Alkaline earth metals Questions in English

(D) . The raw materials used for the manufacture of Portland cement are limestone $(CaCO_3)$,which provides lime,and clay,which provides silica $(SiO_2)$,alumina $(Al_2O_3)$,and iron oxide $(Fe_2O_3)$. Gypsum $(CaSO_4 \cdot 2H_2O)$ is added in small quantities (about $2-3 \%$) to the clinker during grinding to retard the setting time of the cement.

(A) Gypsum is $CaSO_4 \cdot 2H_2O$ and Plaster of Paris is $CaSO_4 \cdot \frac{1}{2}H_2O$.
When Plaster of Paris is mixed with water,it rehydrates to form gypsum: $CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O \rightarrow CaSO_4 \cdot 2H_2O$.
When gypsum is heated at $373 \ K$,it loses water to form Plaster of Paris: $CaSO_4 \cdot 2H_2O \xrightarrow{373 \ K} CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O$.
Comparing the molar masses,the percentage of $Ca$ in gypsum is lower than in Plaster of Paris because of the higher water content in gypsum.

(B) The solubility of alkaline earth metal sulphates decreases from $Be$ to $Ba$ because the hydration energy decreases more rapidly than the lattice energy as the size of the cation increases down the group.
Therefore,the correct option is $B$.

(B) The elements belonging to Group $2$ of the periodic table are known as alkaline earth metals.
The members of this group are:
$1$. Beryllium ($Be$,atomic number $4$)
$2$. Magnesium ($Mg$,atomic number $12$)
$3$. Calcium ($Ca$,atomic number $20$)
$4$. Strontium ($Sr$,atomic number $38$)
$5$. Barium ($Ba$,atomic number $56$)
$6$. Radium ($Ra$,atomic number $88$)
These elements are called alkaline earth metals because their oxides and hydroxides are alkaline in nature and they are found in the earth's crust.
Therefore,the correct set is $Be, Mg, Ca, Sr, Ba$.

(B) The electronic configuration of $Sr$ $(Z=38)$ is $[Kr]\,5s^2$.

(D) is the incorrect statement. $Be$ is the smallest alkaline earth metal and has a high ionization energy and high charge density. Due to its small size and high electronegativity,$Be$ does not react with water,even at elevated temperatures,because of the formation of a protective oxide layer.

(A) Beryllium exhibits anomalous behavior compared to other members of Group-$IIA$ due to the following reasons:
$(i)$ Exceptionally small atomic and ionic size.
$(ii)$ High electronegativity and high ionization energy.
$(iii)$ Absence of $d$-orbitals in its valence shell.
Therefore,the correct option is $A$.

(A) The solubility of alkaline earth metal hydroxides increases down the group as the lattice energy decreases more rapidly than the hydration energy.
The order of solubility is: $Be(OH)_2 < Mg(OH)_2 < Ca(OH)_2 < Sr(OH)_2 < Ba(OH)_2$.
Therefore,$Ba(OH)_2$ is the most soluble in $H_2O$.

(D) For alkaline earth metals,the solubility of hydroxides increases down the group as the atomic number increases.
This is because the lattice energy decreases more rapidly than the hydration energy as the size of the cation increases.
Therefore,the correct option is $D$.

(A) The ionic character of alkaline earth metal compounds increases as we move down the group from $Be$ to $Ba$.
$Be$ has a very small size and high ionization enthalpy,which leads to high polarizing power according to Fajan's rule.
Due to this,$Be$ compounds exhibit significant covalent character compared to other alkaline earth metals.
Therefore,the correct option is $(A)$.

(C) Magnesium $(Mg)$ is an alkaline earth metal. It is less reactive than alkali metals like Sodium $(Na)$.
Magnesium does not react with cold water at room temperature.
However,it reacts with hot water and steam to form Magnesium hydroxide $(Mg(OH)_2)$ or Magnesium oxide $(MgO)$ respectively.

(A) Alkaline earth metals are denser than alkali metals because they have two valence electrons,which leads to stronger metallic bonding.
This allows the atoms to be packed more tightly due to their greater nuclear charge and smaller atomic radii compared to alkali metals.

(B) In alkaline earth metals,as the atomic number increases,the atomic size increases.
For hydroxides,the lattice energy decreases more rapidly than the hydration energy as we move down the group from $Be$ to $Ba$.
This leads to an increase in the solubility of their hydroxides.
Conversely,properties like ionisation energy $(IE)$,electronegativity,and the solubility of their sulphates decrease as the atomic number increases.

(B) Alkaline earth metals are the elements of Group $2$ in the periodic table. Elements of Group $1$ and Group $2$ (s-block) along with Groups $13$ to $17$ (p-block) are collectively known as Representative elements.

(D) The correct answer is $D$.
As we move down the group $2$ in the periodic table,the metallic character increases and the ionization energy decreases.
This leads to an increase in the ionic nature of the metal-oxygen bond in hydroxides,making the release of $OH^-$ ions easier.
Therefore,the basic strength increases down the group: $Be(OH)_2 < Mg(OH)_2 < Ca(OH)_2 < Sr(OH)_2 < Ba(OH)_2$.
Thus,$Ba(OH)_2$ is the strongest base among the given options.

(B) The basic strength of hydroxides increases as we move down a group in the periodic table due to the increase in metallic character and electropositivity.
Among the given elements,$Be$,$Mg$,$Al$,and $Si$,$Mg$ is the most electropositive metal.
Therefore,$Mg(OH)_2$ is the strongest base among the given options.

(C) The chemical formula of limestone is $CaCO_3$.
Quick lime is $CaO$.
Slaked lime is $Ca(OH)_2$.

(C) The reducing power of an element depends on its ability to lose electrons,which is determined by its standard electrode potential $(E^\circ)$.
As we move down the group in alkaline earth metals $(Be$ $\rightarrow Mg$ $\rightarrow Ca$ $\rightarrow Sr$ $\rightarrow Ba)$,the ionization energy decreases.
Consequently,the standard oxidation potential increases,making the metal a better reducing agent.
$Ba$ has the lowest ionization energy and the most negative standard electrode potential among the given options,making it the strongest reducing agent.

(C) Plaster of Paris $(CaSO_4 \cdot \frac{1}{2}H_2O)$ hardens by reacting with water to form gypsum $(CaSO_4 \cdot 2H_2O)$,which is a hard mass.
The reaction is:
$CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O \to CaSO_4 \cdot 2H_2O$

(D) The carbonates of alkaline earth metals ($CaCO_3$,$SrCO_3$,$BaCO_3$) are practically insoluble in water.
This is because their lattice energy is much higher than their hydration energy,which prevents them from dissolving in water.

(A) The solubility of alkaline earth metal hydroxides increases as we move down the group from $Be$ to $Ba$.
$Be(OH)_2$ is amphoteric and has very low solubility in water compared to the other hydroxides in the group.
Therefore,$Be(OH)_2$ is considered insoluble in water.

(D) The metal $M$ is $Be$ (Beryllium).
$1$. $BeSO_4$ is water-soluble.
$2$. $BeO$ (Beryllium oxide) is amphoteric and becomes inert (less reactive) on strong heating.
$3$. $Be(OH)_2$ is amphoteric and dissolves in $NaOH$ to form sodium beryllate,$Na_2[Be(OH)_4]$ or $Na_2BeO_2$.

(A) To reduce the acidity of soil,a basic substance is required to neutralize the excess $H^+$ ions.
Among the given options,$Ca(OH)_2$ (Calcium hydroxide) is a base,while the others are acidic or neutral salts.
Therefore,$Ca(OH)_2$ is used to treat acidic soil.

(A) The alkaline earth metals are the elements of Group $2$ of the periodic table.
These elements have their valence electrons in the $ns$ orbital,which classifies them as $s-$block elements.

(B) The element with atomic number $56$ is Barium $(Ba)$.
Its electronic configuration is $[Xe] 6s^2$.
Since it has two electrons in its outermost $s$-orbital,it belongs to Group $2$,which are known as the alkaline earth metals.

(C) The thermal stability of alkaline earth metal carbonates increases as the size of the metal cation increases down the group.
This is because the larger cation stabilizes the large carbonate anion more effectively.
Therefore,the order of increasing thermal stability is $MgCO_3 < CaCO_3 < SrCO_3 < BaCO_3$.
Conversely,the order of decreasing thermal stability is $BaCO_3 > SrCO_3 > CaCO_3 > MgCO_3$.

(B) The solubility of alkaline earth metal hydroxides increases down the group.
This is because the lattice energy decreases more rapidly than the hydration energy as the size of the cation increases from $Be^{2+}$ to $Ba^{2+}$.
Therefore,$Be(OH)_2$ has a very high lattice energy,making it insoluble,while $Ba(OH)_2$ is highly soluble.

(D) The correct answer is $(D)$.
Beryllium $(Be)$ forms predominantly covalent compounds due to its small atomic size and high ionization energy,which leads to high polarizing power according to Fajan's rule.

(B) The chemical formula for gypsum is $CaSO_4 \cdot 2H_2O$,which contains $2$ molecules of water of crystallization.
The chemical formula for plaster of paris is $CaSO_4 \cdot \frac{1}{2}H_2O$,which contains $\frac{1}{2}$ molecule of water of crystallization.
Therefore,the number of water molecules in gypsum and plaster of paris are $2$ and $1/2$ respectively.

(A) . Plaster of Paris $[(CaSO_4)_2 \cdot H_2O]$ is used in surgery for setting fractured bones,in dentistry,and for the manufacturing of statues and toys.
It is prepared by heating gypsum at $393 \ K$ $(120 \ ^\circ C)$:
$2(CaSO_4 \cdot 2H_2O) \xrightarrow{393 \ K} 2(CaSO_4) \cdot H_2O + 3H_2O$

(B) When $MgCl_2 \cdot 6H_2O$ is heated,it undergoes hydrolysis to form magnesium oxychloride ($MgO \cdot MgCl_2$ or $Mg(OH)Cl$) and hydrogen chloride gas.
$MgCl_2 \cdot 6H_2O \xrightarrow{\Delta} Mg(OH)Cl + HCl + 5H_2O$
However,if the question implies the simple dehydration product or the common industrial observation,it is often associated with the formation of basic magnesium chloride (magnesium oxychloride). Given the options,$B$ is the most chemically accurate product of heating hydrated magnesium chloride.

(D) Sorel's cement is a mixture formed by the reaction of magnesium oxide $(MgO)$ with a concentrated solution of magnesium chloride $(MgCl_2)$.
Its chemical composition is represented as $MgCl_2.5MgO.xH_2O$.

(D) Calcium carbide $(CaC_2)$ is an ionic carbide because it consists of $Ca^{2+}$ and $C_2^{2-}$ ions.
$ZnC$,$TiC$,and $SiC$ are examples of covalent or interstitial carbides.

(B) Superphosphate,which is a mixture of $Ca(H_2PO_4)_2 \cdot H_2O$ and $CaSO_4$,is widely used as a fertilizer in agriculture to provide phosphorus to plants.

(C) The solubility of metal sulphates in water generally decreases down the group for alkaline earth metals,but for heavy metals,the lattice energy and hydration energy play a significant role.
$PbSO_4$ (Lead$(II)$ sulphate) is known to be practically insoluble in water due to its very high lattice energy compared to its hydration energy.
In contrast,$CuSO_4$,$CdSO_4$,and $Bi_2(SO_4)_3$ are generally soluble in water.

(A) The white enamel of our teeth is primarily composed of hydroxyapatite,which is a crystalline form of calcium phosphate,$Ca_3(PO_4)_2$.
However,in the context of dental health and the role of fluoride,the enamel surface can be converted into the harder and more acid-resistant fluorapatite,$CaF_2$,when fluoride ions are present.
Given the standard options provided in many textbooks,$Ca_3(PO_4)_2$ is the primary constituent of the enamel.

(D) The chemical formula of bleaching powder is commonly written as $CaOCl_2$,but its correct structural representation is $CaCl(OCl)$.
In this structure,the calcium ion $(Ca^{2+})$ is bonded to one chloride ion $(Cl^-)$ and one hypochlorite ion $(OCl^-)$.
Thus,the correct representation is $CaCl(OCl)$.

(C) The chemical reaction for the preparation of bleaching powder is as follows:
$Ca(OH)_2 + Cl_2 \to CaOCl_2 + H_2O$
Here,$Ca(OH)_2$ is known as slaked lime. Thus,bleaching powder is obtained by passing chlorine gas over dry slaked lime.

(D) Bleaching powder,which is $CaOCl_2$,is prepared by the action of chlorine gas on dry slaked lime $(Ca(OH)_2)$.
The chemical reaction is:
$Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O$

(A) The correct answer is $A$.
$CaF_2$ has the lowest solubility among the given alkaline earth metal halides.
This is primarily due to its very high lattice energy compared to the hydration energy of its constituent ions,which makes it difficult for the crystal lattice to break apart in water.

(D) The alkaline earth metals have low ionization enthalpies,which allows them to lose their valence electrons easily.
Because they readily lose electrons,they act as strong reducing agents.
Although their reducing power is slightly less than that of the $I^{st}$ group (alkali) metals,they are still considered strong reducing agents.

(B) The reaction between barium sulfate $(BaSO_4)$ and carbon $(C)$ at high temperature is a reduction reaction.
The balanced chemical equation is:
$BaSO_4 + 4C \xrightarrow{\Delta} BaS + 4CO$
Thus,the products formed are barium sulfide $(BaS)$ and carbon monoxide $(CO)$.

(B) . Marble statue contains $CaCO_3$.
$B$. Calcined gypsum is $CaSO_4 \cdot \frac{1}{2}H_2O$ (or $CaSO_4 \cdot 2H_2O$ for gypsum),which does not contain $CaCO_3$.
$C$. Sea shells are primarily composed of $CaCO_3$.
$D$. Dolomite is a double carbonate of calcium and magnesium,$CaCO_3 \cdot MgCO_3$.

(A) The hydration energy decreases from $Be^{2+}$ to $Ba^{2+}$.
Since the hydration energy of the smaller $Be^{2+}$ ion is significantly higher than the lattice energy of $BeSO_4$,it is highly soluble.
As the size of the cation increases down the group,the hydration energy decreases more rapidly than the lattice energy,leading to a decrease in solubility.

(B) The correct answer is $(B)$.
When gypsum $(CaSO_4 \cdot 2H_2O)$ is heated to a temperature of $393 \, K$ $(120 \, ^\circ C)$,it loses three-fourths of its water of crystallization to form Plaster of Paris $(CaSO_4 \cdot \frac{1}{2}H_2O)$.
The reaction is:
$CaSO_4 \cdot 2H_2O \xrightarrow{393 \, K} CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O$

(A) The alkaline earth metals are the elements of Group $2$ of the periodic table.
The first three alkaline earth metals are Beryllium $(Be)$,Magnesium $(Mg)$,and Calcium $(Ca)$.
The third alkaline earth metal is Calcium $(Ca)$,which has an atomic number of $20$.
In a neutral Calcium atom,the number of protons is $20$ and the number of electrons is $20$.
When it forms an ion $(Ca^{2+})$,it loses $2$ electrons.
Therefore,the number of protons remains $20$,and the number of electrons becomes $20 - 2 = 18$.
Thus,the number of electrons is $18$ and the number of protons is $20$.

(A) . In the compounds of alkaline earth metals,all the electrons are paired. Hence,they are diamagnetic in nature.

(A) The flame test is used to identify metal ions based on the characteristic colour they impart to the Bunsen flame.
$Ba^{2+}$ ions,when introduced as $BaCl_2$,impart an apple green colour to the flame.
Therefore,the correct option is $(A)$.

(B) When $CO_2$ is passed through lime water,it forms a white precipitate of calcium carbonate:
$Ca(OH)_2 + CO_2 \xrightarrow{} CaCO_{3(s)} + H_2O$
When excess $CO_2$ is passed,the insoluble $CaCO_3$ reacts with $CO_2$ and water to form water-soluble calcium bicarbonate:
$CaCO_3 + CO_2 + H_2O \xrightarrow{} Ca(HCO_3)_2$
Thus,the milkiness disappears due to the formation of soluble $Ca(HCO_3)_2$.

(B) The flame test is a qualitative analysis used to detect the presence of certain metal ions.
$Sr^{2+}$ ions impart a crimson or bright red colour to the flame.
$Ba^{2+}$ gives an apple green colour,$Ca^{2+}$ gives a brick red colour,and $Cr^{3+}$ does not typically show a characteristic flame colour in this context.
Therefore,the correct option is $(B)$.