Alkali metals Questions in English

(B) $Li, Na$ and $K$ are alkali metals belonging to Group $1$.
They have low ionization energy and possess one electron in their outermost shell.
Due to this,they can easily lose an electron to form a cation $(M^+)$.

(B) The given electronic configuration $1s^2, 2s^2 2p^6, 3s^1$ corresponds to the element $Sodium$ ($Na$,$Z = 11$).
Since the valence shell has $1$ electron $(3s^1)$,the metal $M$ exhibits a valency of $+1$.
Oxygen $(O)$ typically exhibits a valency of $-2$.
To form a neutral oxide,two atoms of $M$ ($+1$ charge each) combine with one atom of $O$ ($-2$ charge).
Therefore,the formula of the oxide is $M_2O$.

(C) To determine the presence of unpaired electrons,we analyze the electronic configuration or total electron count of each species:
$1$. $NO_2^+$: Total electrons = $7 + (2 \times 8) - 1 = 22$ (even,all paired).
$2$. $BaO_2$: Contains the peroxide ion $O_2^{2-}$. Total electrons in $O_2^{2-} = 16 + 2 = 18$ (even,all paired).
$3$. $AlO_2^-$: Total electrons = $13 + (2 \times 8) + 1 = 30$ (even,all paired).
$4$. $KO_2$: Contains the superoxide ion $O_2^-$. Total electrons in $O_2^- = 16 + 1 = 17$ (odd). An odd number of electrons implies at least one unpaired electron.
Therefore,only $KO_2$ contains an unpaired electron.

(D) $Mercury$ $(Hg)$ has very weak interatomic forces compared to other transition metals,which is why it remains in a liquid state at room temperature.

(A) The lattice energy $(U)$ is inversely proportional to the inter-ionic distance $(r^{+} + r^{-})$,where $r^{+}$ and $r^{-}$ are the radii of the cation and anion,respectively.
$U \propto \frac{1}{r^{+} + r^{-}}$
As the size of the alkali metal cation increases from $Li^{+}$ to $Cs^{+}$,the inter-ionic distance increases.
Therefore,the lattice energy decreases as the size of the cation increases.
The correct order is $LiCl > NaCl > KCl > RbCl > CsCl$.

(B) Sodium oxide $(Na_2O)$ reacts with water to form sodium hydroxide $(NaOH)$,which is a strong base. Therefore,sodium oxide is a basic oxide.

(B) The correct answer is $(B)$.
When alkali metals like sodium dissolve in liquid ammonia,they undergo ionization to form ammoniated cations and ammoniated electrons.
The reaction is: $Na + (x + y)NH_3 \to [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$.
The blue colour of the solution is primarily due to the presence of solvated electrons $([e(NH_3)_y]^-)$,which absorb energy in the visible region of the spectrum.

(B) The solution of sodium metal in liquid ammonia is strongly reducing because it contains solvated electrons.
Reduction is defined as the gain of electrons.
In this solution,the solvated electrons act as powerful reducing agents by supplying electrons to other species.
The deep blue color of these solutions is also attributed to the presence of these solvated electrons.

(D) The thermal decomposition of metal nitrates depends on the position of the metal in the electrochemical series.
Nitrates of noble metals like silver $(Ag)$ and mercury $(Hg)$ decompose upon heating to yield the corresponding metal,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$.
The reaction for silver nitrate is:
$2AgNO_3 \xrightarrow{\Delta} 2Ag + 2NO_2 + O_2$
Other nitrates like those of iron,copper,or manganese typically decompose to form their respective metal oxides.

(C) The given electronic configuration $1s^2, 2s^2 2p^6, 3s^1$ corresponds to Sodium $(Na)$,which is an alkali metal.
Since it has a tendency to lose one electron to attain a stable noble gas configuration,it is a monovalent electropositive element.
Alkali metals form basic oxides (e.g.,$Na_2O$).
They have low electron affinity due to their stable octet configuration after losing an electron.
Therefore,the statement that it is a non-metal is incorrect.

(A) The $s$-block elements are those in which the last electron enters the outermost $s$-orbital.
These elements are located in the first two groups of the periodic table,which are designated as group $1$ $(IA)$ and group $2$ $(IIA)$.

(C) Hydrogen has a valence shell configuration of $1s^1$. Like alkali metals (Group $1$),it has a tendency to lose one electron to form a unipositive ion $(H^+)$. Therefore,it resembles alkali metals in this property.

(D) The atomic number $38$ corresponds to the element Strontium $(Sr)$.
Strontium belongs to Group $2$ of the periodic table,which is part of the $s$-block elements.
All elements in the $s$-block (except Hydrogen) are metals.
Therefore,$38$ is the atomic number of a metal.

(D) The $s$-block elements have their last electron entering the $s$-orbital.
Atomic number $3$ corresponds to $Li$ $(1s^2 2s^1)$,which is an $s$-block element.
Atomic number $12$ corresponds to $Mg$ $([Ne] 3s^2)$,which is an $s$-block element.
Therefore,the pair $(3, 12)$ represents $s$-block elements.

(A) The chloride of an element $A$ forms a neutral solution in water if it is a salt of a strong acid and a strong base.
Elements of the $1^{st}$ group (alkali metals) form chlorides like $NaCl$ or $KCl$,which are salts of a strong base $(NaOH/KOH)$ and a strong acid $(HCl)$.
These salts do not undergo hydrolysis,resulting in a neutral solution with $pH = 7$.
Therefore,the element $A$ belongs to the $1^{st}$ group.

(B) The correct answer is $(B)$.
Alkali metals are located at the extreme left of the periodic table.
Due to their large atomic size in their respective periods,the valence electron is loosely held by the nucleus.
Consequently,they have the lowest ionization potential (or ionization energy) in each period.

(A) The lightest metal is $Li$ (Lithium).
It has the lowest density among all metals due to its small atomic size and low atomic mass.

(B) Diagonal relationships exist between certain elements of the second and third periods of the periodic table. $Li$ (Group $1$,Period $2$) shows a diagonal relationship with $Mg$ (Group $2$,Period $3$) due to similarities in their ionic size and charge-to-size ratio.

(A) The electronic configuration of the element with atomic number $55$ is $[Xe] 6s^1$.
Since the last electron enters the $s$-orbital,it belongs to the $s$-block of the periodic table.
The element is caesium $(Cs)$.

(D) is the correct answer.
Potassium $(K)$ has the atomic number $19$.
Its electronic configuration is $[Ar] \, 4s^1$.
Since the last electron enters the $s$-orbital,it is an $s$-block element.

(B) Group $1$ carbonates,except $Li_2CO_3$,are thermally stable and do not decompose on heating.
$Na_2CO_3$ is a stable compound and does not decompose upon heating.
$MgCO_3$ decomposes to $MgO$ and $CO_2$.
$Li_2CO_3$ decomposes to $Li_2O$ and $CO_2$.
$Ca(HCO_3)_2$ decomposes to $CaCO_3$,$H_2O$,and $CO_2$.

(B) $Li$ (Lithium) is a solid metal at room temperature.
$Hg$ (Mercury) is a liquid metal at room temperature.
$Ga$ (Gallium) has a melting point of approximately $29.76 \ ^\circ C$,which is very close to room temperature,and it is often found in a liquid state in warm environments.
$Br$ (Bromine) is a liquid non-metal at room temperature.

(C) The ionic mobility of an ion in an aqueous solution depends on its degree of hydration.
Smaller ions have a higher charge density,which leads to a greater degree of hydration.
Among the alkali metal ions,$Li^{+}$ has the smallest size and therefore the highest degree of hydration.
Due to the large size of the hydrated $Li^{+}$ ion,it moves more slowly through the solution compared to other alkali metal ions.
Thus,$Li^{+}$ has the lowest ionic mobility in aqueous solutions.

(B) Alkali metals have low ionisation energies,which means they lose their valence electrons very easily.
Since they readily lose electrons,they act as strong reducing agents.

(D) The electropositive nature (metallic character) of elements increases as we move down a group in the periodic table.
Since $Li$,$Na$,and $K$ belong to Group $1$ and are arranged in the order $Li < Na < K$ down the group,the correct order of their electropositive nature is $K > Na > Li$.
Therefore,the correct option is $D$.

(A) The reducing nature of elements depends on their ability to lose electrons,which is related to their standard oxidation potential.
$1$. Alkali metals are stronger reducing agents than alkaline earth metals because they have lower ionization enthalpies.
$2$. Among the given options,$K$ (Potassium) is an alkali metal,while $Mg$ (Magnesium),$Al$ (Aluminum),and $Ba$ (Barium) are either alkaline earth metals or post-transition metals.
$3$. Within the alkali metal group,the reducing power increases down the group due to the decrease in ionization enthalpy.
$4$. Comparing the given elements,$K$ has the highest tendency to lose electrons compared to $Mg$,$Al$,and $Ba$. Therefore,$K$ is the strongest reducing agent.

(C) Alkali metals belong to Group $1$ of the periodic table.
These elements have a single valence electron in their outermost $s$-orbital.
The general electronic configuration for the outermost shell of alkali metals is $(n - 1)s^2p^6, ns^1$,where $n$ represents the principal quantum number of the valence shell.

(D) In $CaH_2$,calcium is an alkaline earth metal with an oxidation state of $+2$.
Let the oxidation state of hydrogen be $x$.
For the compound $CaH_2$,the sum of oxidation states is zero: $+2 + 2(x) = 0$.
$2x = -2$,which gives $x = -1$.
Therefore,hydrogen has an oxidation state of $-1$ in $CaH_2$.

(B) Alkali metal hydrides react with water to produce metal hydroxide and $H_2$ gas.
For example: $NaH + H_2O \to NaOH + H_2$.
Since the resulting metal hydroxides are strong bases,the final solution is basic in nature.

(C) Due to plumbosolvency,lead dissolves in water to a small extent to form soluble lead hydroxide,which is poisonous. Therefore,lead pipes are not used for carrying drinking water.

(D) . $Na$ and $K$ are highly reactive and react vigorously with cold water.
$B$. $Pt$ is a noble metal and does not react with water or steam.
$C$. $Fe$ (Iron) does not react with cold water but reacts with steam at high temperatures to produce iron oxide and hydrogen gas: $3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)$.

(A) The lattice energy of all metal nitrates is lower than their solvation energy in water. Therefore,all metal nitrates are soluble in water.

(B) Most metals react with dilute $HNO_3$ to produce nitrogen oxides instead of hydrogen gas because $HNO_3$ is a strong oxidizing agent.
However,very dilute $HNO_3$ (approx. $1\%$ concentration) reacts with very reactive metals like $Mn$ and $Mg$ to evolve hydrogen gas.
The reaction is: $Mn(s) + 2HNO_3(dil.) \to Mn(NO_3)_2(aq) + H_2(g)$.

(C) Alkali metals are highly reactive. They react with alcohol,water,and ammonia,but they do not react with kerosene.
$2C_2H_5OH + 2K \to 2C_2H_5OK + H_2$
$2K + 2H_2O \to 2KOH + H_2$
$K + (x + y)NH_3 \to [K(NH_3)_x]^+ + [e(NH_3)_y]^-$

(B) The correct statement is $(b)$.
After the removal of an electron from an alkali metal atom to form a cation,the effective nuclear charge per electron increases.
This results in a stronger attraction between the nucleus and the remaining electrons,causing the size of the cation to be smaller than that of the neutral atom.

(A) Alkali metals belong to Group $1$ of the periodic table.
Their general valence shell electronic configuration is $ns^1$.
Therefore,they have $1$ valence electron.

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

Element (Group $1$ Ions)	$Li^{+}$	$Na^{+}$	$K^{+}$	$Rb^{+}$	$Cs^{+}$
Ionic Radius $(pm)$	$76$	$102$	$138$	$152$	$167$

Electronegativity,first ionization energy,and melting point all decrease as the atomic number increases down the group.

(C) The reactivity of alkali metals with water increases down the group.
This is because the electropositive character (metallic character) increases as we move down the group from $Li$ to $Cs$.
Greater electropositive character means the metal can lose its valence electron more easily,leading to a faster reaction with water.
Therefore,sodium is more electropositive than lithium.

(A) . Carnallite is an ore of potassium with the chemical formula $KCl \cdot MgCl_2 \cdot 6H_2O$.
$B$. Cryolite is an ore of aluminum with the formula $Na_3AlF_6$.
$C$. Bauxite is a primary ore of aluminum with the formula $Al_2O_3 \cdot 2H_2O$.
$D$. Dolomite is an ore of magnesium and calcium with the formula $MgCO_3 \cdot CaCO_3$.

(C) In the Solvay process,the key step is the precipitation of $NaHCO_3$ by the reaction of $NH_4HCO_3$ with $NaCl$ solution.
$NaCl + NH_4HCO_3 \rightarrow NaHCO_3 \downarrow + NH_4Cl$.
$NaHCO_3$ is relatively less soluble and precipitates out,which can then be heated to form $Na_2CO_3$.
However,in the case of potassium,$KHCO_3$ is much more soluble in water than $NaHCO_3$.
Therefore,$KHCO_3$ does not precipitate out when $CO_2$ is passed through a solution of $KCl$ and $NH_3$,making it impossible to isolate $KHCO_3$ by this method.

(D) In a group of the periodic table, the atomic radius increases as we move down the group due to the addition of new electron shells. The alkali metals belong to Group $1$ and are arranged in the order $Li < Na < K < Rb < Cs$. Therefore, $Li$ has the smallest atomic size among the given options.

Element	Atomic Radius $(pm)$
$Li$	$152$
$Na$	$186$
$K$	$227$
$Rb$	$248$

(B) $Li$ is harder than other alkali metals,whereas other group $I$ metals are soft. Therefore,the statement that $Li$ is much softer than the other group $I$ metals is incorrect and does not illustrate an anomalous property of $Li$.

(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) Sodium is a highly reactive alkali metal. It reacts vigorously with oxygen and moisture present in the air to form oxides and hydroxides,which can lead to fire. Therefore,to prevent these reactions,sodium is stored by immersing it in $Kerosene$ oil,which is inert towards sodium.

(A) The basic character of alkali metal hydroxides increases down the group as the size of the metal cation increases and the bond dissociation energy of the $M-OH$ bond decreases.
The order of basicity is: $LiOH < NaOH < KOH < RbOH$.
Therefore,$RbOH$ is the most basic.

(D) Washing soda is $Na_2CO_3 \cdot 10H_2O$. When it is heated,it loses its water of crystallization.
$Na_2CO_3 \cdot 10H_2O \xrightarrow{\Delta} Na_2CO_3 \cdot H_2O + 9H_2O \uparrow$
On further heating,it becomes anhydrous sodium carbonate $(Na_2CO_3)$.
Therefore,water vapour is released.

(B) The correct statement is that carbonates of alkali metals ($Na_2CO_3,$ $K_2CO_3$) and ammonium carbonate $((NH_4)_2CO_3)$ are soluble in water.
This is because their hydration energy is greater than their lattice energy,which allows them to overcome the crystal lattice forces.

(B) Nitre,also known as saltpeter,is the mineral form of potassium nitrate.
Its chemical formula is $KNO_3$.
It is an ionic salt composed of potassium ions $(K^+)$ and nitrate ions $(NO_3^-)$.

(D) The Nelson cell is an electrolytic cell used for the industrial production of $NaOH$ (Caustic soda) by the electrolysis of brine ($NaCl$ solution).
In this process,$Cl_2$ gas is produced at the anode and $H_2$ gas is produced at the cathode,while $NaOH$ is obtained in the cathode compartment.

(C) The $Le-Blanc$ process was the first industrial process used for the large-scale production of sodium carbonate $(Na_2CO_3)$.
However,it was later replaced by the $Solvay$ process,which is more efficient and environmentally friendly for the industrial manufacturing of sodium carbonate.