Alkali metals Questions in English

(A) The flame test is used to identify metal cations based on the characteristic color they impart to a Bunsen flame.
$Ba^{2+}$ gives an apple-green color,$Ca^{2+}$ gives a brick-red color,and $Pb^{2+}$ gives a pale bluish-white color.
$Mg^{2+}$ does not impart any characteristic color to the Bunsen flame because the energy required to excite its electrons is too high to be provided by the flame.
Therefore,$Mg^{2+}$ is the correct cation.

(D) The metallic lustre is primarily due to the presence of $free \ electrons$ in the metal lattice. When light falls on the surface of the metal,these $free \ electrons$ absorb the energy and get excited. Upon returning to their ground state,they re-emit the absorbed energy as light,which gives the metal its characteristic shiny appearance.

(B) Liquid sodium is used in nuclear reactors as a heat exchanger or coolant.

(B) Sodium is obtained by the electrolytic reduction of its chloride. The melting point of pure $NaCl$ is very high $(1074 \ K)$. To lower the melting point of the electrolyte to a manageable range (around $873 \ K$),$CaCl_2$ is added to the mixture.

(C) Metals that are more reactive than hydrogen (i.e.,have a negative standard reduction potential) can displace hydrogen from dilute acids.
$Fe$ and $Al$ are both more reactive than hydrogen.
However,in the context of standard textbook questions where a single choice is required,$Al$ is a classic example of a metal that reacts with dilute $H_2SO_4$ to liberate $H_2$ gas.
Reaction: $2Al(s) + 3H_2SO_4(dil) \to Al_2(SO_4)_3(aq) + 3H_2(g) \uparrow$.

(D) Most metals do not react with $NaOH$. However,amphoteric metals like $Zn$,$Al$,and $Pb$ react with strong bases like $NaOH$ to evolve $H_2$ gas.
For example,the reaction for $Zn$ is:
$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2 \uparrow$
$Mg$,$Ba$,$Ca$,and $Sr$ are alkaline earth metals and do not react with $NaOH$ to produce $H_2$ gas.

(A) Although $Be$ is placed above $Mg$ in the second group of the periodic table,it is placed below $Mg$ in the electrochemical series.
Since $Be$ has a more positive standard reduction potential (or is less reactive) than $Mg$,it cannot displace $Mg$ from its salt solution.
Therefore,adding $Be$ dust to $MgCl_2$ solution will have no effect.

(A) The reducing power of an element is determined by its standard electrode potential $(E^\circ)$.
$Lithium$ $(Li)$ has the most negative standard reduction potential $(E^\circ = -3.04 \ V)$ among all alkali metals.
This high negative value indicates that $Li$ has the greatest tendency to lose electrons in an aqueous medium,making it the strongest reducing agent.

(C) The chemical formula of Carnallite is $KCl \cdot MgCl_2 \cdot 6H_2O$.
It is a hydrated potassium magnesium chloride mineral,which serves as an important source of $Mg$.

(A) $Na$ (Sodium) is an alkali metal and is highly reactive.
Due to its high reactivity,it readily reacts with other elements to form compounds and is therefore not found in the free state in nature.
In contrast,noble metals like $Au$ (Gold) and $Ag$ (Silver) are less reactive and can be found in the free state.

(B) $(b)$ Copper $(Cu)$ is a reddish-brown metal, whereas Nickel $(Ni)$, Sodium $(Na)$, and Tin $(Sn)$ are silvery-white metals.

(C) . Carnallite is $KCl \cdot MgCl_2 \cdot 6H_2O$.

(A) Chile saltpetre is the common name for sodium nitrate $(NaNO_3)$.
It is called Chile saltpetre because it is a deliquescent crystalline sodium salt that is found chiefly in northern Chile.

(B) Carnallite is an important ore of $Mg$ and $K$. Its chemical formula is $KCl \cdot MgCl_2 \cdot 6H_2O$.

(C) The electrolysis of fused (molten) sodium chloride $(NaCl)$ is known as the $Downs$ process.
The dissociation reaction is: $2NaCl_{(l)} \to 2Na^+ + 2Cl^-$
At the anode (oxidation): $2Cl^- \to Cl_2 + 2e^-$
At the cathode (reduction): $2Na^+ + 2e^- \to 2Na$
Thus,metallic sodium $(Na)$ is produced at the cathode.

(A) Mercury $(Hg)$ does not react with water or steam to produce $H_2$ gas because it is a noble metal with a very high positive standard reduction potential,meaning it is not reactive enough to displace hydrogen from water.

(D) An amalgam is defined as an alloy of $Mercury$ $(Hg)$ with one or more other metals.
Therefore,$Mercury$ is the essential constituent of any amalgam.

(C) Iron is used because mercury does not form an amalgam with iron.
Most other metals like silver,lead,and aluminium readily form amalgams with mercury,which would damage the container.

(D) Among the given options,$HgO$ (mercuric oxide) is a thermally unstable oxide. Upon heating,it decomposes into mercury metal and oxygen gas according to the following reaction:
$2HgO(s) \xrightarrow{\Delta} 2Hg(l) + O_2(g)$
Other oxides like $ZnO$,$Al_2O_3$,and $CuO$ are relatively stable and do not decompose into their constituent elements simply by heating.

(C) Zinc oxide $(ZnO)$ is an amphoteric oxide. When it reacts with a strong base like sodium hydroxide $(NaOH)$,it forms sodium zincate $(Na_2ZnO_2)$ and water.
The balanced chemical equation is:
$ZnO + 2NaOH \to Na_2ZnO_2 + H_2O$

(C) Amphoteric metals like $Zn$ react with hot concentrated alkali to produce hydrogen gas.
The chemical reaction is:
$Zn + 2NaOH \text{ (conc.)} \to Na_2ZnO_2 + H_2 \uparrow$
In this reaction,$Zn$ reacts with sodium hydroxide to form sodium zincate and hydrogen gas is evolved.

(C) $ZnSO_4 \cdot 7H_2O$ is known as white vitriol.
It is a white crystalline solid that is highly soluble in water.
When heated with barium sulphide,it forms a mixture of zinc sulphide and barium sulphate,which is known as lithopone,a white pigment.

(C) Concentrated sulphuric acid acts as an oxidizing agent. Metals like $Cu$ react with concentrated $H_2SO_4$ to produce $SO_2$ gas.
The chemical reaction is: $Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$.

(A) When $Copper$ $(Cu)$ is exposed to moist air containing $CO_2$,it reacts to form a green layer of basic copper carbonate,which is a mixture of $CuCO_3$ and $Cu(OH)_2$.
The chemical reaction is:
$2Cu + H_2O + CO_2 + \frac{1}{2}O_2 \to CuCO_3 \cdot Cu(OH)_2$
Therefore,the correct option is $(A)$.

(B) Concentrated sulphuric acid $(H_2SO_4)$ acts as an oxidizing agent.
Among the given metals,$Ag$ (silver) reacts with hot concentrated $H_2SO_4$ to form silver sulphate,sulphur dioxide,and water.
The chemical reaction is:
$2Ag + 2H_2SO_4 \to Ag_2SO_4 + SO_2 + 2H_2O$
$Au$ (gold) and $Pt$ (platinum) are noble metals and do not react with concentrated $H_2SO_4$.
$Pb$ (lead) forms a protective layer of $PbSO_4$ on its surface,which prevents further reaction.
Therefore,the correct option is $B$.

(C) Upon strong heating,silver nitrate $(AgNO_3)$ decomposes to form metallic silver $(Ag)$,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$. The reaction is: $2AgNO_3(s) \xrightarrow{\Delta} 2Ag(s) + 2NO_2(g) + O_2(g)$. The silver metal remains as a shiny metallic globule.

(D) When $Zn$ reacts with an excess of $NaOH$,it forms $Na_2ZnO_2$ (Sodium zincate) and releases $H_2$ gas.
The balanced chemical equation is:
$Zn + 2NaOH \to Na_2ZnO_2 + H_2$

(D) Amphoteric metals like $Zn$ and $Al$ react with strong bases like $NaOH$ to form soluble complexes and release $H_2$ gas.
The chemical reactions are:
$Zn + 2NaOH \to Na_2ZnO_2 + H_2$
$2Al + 2NaOH + 2H_2O \to 2NaAlO_2 + 3H_2$
Therefore,the correct pair is $Zn$ and $Al$.

(B) $Zn$ is an amphoteric metal. It reacts with hot concentrated $NaOH$ solution to form sodium zincate and hydrogen gas.
$Zn + 2NaOH \xrightarrow{\text{heat}} Na_2ZnO_2 + H_2$

(D) The correct answer is $(D)$.
An amphoteric oxide is one that reacts with both acids and bases to form salt and water.
$ZnO$ is an amphoteric oxide.
Reaction with acid: $ZnO + 2HCl \to ZnCl_2 + H_2O$
Reaction with base: $ZnO + 2NaOH \to Na_2ZnO_2 + H_2O$

(A) . $Zn$ does not react with cold water.
$Zn$ reacts with steam or hot water to yield $H_2$ gas:
$Zn + H_2O(g) \to ZnO + H_2$
$Zn$ reacts with dilute $H_2SO_4$ to produce $H_2$ gas:
$Zn + H_2SO_4(dil) \to ZnSO_4 + H_2$
$Zn$ reacts with dilute $HCl$ to produce $H_2$ gas:
$Zn + 2HCl(dil) \to ZnCl_2 + H_2$
$Zn$ reacts with hot $20 \%$ $NaOH$ to produce $H_2$ gas:
$Zn + 2NaOH(20 \%) \xrightarrow{\Delta} Na_2ZnO_2 + H_2$

(B) The metal that reacts with both acids and strong bases to liberate hydrogen gas is amphoteric in nature.
$Zn + 2HCl \to ZnCl_2 + H_2$
$Zn + 2NaOH \to Na_2ZnO_2 + H_2$
Thus,$Zinc$ is the correct answer.

(A) When solid potassium permanganate $(KMnO_4)$ is heated strongly,it undergoes thermal decomposition to produce oxygen gas,potassium manganate $(K_2MnO_4)$,and manganese dioxide $(MnO_2)$.
The chemical equation is: $2KMnO_4(s) \xrightarrow{\Delta} K_2MnO_4(s) + MnO_2(s) + O_2(g)$.

(A) $Na^+$ and $K^+$ ions play a crucial role in maintaining blood pressure and heart rate.
An excess of $Na^+$ ions in the body leads to an increase in blood pressure $(B.P.)$.

(D) Washing soda is sodium carbonate $(Na_2CO_3)$,which in water undergoes hydrolysis to form a basic solution of sodium hydroxide $(NaOH)$.
$Al$ is an amphoteric metal that reacts with strong bases like $NaOH$ to form soluble sodium aluminate $(Na[Al(OH)_4])$ and releases hydrogen gas.
$2Al + 2NaOH + 6H_2O \rightarrow 2Na[Al(OH)_4] + 3H_2(g)$.
Therefore,washing soda corrodes aluminum utensils by forming soluble aluminates.

(A) When sodium metal dissolves in liquid ammonia,it undergoes ionization: $Na + (x+y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$.
The blue color of the solution is due to the presence of ammoniated (solvated) electrons,which absorb light in the visible region of the spectrum.
The conductivity and paramagnetic nature of the solution are also attributed to these solvated electrons.

(B) The metal $X$ is Magnesium $(Mg)$.
When $Mg$ is heated in nitrogen gas,it forms Magnesium nitride $(Y = Mg_3N_2)$:
$3Mg + N_2 \xrightarrow{\Delta} Mg_3N_2$
When $Mg_3N_2$ reacts with water,it produces ammonia gas $(NH_3)$:
$Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3(g)$
Ammonia gas reacts with $CuSO_4$ solution to form a deep blue complex,tetraamminecopper$(II)$ sulfate:
$Cu^{2+}(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]^{2+}(aq)$ (Deep blue colour)
Therefore,$Y$ is $Mg_3N_2$.

(B) Concentrated $HNO_3$ is a strong oxidizing agent. Metals like $Fe$ and $Zn$ react with it to form their respective nitrates. However,noble metals like $Pt$ (Platinum) are inert and do not react with concentrated $HNO_3$ to form nitrates. $Pb$ and $Ag$ react with concentrated $HNO_3$ to form their respective nitrates ($Pb(NO_3)_2$ and $AgNO_3$). Therefore,among the given options,$Pt$ is the metal that does not form a nitrate. Since the question asks for the metal that does not form a nitrate,and $Pt$ is the only one listed that is inert,the options provided in the question are slightly ambiguous. However,based on chemical reactivity,$Pt$ is the correct answer.

(A) The flame test is used to identify the presence of certain metal ions in a sample.
Barium salts $(Ba^{2+})$ impart a characteristic apple-green color to the flame.
Calcium salts $(Ca^{2+})$ produce a brick-red flame.
Lead salts $(Pb^{2+})$ typically produce a pale bluish-white flame.
Borates are generally tested using the borax bead test,not a simple flame test on a $Pt$ wire for color identification.
Therefore,the correct answer is Barium salt.

(D) Phenolphthalein is a pH indicator that turns pink in an alkaline medium $(pH > 8.2)$.
$MgO$,$CaO$,and $K_2O$ are basic oxides that react with water to form strong bases ($Mg(OH)_2$,$Ca(OH)_2$,and $KOH$ respectively),which turn phenolphthalein pink.
$Fe_2O_3$ is an amphoteric/weakly basic oxide that is practically insoluble in water and does not produce a sufficiently high concentration of $OH^-$ ions to change the color of phenolphthalein.

(C) $Na$ has a very low ionization enthalpy compared to other alkali metals or transition elements.
Therefore,it acts as a strong reducing agent.

(A) The atomic number is $Z = 55$.
The electronic configuration is $[Xe] 6s^1$.
Since the last electron enters the $s-$orbital,the element belongs to the $s-$block (Alkali metal).

(B) The basic strength of metal oxides increases with an increase in the electropositive character (metallic character) of the metal.
$K$ is an alkali metal (Group $1$),$Sr$ is an alkaline earth metal (Group $2$),$Mg$ is an alkaline earth metal (Group $2$),and $Ni$ is a transition metal.
Electropositive character follows the order: $Ni < Mg < Sr < K$.
Therefore,the basic strength order is $NiO < MgO < SrO < K_2O$.

(B) As we move down a group in the periodic table,the atomic size increases due to the addition of new electron shells.
Consequently,the valence electrons are further away from the nucleus and are held less tightly.
Therefore,the ionization energy decreases as we move down Group $1$.

(A) Alkali metals impart characteristic colors to the Bunsen flame because their valence electrons are loosely held due to their $low$ $ionization$ $energy$. When heated in a flame,these electrons absorb energy and get excited to higher energy levels. Upon returning to the ground state,they emit the absorbed energy in the form of visible light.

(A) As we move down the group in alkali metals,the atomic number increases,which leads to an increase in the number of shells. Consequently,the ionic radius increases.

(C) The $Solvay$ process is primarily used for the industrial production of $Na_2CO_3$ (sodium carbonate).

(D) Alkali metals have the lowest ionization enthalpy in their respective periods compared to alkaline earth metals due to their larger atomic size and lower effective nuclear charge.

(B) The reactivity of alkali metals with water increases down the group due to the increase in electropositive character. Since $Na$ is more electropositive than $Li$,it reacts more vigorously with water.

(A) Alkali metals have large atomic sizes due to the presence of an extra shell in each period compared to other elements.
They possess low ionization energy and low electronegativity.
They also have low density.
Therefore,the statement that they have a low atomic size is incorrect.