Alkaline earth metals Questions in English

(B) $MgSO_4 \cdot 7H_2O$ (Epsom salt) is used as a purgative.

(A) In a group,as we move down,the atomic size increases due to the addition of new shells. $Ra$ $(Radium)$ is at the bottom of Group $2$,so it has the largest atomic size among alkaline earth metals. Consequently,it has the lowest ionization energy and is the most electropositive element in the group.

(A) Beryllium nitride $(Be_3N_2)$ is covalent in nature due to the small size and high polarizing power of the $Be^{2+}$ ion. Because of its covalent character,it is volatile. In contrast,other alkaline earth metal nitrides are ionic and exist as crystalline solids.

(A) $BeCl_2$ is covalent in nature due to the small size and high polarizing power of the $Be^{2+}$ ion. Because of its covalent character,it is soluble in organic solvents like ether.

(D) The basic strength of metal hydroxides increases down the group due to the increase in ionic character and the size of the metal cation.
In a period,the basic strength decreases from left to right as the electronegativity of the metal increases,making the $M-OH$ bond more covalent.
$NaOH$ and $KOH$ are strong bases (Group $1$).
$Ca(OH)_2$ is a moderately strong base (Group $2$).
$Be(OH)_2$ is amphoteric in nature and is the weakest base among the given options because $Be^{2+}$ has a high charge density and small size,making the $Be-OH$ bond highly covalent.

(C) The solubility of ionic compounds in water is determined by the balance between lattice energy and hydration energy. For $BeF_2$,the hydration energy of the small $Be^{2+}$ ion is significantly higher than its lattice energy,which overcomes the lattice forces and makes it soluble in water. In contrast,for other alkaline earth metal fluorides,the lattice energy dominates,making them insoluble.

(B) $Zn(OH)_2$ and $Al(OH)_3$ are insoluble in water.
For alkaline earth metals,the solubility of hydroxides increases down the group.
Therefore,$Ba(OH)_2$ is readily soluble in water.

(A) The thermal stability of carbonates of alkali and alkaline earth metals increases as the electropositive character of the metal increases down the group.
$BeCO_3$ is the least stable due to the small size and high polarizing power of $Be^{2+}$ ion.
$MgCO_3$ is more stable than $BeCO_3$.
$CaCO_3$ is more stable than $MgCO_3$.
$K_2CO_3$ (an alkali metal carbonate) is the most stable among the given compounds.
Therefore,the increasing order of thermal stability is: $IV (BeCO_3) < II (MgCO_3) < III (CaCO_3) < I (K_2CO_3)$.

(C) $Be(OH)_2$ and $Al(OH)_3$ are amphoteric in nature,not basic. Therefore,the statement that $Be(OH)_2$ is basic like $Al(OH)_3$ is incorrect.

(A) Due to its smallest atomic size and high charge density,$Be$ (Beryllium) forms complex salts such as $[BeF_3]^-$ and $[BeF_4]^{2-}$.

(B) According to Fajan's rule,smaller cations with higher charge density have greater polarizing power,leading to more covalent character in the bond.
$Be^{2+}$ has the smallest ionic radius among the given elements $(Be, Mg, Ca, Sr)$.
Therefore,$BeCl_2$ has the highest covalent character and the least ionic character.

(C) Due to the high ionization energy of $Mg$,the Bunsen flame cannot provide sufficient energy to excite its electrons to higher energy levels. Therefore,$MgCl_2$ does not give a characteristic flame test.

(B) The solubility of alkaline earth metal sulfates decreases down the group from $Be$ to $Ba$ due to the decrease in hydration energy being more significant than the decrease in lattice energy.
Therefore,$BeSO_4$ is the most soluble and $BaSO_4$ is the least soluble in water.

(A) Alkaline earth metals belong to Group $2$ of the periodic table and have two valence electrons.
They lose these two electrons to achieve a stable noble gas configuration,forming dipositive ions.
The general reaction is: $M \to M^{2+} + 2e^-$.

(A) Alkaline earth metals,such as $Barium$ $(Ba)$,belong to group $2$ of the periodic table and possess a stable electronic configuration of $ns^2$. They typically exhibit a fixed valency of $+2$ and do not show variable valency. In contrast,transition elements like $Titanium$ $(Ti)$ and $Copper$ $(Cu)$,as well as post-transition elements like $Lead$ $(Pb)$,exhibit variable valency due to the involvement of $d$ or $f$ electrons or inert pair effects.

(B) Alkaline earth metals belong to Group $2$ of the periodic table.
They have two electrons in their outermost $s$-orbital.
Therefore,their characteristic outer shell electronic configuration is $ns^2$.

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

(D) Due to the small atomic size and high ionization enthalpy of $Be$,it primarily forms covalent compounds.

(C) $Mg$ burns in air to form $MgO$ and $Mg_3N_2$.
$Mg_3N_2$ reacts with water to produce $NH_3$ gas,which has a characteristic pungent smell.
The reaction is: $Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3$.

(C) $Mg$ can form complex compounds. For example,chlorophyll is a complex of $Mg^{2+}$ ion.

(A) The metal $Be$ forms $BeSO_4$ which is water-soluble.
$Be(OH)_2$ is amphoteric and dissolves in $NaOH$ to form sodium beryllate.
$Be(OH)_2 + 2NaOH \to Na_2BeO_2 + 2H_2O$.
$BeO$ is known to become inert (less reactive) upon strong heating.
Other alkaline earth metals like $Mg, Ca, Sr$ form hydroxides that are not soluble in $NaOH$ as they are basic in nature,whereas $Be(OH)_2$ is amphoteric.

(A) Sprinkling water on cement plasters during curing helps in the hydration of cement compounds,which leads to the development of interlocking needle-like crystals of hydrated silicates. This process provides strength and hardness to the cement structure.

(A) The solubility of alkaline earth metal carbonates decreases down the group because the hydration energy of the cations decreases more rapidly than the lattice energy as the size of the cation increases.

(A) Calcined gypsum is $CaSO_4 \cdot \frac{1}{2}H_2O$ (Plaster of Paris) and does not contain $CaCO_3$.
Sea shells are primarily composed of $CaCO_3$.
Dolomite is $CaCO_3 \cdot MgCO_3$.
$A$ marble statue is made of limestone,which is $CaCO_3$.

(A) Hydration enthalpy is directly proportional to the charge density of the ion.
Charge density is defined as $\frac{\text{charge}}{\text{size}}$.
For $Mg^{2+}$,the charge is $+2$.
Comparing the options:
$Al^{3+}$ has a higher charge $(+3)$ and a smaller ionic radius than $Mg^{2+}$,resulting in a higher charge density.
$Be^{2+}$ also has a higher charge density due to its smaller size.
$Na^+$ has a lower charge $(+1)$ and a larger size,so its hydration enthalpy is lower.
$Mg^{3+}$ is not a stable ion in aqueous solution.
Therefore,both $Al^{3+}$ and $Be^{2+}$ have higher hydration enthalpies than $Mg^{2+}$. Given the standard options,$Al^{3+}$ is the most appropriate answer.

(B) As we move down the group in alkaline earth metals,the atomic number increases.
Due to the decrease in lattice energy being more significant than the decrease in hydration energy,the solubility of their hydroxides increases down the group.
Conversely,ionization energy and electronegativity decrease down the group,and the solubility of their sulfates decreases down the group.

(C) As we move down the group in alkaline earth metals,both the lattice energy and the hydration energy decrease.
However,the decrease in hydration energy is much more significant than the decrease in lattice energy.
Since solubility depends on the balance between these two,the rapid decrease in hydration energy dominates,leading to a decrease in the solubility of carbonates down the group.

(B) The reaction cycle corresponds to the thermal decomposition of $CaCO_3$:
$1$. Heating $X$ $(CaCO_3)$: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2 \uparrow$ (Residue is $CaO$)
$2$. Residue $(CaO)$ with water: $CaO + H_2O \rightarrow Ca(OH)_2$ ($Y$ is $Ca(OH)_2$)
$3$. Excess $CO_2$ through $Y$: $Ca(OH)_2 + 2CO_2 \rightarrow Ca(HCO_3)_2$ ($Z$ is $Ca(HCO_3)_2$)
$4$. Heating $Z$: $Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 + H_2O + CO_2$ (Recovering $X$)
Therefore,$X$ is $CaCO_3$.

(A) The solubility of alkaline earth metal sulfates decreases down the group from $Be$ to $Ba$.
This is because the hydration enthalpy decreases more rapidly than the lattice enthalpy as the size of the cation increases.
For a compound to be highly soluble,its hydration enthalpy must be greater than its lattice enthalpy.
Among the given options,$MgSO_4$ has a relatively small cation $(Mg^{2+})$,which results in a high hydration enthalpy that outweighs the lattice enthalpy,making it the most soluble compound in the list.

(B) The size of the $Be^{2+}$ ion is very small.
Due to its small size,the hydration enthalpy of $BeSO_4$ is significantly higher than its lattice enthalpy,which makes 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.
Therefore,the solubility of their hydroxides in water increases with an increase in atomic number.

(B) $CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g) \uparrow$ (Colorless gas)
$(A) = CaCO_3$,$(B) = CaO$
$CaO(s) + H_2O(l) \to Ca(OH)_2(aq)$
$Ca(OH)_2(aq) + 2CO_2(g) \to Ca(HCO_3)_2(aq)$
$(C) = Ca(HCO_3)_2$
$Ca(HCO_3)_2(s) \xrightarrow{\Delta} CaCO_3(s) + CO_2(g) + H_2O(g)$
Thus,compound $(A)$ is $CaCO_3$.

(B) The chemical formulas for the given compounds are:
$(A)$ Plaster of Paris $= CaSO_4 \cdot \frac{1}{2} H_2O$ (Matches $ii$)
$(B)$ Epsomite $= MgSO_4 \cdot 7H_2O$ (Matches $iii$)
$(C)$ Kieserite $= MgSO_4 \cdot H_2O$ (Matches $iv$)
$(D)$ Gypsum $= CaSO_4 \cdot 2H_2O$ (Matches $i$)
Therefore,the correct matching is $(A)-(ii), (B)-(iii), (C)-(iv), (D)-(i)$.

(A) The enthalpy of hydration is inversely proportional to the size of the ion and directly proportional to the charge density.
For alkaline earth metal ions,the size increases down the group $(Be^{2+} < Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+})$.
Therefore,the enthalpy of hydration decreases down the group.
Among the given options,$Mg^{2+}$ has the smallest ionic radius,resulting in the highest charge density and the highest enthalpy of hydration.

(B) Portland cement consists of several calcium silicates and aluminates. The approximate composition is:
$1.$ Dicalcium silicate $(Ca_2SiO_4)$: $20-30\%$
$2.$ Tricalcium silicate $(Ca_3SiO_5)$: $40-50\%$
$3.$ Tricalcium aluminate $(Ca_3Al_2O_6)$: $10-15\%$
$4.$ Tetracalcium aluminoferrite $(Ca_4Al_2Fe_2O_{10})$: $5-10\%$
Thus,$Ca_3SiO_5$ (Tricalcium silicate) is present in the highest amount.

(C) Helium $(Z = 2)$ has the electronic configuration $1s^2$.
Beryllium $(Z = 4)$ has the electronic configuration $1s^2 \, 2s^2$.
Both elements have an $ns^2$ outer shell configuration,so Statement $A$ is true.
Helium is a noble gas and is chemically inert due to its stable filled shell,whereas Beryllium is a reactive alkaline earth metal that can form compounds by utilizing its $2s$ and $2p$ orbitals. Therefore,Reason $R$ is false.

(A) Assertion $A$ is true because Barium is not essential for human biological processes.
Reason $R$ is true because Barium has an electronic configuration of $[Xe] \, 6s^2$ and exhibits only a $+2$ oxidation state.

(D) $BeF_2$ is predominantly covalent in nature due to the small size and high polarizing power of the $Be^{2+}$ ion.
Because of its covalent character,it has a significantly lower melting point compared to the other alkaline earth metal fluorides,which are primarily ionic.
$BeF_2$ melts at approximately $800 \, ^\circ C$,whereas other alkaline earth metal fluorides like $CaF_2$,$SrF_2$,and $BaF_2$ have much higher melting points (typically above $1200 \, ^\circ C$).

(C) $CaCl_2$ is a very good desiccant,but it cannot be used to dry alcohols or $NH_3$ because it forms addition compounds with them.
For example: $CaCl_2 + 4NH_3 \rightarrow CaCl_2 \cdot 4NH_3$ and $CaCl_2 + 4C_2H_5OH \rightarrow CaCl_2 \cdot 4C_2H_5OH$.
Therefore,Statement $A$ is true,but Reason $R$ is false.

(A) $Be$ can exhibit a maximum covalency of $4$ due to the absence of $d$-orbitals in its valence shell. $Al$ can exhibit a maximum covalency of $6$ because it has vacant $3d$-orbitals available for bonding.

(A) According to Fajan's rule,the smaller the size of the cation,the greater its polarizing power and the higher the covalent character of the compound.
Among the alkaline earth metals,$Be^{2+}$ has the smallest ionic radius.
Therefore,$Be$ compounds exhibit the highest covalent character.

(C) The atomic number $Z = 56$ corresponds to the element Barium $(Ba)$.
The 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 alkaline earth metals.

(C) The atomic number $56$ corresponds to the element Barium $(Ba)$.
Barium is located in Group $2$ of the periodic table.
Elements in Group $2$ are known as alkaline earth metals.
Therefore,the correct option is $C$.

(D) Magnesite is $MgCO_3$.
Calamine is $ZnCO_3$.
Carnallite is $KCl \cdot MgCl_2 \cdot 6H_2O$.
Dolomite is $MgCO_3 \cdot CaCO_3$.
Therefore,dolomite contains both calcium and magnesium.

(A) Dolomite is a common sedimentary rock-forming mineral. Its chemical formula is $CaMg(CO_3)_2$,which can also be represented as $CaCO_3 \cdot MgCO_3$.

(D) Hydration energy is inversely proportional to the ionic size of the cation.
As the size of the cation decreases,the charge density increases,leading to stronger interaction with water molecules.
The ionic radii of the given alkaline earth metal ions follow the order: $Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+}$.
Therefore,$Mg^{2+}$ has the smallest size and the highest charge density,resulting in the maximum hydration energy.

(A) Slaked lime,which is $Ca(OH)_2$,is a key raw material in the production of cement. It is used to adjust the chemical composition of the raw mix during the manufacturing process of Portland cement. The typical composition of Portland cement includes a high percentage of lime $(CaO)$,which is derived from limestone or slaked lime.

\text{Component}	\text{Percentage}
Lime $(CaO)$	$50-60\%$
$SiO_2$	$20-25\%$
Alumina $(Al_2O_3)$	$5-10\%$
$MgO$	$2-3\%$
$Fe_2O_3$	$1-2\%$
$SO_3$	$1-2\%$

(D) The chemical formula of gypsum is $CaSO_4 \cdot 2H_2O$,which contains $2$ water molecules.
The chemical formula of plaster of Paris is $CaSO_4 \cdot \frac{1}{2}H_2O$,which contains $0.5$ water molecules.
The difference in the number of water molecules is $2 - 0.5 = 1.5$ or $1\frac{1}{2}$.

(D) The heat of hydration is inversely proportional to the ionic radius and directly proportional to the charge density of the ion.
$Be^{2+}$ has the smallest ionic radius and the highest charge among the given options,resulting in the highest charge density.
Therefore,the order of decreasing heat of hydration is: $Be^{2+} > Li^{+} > Ba^{2+} > K^{+}$.
Thus,$Be^{2+}$ has the largest heat of hydration.

(D) The correct answer is $(D)$.
$Mg(OH)_2$ is known to sublime upon heating,whereas the hydroxides of alkali metals $(LiOH, KOH, RbOH)$ are generally stable and do not sublime under standard heating conditions.