Chemical Reactivity Questions in English

(A) In the modern periodic table,there are a total of $118$ elements.
Among these,metals constitute the majority.
There are approximately $95$ metals,while non-metals and metalloids make up the remainder.

(C) The electronic configuration of the atom with atomic number $20$ is $2, 8, 8, 2$. It has $2$ valence electrons and tends to lose them to achieve a stable noble gas configuration,forming a cation with a charge of $+2$.
An atom with atomic number $16$ has the electronic configuration $2, 8, 6$. It has $6$ valence electrons and requires $2$ more electrons to complete its octet,forming an anion with a charge of $-2$.
Since the atom with atomic number $20$ can donate $2$ electrons and the atom with atomic number $16$ can accept $2$ electrons,they will combine to form an ionic compound (e.g.,$CaS$).
Therefore,the atom with atomic number $20$ is most likely to combine with the atom having atomic number $16$.

(A) The acidic strength of oxoacids of chlorine depends on the oxidation state of the central chlorine atom.
As the oxidation number of the central atom increases,the electron density on the oxygen atom decreases,which facilitates the release of the $H^+$ ion.
The oxidation states are: $HClO (+1)$,$HClO_2 (+3)$,$HClO_3 (+5)$,and $HClO_4 (+7)$.
Therefore,the correct order of acid strength is $HClO < HClO_2 < HClO_3 < HClO_4$.

(D) The strength of oxoacids of chlorine increases with the increase in the oxidation state of the central chlorine atom.
In $HClO$,the oxidation state of $Cl$ is $+1$.
In $HClO_2$,the oxidation state of $Cl$ is $+3$.
In $HClO_3$,the oxidation state of $Cl$ is $+5$.
In $HClO_4$,the oxidation state of $Cl$ is $+7$.
Since $HClO_4$ has the highest oxidation state $(+7)$,it is the strongest acid among the given options.

(A) The given electronic configuration $1s^2, 2s^2 2p^6, 3s^2 3p^3$ corresponds to the element Phosphorus $(P)$ with an atomic number of $15$.
Phosphorus belongs to Group $15$ of the periodic table.
The element located immediately below Phosphorus in Group $15$ is Arsenic $(As)$.
To find the atomic number of the element below,we add the magic number $18$ to the atomic number of Phosphorus $(15 + 18 = 33)$.
Thus,the atomic number of Arsenic is $33$.

(B) As we move from left to right across a period,the atomic size decreases and the ionization energy increases.
This makes it harder to lose electrons,which decreases the electropositive character and,consequently,the metallic nature of the elements.

(A) The periodic table groups elements based on their valence shell configuration.
$1$. $Mg$ (Magnesium) and $Ba$ (Barium) both belong to Group $2$ (Alkaline Earth Metals).
$2$. $Mg$ is in Group $2$,while $Na$ (Sodium) is in Group $1$.
$3$. $Mg$ is in Group $2$,while $Cu$ (Copper) is in Group $11$.
$4$. $Mg$ is in Group $2$,while $K$ (Potassium) is in Group $1$.
Therefore,the pair $Mg - Ba$ contains members from the same group.

(C) The similarity in the chemical properties of $Li$ (Lithium) and $Mg$ (Magnesium) is due to the diagonal relationship between them in the periodic table.
Diagonal relationships occur between certain elements of the second and third periods (e.g.,$Li$ and $Mg$,$Be$ and $Al$,$B$ and $Si$) due to similar ionic potential (charge/size ratio).

(A) Group $IIA$ (alkaline earth metals) consists of elements like $Be, Mg, Ca, Sr, Ba, Ra$,all of which are metals.
Group $IA$ (alkali metals) contains $H$,which is a non-metal.
Group $IB$ contains $Cu, Ag, Au$,which are metals,but $IIA$ is the most representative group consisting entirely of metals in the context of standard periodic table classification.

(A) Elements in which the last electron enters the $s$ or $p$-orbitals are collectively known as representative elements or common elements.

(D) Aluminium $(Al)$ is diagonally related to Beryllium $(Be)$ in the periodic table.
Both elements exhibit similar chemical properties due to the diagonal relationship,which arises from their similar ionic potential (charge/size ratio).

(C) The electronic configuration of the outermost shell for elements in the same group is identical.
$As$ (Arsenic) and $Bi$ (Bismuth) both belong to Group $15$ ($VA$ group) of the periodic table.
Elements in Group $15$ have a valence shell configuration of $ns^2 np^3$,meaning they all have $5$ electrons in their outermost orbit.

(C) Metallic character is defined by the ability of an element to lose electrons easily.
$X$ has $4$ valence electrons (a metalloid).
$Y$ has $8$ valence electrons (a noble gas).
$Z$ has $1$ valence electron (an alkali metal).
$T$ has $7$ valence electrons (a halogen).
Since $Z$ has only $1$ electron in its outermost shell,it can lose this electron most easily to achieve a stable octet configuration. Therefore,$Z$ is the most metallic element.

(B) The correct answer is $(b)$.
Elements with the same number of valence electrons exhibit similar chemical properties.
Atomic number $3$ corresponds to Lithium $(Li)$,which has the electronic configuration $[He] 2s^1$.
Atomic number $11$ corresponds to Sodium $(Na)$,which has the electronic configuration $[Ne] 3s^1$.
Both elements belong to Group $1$ (alkali metals) and have $1$ valence electron,thus they possess similar chemical properties.

(D) The electronic configuration $2, 8, 2$ corresponds to Magnesium $(Mg)$,which has $2$ valence electrons.
Metallic elements typically have $1, 2,$ or $3$ valence electrons and tend to lose them to form positive ions.
Since $2, 8, 2$ has $2$ valence electrons,it can easily donate them to achieve a stable octet,thus representing a metallic element.

(B) The electronic configuration of atom $A$ is $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^{10}, 4s^2 4p^3$.
The valence shell configuration is $4s^2 4p^3$,which indicates that it belongs to Group $15$ of the periodic table.
Elements in the same group exhibit similar chemical properties.
Nitrogen $(N)$ has the atomic number $7$ with the valence configuration $2s^2 2p^3$,placing it in Group $15$.
Therefore,the chemistry of $A$ is similar to that of Nitrogen.

(B) The given electronic configuration is $1s^2, 2s^2 2p^6, 3s^2 3p^1$.
This corresponds to the element Aluminum $(Al)$ with atomic number $13$.
Since the last electron enters the $p$-orbital,it belongs to the $p$-block.
Elements in $s$-block and $p$-block are collectively known as representative elements.
Therefore,the correct option is $(B)$.

(D) The modern periodic table is arranged based on the atomic number of elements.
Elements belonging to the same group or family possess the same number of valence electrons in their outermost shell,which results in similar chemical properties.

(B) The elements of the third period ($Na$ to $Cl$) are known as typical elements because they represent the properties of their respective groups. Among the given options,$Na$ is the element belonging to the third period.

(D) Metalloids are elements that exhibit properties intermediate between those of metals and non-metals. $Boron$ $(B)$ and $Silicon$ $(Si)$ are well-known examples of metalloids in the periodic table.

(C) Elements of the $s$-block and $p$-block (excluding noble gases) are termed as normal elements or representative elements.
$Li$ is a normal element as it belongs to the $s$-block (Group $1$).

(C) Metalloids are elements that possess properties intermediate between metals and non-metals.
$As$ (Arsenic) is a well-known metalloid.
$Pb$ (Lead) and $Zn$ (Zinc) are metals.

(A) The electronic configuration for the element with atomic number $11$ is $[Ne] \, 3s^1$.
The electronic configuration for the element with atomic number $37$ is $[Kr] \, 5s^1$.
Since both elements have the same valence shell configuration $(ns^1)$,they belong to the same group (Group $1$,the alkali metals).

(D) $Diagonal \ Relationship$ exists between certain pairs of diagonally adjacent elements in the second and third periods of the periodic table. These pairs ($Li$ and $Mg$,$Be$ and $Al$,$B$ and $Si$,etc.) exhibit similar properties.
Such a relationship occurs because crossing and descending the periodic table have opposing effects on atomic properties. On crossing a period,the size of the atoms decreases,and on descending a group,the size of the atoms increases.
Specifically,the chemistry of $Li$ and $Mg$ is similar because the ratio of their charge to size (ionic potential) is nearly the same,which leads to similar polarizing power and chemical behavior.

(A) The tendency to form anions is directly related to the electronegativity and non-metallic character of the elements.
Non-metals have a very strong tendency to gain electrons and form anions.
Among the given sets,$N, O, F$ are the most electronegative elements in their respective periods.
As we move down a group,the metallic character increases and the tendency to form anions decreases.
Therefore,the set $N, O, F$ exhibits the strongest tendency to form anions.

(D) The electronic configuration $2, 8, 1$ indicates that the element has $1$ valence electron,which it can easily lose to achieve a stable octet.
This identifies the element as a metal (Sodium,$Na$).
Metals typically form basic oxides when they react with oxygen.
Therefore,the correct statement is that the element forms a basic oxide.

(A) The similarity in properties between $Li$ (Group $1$) and $Mg$ (Group $2$) is due to their diagonal relationship.
This occurs because they have a similar ionic potential (charge/size ratio).

(B) The electropositive character (metallic character) of an element increases down a group and decreases across a period from left to right.
Among the given options,$[He] \, 2s^1$ is Lithium $(Li)$,$[Xe] \, 6s^1$ is Cesium $(Cs)$,$[He] \, 2s^2$ is Beryllium $(Be)$,and $[Xe] \, 6s^2$ is Barium $(Ba)$.
Cesium $(Cs)$ is an alkali metal located at the bottom of the alkali metal group (Group $1$),making it the most electropositive element among the choices provided.
Therefore,the correct configuration is $[Xe] \, 6s^1$.

(B) The electropositive character (metallic character) of elements decreases as we move from left to right across a period in the periodic table.
Among the given elements,$Mg$ (Magnesium) is in Group $2$,$Al$ (Aluminium) is in Group $13$,$P$ (Phosphorus) is in Group $15$,and $S$ (Sulphur) is in Group $16$.
Since $Mg$ is the leftmost element among these in the $3^{rd}$ period,it has the highest metallic character and is the most electropositive.
Therefore,the correct option is $(B)$.

(B) The electropositive character of an element is defined by its ability to lose electrons easily.
Alkali metals are the most electropositive elements in the periodic table.
Within a group,the electropositive character increases as we move down the group due to the increase in atomic size and decrease in ionization enthalpy.
Among the given options,$Caesium$ $(Cs)$ is an alkali metal located at the bottom of Group $1$,making it the most electropositive element.

(D) The acidic strength of oxides increases as we move from left to right across a period in the periodic table due to the increase in electronegativity and decrease in metallic character.
$Al_2O_3$ is amphoteric.
$SiO_2$ is weakly acidic.
$P_2O_3$ is acidic (anhydride of $H_3PO_3$).
$SO_2$ is more acidic (anhydride of $H_2SO_3$).
Therefore,the correct order of acidic strength is $Al_2O_3 < SiO_2 < P_2O_3 < SO_2$.

(D) The valence shell electronic configuration of $ns^2 np^5$ corresponds to elements having $7$ electrons in their outermost shell.
These elements belong to group $17$ of the periodic table,which are known as the halogens.

(A) The basic nature of oxides depends on the metallic character of the element. As metallic character increases,the electron-donating tendency increases,making the oxide more basic.
$Na_2O$ is strongly basic (alkali metal oxide).
$MgO$ is basic (alkaline earth metal oxide).
$Al_2O_3$ is amphoteric.
$CuO$ is weakly basic (transition metal oxide).
Thus,the decreasing order of basic nature is $Na_2O > MgO > Al_2O_3 > CuO$.

(B) The correct answer is $(B)$.
As we move from left to right along a period,the ionization enthalpy increases and the atomic size decreases,which leads to an increase in non-metallic character.
Consequently,the metallic nature and the basic nature of the oxides decrease across a period.

(A) Non-metallic character decreases as we move down a group and increases from left to right across a period.
Comparing the given elements: $Be$ and $Mg$ are in Group $2$,while $B$ and $Al$ are in Group $13$.
$B$ is in the second period,while $Al$ is in the third period. $Be$ is in the second period,while $Mg$ is in the third period.
Since non-metallic character increases from left to right and decreases down a group,$B$ (Boron) is the most non-metallic among the given options.

(A) The basic character of oxides increases as we move down a group and decreases as we move from left to right across a period in the periodic table.
$Na_2O$ is an oxide of an alkali metal (Group $1$),which is highly basic.
$Al_2O_3$ is amphoteric.
$SiO_2$ and $SO_2$ are acidic oxides.
Therefore,$Na_2O$ is the most basic oxide among the given options.

(B) Metallic character decreases as you move across a period in the periodic table from left to right. This occurs because the effective nuclear charge increases,making it harder to lose electrons.
Metallic character increases as you move down a group in the periodic table. This is because the atomic radius increases,and the valence electrons are further from the nucleus,making them easier to lose.

(B) The basic nature of oxides depends on the metallic character of the element.
As we move from left to right across a period in the periodic table,the metallic character decreases and the non-metallic character increases.
Consequently,the basic nature of the oxides decreases,and the acidic nature increases.
For the given elements $(Mg, Al, S, Cl)$,the order of metallic character is $Mg > Al > S > Cl$.
Therefore,the order of decreasing basic nature of their oxides is $MgO > Al_2O_3 > SO_3 > Cl_2O_7$.

(B) The acidic character of oxides increases as we move from left to right across a period due to an increase in electronegativity and non-metallic character.
$Li_2O$ is strongly basic,$BeO$ is amphoteric,and $B_2O_3$,$CO_2$,and $N_2O_3$ are acidic.
Among the acidic oxides,the acidity increases with the oxidation state and electronegativity of the central atom.
Thus,the correct order of increasing acidic strength is $Li_2O < BeO < B_2O_3 < CO_2 < N_2O_3$.
Therefore,the decreasing order is $N_2O_3 > CO_2 > B_2O_3 > BeO > Li_2O$.

(D) Elements having the same number of electrons in the valence shell exhibit similar chemical properties and are therefore placed in the same group of the periodic table.

(D) As we move down a group in the periodic table,the atomic size increases and the ionization enthalpy decreases.
This makes it easier for the atom to lose electrons,which leads to an increase in the metallic character of the elements.

(C) The tendency to form anions depends on the electron gain enthalpy and electronegativity of the elements.
$N, O$ and $F$ are non-metals located on the right side of the periodic table.
They have high electronegativity and high electron gain enthalpy,which means they have a strong tendency to gain electrons to complete their octet and form anions $(N^{3-}, O^{2-}, F^-)$.

(C) The tendency to lose electrons is a characteristic property of metals,which increases down a group and decreases across a period.
$F$ (Fluorine) is a non-metal with the highest electronegativity,so it has the least tendency to lose electrons.
$S$ (Sulfur) is a non-metal.
Both $Fe$ (Iron) and $Be$ (Beryllium) are metals.
$Be$ has a relatively stable electronic configuration $(1s^2 2s^2)$,making it more difficult to remove an electron compared to $Fe$,which is a transition metal with a larger atomic size and lower ionization energy.
Therefore,$Fe$ has a greater tendency to lose electrons than $Be$.

(A) The basic nature of oxides decreases across a period as metallic character decreases,while the acidic nature of oxides increases.
$Na_2O$ and $CaO$ are strongly basic.
$MgO$ is basic.
$Al_2O_3$ is amphoteric in nature,meaning it acts as both an acid and a base.
Among the given options,$Al_2O_3$ exhibits the most acidic character compared to the others which are basic.

(A) The chemical activity of a metal is related to its standard reduction potential $(E^o)$. $A$ more negative reduction potential indicates a stronger reducing agent and higher chemical activity.
The standard reduction potentials are:
$Na^{+} + e^{-} \to Na$,$E^o = -2.71 \ V$
$Mg^{2+} + 2e^{-} \to Mg$,$E^o = -2.37 \ V$
$Cu^{2+} + 2e^{-} \to Cu$,$E^o = +0.34 \ V$
Comparing the values: $-2.71 \ V < -2.37 \ V < +0.34 \ V$.
Therefore,the order of increasing chemical activity is $Cu < Mg < Na$.

(C) Electropositive character refers to the tendency of an element to lose electrons to form positive ions.
In the periodic table,electropositive character increases down a group and decreases from left to right across a period.
Comparing the elements: $K$ (Group $1$,Period $4$),$Ca$ (Group $2$,Period $4$),$C$ (Group $14$,Period $2$),and $Cl$ (Group $17$,Period $3$).
$K$ is an alkali metal and has the lowest ionization energy among the given options,making it the most electropositive.
The decreasing order of electropositive character is $K > Ca > C > Cl$.
Therefore,the correct option is $(C)$.

(B) The basic character of oxides decreases across a period from left to right in the periodic table.
$MgO$ is a basic oxide because it reacts with water to form a base: $MgO + H_{2}O \to Mg(OH)_{2}$.
$Al_{2}O_{3}$ is amphoteric in nature.
$SiO_{2}$ and $P_{2}O_{5}$ are acidic oxides.
Therefore,$MgO$ is the most basic among the given options.

(C) $Na_2O$,$MgO$,and $CaO$ are basic oxides of $s$-block elements.
$Al_2O_3$ is an amphoteric oxide,but among the given options,it possesses the highest acidic character because acidity increases as we move from left to right across a period in the periodic table.

(D) An amphoteric substance is one that can react with both acids and bases to form salts and water. Since element $A$ dissolves in both acid and alkali,it exhibits amphoteric nature.

(A) The valence shell configuration is $ns^2np^2$.
Total number of valence electrons = $2 + 2 = 4$.
Elements with $4$ valence electrons belong to group $IV$ $A$ (or group $14$ in the modern periodic table).