Chemical Reactivity Questions in English

(C) The similarity in chemical properties between $Li$ (Lithium) and $Mg$ (Magnesium) is due to the diagonal relationship in the periodic table.
Diagonal relationships occur because of similar ionic potential (charge/size ratio) and electronegativity values between elements placed diagonally to each other in the periodic table.
$Li$ is in Group $1$,Period $2$,and $Mg$ is in Group $2$,Period $3$.

(B) The electronic configuration of the ion is $M^{+3} = [Ar]3d^{10}$.
To find the configuration of the neutral atom $M$,we add $3$ electrons back to the $4s$ and $4p$ orbitals.
Thus,$M = [Ar]3d^{10}4s^2 4p^1$.
Since the last electron enters the $p-$orbital,the element $M$ belongs to the $p-$block.

(C) The element with atomic number $Z = 52$ is Tellurium $(Te)$,which belongs to group $16$ and period $5$.
Following the periodic table structure,the element $X$ is located in group $13$ and period $2$,which is Aluminum $(Al)$.
Aluminum oxide $(Al_2O_3)$ is an amphoteric oxide.
Amphoteric oxides react with both acids and bases.
Therefore,$Al_2O_3$ reacts with $HCl$ (acid) to form $AlCl_3$ and water,and it reacts with $NaOH$ (base) to form sodium aluminate $(NaAlO_2)$ and water.
Thus,the correct statement is that it reacts with both $HCl$ and $NaOH$.

(B) The acidic strength of oxides increases as the non-metallic character of the central atom increases across a period from left to right.
As the electronegativity $(EN)$ of the central atom increases,the acidic strength of its oxide increases.
The order of electronegativity for the given elements is $Al < Si < P < S$.
Therefore,the correct order of acidic strength is $Al_2O_3 < SiO_2 < P_2O_5 < SO_3$.

(C) The acidic strength of oxides generally increases from left to right across a period and decreases from top to bottom down a group.
Comparing the oxides:
$1$. Across a period (left to right): $SiO_2 < P_4O_{10} < CO_2 < N_2O_5$ (Acidic strength increases).
$2$. Down a group: Acidic strength decreases.
Looking at the cycle:
- $CO_2 \rightarrow N_2O_5$: Increase (Left to right)
- $N_2O_5 \rightarrow P_4O_{10}$: Decrease (Top to bottom)
- $P_4O_{10} \rightarrow SiO_2$: Decrease (Right to left)
- $SiO_2 \rightarrow CO_2$: Increase (Bottom to top)
This matches the diagram in option $C$.

(A) $Li$ and $Mg$ exhibit a diagonal relationship because they possess a similar ionic potential (also known as ionic potential or charge-to-size ratio,$\phi = \frac{\text{Charge}}{\text{Radius}}$).
This similarity in charge density leads to comparable polarizing power and chemical properties.

(B) Carbon can reduce $XO$ and $YO$ to their respective metals,which means $X$ and $Y$ are less reactive than carbon.
Since carbon cannot remove oxygen from $ZO$,$Z$ is more reactive than carbon.
Therefore,$Z$ is the most reactive metal.
Given that $Y$ can remove oxygen from $XO$,$Y$ is more reactive than $X$.
Combining these observations,the order of reactivity is $Z > Y > X$.
Thus,the correct order is $ZYX$.

(D) $Be$ and $Al$ show a diagonal relationship in the periodic table.
Both elements exhibit similar properties due to their similar ionic potential (charge/size ratio).
$(1)$ Both form a protective oxide layer when treated with $HNO_3$,making them passive.
$(2)$ Both $BeCl_2$ and $AlCl_3$ act as Lewis acids due to their electron-deficient nature.
$(3)$ Both $Be(OH)_2$ and $Al(OH)_3$ are amphoteric,meaning they dissolve in both acids and bases.

(D) Elements in the same group of the Periodic table exhibit similar chemical properties because they possess the same number of valence electrons.
These valence electrons determine the chemical reactivity and bonding behavior of the atoms.
While the atomic number determines the position of an element,the specific configuration of valence electrons is the direct cause of the similarity in chemical properties within a group.

(A) Elements with atomic numbers $40$ $(Zr)$ and $72$ $(Hf)$ belong to the same group $(Group \ 4)$ in the periodic table.
Elements in the same group exhibit similar chemical properties due to having the same valence shell electronic configuration.
Therefore,the pair $(40, 72)$ is expected to have similar properties.

(B) The electronic configuration of the element with $Z = 120$ is $[Og]_{118} \, 8s^2$,where $Og$ is Oganesson.
Since the valence shell configuration is $ns^2$,it belongs to Group $2$ of the periodic table.
Elements in Group $2$ are known as alkaline earth metals.

(A) $1$. From reaction $(I)$,$B + ANO_3 \longrightarrow BNO_3 + A$,metal $B$ displaces $A$ from its salt solution,so $B$ is more reactive than $A$ $(B > A)$.
$2$. From reaction $(II)$,$A + HCl \longrightarrow ACl + \frac{1}{2} H_2$,metal $A$ reacts with $HCl$ to evolve $H_2$ gas,meaning $A$ is more reactive than $H$ $(A > H)$.
$3$. From reaction $(III)$,$D + ECl \longrightarrow DCl + E$,metal $D$ displaces $E$ from its salt solution,so $D$ is more reactive than $E$ $(D > E)$.
$4$. From reaction $(IV)$,$D + HNO_3$ does not evolve $H_2$ gas,which is typical for metals less reactive than $H$ (or metals that act as reducing agents for $HNO_3$),implying $D < H$.
$5$. Combining these: $B > A > H > D > E$. Thus,the decreasing order of reactivity is $B > A > D > E$.

(A) The basic character of oxides depends on the nature of the element forming the oxide.
$1$. $NO_2$ is a non-metallic oxide,which is acidic in nature.
$2$. $ZnO$ is an amphoteric oxide,which shows both acidic and basic properties.
$3$. $K_2O$ is a metallic oxide,which is strongly basic in nature.
Therefore,the order of increasing basic character is: $\text{Acidic oxide} < \text{Amphoteric oxide} < \text{Basic oxide}$.
Thus,the correct order is: $NO_2 < ZnO < K_2O$.

(A) The acidic strength of oxides is directly proportional to the electronegativity $(EN)$ of the central atom and the oxidation state of the central atom.
For option $A$,the central atoms are $Cl$ $(+7)$,$S$ $(+6)$,and $P$ $(+5)$. Since electronegativity increases from $P$ to $S$ to $Cl$ and the oxidation state also increases,the acidic strength follows the order $Cl_2O_7 > SO_3 > P_4O_{10}$.
Thus,the correct option is $A$.

(C) Electropositive character increases down a group and decreases across a period from left to right.
Cesium $(Cs)$ is the most electropositive stable element in the periodic table (excluding radioactive Francium).
The electronic configuration of Cesium ($Cs$,atomic number $55$) is $[Xe] \ 6s^1$.
Therefore,the element with the configuration $[Xe] \ 6s^1$ is the most electropositive.

(D) The metallic element that is liquid at room temperature is Mercury $(Hg)$,which has an atomic number of $80$.
The non-metallic element that is liquid at room temperature is Bromine $(Br)$,which has an atomic number of $35$.
Therefore,the atomic numbers of the metallic and non-metallic elements are $80$ and $35$ respectively.

(B) Metallic character is defined by the ease with which an atom can lose electrons.
Across a period,from left to right,the effective nuclear charge increases and the atomic size decreases,making it harder to lose electrons; thus,metallic character decreases.
Down a group,the atomic size increases and the valence electrons are further from the nucleus,making it easier to lose electrons; thus,metallic character increases.

(B) The electropositive character of an element increases down a group and decreases across a period from left to right in the periodic table.
$Be$ is in group $2$,period $2$.
$Rb$ is in group $1$,period $5$.
$Mn$ is a transition metal in period $4$.
$Tl$ is in group $13$,period $6$.
Among the given options,$Rb$ (Rubidium) is an alkali metal belonging to group $1$,which are the most electropositive elements in the periodic table. Therefore,$Rb$ is the most electropositive.

(A) The acidic nature of non-metal oxides increases as the oxidation state of the central non-metal atom increases.
Comparing the oxidation states:
$Cl_2O_7$: $Cl$ is in $+7$ state.
$P_4O_{10}$: $P$ is in $+5$ state.
$SO_3$: $S$ is in $+6$ state.
$B_2O_3$: $B$ is in $+3$ state.
Since $Cl$ has the highest oxidation state of $+7$,$Cl_2O_7$ is the most acidic oxide among the given options.

(D) Given that $X$ is in the $16^{th}$ group and $3^{rd}$ period,$X$ is Sulphur $(S)$.
Based on the relative positions:
$Y$ is above $X$ in the same group ($16^{th}$ group,$2^{nd}$ period),so $Y$ is Oxygen $(O)$.
$W$ is to the left of $X$ in the same period ($15^{th}$ group,$3^{rd}$ period),so $W$ is Phosphorus $(P)$.
$Z$ is to the right of $X$ in the same period ($17^{th}$ group,$3^{rd}$ period),so $Z$ is Chlorine $(Cl)$.
Evaluating the statements:
$(a)$ Electronegativity order: $O > Cl > S > P$. Thus,$Y$ $(O)$ has the maximum electronegativity. (Correct)
$(b)$ Catenation property order: $S > P > O > Cl$. Thus,$X$ $(S)$ has the maximum catenation property. (Correct)
$(c)$ Electron affinity order: $Cl > S > O > P$. Note: $Cl$ has higher electron affinity than $O$. Thus,$Z$ $(Cl)$ has the maximum electron affinity. (Correct)
$(d)$ Oxygen $(Y)$ has no $d$-orbitals and can only exhibit a covalency of $2$. It does not exhibit variable covalency. (Incorrect)
Therefore,the incorrect statement is $(d)$.

(D) $I.$ $[Kr]\, 5s^1$ is Rubidium $(Rb)$,an alkali metal,which shows only a single oxidation state of $+1$. Statement $(i)$ is False.
$II.$ $[Rn]\, 5f^{14}\, 6d^1\, 7s^2$ is Lawrencium ($Lr$,$Z=103$),which is an $f$-block element. Statement $(ii)$ is False.
$III.$ $I$ is a metal $(Rb)$ and $III$ is a non-metal (Bromine,$Br$). The compound formed between them $(RbBr)$ is ionic. Statement $(iii)$ is False.
$IV.$ $[Ar]\, 3d^6\, 4s^2$ is Iron ($Fe$,$Z=26$),which is a transition metal and shows variable oxidation states $(+2, +3)$. Statement $(iv)$ is False.
Therefore,all statements are False $(FFFF)$.

(B) metal with higher reactivity can reduce the oxide of a metal with lower reactivity.
Given that metal $(P)$ can reduce the oxide of $(R)$,it implies that $(P)$ is more reactive than $(R)$,i.e.,$P > R$.
Given that metal $(R)$ can reduce the oxide of $(Q)$,it implies that $(R)$ is more reactive than $(Q)$,i.e.,$R > Q$.
Combining these two relationships,we get the order of reactivity as $P > R > Q$.

(B) For hydrohalic acids $(HI, HBr, HCl, HF)$,the acidic strength increases down the group due to the decrease in bond dissociation energy: $HI > HBr > HCl > HF$.
For oxoacids of halogens,the acidic strength is directly proportional to the electronegativity of the halogen atom when the oxidation state is the same.
In $HIO_4, HBrO_4,$ and $HClO_4$,the central atoms are in the same oxidation state $(+7)$. Since electronegativity follows the order $Cl > Br > I$,the correct order of acidic strength is $HClO_4 > HBrO_4 > HIO_4$.
Therefore,the order given in option $B$ is incorrect.

(C) Metallic oxides are generally basic in nature. As we move from left to right across a period in the periodic table,the acidic character of the oxides increases. Among the given options,$Na_2O$,$MgO$,and $CaO$ are basic oxides,while $Al_2O_3$ is amphoteric. However,in the context of comparing these specific oxides,$Al_2O_3$ exhibits the highest acidic character (or least basic character) compared to the others.

(D) The acidic strength of oxides in a period increases as we move from left to right.
$Al_2O_3$ is amphoteric,$SiO_2$ is weakly acidic,$P_2O_3$ is acidic,and $SO_2$ is strongly acidic.
Therefore,the correct order of acidic strength is $Al_2O_3 < SiO_2 < P_2O_3 < SO_2$.

(A) $CaO$ and $CuO$ are metallic oxides and are basic in nature. $CaO$ is more basic than $CuO$.
$H_2O$ is neutral.
$CO_2$ is a non-metallic oxide and is acidic in nature.
Therefore,the correct order of acidic strength is $CaO < CuO < H_2O < CO_2$.

(A) more electronegative halogen can displace a less electronegative halogen from its salt solution.
Since $Cl$ is more electronegative than $Br$,$Cl_2$ can displace $Br^-$ from the solution to form $Br_2$ and $Cl^-$.
Therefore,$X_2$ is $Cl_2$.

(C) Non-metals have a high tendency to form anions by gaining electrons.
Among the given sets,$N, O,$ and $F$ are non-metals with high electronegativity.
Due to their high electronegativity,they have the greatest tendency to gain electrons and form anions compared to the other sets,which consist mostly of metals.

(C) more electronegative halogen displaces a less electronegative halogen from its salt solution.
Since chlorine is more electronegative than bromine,it displaces bromine from potassium bromide:
$2KBr + Cl_2 \to 2KCl + Br_2$

(D) The oxidizing power of halogens decreases down the group $(F_2 > Cl_2 > Br_2 > I_2)$.
Since $I_2$ is the weakest oxidizing agent among the halogens,it cannot displace $Cl^-, Br^-,$ or $F^-$ ions from their respective salt solutions.
Therefore,no reaction occurs.

(B) more electronegative element can displace a less electronegative element from its salt solution. The order of electronegativity/oxidizing power is $F_2 > Cl_2 > Br_2 > I_2$. Since chlorine is more electronegative than bromine,it will displace bromine from $KBr$ solution according to the reaction: $Cl_2 + 2KBr \rightarrow 2KCl + Br_2$.

(C) The acidic strength of oxyacids depends on the oxidation state of the central atom and the electronegativity of the central atom.
For the given acids,the $pK_a$ values are:
$HClO_3$ $(pK_a \approx -3)$
$H_2SO_4$ ($pK_a \approx -3$ for the first dissociation,but generally considered stronger than $H_3PO_4$)
$H_3PO_4$ $(pK_a \approx 2.12)$
Comparing the acid dissociation constants,the order of acidic strength is $HClO_3 > H_2SO_4 > H_3PO_4$,which corresponds to $III > I > II$.

(A) The acidic strength of oxyacids depends on the electronegativity of the central atom.
As the electronegativity of the halogen atom decreases in the order $Cl > Br > I$,the electron-withdrawing effect on the $O-H$ bond decreases.
Consequently,the ability to release $H^+$ ions decreases.
Therefore,the correct order of acidic strength is $Cl(OH) > Br(OH) > I(OH)$,which corresponds to $I > II > III$.

(A) The acidic strength of oxides increases as we move from left to right across a period in the periodic table.
Comparing the non-metallic character and oxidation states,the acidic strength order is $Cl_2O_7 > SO_2 > P_4O_{10}$ because chlorine is more electronegative and has a higher oxidation state compared to sulfur and phosphorus in these oxides.

(D) The acidic strength of oxoacids of halogens increases with an increase in the oxidation state of the central halogen atom.
The oxidation states are as follows:
$HOCl$ $(+1)$
$HOClO$ $(+3)$
$HOClO_2$ $(+5)$
$HOClO_3$ $(+7)$
Therefore,the correct order of increasing acidic strength is: $HOCl < HOClO < HOClO_2 < HOClO_3$.

(D) Element $B$ has the valence shell configuration $3s^2, 3p^3$,which means it has $5$ valence electrons and belongs to Group $15$ (valency $3$).
Element $C$ has the valence shell configuration $3s^2, 3p^5$,which means it has $7$ valence electrons and belongs to Group $17$ (valency $1$).
To form a stable compound,$B$ shares its $3$ electrons with $3$ atoms of $C$.
Therefore,the molecular formula of the compound formed is $BC_3$.

(C) $1$. In molten state,conductivity depends on the mobility of ions. As the size of the cation increases $(Mg^{+2} < Ca^{+2} < Sr^{+2} < Ba^{+2})$,the degree of hydration decreases (in aqueous solution) or the ionic mobility increases,making the order $Mg^{+2} < Ca^{+2} < Sr^{+2} < Ba^{+2}$ correct for conductivity.
$2$. Hydration energy is inversely proportional to the size of the ion. Since the size increases from $Li^{+1}$ to $Cs^{+1}$,the hydration energy decreases. Thus,the order $Li^{+1} > Na^{+1} > K^{+1} > Rb^{+1} > Cs^{+1}$ is correct,making option $B$ incorrect.
$3$. Reactivity of alkali metals increases down the group due to the decrease in ionization enthalpy. Thus,$Li < Na < K < Rb < Cs$ is the correct order of reactivity.
$4$. Since option $B$ is incorrect,the correct answer is $C$.

(D) The acidic strength of oxides increases as the non-metallic character increases across a period and decreases down a group.
For non-metallic oxides,the acidic strength is directly proportional to the oxidation state of the central atom.
Comparing the oxides: $SO_3$ (sulfur in $+6$ state),$CO_2$ (carbon in $+4$ state),and $B_2O_3$ (boron in $+3$ state).
Since the oxidation state decreases from $S$ to $C$ to $B$,the acidic strength follows the order $SO_3 > CO_2 > B_2O_3$.

(B) Metalloids are elements that exhibit properties intermediate between those of metals and non-metals.
In the periodic table,the metalloids are $B, Si, Ge, As, Sb, Te,$ and $Po$.
Looking at the options:
$A) Be$ is a metal,$B$ is a metalloid.
$B) Si$ and $Ge$ are both metalloids.
$C) B$ is a metalloid,$C$ is a non-metal.
$D) Sn$ and $Pb$ are metals.
Therefore,the correct pair is $Si$ and $Ge$.

(D) The acidic strength of oxides depends on the non-metallic character and the oxidation state of the central atom.
Acidic character increases as the non-metallic character increases across a period and decreases down a group.
Comparing the oxides:
$SO_3$ (Sulfur is in $+6$ oxidation state),$CO_2$ (Carbon is in $+4$ oxidation state),and $B_2O_3$ (Boron is in $+3$ oxidation state).
Since the non-metallic character follows the order $S > C > B$,the acidic strength of their oxides follows the order $SO_3 > CO_2 > B_2O_3$.
Therefore,option $D$ is correct.

(A) pseudo-noble gas configuration is defined as an electronic configuration where the outermost shell contains $18$ electrons $(ns^2 np^6 nd^{10})$.
For $Cu$ $(Z=29)$,the ground state configuration is $[Ar] 3d^{10} 4s^1$.
When $Cu$ loses one electron to form $Cu^{+}$,the configuration becomes $[Ar] 3d^{10}$.
This configuration has $18$ electrons in the outermost shell $(3s^2 3p^6 3d^{10})$,which is characteristic of a pseudo-noble gas configuration.
Therefore,$Cu^{+}$ is the correct ion.

(A) Representative elements are those elements that belong to the $s$-block and $p$-block of the periodic table (excluding noble gases in some definitions,but generally including groups $1, 2, 13-17$).
$Al$ (Aluminum) is in group $13$ ($p$-block) and $Mg$ (Magnesium) is in group $2$ ($s$-block).
Both $Al$ and $Mg$ are representative elements.
$Cr$ and $Zn$ are transition elements ($d$-block).
$Ag$ is a transition element and $At$ is a halogen (representative).
$La$ and $Th$ are inner transition elements ($f$-block).
Therefore,the correct pair is $Al$ and $Mg$.

(A) The basicity of oxides depends on the metallic character of the element. As we move from left to right across a period,the metallic character decreases and the non-metallic character increases,leading to a decrease in the basicity of the oxides.
$Na_2O$ is an oxide of an alkali metal (Group $1$),which is strongly basic.
$MgO$ is an oxide of an alkaline earth metal (Group $2$),which is basic.
$CuO$ is a transition metal oxide,which is amphoteric to weakly basic.
$SiO_2$ is a non-metal oxide (Group $14$),which is acidic.
Therefore,the correct order of decreasing basicity is $Na_2O > MgO > CuO > SiO_2$.

(C) In a group,as the atomic number increases,the atomic mass increases significantly more than the atomic volume. Since density is defined as $\text{density} = \frac{\text{mass}}{\text{volume}}$,the increase in mass dominates,leading to an overall increase in atomic density as we move down the group.

(B) The reaction $2KBr + I_2 \to 2KI + Br_2$ is not feasible.
In this reaction,$I_2$ would need to act as an oxidizing agent to oxidize $Br^-$ to $Br_2$.
However,the standard reduction potential of $Br_2$ is higher than that of $I_2$,meaning $Br_2$ is a stronger oxidizing agent than $I_2$.
Therefore,$I_2$ cannot displace $Br^-$ from its salt solution,making the reaction non-spontaneous.

(C) In the context of chemical reactivity and displacement in gaseous mixtures,$CO$ (carbon monoxide) is highly reactive due to its electronic structure and the presence of a lone pair on the carbon atom.
It acts as a strong ligand and reducing agent,allowing it to displace other gases like $O_2$,$N_2$,and $H_2$ in specific chemical environments or catalytic processes.

(D) The element with atomic number $Z = 117$ belongs to group $17$ of the periodic table.
Group $17$ elements are known as halogens.
These elements have a general valence shell electronic configuration of $ns^2 np^5$.
For $Z = 117$,the valence shell is $n = 7$,so the configuration is $7s^2 7p^5$.
This configuration indicates that the element has $7$ valence electrons $(2 + 5 = 7)$.
Therefore,all the given statements are correct.

(D) The acidic nature of non-metallic oxides is directly proportional to the oxidation state of the central atom.
Calculating the oxidation states:
$N_2O_5$: $N$ is $+5$
$SO_2$: $S$ is $+4$
$CO_2$: $C$ is $+4$
$CO$: $C$ is $+2$
Comparing the oxidation states: $+5 > +4 > +4 > +2$.
For oxides with the same oxidation state (like $SO_2$ and $CO_2$),the acidic strength depends on the electronegativity of the central atom. Since $S$ is more electronegative than $C$ in this context,$SO_2$ is more acidic than $CO_2$.
Thus,the correct order is $N_2O_5 > SO_2 > CO_2 > CO$.

(C) The basic nature of oxides depends on the electropositivity of the metal.
As we move from left to right in the periodic table,the metallic character decreases,and thus the basic nature of oxides decreases.
$Al_2O_3$ is amphoteric,$MgO$ is basic,$Na_2O$ is more basic,and $K_2O$ is the most basic among these due to the highest electropositivity of $K$.
The correct order of increasing basic nature is: $Al_2O_3 < MgO < Na_2O < K_2O$.