Alkaline earth metals Questions in English

(D) The solubility product $(K_{sp})$ values for the oxalates of alkaline earth metals are very low.
Specifically,$BaC_2O_4$,$SrC_2O_4$,and $CaC_2O_4$ are all sparingly soluble in water.
Therefore,when an oxalate solution is added to a mixture containing $Ba^{2+}$,$Sr^{2+}$,and $Ca^{2+}$,all three ions will form precipitates of their respective oxalates.

(B) is the correct answer.
$ZnCO_3 \xrightarrow{\Delta} ZnO + CO_2 \uparrow$
$CO_2$ gas turns lime water milky.
$ZnO$ is yellow when hot and white when cold.

(C) The chemical formulas for the given ores are as follows:

Dolomite	$MgCO_3 \cdot CaCO_3$
Magnesite	$MgCO_3$
Carnallite	$KCl \cdot MgCl_2 \cdot 6H_2O$

As observed from the formulas,magnesium $(Mg)$ is present in all three ores. Therefore,$Mg$ is the metal that can be extracted from all of them.

(C) Electric furnaces are lined with magnesia $(MgO)$ because it has a very high melting point $(2852 \ ^{\circ}C)$.
This property allows the furnace lining to remain stable and intact even when the furnace operates at extremely high temperatures.

(B) The cheap compound having a high melting point used in a furnace is $CaO$ (calcium oxide). It is commonly used as a flux or refractory material in metallurgical processes due to its high thermal stability and low cost.

(A) When $ZnO$ (zinc oxide) is heated with $BaO$ (barium oxide) at $1100\,^{\circ}C$,they react to form a mixed oxide known as barium zincate.
The chemical equation for the reaction is: $ZnO + BaO \xrightarrow{1100\,^{\circ}C} BaZnO_2$.

(A) The flame test is used to identify metal ions based on the characteristic color they impart to a non-luminous flame.
For alkaline earth metals,the flame colors are:
$Ca^{2+}$: Brick red
$Sr^{2+}$: Crimson red
$Ba^{2+}$: Apple green
$Mg^{2+}$: Does not impart any characteristic color to the flame.
Therefore,$BaCl_2$ (Barium chloride) imparts an apple green color to the flame.

(A) In a Bunsen burner flame test,$Barium$ $(Ba^{2+})$ ions impart a characteristic apple-green color to the flame.
Therefore,the metal ion present in the salt is $Barium$.

(A) The flame test is a qualitative analysis used to detect the presence of certain elements,primarily metal ions,based on each element's characteristic emission spectrum.
Barium $(Ba^{2+})$ salts are known to impart a characteristic apple-green color to the Bunsen burner flame.
Among the given options,$Ba(OMe)_2$ (Barium methoxide) contains the $Ba^{2+}$ ion,which is responsible for the green flame coloration.

(B) The flame test is based on the excitation of electrons to higher energy levels by the heat of the flame.
$Be^{2+}$ and $Mg^{2+}$ ions have very high ionization energies and their electrons are too strongly bound to be excited by the energy available in a Bunsen flame.
Therefore,$Be^{2+}$ does not impart any characteristic color to the flame.

(B) When $CO_2$ is passed through lime water $(Ca(OH)_2)$,it forms a white precipitate of calcium carbonate $(CaCO_3)$,which makes the solution milky:
$Ca(OH)_2(aq) + CO_2(g) \rightarrow CaCO_3(s) + H_2O(l)$
When excess $CO_2$ is passed,the insoluble $CaCO_3$ reacts with $CO_2$ and water to form calcium bicarbonate $(Ca(HCO_3)_2)$,which is soluble in water,causing the milkiness to disappear:
$CaCO_3(s) + H_2O(l) + CO_2(g) \rightarrow Ca(HCO_3)_2(aq)$

(D) $Ca$ and $Zn$ are chemically dissimilar because $Ca$ belongs to the $s$-block (alkaline earth metal) with valence configuration $ns^2$,while $Zn$ belongs to the $d$-block (transition metal) with valence configuration $(n-1)d^{10}ns^2$. $Na$ and $K$ are both alkali metals,$Ba$ and $Sr$ are both alkaline earth metals,and $Zr$ and $Hf$ exhibit lanthanoid contraction,making them chemically very similar.

(C) The basic strength of alkaline earth metal hydroxides increases as we move down the group from $Be$ to $Ba$. \\ This is because the metallic character and the electropositive nature of the elements increase down the group,making the $M-OH$ bond more ionic and easier to dissociate into $M^{2+}$ and $OH^-$ ions. \\ Therefore,the order of basic strength is: $Be(OH)_2 < Mg(OH)_2 < Ca(OH)_2 < Sr(OH)_2 < Ba(OH)_2$.

(A) Carbonates of alkali metals (except $Li_2CO_3$) are thermally stable and do not decompose upon heating.
$MgCO_3$ is an alkaline earth metal carbonate,which decomposes on heating as follows:
$MgCO_3 \xrightarrow{\Delta} MgO + CO_2$

(D) $Be$ (Beryllium) shows anomalous behavior compared to other alkaline earth metals due to its small size and high polarizing power.
It exhibits a diagonal relationship with $Al$ (Aluminum),which includes having a similar electronegativity value (approximately $1.5$ on the Pauling scale).

(A) Alkali metal carbonates (except $Li_2CO_3$) are thermally stable and do not decompose upon heating.
$MgCO_3$ is an alkaline earth metal carbonate,which decomposes on heating to form $MgO$ and $CO_2$ according to the reaction: $MgCO_3 \xrightarrow{\Delta} MgO + CO_2$.

(A) The standard electrode potential $(E^\circ)$ of alkaline earth metals becomes more negative as we move down the group from $Be$ to $Ba$.
Since $Be$ has the least negative (or most positive) value among the given options,it has the highest electrode potential.

(B) The basic character of oxides of Group-$2$ elements increases as we move down the group.
$BeO$ is amphoteric,while $MgO$,$CaO$,and $BaO$ are basic.
As the metallic character increases down the group,the basic strength of the oxides increases in the order: $BeO < MgO < CaO < BaO$.

(D) In a group,as we move down,the ionization energy decreases,which leads to an increase in the electropositive character.
Therefore,$Barium$ $(Ba)$ is the most electropositive element among the given alkaline earth metals.

(C) Both $Ca$ and $CaH_2$ react with water to produce $H_2$ gas.
$Ca + 2H_2O \to Ca(OH)_2 + H_2 \uparrow$
$CaH_2 + 2H_2O \to Ca(OH)_2 + 2H_2 \uparrow$

(C) The reducing power of alkaline earth metals increases as we move down the group.
Since $Ba$ is at the bottom of the given elements in Group $2$,it has the lowest ionization enthalpy and thus acts as the strongest reducing agent.

(A) Alkaline earth metals have a general valence shell configuration of $ns^2$.
In their compounds,they exist as dipositive ions $(M^{2+})$ by losing two electrons.
These $M^{2+}$ ions have a noble gas configuration where all electrons are paired.
Due to the presence of all paired electrons,the compounds of alkaline earth metals are diamagnetic in nature.

(D) Alkaline earth metals have low ionization enthalpies and can easily lose their valence electrons to form $M^{2+}$ ions. Therefore,they act as strong reducing agents.

(D) The fluorides of alkaline earth metals (except $BeF_2$) are generally insoluble in water. $CaF_2$ is known to be sparingly soluble or insoluble in water due to its high lattice energy.

(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 increases the ionic character of the metal-oxygen bond in the carbonate,making it more stable.
Therefore,the correct decreasing order of thermal stability is $BaCO_3 > SrCO_3 > CaCO_3 > MgCO_3$.

(D) Alkaline earth metals belong to Group $2$ of the periodic table.
They have two valence electrons in their outermost shell $(ns^2)$.
They are monoatomic in nature and readily lose two electrons to form $M^{2+}$ ions to achieve a stable noble gas configuration.

(D) In Group $2$,the basic strength of hydroxides increases as we move down the group because the metallic character increases.
Therefore,$Ba(OH)_2$ is more basic than $Be(OH)_2$.
Statement $D$ is incorrect because $Be(OH)_2$ is amphoteric and less basic than $Ba(OH)_2$.

(A) The solubility of alkaline earth metal sulfates decreases down the group.
This is primarily because the hydration enthalpy of the metal ions decreases significantly from $Be^{2+}$ to $Ba^{2+}$ as the ionic size increases.
Since the sulfate ion is large,the lattice energy does not decrease as rapidly as the hydration enthalpy,leading to a decrease in overall solubility.

(C) Both $Mg$ (Magnesium) and $Ba$ (Barium) are alkaline earth metals belonging to group $2$ of the periodic table.

(C) The green color in fireworks is produced by the presence of $Ba$ (Barium) salts. When $Ba^{2+}$ ions are heated in a flame,they emit light in the green region of the spectrum.

(C) Plaster of Paris $(CaSO_4 \cdot \frac{1}{2}H_2O)$ sets into a hard mass by combining with water to form Gypsum $(CaSO_4 \cdot 2H_2O)$.
The reaction is: $CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O \rightarrow CaSO_4 \cdot 2H_2O$.

(A) Epsom salt is the heptahydrate form of magnesium sulfate,with the chemical formula $MgSO_4 \cdot 7H_2O$.

(B) When gypsum $(CaSO_4 \cdot 2H_2O)$ is heated at $300\,^oC$,it undergoes complete dehydration to form anhydrous calcium sulfate,also known as dead burnt plaster.
The reaction is: $CaSO_4 \cdot 2H_2O \xrightarrow{300\,^oC} CaSO_4 + 2H_2O$.

(A) The chemical formula of gypsum is $CaSO_4 \cdot 2H_2O$ and plaster of Paris is $CaSO_4 \cdot \frac{1}{2}H_2O$.
Since gypsum has more water molecules per formula unit,the percentage by mass of calcium in gypsum is lower than in plaster of Paris.

(D) The chemical formula for gypsum is $CaSO_4 \cdot 2H_2O$.
The chemical formula for plaster of Paris is $CaSO_4 \cdot \frac{1}{2}H_2O$.
The difference in the number of water molecules is $2 - \frac{1}{2} = \frac{3}{2}$ or $1.5$.

(C) $Ca(OH)_2$ is used to remove temporary hardness of water by reacting with calcium bicarbonate:
$Ca(OH)_2 + Ca(HCO_3)_2 \to 2CaCO_3 + 2H_2O$
It reacts with sodium carbonate to produce caustic soda $(NaOH)$:
$Ca(OH)_2 + Na_2CO_3 \to 2NaOH + CaCO_3$
It reacts with $CO_2$ to form calcium carbonate,which makes the solution milky:
$Ca(OH)_2 + CO_2 \to CaCO_3 \downarrow + H_2O$
Therefore,$A$ is $Ca(OH)_2$.

(A) Plaster of Paris,chemically $CaSO_4 \cdot \frac{1}{2}H_2O$,is widely used in the construction industry for plastering walls and ceilings,as well as in medicine for setting fractured bones.

(D) When quicklime $(CaO)$ is heated with coke $(C)$ in an electric furnace,it produces calcium carbide $(CaC_2)$ and carbon monoxide $(CO)$.
The chemical reaction is: $CaO + 3C \xrightarrow{\Delta} CaC_2 + CO$.

(D) The setting of Plaster of Paris involves hydration to form a dihydrate (gypsum).
$CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O \rightarrow CaSO_4 \cdot 2H_2O$
(Plaster of Paris) $\rightarrow$ (Gypsum)

(C) In Group $2$ (alkaline earth metals),the ionization enthalpy decreases as we move down the group from $Be$ to $Ba$.
Since the tendency to lose electrons increases down the group,the chemical reactivity increases from $Be$ to $Ba$.
Therefore,$Ba$ is the most chemically reactive element among the given options.

(D) The raw materials used in the manufacture of Portland cement are limestone $(CaCO_3)$,clay ($SiO_2$,$Al_2O_3$,$Fe_2O_3$),and gypsum $(CaSO_4 \cdot 2H_2O)$.

(A) The solubility of alkaline earth metal sulfates decreases down the group. This is because the hydration enthalpy decreases more rapidly than the lattice enthalpy as the size of the cation increases. Therefore,$BeSO_4$ has the highest solubility among the given options.

(C) The chemical formula of Plaster of Paris is $CaSO_4 \cdot \frac{1}{2}H_2O$. It is a hemihydrate of calcium sulfate.

(D) The basic character of alkaline earth metal hydroxides increases as we move down the group due to the increase in ionic character and size of the metal cation.
$Ba(OH)_2$ is the strongest base among the given options because $Ba$ is the largest alkaline earth metal listed.
$Al(OH)_3$ is amphoteric in nature.

(A) The term 'blanc fixe' refers to finely powdered $BaSO_4$ (barium sulfate),which is commonly used as a filler or pigment extender in paints.

(D) $Mg$ metal forms carbide $MgC_2$,which upon heating converts to $Mg_2C_3$. $Mg_2C_3$ reacts with water to produce propyne (also known as allylene).
$Mg_2C_3 + 4H_2O \to 2Mg(OH)_2 + CH_3-C \equiv CH$
(Propyne)

(D) The tendency to form peroxide increases as the size of the alkaline earth metal cation increases down the group.
Among the given options,$Radium$ $(Ra)$ has the largest atomic size,which stabilizes the large peroxide ion $(O_2^{2-})$.
Therefore,$Radium$ shows the highest tendency to form peroxide.

(B) The thermal decomposition of $CaCO_3$ is given by the reaction:
$CaCO_3 \xrightarrow{\Delta} CaO + CO_2$
Here,$CaO$ is a basic oxide and $CO_2$ is an acidic oxide.