Alkaline earth metals Questions in English

(A) In the polymeric structure of beryllium chloride,each $Be$ atom is bonded to four $Cl$ atoms.
Each bridging $Cl$ atom is bonded to two $Be$ atoms,one via a normal covalent bond and the other via a coordinate (dative) bond.
This bridging unit involves three atoms $(Be-Cl-Be)$ sharing four electrons (two from the covalent bond and two from the lone pair of $Cl$ used in the dative bond).
Therefore,these are referred to as three-centre four-electron bonds.

(C) When a metal is burnt in air,it reacts with both $O_2$ and $N_2$ to form its oxide and nitride respectively.
For magnesium $(Mg)$,the reactions are:
$3 Mg + O_2 \rightarrow 2 MgO$
$3 Mg + N_2 \rightarrow Mg_3N_2$
The resulting ash contains $Mg_3N_2$. Upon moistening,the magnesium nitride undergoes hydrolysis to release ammonia gas:
$Mg_3N_2 + 6 H_2O \rightarrow 3 Mg(OH)_2 + 2 NH_3 \uparrow$
The smell of $NH_3$ confirms the presence of the nitride.

(C) . $BeCO_3$ has the lowest thermal stability among alkaline earth metal carbonates. It decomposes to give $BeO$ and $CO_2$. Hence,it is stored in an atmosphere of $CO_2$ to prevent decomposition. This statement is correct.
$B$. $Be$ is amphoteric and dissolves in alkali to form the beryllate ion: $Be + 2OH^- + 2H_2O \rightarrow [Be(OH)_4]^{2-} + H_2$. This statement is correct.
$C$ and $D$. $BeF_2$ reacts with $NaF$ to form the complex $Na_2[BeF_4]$. In this complex,$Be$ is part of the complex anion $[BeF_4]^{2-}$. Therefore,the statement that $Be$ goes with the cation is incorrect,and the statement that $Be$ goes with the anion is correct.

(A) $CaCO_3 \stackrel{\Delta}{\longrightarrow} CaO + CO_2$ (colourless gas)
$CaO + H_2O \longrightarrow Ca(OH)_2$ $(Y)$
$Ca(OH)_2 + 2CO_2 \text{ (excess)} \longrightarrow Ca(HCO_3)_2$ $(Z)$
$Ca(HCO_3)_2 \stackrel{\Delta}{\longrightarrow} CaCO_3 + CO_2 + H_2O$ $(X)$
Therefore,the compound $X$ is $CaCO_3$.

(B) The thermal decomposition of metal nitrates follows specific patterns. Zinc nitrate,$Zn(NO_3)_2$,decomposes upon heating to produce zinc oxide $(ZnO)$,nitrogen dioxide $(NO_2)$,and oxygen gas $(O_2)$.
The balanced chemical equation is:
$2Zn(NO_3)_2(s) \xrightarrow{\Delta} 2ZnO(s) + 4NO_2(g) + O_2(g)$.

(B) The reaction between calcium hydroxide and sulfur dioxide is as follows:
$Ca(OH)_2 + SO_2 \to CaSO_3 + H_2O$
Calcium sulfite $(CaSO_3)$ is formed as a white precipitate,which turns lime water milky.

(A) The thermal stability of alkaline earth metal carbonates $(MCO_3)$ increases as the size of the metal cation $(M^{2+})$ increases down the group.
This is because the larger cation polarizes the large carbonate ion $(CO_3^{2-})$ less effectively,making the $M-O$ bond stronger and the carbonate ion more stable.
The order of ionic radii is $Be^{2+} < Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+}$.
Therefore,the thermal stability order is $BeCO_3 < MgCO_3 < CaCO_3 < SrCO_3 < BaCO_3$.

(C) The reaction sequence is as follows:
$1$. $CaO + CO_2 \to CaCO_3$ $(A = CaCO_3)$
$2$. $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$ $(B = CaO)$
$3$. $CaO + H_2O \to Ca(OH)_2$ $(C = Ca(OH)_2)$
$4$. $Ca(OH)_2 + CO_2 \to CaCO_3 + H_2O$ $(D = CaCO_3)$
$5$. $CaCO_3 + H_2O + CO_2 \to Ca(HCO_3)_2$ $(E = Ca(HCO_3)_2)$
Thus,$D$ is $CaCO_3$ and $E$ is $Ca(HCO_3)_2$.

(B) The basic nature of hydroxides depends on the ease with which the $M-OH$ bond breaks,which is related to the ionic character of the bond.
As we move from left to right in the periodic table,the electronegativity of the central atom increases,and the ionic character of the $M-OH$ bond decreases.
$Be(OH)_2$ is amphoteric,$Mg(OH)_2$ is basic,$Al(OH)_3$ is amphoteric,and $Si(OH)_4$ is acidic.
Among the given options,$Mg(OH)_2$ is the most basic because it is an alkaline earth metal hydroxide with significant ionic character compared to the others.

(D) Hydrolith is the common name for calcium hydride,which has the chemical formula $CaH_2$.
It is produced by the reaction of calcium metal with dihydrogen gas:
$Ca(s) + H_2(g) \rightarrow CaH_2(s)$.

(C) White vitriol is the common name for zinc sulfate heptahydrate,which has the chemical formula $ZnSO_4 \cdot 7H_2O$.
$CuSO_4 \cdot 5H_2O$ is known as blue vitriol.
$FeSO_4 \cdot 7H_2O$ is known as green vitriol.
Therefore,the correct option is $C$.

(A) $MgO$ (magnesium oxide) has a very high melting point of $2852 \ ^{\circ}C$.
Due to this high melting point and its chemical stability at elevated temperatures,it is widely used as a refractory material for lining furnaces.

(B) The complex-forming tendency of an ion is inversely proportional to its size and directly proportional to its charge density.
Among the given ions,$Be^{2+}$ has the smallest ionic radius and a high charge,resulting in the highest charge density.
Therefore,$Be^{2+}$ exhibits the maximum complex-forming tendency.

(A) $MgO$ is a basic oxide because it is an oxide of an alkaline earth metal (Group $2$).
$SnO$,$ZnO$,and $PbO$ are amphoteric oxides,meaning they react with both acids and bases.
Therefore,$MgO$ is different from the other three.

(A) Milk of magnesia is a suspension of magnesium hydroxide in water. Its chemical formula is $Mg(OH)_2$. It is commonly used as an antacid to treat indigestion and heartburn.

(C) The solubility of group $2$ metal carbonates is generally low in water.
All group $1$ metal salts (like $KNO_3$ and $NaNO_3$) are highly soluble in water.
$Ba(OH)_2$ is moderately soluble in water.
$BaCO_3$ is an alkaline earth metal carbonate and is insoluble in water due to its high lattice energy compared to its hydration energy.
Therefore,the correct option is $BaCO_3$.

(B) The raw materials for cement are limestone,clay,and gypsum. Cement is a greyish heavy powder containing calcium aluminates and silicates. Gypsum $(CaSO_4 \cdot 2H_2O)$ is added to the cement clinker during the grinding process to increase the setting time of cement,which allows it to be placed and hardened properly. The setting of cement is an exothermic process involving the hydration of calcium aluminates and silicates. Adding gypsum slows down the initial hydration reaction.

(A) . $Mg(NO_3)_2 \xrightarrow{\Delta} MgO + 2NO_2 \uparrow + \frac{1}{2}O_2 \uparrow$. $NO_2$ is a paramagnetic acidic gas.
$B$. $Fe_2(SO_4)_3 \xrightarrow{\Delta} Fe_2O_3 + 3SO_3 \uparrow$. $SO_3$ is diamagnetic.
$C$. $FeCO_3 \xrightarrow{\Delta} FeO + CO_2 \uparrow$. $CO_2$ is diamagnetic.
$D$. $HgC_2O_4 \xrightarrow{\Delta} Hg + CO_2 \uparrow + CO \uparrow$. Both $CO_2$ and $CO$ are diamagnetic.

(B) $2NH_4ClO_4 \xrightarrow{\Delta} N_2 \uparrow + Cl_2 \uparrow + 2O_2 \uparrow + 4H_2O$
$CaCO_3 \xrightarrow{\Delta} \underbrace{CaO}_{\text{Basic}} + \underbrace{CO_2 \uparrow}_{\text{Acidic}}$
$NH_4NO_2 \xrightarrow{\Delta} N_2 \uparrow + 2H_2O$
$NaNO_3 \xrightarrow{\text{below } 500^{\circ}C} NaNO_2 + \frac{1}{2}O_2 \uparrow$
$2NaNO_3 \xrightarrow{800^{\circ}C} Na_2O + N_2 \uparrow + \frac{5}{2}O_2 \uparrow$
Thus,$CaCO_3$ decomposes to give a basic oxide $(CaO)$ and an acidic oxide $(CO_2)$.

(D) $1$. For $Group \ 2$ hydroxides,solubility increases down the group due to the decrease in lattice energy being more significant than the decrease in hydration energy. Thus,$Ca(OH)_2 < Sr(OH)_2 < Ba(OH)_2$ is correct.
$2$. For alkali metal carbonates,solubility increases down the group as the lattice energy decreases significantly. Thus,$Li_2CO_3 < Na_2CO_3 < K_2CO_3$ is correct.
$3$. For nitrates of alkali metals,solubility decreases down the group as the hydration energy decreases more rapidly than the lattice energy. Therefore,the correct order is $CsNO_3 < RbNO_3 < KNO_3 < NaNO_3 < LiNO_3$. Thus,$CsNO_3 < RbNO_3 < KNO_3$ is correct.
$4$. For alkaline earth metal thiosulfates,solubility generally decreases down the group due to the dominance of lattice energy over hydration energy. Therefore,the order $BeS_2O_3 < MgS_2O_3 < CaS_2O_3$ is incorrect; it should be $BeS_2O_3 > MgS_2O_3 > CaS_2O_3$.

(A) Due to the very small size and high ionization enthalpy of $Be^{2+}$ ion,it has a high polarizing power according to Fajan's rule.
This results in significant covalent character in its compounds,unlike other alkaline earth metals which primarily form ionic compounds.

(C) The flame test is used to identify metal ions based on the characteristic color they impart to a flame.
$Ba^{2+}$ ions impart an apple-green color to the flame.
$Na^{+}$ ions impart a golden yellow color.
$Ca^{2+}$ ions impart a brick-red color.
$Mg^{2+}$ ions do not give a characteristic color in the flame test because their excitation energy is very high.

(A) Beryllium $(Be)$ is an alkaline earth metal,but it shows anomalous behavior compared to other members of the group.
$Be$ does not react directly with $H_2$ to form $BeH_2$ because of its high ionization energy and small size.
$BeH_2$ is typically prepared by the reaction of $BeCl_2$ with $LiAlH_4$.
Therefore,the reaction $Be + H_2 \longrightarrow BeH_2$ is incorrect.

(D) Sorel's cement is a mixture formed by the reaction of magnesium oxide $(MgO)$ with a concentrated solution of magnesium chloride $(MgCl_2)$.
The chemical composition is represented as $MgCl_2 \cdot 5MgO \cdot xH_2O$ (where $x$ is typically $13$).
It is a hard,stone-like material used in dental fillings and flooring.

(D) The alkaline earth metals $Be$ and $Mg$ have very high ionization enthalpies.
When heated in a flame,the energy provided is not sufficient to excite their electrons to higher energy levels.
Therefore,they do not impart any characteristic colour to the flame.

(D) An amphoteric oxide is a substance that can act as both an acid and a base.
$BeO$ (Beryllium oxide),$SnO$ (Tin$(II)$ oxide),and $ZnO$ (Zinc oxide) are all well-known examples of amphoteric oxides.
They react with both acids and bases to form salts and water.
Therefore,the correct answer is $D$.

(D) For $LiF < LiCl < LiBr < LiI$,the solubility increases due to the decrease in lattice energy as the size of the anion increases. This is correct.
For $Li_2CO_3 < Na_2CO_3 < K_2CO_3 < Rb_2CO_3$,the solubility of alkali metal carbonates increases down the group as the hydration energy decreases less rapidly than the lattice energy. This is correct.
For $Be(OH)_2 < Mg(OH)_2 < Ca(OH)_2 < Sr(OH)_2$,the solubility of alkaline earth metal hydroxides increases down the group because the lattice energy decreases more significantly than the hydration energy. This is correct.
For $BeF_2 < MgF_2 < CaF_2 < SrF_2$,the solubility of alkaline earth metal fluorides decreases down the group because the hydration energy decreases more rapidly than the lattice energy. Therefore,the correct order should be $BeF_2 > MgF_2 > CaF_2 > SrF_2$. Thus,option $D$ is the incorrect order.

(A) $CaCO_3$ $(x) \xrightarrow{\Delta} CaO_{(s)} + CO_2 \uparrow$ (colourless gas).
$CaO_{(s)} + H_2O \rightarrow Ca(OH)_2$ $(y)$.
$Ca(OH)_2$ $(y) + 2CO_2 \rightarrow Ca(HCO_3)_2$ $(z)$.
$Ca(HCO_3)_2$ $(z) \xrightarrow{\Delta} CaCO_3$ $(x) + H_2O + CO_2$.

(B) The element $P$ with atomic number $56$ is Barium $(Ba)$,which belongs to Group $2$ of the periodic table.
Group $2$ elements have $2$ valence electrons and tend to lose these electrons to form a cation with a charge of $+2$,i.e.,$P^{2+}$.
Halogens $(X)$ belong to Group $17$ and gain $1$ electron to form an anion with a charge of $-1$,i.e.,$X^-$.
To form a neutral compound,one $P^{2+}$ ion combines with two $X^-$ ions.
Therefore,the formula of the halide is $PX_2$.

(A) The solubility of alkaline earth metal sulphates depends on the balance between lattice energy and hydration energy.
As we move down the group,the size of the cation increases.
The hydration energy decreases significantly as the size of the cation increases.
Since the decrease in hydration energy is more pronounced than the decrease in lattice energy,the overall solubility decreases down the group.

(C) The substance $A$ is calcium hydroxide,$Ca(OH)_2$,also known as slaked lime.
$1$. $Ca(OH)_2$ is used to remove temporary hardness of water by precipitating calcium bicarbonate as calcium carbonate: $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3(s) + 2H_2O$.
$2$. It reacts with sodium carbonate $(Na_2CO_3)$ to produce caustic soda $(NaOH)$: $Ca(OH)_2 + Na_2CO_3 \rightarrow CaCO_3 + 2NaOH$.
$3$. When $CO_2$ is passed through lime water $(Ca(OH)_2)$,it turns milky due to the formation of insoluble $CaCO_3$: $Ca(OH)_2 + CO_2 \rightarrow CaCO_3(s) + H_2O$.

(C) Amphoteric hydroxides are those that can react with both acids and bases.
$Al(OH)_3$ is amphoteric because it reacts with $HCl$ (acid) to form $AlCl_3$ and with $NaOH$ (base) to form $Na[Al(OH)_4]$.
$Be(OH)_2$ is also amphoteric as it reacts with $HCl$ to form $BeCl_2$ and with $NaOH$ to form $Na_2[Be(OH)_4]$.
Therefore,the pair $Al(OH)_3$ and $Be(OH)_2$ consists of amphoteric hydroxides.

(C) The solubility of sulfates of alkaline earth metals and heavy metals depends on the lattice energy and hydration energy.
$PbSO_4$ (Lead$(II)$ sulfate) is known to be insoluble in water due to its very high lattice energy compared to its hydration energy.
In contrast,$CuSO_4$,$CdSO_4$,and $Al_2(SO_4)_3$ are soluble in water.

(D) The production of bleaching powder involves the reaction of chlorine gas with dry slaked lime $(Ca(OH)_2)$.
The chemical equation is: $Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O$.
Therefore,the correct option is $D$.

(D) Magnesium $(Mg)$ belongs to Group $2$ (alkaline earth metals).
$1$. Sodium $(Na)$ is in Group $1$ and is more electropositive than $Mg$.
$2$. Electrolysis of aqueous $MgCl_2$ produces $H_2$ gas at the cathode instead of $Mg$ because $H^+$ ions are reduced more easily than $Mg^{2+}$ ions. $Mg$ is obtained by the electrolysis of fused (molten) $MgCl_2$.
$3$. $Mg$ is a reducing agent,but not as strong as alkali metals like $Na$.
$4$. Magnesium shows a diagonal relationship with Lithium $(Li)$,not Boron $(B)$.
Wait,re-evaluating the options: The question asks for the true statement. Upon review,none of the provided options are strictly correct based on standard chemistry. However,if we must choose the most chemically relevant concept often tested,the diagonal relationship is between $Mg$ and $Li$. Given the constraints,if this is a multiple-choice question where one must be selected,there might be an error in the question source. Assuming the intended answer is related to the diagonal relationship (even if the element is incorrect),or if the question implies a specific context,$I$ will mark the most plausible intended answer if applicable. Since all are factually incorrect,$I$ will flag this as an error.

(D) Alkaline earth metals have low ionization enthalpies and can easily lose their valence electrons to form $M^{2+}$ ions.
Since they readily lose electrons,they act as good electron donors.
Substances that donate electrons are known as reducing agents.
Because they lose electrons easily,they are considered strong reducing agents.

(D) The solubility of alkaline earth metal halides decreases down the group due to the increase in lattice energy being more significant than the hydration energy.
$CaF_2$ (calcium fluoride) is known to be sparingly soluble or insoluble in water due to its very high lattice energy compared to its hydration energy.
In contrast,nitrates like $NaNO_3$ and $Ca(NO_3)_2$ are highly soluble in water.
$HgCl_2$ is also soluble in water.
Therefore,the correct option is $D$.

(D) Alkaline earth metals belong to Group $2$ of the periodic table.
They have a general electronic configuration of $ns^2$.
Due to the presence of two electrons in their outermost shell,they readily lose these two electrons to achieve a stable noble gas configuration,forming $M^{2+}$ ions.
They exist as monatomic species in their metallic state.
Therefore,they are monatomic and form $M^{2+}$ ions.

(C) $1$. $Be$ is the first element of Group $2$ and has a small size and high ionization energy,making its hydroxide $Be(OH)_2$ amphoteric in nature.
$2$. As we move down the group,the metallic character increases,and the basic character of hydroxides increases. Thus,$Ba(OH)_2$ is strongly basic,while $Be(OH)_2$ is amphoteric.
$3$. Therefore,the statement that $Be(OH)_2$ is more basic than $Mg(OH)_2$ is incorrect,as $Mg(OH)_2$ is more basic than $Be(OH)_2$.

(A) The solubility of alkaline earth metal sulfates in water decreases down the group from $Be$ to $Ba$.
This is because the hydration enthalpy of the metal ions decreases much more rapidly than the lattice energy as the size of the cation increases down the group.
For small ions like $Be^{2+}$,the hydration enthalpy is very high,which compensates for the lattice energy,making the sulfate highly soluble in water.

(C) The characteristic flame colors of alkaline earth metals are due to the excitation of electrons to higher energy levels.
When these elements or their salts are heated in a flame,they emit specific wavelengths of light.
$Ba$ (Barium) salts impart a characteristic apple-green color to the flame.
$Na$ gives a golden yellow flame,$K$ gives a violet flame,and $Ca$ gives a brick-red flame.
Therefore,the green flame is due to the presence of $Ba$.

(C) Plaster of Paris is $CaSO_4 \cdot \frac{1}{2}H_2O$.
When it is mixed with water,it undergoes hydration to form Gypsum $(CaSO_4 \cdot 2H_2O)$,which is a hard,solid mass.
The reaction is: $CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O \rightarrow CaSO_4 \cdot 2H_2O$.

(A) Plaster of Paris $(CaSO_4 \cdot \frac{1}{2}H_2O)$ sets into a hard mass by hydration to form Gypsum $(CaSO_4 \cdot 2H_2O)$.
Common salt $(NaCl)$ acts as an accelerator for this setting process,thereby increasing the rate of setting.

(A) Calcium is an alkaline earth metal with a high negative reduction potential.
It cannot be obtained by the reduction of its salts with carbon or by the electrolysis of its aqueous solution because hydrogen gas would be evolved at the cathode instead of calcium.
Therefore,calcium is commercially prepared by the electrolysis of fused (molten) $CaCl_2$ containing a small amount of $CaF_2$ to lower the melting point.

(D) 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 $0.5$ molecules of water of crystallization.
The difference in the number of water molecules is $2 - 0.5 = 1.5$.

(B) Plaster of Paris,$CaSO_4 \cdot \frac{1}{2}H_2O$,is widely used in medicine for setting fractured bones in a cast.
It is also used in making toys,statues,and for plastering walls.

(D) Plaster of Paris is $CaSO_4 \cdot \frac{1}{2}H_2O$.
When it is mixed with water,it undergoes hydration to form a hard solid mass known as gypsum $(CaSO_4 \cdot 2H_2O)$.
The reaction is: $CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O \rightarrow CaSO_4 \cdot 2H_2O$.
Thus,the setting of Plaster of Paris is due to the formation of another hydrate by hydration.

(C) The chemical reactivity of alkaline earth metals $(Group \ 2)$ increases as we move down the group.
This is because the ionization enthalpy decreases down the group,making it easier for the atom to lose electrons.
The order of reactivity is $Be < Mg < Ca < Sr < Ba$.
Therefore,$Ba$ (Barium) is the most reactive among the given options.

(C) Plaster of Paris is chemically known as calcium sulphate hemihydrate.
Its chemical formula is $CaSO_4 \cdot 0.5H_2O$ or $CaSO_4 \cdot \frac{1}{2}H_2O$.
It is obtained by heating gypsum $(CaSO_4 \cdot 2H_2O)$ at $393 \ K$.

(B) The melting points of alkaline earth metals $(Group \ 2)$ do not show a regular trend due to differences in their crystal structures.
However,$Mg$ $(Magnesium)$ has the lowest melting point among the alkaline earth metals $(Be, Mg, Ca, Sr, Ba)$.
$Be$ has a high melting point due to its small size and strong metallic bonding.
$Mg$ has a relatively lower melting point of approximately $923 \ K$ compared to the others in the group.