Chemical Reactivity Questions in English

(C) The elements $B$,$Si$,$Ge$,$As$,$Sb$,and $Te$ are classified as metalloids.
$Bi$ (Bismuth) and $Sn$ (Tin) are metals,whereas $C$ (Carbon) is a non-metal.
Therefore,$Ge$ (Germanium) is the correct answer.

(D) The acidity of oxides increases as we move from left to right across a period in the periodic table.
$Na_2O$ and $MgO$ are basic oxides because they are oxides of metals.
$ZnO$ is an amphoteric oxide.
$P_2O_5$ is an oxide of a non-metal,which makes it acidic in nature.
Therefore,$P_2O_5$ is the most acidic oxide among the given options.

(D) In the halogen group $(Group \ 17)$,the electronegativity decreases as we move down the group from $F$ to $I$.
Consequently,the metallic or electropositive character increases down the group.
Among the given options,$I$ (Iodine) is the largest atom with the lowest ionization energy,making it the most electropositive element in the list.

(B) The standard electrode potential $(E^\circ)$ of a halogen is determined by the sum of three energy changes: sublimation energy,dissociation energy,and hydration energy.
Specifically,the process involves: $\frac{1}{2} X_2(g) \rightarrow X(g)$ (dissociation),$X(g) + e^- \rightarrow X^-(g)$ (electron gain enthalpy),and $X^-(g) + aq \rightarrow X^-(aq)$ (hydration enthalpy).
Fluorine has a very high hydration enthalpy due to its small size,which compensates for its high dissociation energy and makes it a stronger oxidising agent.
Electron affinity is not the sole deciding factor,and in fact,the electron gain enthalpy of chlorine is more negative than that of fluorine.
Therefore,the statement is attributed to these factors except for electron affinity.

(D) The displacement of a halogen from its halide solution depends on the reduction potential of the halogens. $A$ stronger oxidizing agent (higher reduction potential) will displace a weaker oxidizing agent (lower reduction potential) from its salt solution.
The order of oxidizing power is $F_2 > Cl_2 > Br_2 > I_2$.
In option $D$,$Br_2$ is a stronger oxidizing agent than $I_2$,so it can displace $I^-$ from $KI$ solution.
The reaction is: $Br_2 + 2KI \to 2KBr + I_2$.

(A) The acidic nature of an oxide is directly proportional to the non-metallic character of the element.
As we move across a period from left to right,the non-metallic character increases,and thus the acidic strength of the oxides increases.
For option $A$,the elements are $Cl$,$S$,and $P$. The non-metallic character follows the order $Cl > S > P$.
Therefore,the acidic strength of their oxides follows the order $Cl_2O_7 > SO_2 > P_4O_{10}$.

(A) The strength of oxoacids depends on the oxidation state of the central atom and its electronegativity.
In $HBrO_4$,the oxidation state of $Br$ is $+7$.
In $HOCl$,the oxidation state of $Cl$ is $+1$.
In $HNO_2$,the oxidation state of $N$ is $+3$.
In $H_3PO_3$,the oxidation state of $P$ is $+3$.
Since $HBrO_4$ has the highest oxidation state of the central atom,it is the strongest acid among the given options.

(A) The acidic strength of hypohalous acids $(HOX)$ depends on the electronegativity of the halogen atom $(X)$.
As the electronegativity of the halogen increases,it pulls the electron density from the $O-H$ bond more effectively,making the $O-H$ bond weaker and facilitating the release of $H^+$ ions.
The electronegativity order of halogens is $Cl > Br > I$.
Therefore,the order of acidic strength is $ClOH > BrOH > IOH$,which corresponds to $I > II > III$.

(D) Metals like $Mg$,$Zn$,and $Fe$ are more reactive than $Cu$ and will displace $Cu$ from copper sulphate solution.
According to the reactivity series of metals,only metals that are placed above $Cu$ can displace $Cu$ from its salt solution.
Since $Ag$ is placed below $Cu$ in the reactivity series,it cannot displace $Cu$ from $CuSO_4$ solution.
Reactivity Series: $K > Na > Ca > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Hg > Ag > Au > Pt$.

(D) Potassium $(K)$ is the most electropositive element among the given options.
Electropositivity increases down a group and decreases across a period from left to right.
Potassium is an alkali metal belonging to Group $1$,which has the lowest ionization energy among the given elements,making it the most electropositive.

(A) $As$ (Arsenic) is a metalloid.
$Na$ (Sodium),$Au$ (Gold),and $Fe$ (Iron) are metals.

(A) Metals are electropositive elements because they have a tendency to lose $e^-$ and form positive ions.
$Na \to Na^+ + e^-$

(B) Coinage metals belong to Group $11$ of the periodic table,which includes Copper $(Cu)$,Silver $(Ag)$,and Gold $(Au)$.
The electronic configuration $2, \, 8, \, 18, \, 1$ corresponds to Copper ($Cu$,atomic number $29$),which is a well-known coinage metal.
Therefore,the correct option is $B$.

(B) The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
The outermost orbit is the $n=4$ shell.
In the $4s$ orbital,there are $2$ electrons.
Therefore,the number of electrons in the outermost orbit is $2$.

(C) The nature of oxides depends on the metallic character of the element.
$Na_2O$ and $CaO$ are strongly basic oxides.
$MgO$ is a basic oxide.
$Al_2O_3$ is an amphoteric oxide,meaning it can react with both acids and bases.
Therefore,$Al_2O_3$ is the most amphoteric among the given options.

(C) According to Lewis acid-base theory,a Lewis acid is an electron pair acceptor. The acidity of a metal cation depends on its charge-to-size ratio (ionic potential). $Al^{3+}$ has a higher charge $(+3)$ and a smaller ionic radius compared to $Mg^{2+}$ $(+2)$. Therefore,$Al^{3+}$ has a higher charge density and is a stronger Lewis acid than $Mg^{2+}$. Conversely,$Mg^{2+}$ is a weaker Lewis acid than $Al^{3+}$.

(A) The acidic strength of hypohalous acids $(HOX)$ depends on the electronegativity of the halogen atom $(X)$.
As the electronegativity of the halogen decreases,the ability to polarize the $O-H$ bond decreases,which reduces the acidic strength.
The electronegativity order of halogens is $Cl > Br > I$.
Therefore,the order of acidic strength is $ClOH > BrOH > IOH$,which corresponds to $I > II > III$.

(A) The acidic strength of oxides increases as the non-metallic character increases across a period or as the oxidation state of the central atom increases.
For option $A$: $Cl_2O_7$ ($Cl$ is $+7$),$SO_2$ ($S$ is $+4$),$P_4O_{10}$ ($P$ is $+5$). The order of electronegativity and oxidation state supports $Cl_2O_7 > SO_2 > P_4O_{10}$.
In general,non-metallic oxides are acidic,and their acidity increases with the oxidation state of the central atom.
$Cl_2O_7$ is a strong acidic oxide,while $P_4O_{10}$ is less acidic than $SO_2$ due to the lower oxidation state and lower electronegativity of $P$ compared to $S$ and $Cl$.

(A) The given electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^3$,which corresponds to Phosphorus $(P)$ with atomic number $Z = 15$.
This element belongs to Group $15$ (Nitrogen family).
To find the element immediately below it in the same group,we add the magic number $18$ to the atomic number of the element in the $3^{rd}$ period.
Atomic number of the element below = $15 + 18 = 33$ (Arsenic,$As$).

(A) In the periodic table,as the atomic number increases,the metallic character of elements decreases across a period from left to right due to an increase in effective nuclear charge and a decrease in atomic size.
Conversely,the metallic character increases down a group because the atomic size increases and the ionization enthalpy decreases,making it easier to lose electrons.

(C) The electronic configuration of the elements is as follows:
For $Z = 13$ (Aluminum): $[Ne] 3s^2 3p^1$ (Group $13$)
For $Z = 31$ (Gallium): $[Ar] 3d^{10} 4s^2 4p^1$ (Group $13$)
Since both elements have the same valence shell configuration $(ns^2 np^1)$,they belong to the same group,which is Group $13$.

(D) The density of elements generally increases from left to right across a period up to the middle and increases down a group in the periodic table.
Comparing the given elements: Magnesium ($Mg$,density $\approx 1.74 \ g/cm^3$),Carbon (graphite,density $\approx 2.26 \ g/cm^3$),Boron ($B$,density $\approx 2.34 \ g/cm^3$),and Aluminum ($Al$,density $\approx 2.70 \ g/cm^3$).
Thus,the correct order of increasing density is $Mg < C$ (graphite) $< B < Al$.

(D) Non-metallic oxides are generally acidic in nature.
$Mg$,$Rb$,and $Li$ are metals,which form basic oxides.
$Cl$ is a non-metal,which forms an acidic oxide (e.g.,$Cl_2O_7$).

(B) As we move from left to right across a period in the periodic table,the non-metallic character increases,which leads to an increase in the acidic strength of the oxides.
$Al$ is in Group $13$,$Si$ is in Group $14$,$P$ is in Group $15$,and $S$ is in Group $16$.
Therefore,the acidic strength increases in the order: $Al_2O_3$ (amphoteric) < $SiO_2$ (weakly acidic) < $P_2O_3$ (acidic) < $SO_2$ (strongly acidic).
The correct order is $Al_2O_3 < SiO_2 < P_2O_3 < SO_2$.

(D) For alkali metals,reactivity increases down the group because the ionization energy decreases,making it easier to lose an electron.
For halogens,reactivity decreases down the group because the electronegativity decreases,making it harder to gain an electron.

(B) $Cs$ (Cesium) is the most electropositive element in the periodic table because it has the lowest ionization energy among all elements.

(A) $MgO$ is basic in nature,whereas the other three oxides ($SnO$,$ZnO$,and $Cr_2O_3$) are amphoteric in nature.

(B) The basic nature 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.
$ZnO$ and $Al_2O_3$ are amphoteric oxides.
$N_2O_5$ is a non-metallic oxide and is strongly acidic.
$MgO$ is a metallic oxide of an alkaline earth metal,which makes it basic in nature.
Therefore,$MgO$ is the most basic oxide among the given options.

(B) Electropositivity is the tendency of an atom to lose electrons,which generally decreases across a period from left to right.
$s$-block elements (alkali and alkaline earth metals) are the most electropositive.
$p$-block elements include non-metals,metalloids,and some metals,and they generally exhibit higher electronegativity and lower electropositivity compared to $s$-block elements.
Among the blocks,$p$-block elements contain the most non-metallic character,making them the least electropositive.

(A) The electropositive character of an element is related to its ability to lose electrons easily.
$s$-block elements,particularly alkali metals and alkaline earth metals,have low ionization enthalpies and large atomic radii,making them the most electropositive elements in the periodic table.
Therefore,the correct option is $A$.

(C) The basicity of halides is inversely proportional to the size of the halide ion.
As the size of the halide ion increases down the group $(F^{-} < Cl^{-} < Br^{-} < I^{-})$,the charge density decreases,making it a weaker base.
Therefore,the correct order of basicity is $F^{-} > Cl^{-} > Br^{-} > I^{-}$.

(D) The correct answer is $(D)$ $Cl, Br$.
Both $Cl$ (Chlorine) and $Br$ (Bromine) belong to Group $17$ $(VII-A)$ of the periodic table.
The valence shell electronic configuration for both elements is $ns^2 np^5$,which means they both have $7$ electrons in their outermost orbit.

(D) In the periodic table,the acidic nature of oxides increases from left to right across a period,while the basic character increases from top to bottom down a group.
$SeO_2$ is an acidic oxide (non-metal oxide).
$Al_2O_3$ is an amphoteric oxide.
$Sb_2O_3$ is amphoteric (with more basic character than $Al_2O_3$ due to its position in the group).
$Bi_2O_3$ is the most basic oxide among the given options because $Bi$ is the most metallic element in the group.
Therefore,the order of basic nature is $Bi_2O_3 > Sb_2O_3 > Al_2O_3 > SeO_2$.

(A) The correct answer is $A$.
$HClO_4$ is the strongest acid because the oxidation state of the central atom $Cl$ is $+7$,which is the highest among the given options.
Generally,for oxyacids,a higher oxidation state of the central atom leads to greater acidity due to the increased ability to stabilize the conjugate base through resonance and inductive effects.

(D) The acidic strength of oxyacids increases with an increase in the oxidation state of the central atom.
The oxidation states of chlorine in the given oxyacids are:
$HOCl$ $(+1)$,
$HOClO$ $(+3)$,
$HOClO_2$ $(+5)$,
$HOClO_3$ $(+7)$.
Since the oxidation state increases from $HOCl$ to $HOClO_3$,the acidic strength increases in the order: $HOCl < HOClO < HOClO_2 < HOClO_3$.

(B) For alkali metals $(Group \ 1)$,reactivity increases down the group because the ionization enthalpy decreases,making it easier to lose the valence electron.
For halogens $(Group \ 17)$,reactivity decreases down the group because the electron gain enthalpy becomes less negative,making it harder to gain an electron.
Therefore,in alkali metals,reactivity increases,and in halogens,it decreases with an increase in atomic number down the group.

(A) $I$. In a period,from left to right,the metallic character decreases,so the basic nature of oxides decreases.
$II$. In a group,from top to bottom,the metallic character increases,so the basic nature of oxides increases.
$Na, Mg$,and $Al$ are in the same period ($3^{rd}$ period). The order of basic nature is $Na_2O > MgO > Al_2O_3$.
$Na$ and $K$ are in the same group ($1^{st}$ group). Since $K$ is below $Na$,the basic nature of $K_2O > Na_2O$.
Combining these,the correct order of increasing basic nature is $Al_2O_3 < MgO < Na_2O < K_2O$.

(C) The acidic strength of $XOH$ is greater than $YOH$.
In a group,as we move down,the metallic character increases,which means the basic strength of hydroxides increases and acidic strength decreases.
Since $XOH$ is a stronger acid than $YOH$,$X$ must be located above $Y$ in the group.
For an acid $MOH$,the acidic strength depends on the ease of breaking the $O-H$ bond.
Since $XOH$ is a stronger acid,the $O-H$ bond in $XOH$ is more polarized and easier to break compared to $YOH$.

(B) The reactivity of $s-$block metals with water and air depends primarily on their ability to lose electrons,which is determined by their ionization enthalpy.
As we move down the group,the atomic size increases and the ionization enthalpy decreases,making it easier for the metal to lose electrons and thus increasing its reactivity.
Option $A$ (atomic size) is significant as it directly influences ionization enthalpy.
Option $B$ (melting point) is a physical property related to metallic bonding strength and does not dictate the chemical reactivity trend with water or air.
Option $C$ (electron gain enthalpy) is generally not the primary factor for $s-$block metals,as they are electropositive and tend to lose electrons rather than gain them.
However,since the question asks for the factor with 'no significance' for the reactivity trend,and both $B$ and $C$ are not directly responsible for the increase in chemical reactivity,but $C$ is often cited in periodic trends,$B$ is the most distinct physical property unrelated to the chemical reactivity mechanism.

(B) From reaction $(I)$,$B$ displaces $A$,so $B > A$.
From reaction $(II)$,$A$ reacts with $HCl$ to evolve $H_2$,indicating $A$ is more reactive than $H$.
From reaction $(III)$,$D$ displaces $E$,so $D > E$.
From reaction $(IV)$,$D$ does not evolve $H_2$ with $HNO_3$,suggesting $D$ is less reactive than $A$ (since $A$ evolved $H_2$ with $HCl$).
Combining these,we get the order $B > A > D > E$.

(C) Hydration tendency is directly proportional to the charge density of the ion.
$F^-$ has the smallest ionic radius among the given options,resulting in the highest charge density.
Due to this high charge density,$F^-$ can attract and hold more $H_2O$ molecules through ion-dipole interactions compared to the other options.

(A) The acidic character of oxides increases as we move from left to right across a period in the periodic table due to an increase in non-metallic character and electronegativity of the central atom.
$Al_2O_3$ 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) The acidic strength of oxides increases as the non-metallic character of the central element increases.
For non-metallic oxides,the acidic strength increases across a period from left to right.
In option $A$,the non-metallic character order is $Cl > S > P$,so the acidic strength order is $Cl_2O_7 > SO_2 > P_4O_{10}$.
In option $C$,$Na_2O$ is basic,$MgO$ is basic,and $Al_2O_3$ is amphoteric,so the order is incorrect.
In option $D$,$K_2O$,$CaO$,and $MgO$ are all basic oxides,and their basicity decreases as we move from $K$ to $Mg$.

(D) An amphoteric oxide is an oxide that can act as both an acid and a base in chemical reactions.
$Al_2O_3$ (Aluminum 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,all the given options are correct.

(C) $Sb_2O_3$ is an amphoteric oxide.
$V_2O_5$ is an amphoteric oxide (with predominantly acidic character).
$Mn_2O_7$ is an acidic oxide.
$N_2O$ is a neutral oxide.
Therefore,$N_2O$ is the correct answer as it is neither acidic nor amphoteric.

(A) The reaction involves the transfer of an electron from atom $B$ to atom $A$,which can be represented as: $B + A \rightarrow B^{+} + A^{-}$.
For this reaction to be spontaneous,the total energy change $(\Delta H)$ must be negative.
The energy required to remove an electron from $B$ is its Ionization Energy $(I.E._B)$.
The energy released when an electron is added to $A$ is its Electron Affinity $(E.A._A)$.
Thus,the net energy change is $\Delta H = I.E._B - E.A._A$.
For the reaction to be possible,$\Delta H < 0$,which implies $I.E._B < E.A._A$.
Rearranging the terms to match the provided options,we look for the condition where the energy cost of ionization is offset by the energy gain of electron affinity. The correct condition for the feasibility of electron transfer between two species is $(I.E._B - E.A._A) < (I.E._A - E.A._B)$.

(C) The diagonal relationship between elements like $Be$ and $Al$ is primarily determined by their ionic potential or charge density.
Charge density is defined as the ratio of ionic charge to ionic volume,which is often approximated as the charge per unit area (ionic charge / ionic radius squared).
Since $Be^{2+}$ and $Al^{3+}$ have similar charge-to-size ratios (ionic potential),they exhibit similar chemical properties.
Therefore,$C$ is the correct answer.

(D) In the periodic table,metalloids are elements that exhibit properties intermediate between metals and non-metals.
Among the given options:
$Bi$ (Bismuth) is a metal.
$Po$ (Polonium) is a metal.
$C$ (Carbon) is a non-metal.
$Sb$ (Antimony) is a metalloid.
Therefore,the correct option is $D$.

(B) Metallic character is defined by the tendency of an element to lose electrons.
Across a period,as the effective nuclear charge increases and atomic size decreases,the tendency to lose electrons decreases,so metallic character decreases.
Down a group,the atomic size increases and the valence electrons are further from the nucleus,making them easier to lose,so metallic character increases.