Alkaline earth metals Questions in English

(B) When alkaline earth metals $(M)$ are dissolved in liquid ammonia,they form ammoniated metal cations and solvated electrons,represented by the reaction: $M + (x+y)NH_3 \rightarrow [M(NH_3)_x]^{2+} + 2[e(NH_3)_y]^-$.
These solvated electrons are responsible for the characteristic deep blue colour of the solution.
Therefore,the correct option is $B$.

(B) Barium tests are used to examine the digestive tract using a white powder called barium sulphate $(BaSO_4)$.
This powder is insoluble in water and is not absorbed by the body,making it safe for internal use.
For a barium meal,the barium sulphate powder is mixed with water and swallowed.
Since barium is a heavy element,it absorbs $X$-rays effectively,allowing the digestive tract to be clearly visible on $X$-ray images.

(D) $Be$ (Beryllium) does not react with water even at high temperatures because of the formation of a stable,protective oxide layer on its surface.

(C) Beryllium $(Be)$ is the only alkaline earth metal that does not react with water or steam,even at red heat.
This is due to its exceptionally small atomic size and high ionization energy compared to other elements in the group $(Group \ 2)$.

(B) The $second$ group consists of alkaline earth metals,which are located in Group $2$ of the periodic table.
The $fifth$ period corresponds to the row where the element is located.
Strontium $(Sr)$ has an atomic number of $38$ and its electronic configuration is $[Kr] 5s^2$.
Therefore,it belongs to Group $2$ and Period $5$ of the periodic table.

(D) Beryllium oxide $(BeO)$ is amphoteric in nature,meaning it reacts with both acids and bases.
$1$. Reaction with aqueous $HCl$:
$BeO + 2HCl_{(aq)} \longrightarrow BeCl_2 + H_2O$
The products are beryllium chloride $(BeCl_2)$ and water.
$2$. Reaction with aqueous $NaOH$:
$BeO + 2NaOH_{(aq)} \longrightarrow Na_2BeO_2 + H_2O$
The products are sodium beryllate $(Na_2BeO_2)$ and water.
Therefore,the products obtained are $BeCl_2$ and $Na_2BeO_2$.

(B) The alkaline earth metals are the elements belonging to Group $2$ of the periodic table.
These elements have a general valence shell electronic configuration of $ns^2$.
Among the given options:
$Rb$ $(Rubidium)$ and $Cs$ $(Cesium)$ belong to Group $1$ (Alkali metals).
$Fr$ $(Francium)$ also belongs to Group $1$ (Alkali metals).
$Sr$ $(Strontium)$ belongs to Group $2$ (Alkaline earth metals).
Therefore,the correct option is $B$.

(C) In nuclear reactors,$Be$ (Beryllium) is used as a moderator and reflector because of its low atomic mass and high scattering cross-section for neutrons,which helps in slowing down fast-moving neutrons.

(C) $Be$ (Beryllium) is inert towards water even at high temperatures due to its high ionization enthalpy and small atomic size. The other alkaline earth metals $(Mg, Ca, Sr)$ react with water to form their respective hydroxides and release hydrogen gas.

(B) Alkali metals have lower ionization energies compared to alkaline earth metals,making them more electropositive.
Alkaline earth metals have two valence electrons and form $M^{2+}$ ions with a stable inert gas configuration.
Most compounds of alkaline earth metals are diamagnetic and colourless because they do not have unpaired electrons.
Therefore,the statement that alkaline earth metals are more electropositive than alkali metals is incorrect.

(D) Beryllium $(Be)$ is the only alkaline earth metal that does not react with water or steam even at high temperatures.
This is due to its very small atomic size and high ionization enthalpy compared to other elements in the group.

(D) Magnesium is found in minerals such as magnesite $(MgCO_3)$,dolomite $(MgCO_3 \cdot CaCO_3)$,brucite $(Mg(OH)_2)$,and serpentinite.
Among the given options,dolomite is an important ore of magnesium.
Siderite is an ore of iron $(FeCO_3)$,calamine is an ore of zinc $(ZnCO_3)$,and limonite is an ore of iron $(FeO(OH) \cdot nH_2O)$.

(D) The reactivity of alkaline earth metals with water increases as we move down the group from $Be$ to $Ba$.
This is because the atomic size increases and the ionization enthalpy decreases down the group,making it easier to lose valence electrons.
The order of reactivity is $Be < Mg < Ca < Sr < Ba$.
Therefore,the correct increasing order is $Mg < Ca < Sr < Ba$.

(D) Hydrolysis of metal chlorides depends on the ionic character of the $M-Cl$ bond.
As we move down the group from $Be$ to $Ba$,the ionic character of the metal chloride increases due to the decrease in polarizing power of the cation.
$BeCl_{2}$ is covalent and undergoes hydrolysis,whereas $BaCl_{2}$ is highly ionic and is the least susceptible to hydrolysis among the given options.
Therefore,$BaCl_{2}$ is the least hydrolysed.

(C) Quicklime $(CaO)$ is prepared by the thermal decomposition of limestone $(CaCO_3)$.
The chemical reaction is as follows:
$CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)$
Therefore,calcium carbonate is the correct compound used for the preparation of quicklime.

(B) Gypsum is a naturally occurring mineral with the chemical formula $CaSO_{4} \cdot 2H_{2}O$.

(D) Beryllium $(Be)$ of the second period (Group $2$) shows a diagonal relationship with Aluminum $(Al)$ of the third period (Group $13$).
This is due to their similar ionic sizes and charge-to-radius ratios (ionic potential).

(B) All the metals of group $2$,except beryllium,when heated with hydrogen form $MH_2$ type hydrides. Hence,statement $(B)$ is correct.
Group $2$ elements exhibit an oxidation state of $+2$.
Elements at the top of group $2$ do not catch fire when kept on water.
The reducing power of group $2$ elements is less than that of group $1$ elements due to higher ionization enthalpy. Hence,statements $(A)$,$(C)$,and $(D)$ are incorrect.

(C) The Nicol prism is an optical device made from a crystal of $CaCO_3$ (calcite).
It is used to produce and analyze plane-polarized light due to its property of double refraction.

(A) In group-$2$,the first element,$Be$ (Beryllium),exhibits anomalous properties compared to the rest of the group members.
This is due to its small atomic size,high electronegativity,and high ionization enthalpy compared to other alkaline earth metals.

(B) The correct answer is $CaCO_{3}$.
Nicol prism is a type of optical device used to produce plane-polarized light.
It is constructed from a crystal of calcite,which is a naturally occurring form of calcium carbonate $(CaCO_{3})$,also known as Iceland Spar.

(D) $BeO$,$Al_2O_3$,and $ZnO$ are amphoteric in nature.
$MgO$ and $Li_2O$ are basic in nature,whereas $B_2O_3$ is acidic.
Thus,the pair of amphoteric oxides is $BeO$ and $ZnO$.

(C) The solubility of ionic compounds,such as the hydroxides of alkaline earth metals,depends on two factors: $(i)$ lattice energy and $(ii)$ hydration energy.
As we move down the group,the lattice energy decreases more significantly compared to the hydration energy.
Consequently,the solubility of alkaline earth metal hydroxides increases with an increase in their atomic number.
In contrast,other properties such as ionisation enthalpy,electronegativity,and the solubility of their sulphates in water decrease as the atomic number increases.

(A) When gypsum is heated at a temperature above $393 \ K$,all water of crystallization is removed,and anhydrous calcium sulphate $(CaSO_4)$ is formed. This is known as dead burnt plaster.
$CaSO_4 \cdot 2 H_2O \rightarrow CaSO_4 + 2 H_2O$

(A) Plaster of Paris is a hemihydrate of calcium sulfate.
It is chemically represented as $CaSO_4 \cdot \frac{1}{2} H_2O$.

(D) . $LiOH$ is a weaker base than $NaOH$ because $Li^+$ has a higher charge density than $Na^+$,which increases the covalent character of the $Li-OH$ bond,making it less ionic and thus a weaker base. This statement is correct.
$B$. Salts of $Be$ undergo hydrolysis due to the high charge density and small size of $Be^{2+}$ ion,which polarizes the water molecules. This statement is correct.
$C$. $Ca(HCO_3)_2$ is soluble in water. This statement is correct.
$D$. Hydrolysis of beryllium carbide $(Be_2C)$ gives methane $(CH_4)$,not acetylene $(C_2H_2)$. The reaction is: $Be_2C + 4H_2O \rightarrow CH_4 + 2Be(OH)_2$. Therefore,this statement is incorrect.

(C) The electronic configuration of $He$ is $1s^2$ and $Be$ is $1s^2 2s^2$.
Both have $ns^2$ configuration,but helium is a noble gas and is chemically inert because its valence shell $(1s)$ is completely filled.
Beryllium is an alkaline earth metal and is reactive.
Therefore,$(A)$ is correct but $(R)$ is incorrect.

(A) The common hydrate forms for these alkaline earth metal chlorides are $BaCl_2 \cdot 2 H_2 O$,$CaCl_2 \cdot 6 H_2 O$,and $SrCl_2 \cdot 6 H_2 O$.
Comparing these with the given formulas $BaCl_2 \cdot x H_2 O$,$CaCl_2 \cdot y H_2 O$,and $SrCl_2 \cdot z H_2 O$,we get $x = 2$,$y = 6$,and $z = 6$.

(C) Enamel provides the hard surface to the teeth.
It consists of carbonate-substituted hydroxyapatite crystallites.
It is primarily composed of calcium phosphate and calcium hydroxide.
Therefore,the main constituent of enamel on the surface of teeth is $[3 Ca_3(PO_4)_2 \cdot Ca(OH)_2]$.

(A) For a good quality cement,the chemical composition is strictly controlled to ensure proper setting and strength.
The ratio of silica $(SiO_2)$ to alumina $(Al_2O_3)$ is maintained in the range of $2.5-4.0$.
Additionally,the ratio of lime $(CaO)$ to the sum of silica,alumina,and iron oxide $(SiO_2 + Al_2O_3 + Fe_2O_3)$ is kept between $2.0$ and $3.0$.

(B) The correct matches are:
$I$. $CaSO_4 \cdot 2H_2O$ is Gypsum $(B)$.
$II$. $CaSO_4 \cdot 0.5H_2O$ is Plaster of Paris $(C)$.
$III$. $CaSO_4$ is Dead burnt plaster $(D)$.
Therefore,the correct sequence is $I-B, II-C, III-D$.

(D) The correct answer is $D$.
$1$. The solubility of alkaline earth metal carbonates decreases down the group due to the decrease in hydration energy being more significant than the decrease in lattice energy.
$2$. Alkaline earth metal sulphates are generally thermally stable.
$3$. All nitrates of alkaline earth metals decompose upon heating to form oxides,$NO_2$,and $O_2$.
$4$. While most halides of alkaline earth metals are ionic,$BeCl_2$ is covalent in nature due to the small size and high polarizing power of the $Be^{2+}$ ion. Therefore,the statement that all halides are ionic is incorrect.

(A) The density of alkaline earth metals generally increases down the group as the atomic mass increases more significantly than the atomic volume.
However,there is an anomaly in the case of $Mg$ and $Ca$.
The densities of the elements are: $Be$ $(1.85 \ g/cm^3)$,$Mg$ $(1.74 \ g/cm^3)$,$Ca$ $(1.55 \ g/cm^3)$,and $Sr$ $(2.63 \ g/cm^3)$.
Comparing these values,the correct order is $Sr > Be > Mg > Ca$.

(A) Statement-$I$ is correct: The solubility of alkaline earth metal sulfates decreases down the group. $BeSO_4$ and $MgSO_4$ have high hydration enthalpies which overcome the lattice enthalpy,making them readily soluble in water.
Statement-$II$ is correct: Alkaline earth metal nitrates decompose on heating. $Be(NO_3)_2$ decomposes to give $BeO$,$NO_2$,and $O_2$. Other nitrates like $Ca(NO_3)_2$,$Sr(NO_3)_2$,and $Ba(NO_3)_2$ decompose to form their respective nitrites $(M(NO_2)_2)$ and $O_2$.

(B) $I$. All alkaline earth metals (except $Be$) react with hydrogen on heating to form hydrides $(MH_2)$. $Be$ does not react directly with hydrogen. Thus,statement $I$ is incorrect.
$II$. Calcium hydroxide $(Ca(OH)_2)$,also known as slaked lime,is used in the purification of sugar by precipitating impurities as insoluble calcium salts. Thus,statement $II$ is correct.
$III$. In the gaseous phase,$BeCl_2$ exists as a dimer $(Be_2Cl_4)$ at lower temperatures and as a monomer at higher temperatures. It is generally considered a dimer in the vapor phase. Thus,statement $III$ is correct.
Therefore,statements $II$ and $III$ are correct.

(A) The thermal stability of alkaline earth metal carbonates increases as the size of the cation increases down the group.
As the size of the cation increases,the polarizing power of the cation decreases,making the $CO_3^{2-}$ ion more stable.
The ionic radii order is $Mg^{2+} < Ca^{2+} < Ba^{2+}$.
Therefore,the thermal stability order is $MgCO_3 < CaCO_3 < BaCO_3$,which corresponds to $X < Z < Y$ or $Y > Z > X$.

(A) The densities of the alkaline earth metals are as follows:
$Be: 1.84 \ g \ cm^{-3}$
$Mg: 1.74 \ g \ cm^{-3}$
$Ca: 1.55 \ g \ cm^{-3}$
$Sr: 2.63 \ g \ cm^{-3}$
Comparing these values,the order of density is $Sr (2.63) > Be (1.84) > Mg (1.74) > Ca (1.55)$.
Therefore,the correct order is $Sr > Be > Mg > Ca$.

(B) $Be(OH)_2$ is amphoteric,but the hydroxides of $Mg$,$Ca$,$Sr$ and $Ba$ are basic. The strength increases from $Mg$ to $Ba$.
Carbonates of group $2$ elements become more thermally stable down the group. The larger compounds require more heat than the lighter compounds in order to decompose.

$BeCO_3$	$< 100 \ ^\circ C$
$MgCO_3$	$540 \ ^\circ C$
$CaCO_3$	$900 \ ^\circ C$
$SrCO_3$	$1290 \ ^\circ C$
$BaCO_3$	$1360 \ ^\circ C$

The solubility of the sulphates in water decreases down the group. $Be > Mg \gg Ca > Sr > Ba$. The higher solubilities of $BeSO_4$ and $MgSO_4$ are due to the high enthalpy of solvation of the smaller $Be^{2+}$ and $Mg^{2+}$ ions.
On heating,the alkaline earth metal nitrates decompose to give metal oxide,nitrogen dioxide and oxygen.
$2 M(NO_3)_2 \xrightarrow{\Delta} 2 MO + 4 NO_2 + O_2$ $(M = Be, Mg, Ca, Sr, Ba)$

(A) Among the alkaline earth metals,$Be$ (Beryllium) is the only element that does not react directly with hydrogen to form its hydride $(BeH_2)$.
This is due to the high ionization energy and small size of the $Be^{2+}$ ion,which makes the formation of the ionic hydride energetically unfavorable.
Other alkaline earth metals like $Mg$,$Ca$,$Sr$,and $Ba$ react directly with hydrogen upon heating to form their respective ionic hydrides $(MH_2)$.

(A) The densities of alkaline earth metals are as follows:
$Be: 1.84 \ g/cm^3$ $(II)$
$Mg: 1.74 \ g/cm^3$ $(I)$
$Ca: 1.55 \ g/cm^3$ $(IV)$
$Sr: 2.63 \ g/cm^3$ $(III)$
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(B) $4 LiNO_3 \xrightarrow{\Delta} 2 Li_2O + 4 NO_2 + O_2$ (anomalous behavior).
$2 Ba(NO_3)_2 \xrightarrow{\Delta} 2 BaO + 4 NO_2 + O_2$.
Thus,statement $(i)$ is correct.
The solubility of alkaline earth metal sulphates decreases down the group due to a decrease in hydration enthalpy compared to the lattice enthalpy.
Thus,$BaSO_4$ is less soluble in water than $CaSO_4$. Thus,statement $(ii)$ is correct.
Alkaline earth metals dissolve in liquid ammonia to form a deep blue coloured solution.
$M + (x + y) NH_3 \rightarrow [M(NH_3)_x]^{2+} + 2[e(NH_3)_y]^{-}$.
Thus,statement $(iii)$ is incorrect.

(C) The density of alkaline earth metals generally increases down the group,but there are irregularities due to differences in atomic packing and atomic volume.
The density order for alkaline earth metals is: $Ca < Mg < Be < Sr < Ba < Ra$.
Among the given options,$Ca$ has the lowest density due to its relatively large atomic volume compared to $Be$ and $Mg$.

(A) For a compound to be soluble in water,the hydration enthalpy must be greater than the lattice enthalpy.
In $MgSO_{4}$,the $Mg^{2+}$ ion has a relatively small size,which results in a high hydration enthalpy.
This high hydration enthalpy of $Mg^{2+}$ ions is sufficient to overcome the lattice enthalpy of $MgSO_{4}$.
Therefore,$MgSO_{4}$ is readily soluble in water.
Thus,both $A$ and $R$ are correct and $R$ is the correct explanation of $A$.

(C) When alkaline earth metals like magnesium $(Mg)$ and calcium $(Ca)$ are burnt in air,they react with both oxygen and nitrogen present in the air.
Reaction with oxygen: $2 Mg + O_2 \longrightarrow 2 MgO$ and $2 Ca + O_2 \longrightarrow 2 CaO$.
Reaction with nitrogen: $3 Mg + N_2 \longrightarrow Mg_3N_2$ and $3 Ca + N_2 \longrightarrow Ca_3N_2$.
Thus,the pair of elements that form both oxides and nitrides is $Mg$ and $Ca$.

(B) $BeSO_4$ is soluble in water due to the high hydration energy of the $Be^{2+}$ ion,which overcomes the lattice energy. Thus,the statement is correct.
$(B)$ $BeO$ is an amphoteric oxide because it reacts with both acids and bases. Thus,the statement is correct.
$(C)$ Heating $CaCO_3$ at $1070-1270 \ K$ yields $CaO$ and $CO_2$,not $CO$. The reaction is: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2 \uparrow$. Thus,the statement is incorrect.
Therefore,the correct statements are $A$ and $B$.

(D) The compound $BeCl_2$ exists as a polymeric chain in the solid state.
In the vapour phase at lower temperatures,it exists as a chloro-bridged dimer.
At high temperatures (around $1200 \ K$),the dimer dissociates into a linear monomeric molecule,$Cl-Be-Cl$.
Therefore,the metal $M$ is $Be$ (Beryllium).

(D) The tendency to form hydrates decreases as the size of the alkaline earth metal cation increases down the group.
Smaller cations have higher charge density,which allows them to polarize water molecules more effectively,leading to stronger hydration.
The ionic radii follow the order: $Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+}$.
Therefore,the tendency to form hydrates follows the order: $Mg^{2+} > Ca^{2+} > Sr^{2+} > Ba^{2+}$.
Thus,the correct order is $Mg > Ca > Sr > Ba$.

(D) All the given statements are correct:
$(i)$ Beryllium oxide $(BeO)$ is amphoteric,meaning it reacts with both acids and bases.
$(ii)$ Group $II$ elements dissolve in liquid ammonia to form deep blue-black solutions due to the presence of ammoniated electrons,which absorb energy in the visible region.
$(iii)$ Hydration enthalpy is inversely proportional to the size of the ion. As the size of the cation increases from $Be^{2+}$ to $Ba^{2+}$,the hydration enthalpy decreases.

(D) As we move down group $2$ from $Mg(OH)_2$ to $Ba(OH)_2$:
$1$. The basic character increases because the metallic character and the ease of releasing $OH^-$ ions increase due to the increase in atomic size.
$2$. The chemical stability (thermal stability) increases as the electropositive character of the metal increases,making the $M-O$ bond stronger relative to the $O-H$ bond.
$3$. The solubility in water increases because the lattice energy decreases more rapidly than the hydration energy as the size of the cation increases.
Therefore,all three properties increase down the group.