical equation is:
$2(CaSO_4 \cdot 2H_2O) \xrightarrow{120\ ^\circ C} 2CaSO_4 \cdot H_2O + 3H_2O$

(D) $1$. $Mg$ is the central metal ion in the chlorophyll molecule,which is correct.
$2$. Alkaline earth metals (Group $2$) form oxides $(MO)$ and peroxides $(MO_2)$ with oxygen,but they do not form superoxides $(MO_2)$ like alkali metals,which is correct.
$3$. $NaHCO_3$ (Sodium bicarbonate) is commonly known as baking soda,which is correct.
$4$. Permanent hardness of water is caused by the presence of dissolved chlorides and sulfates of calcium and magnesium. Boiling only removes temporary hardness (caused by bicarbonates). Therefore,this statement is $\text{Incorrect}$.

(A) The given reaction is the reduction of barium sulfate with carbon at high temperatures.
The balanced chemical equation is:
$BaSO_4 + 4C \rightarrow BaS + 4CO$
Here,$BaSO_4$ is reduced to $BaS$ (barium sulfide) by carbon,which acts as a reducing agent.
Therefore,the product $A$ is $BaS$.

(D) Beryllium $(Be)$ is the first element of Group $2$. Due to its small size and high ionization energy,it shows diagonal relationship with Aluminum $(Al)$.
$BeO$ (Beryllium oxide) and $Be(OH)_2$ (Beryllium hydroxide) are amphoteric in nature,meaning they react with both acids and bases.
For example,$BeO + 2HCl \rightarrow BeCl_2 + H_2O$ and $BeO + 2NaOH \rightarrow Na_2BeO_2 + H_2O$.
Similarly,$Be(OH)_2$ also reacts with both acids and bases.
Therefore,both $(A)$ and $(C)$ are correct.

(C) The alkaline earth metal $(M)$ is $Ba$ (Barium).
$1$. The salt with chlorine is barium chloride $(BaCl_2)$,which is known to be less soluble in cold water compared to hot water.
$2$. The insoluble sulphate is barium sulphate $(BaSO_4)$.
$3$. 'Lithopone' is a white pigment consisting of a mixture of barium sulphate $(BaSO_4)$ and zinc sulphide $(ZnS)$. It is produced by the reaction: $BaS + ZnSO_4 \rightarrow BaSO_4 + ZnS$.

(D) For alkaline earth metal hydroxides,as we move down the group from $Be$ to $Ba$,the lattice enthalpy decreases much faster than the hydration enthalpy.
Consequently,the solubility of these hydroxides increases down the group.
Therefore,$Be(OH)_2$ is the least soluble among them and possesses the lowest value of solubility product $(K_{sp})$.

(A) Milk of Magnesia is a suspension of magnesium hydroxide,$Mg(OH)_2$,in water.
It acts as an antacid by neutralizing excess stomach acid $(HCl)$ through the reaction: $Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O$.
Therefore,the chemical composition of milk of magnesia is $Mg(OH)_2$.

(A) The tendency to form complexes depends on the charge density of the metal ion.
As we move down the group,the atomic size of alkaline earth metals increases.
This leads to a decrease in the charge-to-size ratio (ionic potential),which reduces the electrostatic attraction towards ligands.
Therefore,the complex formation tendency decreases down the group.

(A) Alkaline earth metals have high ionization enthalpy.
When heated in a Bunsen flame,the energy of the flame is not sufficient to excite the electrons to higher energy levels in the case of $Be$ and $Mg$.
Therefore,$Be$ and $Mg$ do not impart any characteristic colour to the flame.
Other alkaline earth metals like $Ca$,$Sr$,and $Ba$ impart characteristic colours to the flame due to the excitation of electrons.

(A) When $CaSO_4 \cdot 2H_2O$ (gypsum) is heated at $120^\circ C$ $(393 \ K)$,it loses $1.5$ molecules of water to form plaster of paris $(CaSO_4 \cdot \frac{1}{2}H_2O)$.
Thus,$X = CaSO_4 \cdot \frac{1}{2}H_2O$ (plaster of paris).
When $CaSO_4 \cdot 2H_2O$ is heated at $205^\circ C$ $(478 \ K)$,it loses all its water of crystallization to form anhydrous calcium sulfate $(CaSO_4)$,which is known as dead burnt plaster.
Thus,$Y = CaSO_4$ (dead burnt plaster).
Therefore,$X$ is plaster of paris and $Y$ is dead burnt plaster.

(B) The alkaline earth metal $Be$ (Beryllium) exhibits anomalous properties compared to other group $2$ elements.
$1$. $BeSO_4$ is water-soluble due to the high hydration enthalpy of the small $Be^{2+}$ ion.
$2$. $Be(OH)_2$ is water-insoluble and is amphoteric in nature,meaning it reacts with both acids and bases (e.g.,$Be(OH)_2 + 2NaOH \rightarrow Na_2[Be(OH)_4]$).
$3$. $BeO$ is a hard,covalent solid with a very high melting point and is amphoteric.
Therefore,the metal $M$ is $Be$.

(B) The basic strength of oxides of alkaline earth metals increases down the group as the metallic character increases and the ionization energy decreases.
As we move down the group from $Be$ to $Ba$,the size of the metal atom increases,which makes the $M-O$ bond more ionic.
Therefore,the basic character of the oxides follows the order: $BeO < MgO < CaO < SrO < BaO$.
Thus,the correct order is $SrO > CaO > MgO > BeO$.

(B) The given reaction is the industrial preparation of beryllium chloride:
$BeO + C + Cl_2 \xrightarrow{1000 \ K} BeCl_2 + CO$
Beryllium chloride $(BeCl_2)$ reacts with water as follows:
$BeCl_2 + 2H_2O \rightarrow Be(OH)_2 + 2HCl$
$BeCl_2$ exists as a polymeric chain in the solid state and is an electron-deficient molecule (Lewis acid). Among alkaline earth metals,only $Be$ compounds exhibit such polymeric structures due to the small size and high polarizing power of the $Be^{2+}$ ion.

(A) The thermal stability of alkaline earth metal carbonates increases as we move down the group from $Be$ to $Ba$.
This is because the electropositive character of the metal increases down the group,which leads to a stronger ionic bond between the metal cation and the carbonate anion.
Therefore,the order of thermal stability is $MgCO_3 < CaCO_3 < SrCO_3 < BaCO_3$,which is equivalent to $BaCO_3 > SrCO_3 > CaCO_3 > MgCO_3$.

(C) The correct option is $(C)$.
Explanation:
$Ca$ and $CaH_2$ both react with water to produce calcium hydroxide $(Ca(OH)_2)$ and hydrogen gas $(H_2)$.
Reaction for $Ca$:
$Ca + 2 H_2O \rightarrow Ca(OH)_2 + H_2$
Reaction for $CaH_2$:
$CaH_2 + 2 H_2O \rightarrow Ca(OH)_2 + 2 H_2$
Note: While the stoichiometry of $H_2$ produced differs,the chemical species produced ($Ca(OH)_2$ and $H_2$) are the same.

(A) The statement '$Mg$ cannot form complexes' is incorrect because $Mg$ can form complexes. For example,chlorophyll contains a $Mg^{2+}$ ion coordinated to a porphyrin ring,and $Mg$ also forms complexes with $EDTA^{4-}$.

(B) Elements belonging to the same group in the periodic table exhibit similar chemical properties because they have the same valence shell electronic configuration.
$Mg$,$Sr$,and $Ba$ all belong to Group $2$ (alkaline earth metals).
Therefore,they possess similar chemical properties.

(D) Barium $(Ba)$ is an alkaline earth metal.
$1$. It is a silvery-white metal.
$2$. It forms $Ba(NO_3)_2$,which imparts a characteristic apple-green color to flames,used in fireworks.
$3$. Ionization potential decreases down the group. Since $Ba$ is above $Ra$ in Group $2$,$Ba$ has a higher ionization potential than $Ra$. Therefore,the statement that its ionization potential is lower than radium is incorrect.
$4$. Barium does not exhibit significant photoelectric effect under visible light compared to more reactive elements like cesium.

(C) $Be$ (Beryllium) and $Al$ (Aluminium) show a diagonal relationship. Both form amphoteric oxides,polymeric hydrides,and covalent halides. However,they differ in their maximum covalency. The maximum covalency of $Be$ is $4$ (due to the presence of only $2s$ and $2p$ orbitals),whereas the maximum covalency of $Al$ is $6$ (due to the availability of $3d$ orbitals).

(D) In group $2$,the solubility of alkaline earth metal sulphates decreases down the group due to the decrease in hydration energy being more significant than the decrease in lattice energy.
The order of solubility is: $BeSO_4 > MgSO_4 > CaSO_4 > SrSO_4 > BaSO_4$.
Therefore,$SrSO_4$ has greater solubility than $BaSO_4$.

(B) Portland cement consists of $C_3A$ (tricalcium aluminate),which reacts very rapidly with water,causing the cement to set almost immediately.
To prevent this flash setting and to allow sufficient time for mixing and placing the concrete,a small amount of gypsum $(CaSO_4 \cdot 2H_2O)$ is added.
Gypsum acts as a retarder,which slows down the hydration of $C_3A$,thereby delaying the setting process of the cement.

(C) The compound $(A)$ is $MgSO_{4} \cdot 7H_{2}O$ (Epsom salt),which is used as a purgative.
When treated with ammonium phosphate,it forms a white precipitate of magnesium ammonium phosphate or magnesium phosphate: $MgSO_{4} + (NH_{4})_{3}PO_{4} \rightarrow Mg_{3}(PO_{4})_{2} \downarrow + 3(NH_{4})_{2}SO_{4}$.
In the cobalt nitrate test,magnesium salts form a pink-colored residue $(MgO \cdot CoO)$ upon heating.

(D) Portland cement is a complex mixture of calcium silicates and aluminates.
Its primary constituents are Dicalcium silicate $(Ca_2SiO_4)$,Tricalcium silicate $(Ca_3SiO_5)$,and Tricalcium aluminate $(Ca_3Al_2O_6)$.
Therefore,all three compounds listed are present in Portland cement.

(B) When a mixture of beryllium oxide $(X)$,carbon,and chlorine is heated at high temperature,beryllium dichloride $(Y)$ and carbon monoxide are obtained:
$BeO + C + Cl_2 \to BeCl_2 + CO$
The hydrolysis of beryllium dichloride $(Y)$ gives beryllium hydroxide $(Z)$ and $HCl$:
$BeCl_2 + 2H_2O \to Be(OH)_2 + 2HCl$
Beryllium dichloride $(BeCl_2)$ is an electron-deficient molecule and exists in a polymeric chain structure in the solid state.

(D) The melting point of ionic compounds depends on the lattice energy and the ionic character of the bond.
According to Fajan's rule,as the size of the cation increases,the polarizing power decreases,leading to an increase in the ionic character of the metal-chloride bond.
Among the given alkaline earth metal chlorides $(BeCl_2, MgCl_2, CaCl_2, BaCl_2)$,$Ba^{2+}$ has the largest ionic radius.
Therefore,$BaCl_2$ possesses the highest ionic character and the strongest electrostatic forces of attraction,resulting in the highest melting point.

(C) $BeF_2$ is predominantly covalent in nature,which results in a significantly lower melting point $\left(800^{\circ} C\right)$ compared to other alkaline earth metal fluorides,which are ionic and melt at temperatures above $1200^{\circ} C$.
Furthermore,$BeF_2$ is the most soluble fluoride in water among group $2$ elements because the high hydration energy of the small $Be^{2+}$ ion outweighs its lattice energy.

(A) The reactivity of alkaline earth metals with water increases down the group.
$Be$ (Beryllium) is the first member of the group and has a very high ionization enthalpy and a small atomic size.
Due to its high hydration energy and strong covalent character in its compounds,$Be$ does not react with water or steam even at red heat.
In contrast,$Na$,$Ca$,and $K$ are highly reactive and react readily with water to form their respective hydroxides and release $H_2$ gas.

(A) Option $A$ is correct.
Elements in the same group have the same valence shell electronic configuration.
For option $A$ $(Z = 4, 12, 38, 88)$:
$Z = 4$ is $Be$ $([He] 2s^2)$
$Z = 12$ is $Mg$ $([Ne] 3s^2)$
$Z = 38$ is $Sr$ $([Kr] 5s^2)$
$Z = 88$ is $Ra$ $([Rn] 7s^2)$
All these elements belong to Group $2$ (Alkaline Earth Metals) as they all have $ns^2$ valence configuration.

(C) The solubility of alkaline earth metal fluorides is governed by the balance between lattice energy and hydration energy.
$BeF_2$ is highly soluble in water due to the very high hydration energy of the small $Be^{2+}$ ion,which compensates for its lattice energy.
As the size of the cation increases from $Be^{2+}$ to $Ba^{2+}$,the hydration energy decreases significantly,leading to a decrease in solubility.
Therefore,the correct order is $MgF_2 < CaF_2 < SrF_2 < BeF_2$.

(C) Both $Be$ and $Al$ react with concentrated nitric acid $(HNO_3)$ to form a thin,non-porous,and stable protective oxide layer on their surfaces.
This layer prevents further reaction of the metal with the acid,rendering the metals passive.

(C) The chemical composition of bleaching powder is represented as $CaCl_{2} \cdot Ca(OCl)_{2} \cdot Ca(OH)_{2} \cdot 2H_{2}O$.
In aqueous solution,it dissociates to provide $Ca^{2+}$,$Cl^{-}$,and $OCl^{-}$ ions.

(D) Methanides are carbides that contain the methanide ion,which is the $C^{4-}$ ion.
Upon hydrolysis,these carbides yield methane $(CH_4)$ gas.
Examples of methanides include $Be_2C$ and $Al_4C_3$.
$Be_2C + 4H_2O \rightarrow 2Be(OH)_2 + CH_4$
$Al_4C_3 + 12H_2O \rightarrow 4Al(OH)_3 + 3CH_4$

(B) The hydroxide with the lowest solubility product $(K_{sp})$ is $Be(OH)_2$,which has a $K_{sp}$ value in the $10^{-22}$ range.
This is because the $Be^{2+}$ ion is significantly smaller than the other alkaline earth metal ions,leading to a much stronger lattice energy and a tighter bond with $OH^{-}$ ions,which results in very low solubility.
As we move down group $2A$,the size of the metal ions increases,which weakens the lattice energy and the bond between the metal and $OH^{-}$ ions,causing the $K_{sp}$ values to increase.

(D) The density of alkaline earth metals ($IIA$ group) generally increases down the group as atomic mass increases faster than atomic volume.
However,$Mg$ is an exception due to its crystal structure.
The densities of the elements are: $Be (1.85 \ g/cm^3)$,$Mg (1.55 \ g/cm^3)$,$Ca (1.55 \ g/cm^3)$,$Sr (2.63 \ g/cm^3)$,$Ba (3.59 \ g/cm^3)$.
Comparing the given options,the order $Mg < Be < Ca$ is the most accurate representation of the relative densities among these three elements.