Chemical Reactivity Questions in English

(B) $Cl_2O_7$ is an acidic oxide because it is a non-metal oxide with a high oxidation state.
$Na_2O$ is a basic oxide as it is an alkali metal oxide.
$Al_2O_3$ is an amphoteric oxide as it reacts with both acids and bases.
$N_2O$ is a neutral oxide.
Therefore,the correct matching is: $A-IV, B-II, C-I, D-III$.

(A) Metallic character is defined by the ease with which an element can lose electrons.
Across a period from left to right,the effective nuclear charge increases and atomic size decreases,making it harder to lose electrons.
Therefore,metallic character decreases across a period.
For the given elements: $Na$ (Group $1$),$Mg$ (Group $2$),$Be$ (Group $2$),$Si$ (Group $14$),and $P$ (Group $15$).
Comparing their positions: $Na$ is the most metallic,followed by $Mg$,then $Be$,then $Si$,and finally $P$ is the least metallic.
The correct decreasing order is $Na > Mg > Be > Si > P$.

(B) The reactivity of oxides towards water is directly proportional to their acidic strength.
As the electronegativity of the central atom increases,the acidic strength of the oxide increases.
The electronegativity order of the central atoms is $B < P < Cl$.
Consequently,the acidic strength and reactivity towards water follow the order: $B_2O_3 < P_2O_5 < Cl_2O_7$.
Thus,the correct option is $B$.

(B)
The reactivity of metals with water is determined by their position in the reactivity series.
Potassium $(K)$ is a highly reactive alkali metal that reacts vigorously with cold water.
Magnesium $(Mg)$ reacts with hot water.
Zinc $(Zn)$ reacts with steam.
Gold $(Au)$ is a noble metal and does not react with water.
Therefore,the decreasing order of reactivity is $K > Mg > Zn > Au$.

(D) The correct option is $D$.
In the periodic table,as we move from left to right across a period,the metallic character decreases and the non-metallic character increases.
Consequently,the basic nature of the oxides decreases,and the acidic nature increases.
Comparing the given oxides: $Na_2O$ (alkali metal oxide),$Al_2O_3$ (amphoteric oxide),$SiO_2$ (acidic oxide),and $P_2O_5$ (acidic oxide).
$Na_2O$ is the most basic oxide because $Na$ is an alkali metal.
The order of basicity is $P_2O_5 < SiO_2 < Al_2O_3 < Na_2O$.

(A) Metallic character is defined as the tendency of an element to lose electrons.
It increases down a group and decreases across a period from left to right.
The positions of the elements in the periodic table are: $Be$ (Group $2$,Period $2$),$Mg$ (Group $2$,Period $3$),$K$ (Group $1$,Period $4$),and $Si$ (Group $14$,Period $3$).
Comparing these:
$1$. $Be$ and $Mg$ are in Group $2$,so $Mg > Be$.
$2$. $K$ is in Group $1$,which is more metallic than Group $2$ elements.
$3$. $Si$ is a metalloid and is the least metallic among these.
Thus,the order of increasing metallic character is $Si < Be < Mg < K$.

(A) Statement $I$ is true: Chlorine oxides like $Cl_2O$,$ClO_2$,$Cl_2O_6$,and $Cl_2O_7$ are highly reactive oxidizing agents and are known to be unstable and prone to explosion.
Statement $II$ is true: The chemical reactivity of elements is often assessed by observing their ability to react with oxygen and halogens,which helps in determining their oxidation states and bonding characteristics.
Therefore,both statements are correct.

(C) Metallic character increases down a group and decreases across a period from left to right.
$K$ (Potassium) is in Group $1$ and Period $4$.
$Ca$ (Calcium) is in Group $2$ and Period $4$.
$Be$ (Beryllium) is in Group $2$ and Period $2$.
Comparing these,$K$ is the most metallic as it is in Group $1$. Between $Ca$ and $Be$,$Ca$ is more metallic because it is below $Be$ in Group $2$.
Therefore,the correct order is $K > Ca > Be$.

(C) Chemical reactivity of elements generally decreases from group $1$ to group $17$ and is lowest for noble gases (group $18$),so Statement-$I$ is false.
Group-$1$ elements are alkali metals and form basic oxides (e.g.,$Na_2O$),whereas group-$17$ elements are halogens and form acidic oxides (e.g.,$Cl_2O_7$). Thus,Statement-$II$ is true.

(B) $I$. In $p-$block,both metals and non-metals are present,but in $d-$block,only metals are present. Therefore,Statement $I$ is false.
$II$. Non-metals generally have higher ionisation enthalpy $(IE)$ and higher electronegativity $(EN)$ compared to metals due to their smaller atomic size and higher effective nuclear charge. Therefore,Statement $II$ is true.
Conclusion: Statement $I$ is false and Statement $II$ is true.

(D) The nuclear reaction is given by: ${}_{29}^{83}Cu + {}_{1}^{1}H \rightarrow 6{}_{0}^{1}n + {}_{2}^{4}\alpha + 2{}_{1}^{1}H + {}_{Z}^{A}X$
Applying the law of conservation of mass number: $83 + 1 = 6(1) + 4 + 2(1) + A$ $\Rightarrow 84 = 12 + A$ $\Rightarrow A = 72$
Applying the law of conservation of atomic number: $29 + 1 = 6(0) + 2 + 2(1) + Z$ $\Rightarrow 30 = 4 + Z$ $\Rightarrow Z = 26$
The element with atomic number $Z = 26$ is Iron $(Fe)$.
Iron $(Fe)$ belongs to group $8$ of the periodic table.

(A) Statement-$I$ is false because elements on the extreme left of the periodic table are alkali and alkaline earth metals,which form basic oxides.
Statement-$II$ is true because non-metals on the extreme right of the periodic table form acidic oxides,which react with water to form acids (e.g.,$SO_3 + H_2O \rightarrow H_2SO_4$).
As we move from left to right in the periodic table,the non-metallic character increases,leading to an increase in the acidic strength of the oxides.

(A) The atomic radii of the given elements decrease across a period and increase down a group.
Comparing the elements $Li$,$Na$,$Be$,$Mg$,$B$,and $Al$:
$B$ (Boron) is in the $2^{nd}$ period and group $13$.
$Be$ (Beryllium) is in the $2^{nd}$ period and group $2$.
$Li$ (Lithium) is in the $2^{nd}$ period and group $1$.
Across the $2^{nd}$ period,the atomic radius decreases from $Li$ to $Be$ to $B$.
Thus,$B$ has the smallest atomic radius among the given elements.
Boron $(B)$ belongs to group $13$,which has a valence of $+3$.
Therefore,it forms an oxide of the type $B_2O_3$,which corresponds to the general formula $A_2O_3$.

(B) The valence shell configuration of elements in the same group is identical.
$Se$ (Selenium) and $Te$ (Tellurium) both belong to Group $16$ (Oxygen family).
Their valence shell electronic configuration is $ns^2 np^4$,meaning both have $6$ valence electrons.
$Na$ ($1$ valence electron) and $Sr$ ($2$ valence electrons) belong to different groups.
$Mn$ and $Fe$ are transition metals with varying valence electron counts.
$He$ has $2$ valence electrons,while $Ne$ has $8$ valence electrons.

(A) To determine the acidic strength,we look at the nature of the oxides and the compound:
$1$. $CaO$ is a strong basic oxide (alkaline earth metal oxide).
$2$. $CuO$ is a basic oxide (transition metal oxide),but less basic than $CaO$.
$3$. $H_2O$ is amphoteric/neutral,acting as a very weak acid.
$4$. $CO_2$ is an acidic oxide (non-metal oxide) which forms carbonic acid $(H_2CO_3)$ in water.
Therefore,the order of increasing acidic strength is: $CaO < CuO < H_2O < CO_2$.

(B) $1$. Basic character of oxides increases down the group as metallic character increases. Thus,$Li_2O < Na_2O < K_2O < Cs_2O$ is correct.
$2$. Acidic character of oxides increases with the increase in oxidation state of the non-metal. For $Al_2O_3$ (amphoteric),$SiO_2$ (weakly acidic),$P_2O_5$ (acidic),and $Cl_2O_7$ (strongly acidic),the order $Al_2O_3 < SiO_2 < P_2O_5 < Cl_2O_7$ is correct.
$3$. Acidic strength of hydrohalic acids increases down the group due to decrease in bond dissociation enthalpy. Thus,$HF < HCl < HBr < HI$ is correct.
$4$. Acidic character of nitrogen oxides increases with the oxidation state of nitrogen: $N_2O (+1) < N_2O_3 (+3) < NO_2 (+4) < N_2O_5 (+5)$.
Since the question asks for the correct order,and multiple options provided in the original text were chemically incorrect,option $B$ represents the standard periodic trend for acidic oxides.

(B) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] \ 3d^5 \ 4s^2$.
In this configuration,the $d$-orbital is half-filled $(d^5)$.
Therefore,$Mn$ possesses half-filled $d$-orbitals.

(D) $SO_3$ is an acidic oxide.
$Na_2O$ is a basic oxide.
$N_2O$ is a neutral oxide.
$Al_2O_3$ is an amphoteric oxide,as it reacts with both acids and bases.

(C) An amphoteric oxide is one that can react with both acids and bases to form salt and water.
$Al_{2}O_{3}$ is a well-known amphoteric oxide.
It reacts with acids: $Al_{2}O_{3} + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}O$.
It reacts with bases: $Al_{2}O_{3} + 2NaOH + 3H_{2}O \rightarrow 2Na[Al(OH)_{4}]$.
$SO_{3}$ and $P_{4}O_{10}$ are acidic oxides,while $MgO$ is a basic oxide.

(C) The elements of group $16$ (such as $O, S$,etc.) have a valence shell configuration of $ns^2 np^4$.
They require $2$ more electrons to complete their octet $(ns^2 np^6)$,which corresponds to the stable electronic configuration of the nearest noble gas.
Therefore,group $16$ elements achieve noble gas configuration by gaining $2$ electrons.

(C) $Be$ is the first element of group $2$ which shows a diagonal relationship with $Al$,the second element of group $13$.

(D) The diagonal relationship is observed between certain elements of the second and third periods. $Li$ (Group $1$,Period $2$) shows a diagonal relationship with $Mg$ (Group $2$,Period $3$) due to their similar ionic radii and charge-to-size ratios.

(C) The diagonal relationship between $Be$ and $Al$ is due to their similar ionic size and charge/radius ratio (ionic potential).
Diagonal relationships occur because the increase in ionic charge/radius ratio when moving across a period is compensated by the decrease in ionic charge/radius ratio when moving down a group.
This results in similar polarizing power,leading to similar chemical properties.

(C) Metal oxides are generally basic in nature.
$Na_2O$,$MgO$,and $CaO$ are basic oxides.
$As_2O_3$ and $ZnO$ are amphoteric in nature.
$Mn_2O_3$ is basic,but $N_2O$ and $N_2O_5$ are acidic oxides.
However,in the context of standard chemistry problems,$Na_2O$,$MgO$,and $CaO$ are the primary basic oxides listed.
Thus,there are $3$ basic oxides.

(B) Metalloids are elements that exhibit properties intermediate between metals and non-metals. In the given list,the metalloids are $B$ (Boron),$Sb$ (Antimony),$As$ (Arsenic),$Ge$ (Germanium),and $Si$ (Silicon).
Thus,the total number of metalloids is $5$.
Note: $Si$ is classified as a metalloid in most standard chemistry textbooks.

(A) The acidic nature of oxides depends on the electronegativity of the central atom. Higher electronegativity leads to more acidic oxides.
As we move from left to right across a period,electronegativity increases,so the acidic nature of oxides increases.
As we move from top to bottom down a group,electronegativity decreases due to an increase in atomic size,so the acidic nature of oxides decreases.
Therefore,the acidic nature of oxides increases from left to right and decreases from top to bottom.

(C) Non-metallic character increases across a period from left to right and decreases down a group.
In the given elements,$F, N, C, B$ are in the second period,and their non-metallic character order is $F > N > C > B$.
$Si$ is in the third period,below $C$. Since non-metallic character decreases down a group,$C > Si$.
Combining these,the overall order of non-metallic character is $F > N > C > B > Si$.

(B) Metallic nature decreases in a period from left to right due to an increase in the effective nuclear charge.
Across the third period,the elements are $Na$,$Mg$,$Al$,and $Si$.
Since metallic character decreases from left to right,the correct order is $Na > Mg > Al > Si$.

(C) Based on the given electronic configurations:
$X$ is $Al$ $([Ne] 3s^2 3p^1)$,
$Y$ is $Cl$ $([Ne] 3s^2 3p^5)$,
$Z$ is $Mg$ $([Ne] 3s^2)$.
All three elements belong to the $3^{rd}$ period of the periodic table.
Across a period from left to right,the metallic character decreases and the non-metallic character increases,which leads to an increase in the acidic nature of their oxides.
The order of elements in the $3^{rd}$ period is $Mg (Z) < Al (X) < Cl (Y)$.
Therefore,the increasing order of the acidic nature of their oxides is $Z < X < Y$.

(A) Metallic character is defined as the tendency of an element to lose electrons.
It decreases along a period from left to right and increases down a group.
Comparing the positions in the periodic table:
$B$ (Boron) is in period $2$,group $13$.
$Al$ (Aluminum) is in period $3$,group $13$.
$Mg$ (Magnesium) is in period $3$,group $2$.
$K$ (Potassium) is in period $4$,group $1$.
Since metallic character increases down a group and decreases across a period,the order of increasing metallic character is $B < Al < Mg < K$.

(A) The oxidizing property of an element is determined by its ability to gain electrons,which is related to its electron gain enthalpy and electronegativity.
Fluorine $(F)$ is the most electronegative element and has the highest oxidizing power.
Oxygen $(O)$ follows fluorine.
Chlorine $(Cl)$ is less oxidizing than oxygen due to its larger size.
Nitrogen $(N)$ has a lower oxidizing tendency compared to the others.
Thus,the correct order of oxidizing property is $F > O > Cl > N$.

(D) Acidic oxides $\rightarrow Mn_2O_7, CrO_3, V_2O_5$ (due to higher oxidation state of the central metal).
Basic oxides $\rightarrow Na_2O, CaO, BaO$ (due to lower oxidation state of the central metal).
Alkali and alkaline earth metal oxides are generally basic.
Neutral oxides $\rightarrow CO, NO, N_2O$.
Amphoteric oxides $\rightarrow ZnO, PbO, BeO$ (exhibit both acidic and basic properties).
Hence,option $(D)$ is correct.

(C) The basic nature of oxides increases as the metallic character of the element increases.
In a period,the basic nature of oxides decreases from left to right,while in a group,it increases from top to bottom.
$P_2O_5$ is an acidic oxide.
$Al_2O_3$ is an amphoteric oxide.
$MgO$ is a basic oxide.
$K_2O$ is a strongly basic oxide.
Thus,the increasing order of basic nature is $P_2O_5 < Al_2O_3 < MgO < K_2O$,which corresponds to $c < a < d < b$.

(B) Amphoteric oxides are those that react with both acids and bases to form salt and water.
$ZnO$,$PbO$,and $SnO_2$ are well-known amphoteric oxides.
$Tl_2O_3$ and $In_2O_3$ are basic in nature.
$B_2O_3$ is acidic in nature.
Therefore,the correct set of amphoteric oxides is $ZnO, PbO, SnO_2$.

(D) In the periodic table,metals of groups $7$,$8$,and $9$ do not form hydrides. This region is known as the hydride gap.
However,$Cr$ (group $6$) is an exception that forms an interstitial hydride.
In these compounds,hydrogen atoms occupy the interstitial sites (voids) in the metal crystal lattice,which is why they are called interstitial hydrides.

(D) $(i)$ $ZnO, PbO, Sb_2O_3$ are amphoteric oxides,not neutral. They react with both acids and bases.
(ii) $CO$ and $NO$ are neutral oxides,not amphoteric.
(iii) $CrO_3, Mn_2O_7, V_2O_5$ are acidic oxides,not basic. They are oxides of transition metals in high oxidation states.

(A) An amphoteric oxide is one that reacts with both acids and bases,such as $SnO$ or $ZnO$. $A$ neutral oxide is one that does not react with either acids or bases,such as $N_2O$,$NO$,or $CO$.
In the given options,$SnO$ is amphoteric and $N_2O$ is neutral. However,looking at the options provided,$SnO_2$ is acidic/amphoteric and $N_2O$ is neutral. Given the standard classification,$SnO$ is amphoteric. If we evaluate the options,$SnO_2$ is often considered weakly amphoteric,and $N_2O$ is neutral. Thus,$SnO_2, N_2O$ is the most appropriate pair.

(B) The acidity of oxoacids depends on the oxidation state and electronegativity of the central atom.
$1$. The oxidation state of $Cl$ in $HClO_4$ is $+7$,while that of $S$ in $H_2SO_4$ is $+6$. $A$ higher oxidation state leads to stronger electron-withdrawing effects,making the $O-H$ bond more polar and easier to break.
$2$. $Cl$ is more electronegative than $S$,which further increases the electron-withdrawing effect on the $O-H$ bond.
$4$. Due to the higher oxidation state and electronegativity of $Cl$,the $ClO_3$ group exerts a stronger inductive effect than the $SO_3$ group,facilitating the release of $H^+$.
Statement $3$ is a property of these acids but does not explain why $HClO_4$ is a stronger acid than $H_2SO_4$.
Therefore,statements $1$,$2$ and $4$ are the correct reasons.

(B) As we move down the group in the periodic table,the atomic size of the halogen increases significantly.
This increase in size leads to a decrease in the bond dissociation enthalpy of the $H-X$ bond.
Consequently,the bond becomes weaker and it becomes easier to release the $H^+$ ion.
Therefore,the acidic strength increases down the group.
The correct order is $HF < HCl < HBr < HI$.

(C) 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 electronegativity and oxidation state of the central atom.
Comparing the central atoms: $S$ (Group $16$),$P$ (Group $15$),$Cl$ (Group $17$),and $N$ (Group $15$).
Among these,$Cl$ is the most electronegative and exhibits the highest oxidation state $(+7)$ in $Cl_2O_7$.
Therefore,$Cl_2O_7$ is the most acidic oxide.
The order of acidic strength is: $SO_3 < P_4O_{10} < N_2O_5 < Cl_2O_7$.

(A) Metallic character is defined by the ease with which an element can lose electrons.
Across a period (from left to right),the metallic character decreases due to an increase in effective nuclear charge.
Down a group,the metallic character increases due to an increase in atomic size and decrease in ionization enthalpy.
Comparing the given elements:
$K$ (Potassium) is in Group $1$,Period $4$.
$Na$ (Sodium) is in Group $1$,Period $3$.
$Al$ (Aluminium) is in Group $13$,Period $3$.
$Be$ (Beryllium) is in Group $2$,Period $2$.
Since $K$ is below $Na$,$K > Na$.
Since $Na$ is to the left of $Al$ in the same period,$Na > Al$.
Since $Al$ is in the $3^{rd}$ period and $Be$ is in the $2^{nd}$ period,$Al > Be$.
Thus,the correct order of metallic character is $K > Na > Al > Be$.

(A) The oxides and hydroxides of both $Be$ and $Al$ are amphoteric in nature,not basic.
Both $Be$ and $Al$ exhibit diagonal relationship due to similar ionic potential ($charge/size$ ratio).
$Be$ and $Al$ both form complexes (e.g.,$[BeF_4]^{2-}$ and $[AlF_6]^{3-}$).
Their chlorides ($BeCl_2$ and $Al_2Cl_6$) exist as bridged structures in the vapour phase and are covalent,making them soluble in organic solvents.
Therefore,the statement that they form basic oxides and hydroxides is incorrect as they are amphoteric.

(A) $Be$ and $Al$ exhibit a diagonal relationship,leading to many similar properties.
Both metals form covalent compounds and their chlorides act as Lewis acids.
Both metals dissolve in alkalies to form soluble complexes (beryllates and aluminates).
The oxides and hydroxides of both $Be$ and $Al$ are amphoteric in nature,not basic.
Therefore,the statement that both form basic oxides and hydroxides is incorrect.

(D) $Na_2O$ is a metal oxide,which is basic in nature. Upon hydrolysis,it forms $NaOH$,which is a strong base.
$Al_2O_3$ is an amphoteric oxide,meaning it reacts with both acids and bases.
$Cl_2O_7$ is a non-metal oxide,which is acidic in nature.
$CO$ is a neutral oxide.

(A) The electronic configurations given in options $B$,$C$,and $D$ are $[ Ar ] 3 d^{10} 4 s^2$,$[ Kr ] 4 d^{10} 5 s^2$,and $[ Xe ] 4 f^{14} 5 d^{10} 6 s^2$ respectively.
These configurations correspond to the general valence shell configuration of $(n-1)d^{10} ns^2$,which is characteristic of Group $12$ elements (Zinc,Cadmium,and Mercury).
Option $A$ has the configuration $[ Ne ] 3 s^2 3 p^5$,which corresponds to Chlorine (Group $17$,Halogen family).
Therefore,the element in option $A$ does not belong to the same family as the others.

(C) $Si$,$Ge$,and $Sn$ belong to the $3^{rd}$,$4^{th}$,and $5^{th}$ periods of group $14$ of the periodic table.
Moving down the group,the atomic size increases,which leads to an increase in bond length.
Therefore,the order of bond length is $Si-Si < Ge-Ge < Sn-Sn$.
Since bond enthalpy is inversely proportional to bond length,the order of bond enthalpy is $Si-Si > Ge-Ge > Sn-Sn$.
Given the value for $Ge-Ge$ is $260 \ kJ \ mol^{-1}$,the values for $Si-Si$ and $Sn-Sn$ are $297 \ kJ \ mol^{-1}$ and $240 \ kJ \ mol^{-1}$ respectively.

(A) The reactivity of a metal is determined by its ability to lose electrons,which is inversely proportional to its first ionisation enthalpy $( \Delta_i H_1 )$.
Lower $ \Delta_i H_1 $ values indicate that the metal can lose its valence electron more easily,making it more reactive.
Comparing the $ \Delta_i H_1 $ values:
$ I = 520 \ kJ \ mol^{-1} $
$ II = 490 \ kJ \ mol^{-1} $
$ III = 1681 \ kJ \ mol^{-1} $
$ IV = 2372 \ kJ \ mol^{-1} $
Element $ II $ has the lowest first ionisation enthalpy $( 490 \ kJ \ mol^{-1} )$,which indicates it is the most reactive metal among the given options.

(C) Across a period from left to right,the electronegativity of the element increases,which leads to an increase in the acidic strength of its oxides.
The elements $Al, Si, P$ and $S$ belong to the same period (Period $3$).
The order of electronegativity is $Al < Si < P < S$.
Thus,the correct order of acidic strength is $Al_2O_3 < SiO_2 < P_2O_3 < SO_2$.

(A) The elements $Sr, Zr, Cd,$ and $Sn$ all belong to the $5^{th}$ period of the periodic table.
Their arrangement from left to right in the $5^{th}$ period is $Sr$ (Group $2$),$Zr$ (Group $4$),$Cd$ (Group $12$),and $Sn$ (Group $14$).
Metallic property decreases as we move from left to right across a period because the ionization enthalpy increases and the tendency to lose electrons decreases.
Therefore,the correct order of metallic property is $Sn < Cd < Zr < Sr$.

(D) The acidic nature of oxides depends on the electronegativity of the central atom. As we move from left to right across a period,the electronegativity of elements increases,which leads to an increase in the acidic nature of their corresponding oxides.
For the elements of period $2$,the order of electronegativity is: $Li < Be < B < C < N$.
Consequently,the acidic nature of their oxides increases in the same order: $Li_2O < BeO < B_2O_3 < CO_2 < N_2O_3$.
Therefore,the correct order of decreasing acidic nature is: $N_2O_3 > CO_2 > B_2O_3 > BeO > Li_2O$.