Chemical properties Questions in English

(C) Among the given compounds,ferric chloride $(FeCl_3)$ is the most acidic in water.
This is due to the higher charge density and effective nuclear charge of the $Fe^{3+}$ ion compared to $Al^{3+}$ and $Be^{2+}$.
As the charge density of the central metal cation increases,its ability to polarize the $O-H$ bond in coordinated water molecules increases,leading to greater acidity.
All these compounds undergo hydrolysis in water to liberate $HCl$ gas:
$BeCl_2 + 2 H_2O \rightarrow Be(OH)_2 + 2 HCl \uparrow$
$AlCl_3 + 3 H_2O \rightarrow Al(OH)_3 + 3 HCl \uparrow$
$FeCl_3 + 3 H_2O \rightarrow Fe(OH)_3 + 3 HCl \uparrow$
$Fe^{3+}$ has a higher charge-to-size ratio,making it the strongest Lewis acid among the choices.

(B) To find the oxidation state of $Mn$ in each compound:
$1$. In $K_2MnO_4$: $2(+1) + x + 4(-2) = 0 \implies x = +6$.
$2$. In $KMnO_4$: $1(+1) + x + 4(-2) = 0 \implies x = +7$.
$3$. In $MnO_2$: $x + 2(-2) = 0 \implies x = +4$.
$4$. In $Mn_2O_3$: $2x + 3(-2) = 0 \implies x = +3$.
Therefore,the highest oxidation state of $Mn$ is $+7$,which is shown by $KMnO_4$.

(C) binary compound is a chemical compound composed of two different elements.
Elements that exhibit variable oxidation states can form more than one type of binary compound with a given non-metal.
$K$ (potassium) is an alkali metal with a fixed oxidation state of $+1$,forming only $KCl$.
$Ca$ (calcium) is an alkaline earth metal with a fixed oxidation state of $+2$,forming only $CaCl_2$.
$Zn$ (zinc) is a transition metal but exhibits a fixed oxidation state of $+2$ in its compounds,forming only $ZnCl_2$.
$Fe$ (iron) is a transition metal that exhibits variable oxidation states of $+2$ and $+3$.
Therefore,$Fe$ can form two types of binary compounds with chlorine: $FeCl_2$ (iron$(II)$ chloride) and $FeCl_3$ (iron$(III)$ chloride).
Thus,the correct option is $C$.

(C) When a mixture of $Cr^{3+}$ and $Fe^{3+}$ is treated with concentrated $NaOH$,$Cr^{3+}$ forms a soluble complex,while $Fe^{3+}$ forms an insoluble precipitate.
For $Cr^{3+}$:
$Cr^{3+} + 3OH^{-} \longrightarrow Cr(OH)_3 \downarrow$
$Cr(OH)_3 + OH^{-} \longrightarrow [Cr(OH)_4]^{-} \text{ (soluble green complex)}$
For $Fe^{3+}$:
$Fe^{3+} + 3OH^{-} \longrightarrow Fe(OH)_3 \downarrow \text{ (brown precipitate)}$
Since $Cr^{3+}$ remains in the solution as a complex and $Fe^{3+}$ precipitates out as $Fe(OH)_3$,they can be separated by filtration.

(C) The reduction of $Cu^{2+}$ to $Cu^{+}$ occurs when $Cu^{2+}$ reacts with iodide ions $(I^-)$ because $CuI$ is insoluble and the reaction is thermodynamically favorable.
The balanced chemical equation is: $2Cu^{2+} + 4I^- \to 2CuI(s) + I_2(s)$.

(C) The ease of oxidation $(M^{2+} \to M^{3+} + e^-)$ is determined by the oxidation potential,which is the negative of the reduction potential $({E^0}_{M^{2+}/M^{3+}} = -{E^0}_{M^{3+}/M^{2+}})$. $A$ higher (more positive) oxidation potential indicates that the oxidation process is easier.

Metal	Oxidation Potential $({E^0}_{M^{2+}/M^{3+}})$
$Cr$	$+0.41 \ V$
$Mn$	$-1.57 \ V$
$Fe$	$-0.77 \ V$
$Co$	$-1.97 \ V$

Since $Cr$ has the highest oxidation potential $(+0.41 \ V)$,the change in oxidation state from $+2$ to $+3$ is easiest for $Cr$.

(C) An amphoteric substance is one that can react with both acids and bases.
$Al^{3+}$,$Cr^{3+}$,and $Zn^{2+}$ form amphoteric oxides or hydroxides.
$Fe^{3+}$ forms $Fe_2O_3$ and $Fe(OH)_3$,which are basic in nature and do not show amphoteric character.

(C) $Fe$,$Co$,and $Ni$ are transition metals that act as effective catalysts in hydrogenation processes. This is because they have a high tendency to adsorb (occlude) hydrogen gas on their free metal surfaces,which facilitates the reaction.

(A) The reactivity of metals is determined by their standard oxidation potential or their tendency to lose electrons. Among the given transition metals,$Fe$ (Iron) is the most reactive.
$Fe$ readily undergoes oxidation in the presence of moist air to form hydrated iron$(III)$ oxide,commonly known as rust:
$4Fe(s) + 3O_2(g) + xH_2O(l) \rightarrow 2Fe_2O_3 \cdot xH_2O(s)$
In contrast,$Pt$ (Platinum) is a noble metal and is highly unreactive. $Ni$ and $Co$ are less reactive than $Fe$ under standard atmospheric conditions.

(A) Both $Zn$ and $Hg$ belong to Group $12$ of the periodic table.
$Zn$ is a reactive metal that forms $ZnO$ readily upon heating in air.
$Hg$ also forms $HgO$ when heated in air at temperatures between $300^{\circ}C$ and $350^{\circ}C$.
Therefore,the property shared by both is that they form oxides readily.

(A) The correct option is $A$.
Manganese exhibits its most powerful oxidizing property in the $+7$ oxidation state.
For example,in $KMnO_4$,manganese is in the $+7$ oxidation state,which acts as a strong oxidizing agent in acidic,neutral,and alkaline media.

(D) The correct answer is $(D)$. $Ta$ (Tantalum) is used for surgical implants like plates,screws,and wires because it is highly biocompatible and non-corrosive in the human body environment.

(A) The ionic character of an oxide depends on the oxidation state of the metal.
Lower oxidation states of metals generally result in ionic oxides,while higher oxidation states lead to covalent character.
In $MnO$,the oxidation state of $Mn$ is $+2$.
In $Mn_2O_7$,the oxidation state of $Mn$ is $+7$.
In $CrO_3$,the oxidation state of $Cr$ is $+6$.
$P_2O_5$ is an oxide of a non-metal and is covalent.
Therefore,$MnO$ is the most ionic oxide among the given options.

(C) When $KI$ is added to a solution of $CuSO_4$,$Cu^{2+}$ ions are reduced to $Cu^+$ ions by $I^-$ ions.
The reaction is: $2CuSO_4 + 4KI \to Cu_2I_2 + 2K_2SO_4 + I_2$.
Here,$Cu_2I_2$ (Cuprous iodide) is formed as a white precipitate,and iodine $(I_2)$ is liberated,which turns the solution brown.

(C) When $KI$ is added to $CuSO_4$,the initial reaction is:
$2KI + CuSO_4 \to CuI_2 + K_2SO_4$
$CuI_2$ is unstable and immediately undergoes reduction to form $Cu_2I_2$ and $I_2$:
$2CuI_2 \to Cu_2I_2 + I_2$
Combining these,the overall reaction is:
$2CuSO_4 + 4KI \to Cu_2I_2 + 2K_2SO_4 + I_2$
Thus,the final mixture contains $K_2SO_4$,$Cu_2I_2$,and $I_2$.

(C) $V_2O_3$ (Vanadium $(III)$ oxide) acts as a strong reducing agent.
This is because vanadium is an electropositive transition metal.
In its lower oxidation states,it tends to lose electrons to reach a more stable higher oxidation state,thereby facilitating the reduction of other substances.

(A) $Cu^{2+}$ ions act as an oxidizing agent and oxidize $I^-$ ions to $I_2$,while $Cu^{2+}$ is reduced to $Cu^+$. The reaction is as follows:
$2CuSO_4 + 4KI \to Cu_2I_2 + I_2 + 2K_2SO_4$
Thus,iodine is formed when $KI$ reacts with a solution of $CuSO_4$.

(D) Iron becomes passive when treated with concentrated $HNO_3$ due to the formation of a thin,protective,and non-porous oxide film $(Fe_2O_3)$ on its surface.
This layer prevents further reaction of the metal with the acid.
Other oxidizing agents like $K_2Cr_2O_7$,$KMnO_4$,chloric acid,and chromic acid also render iron passive.
Passive iron can be made active again by scratching the surface mechanically or by chemical reduction.

(B) The most important oxidation state of copper is $+2$.
Although copper exhibits both $+1$ and $+2$ oxidation states,the $+2$ state is more stable in aqueous solution due to its higher hydration enthalpy,which compensates for the second ionization energy.
Standard electrode potentials are:
$Cu^{+} + e^{-} \to Cu$; $E^{0} = +0.52 \ V$
$Cu^{2+} + 2e^{-} \to Cu$; $E^{0} = +0.34 \ V$

(D) Transition metals with low oxidation numbers have a high tendency to lose electrons to reach a more stable state.
Therefore,they act as reducing agents.
However,the provided option $(C)$ is incorrect based on chemical principles.
Transition metals with high oxidation states act as oxidizing agents,while those with low oxidation states act as reducing agents.
Since the question asks for the behavior of low oxidation state metals and the options do not include 'reducing agent',the correct choice is $D$.

(B) The correct answer is $B$. Iron $(Fe)$ reacts with steam at high temperatures to produce iron oxide and hydrogen gas.
$3Fe(s) + 4H_2O(g) \to Fe_3O_4(s) + 4H_2(g)$

(C) The yellow color is due to the formation of the chromate ion,$(CrO_4^{2-})$.
The chemical reaction is:
$2Cr(OH)_3 + 4NaOH + 3Na_2O_2 \rightarrow 2Na_2CrO_4 + 8H_2O$
Here,the chromium in $+3$ oxidation state is oxidized to $+6$ oxidation state,forming sodium chromate,which is yellow in color.

(B) In $CrO_2Cl_2$: Let the oxidation state of $Cr$ be $x$. $x + 2(-2) + 2(-1) = 0 \implies x - 4 - 2 = 0 \implies x = +6$. This is the maximum oxidation state for $Cr$ $(3d^5 4s^1)$.
In $MnO_4^-$: Let the oxidation state of $Mn$ be $x$. $x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7$. This is the maximum oxidation state for $Mn$ $(3d^5 4s^2)$.

(C) The change from $+2$ to $+3$ oxidation state involves the oxidation of the metal ion: $M^{2+} \rightarrow M^{3+} + e^-$.
The standard electrode potential given is for reduction: $M^{3+} + e^- \rightarrow M^{2+}$.
The oxidation potential is the negative of the reduction potential: $E^{0}_{M^{2+}/M^{3+}} = -E^{0}_{M^{3+}/M^{2+}}$.
For $Cr$: $E^{0}_{Cr^{2+}/Cr^{3+}} = -(-0.41 \ V) = +0.41 \ V$.
For $Mn$: $E^{0}_{Mn^{2+}/Mn^{3+}} = -(+1.57 \ V) = -1.57 \ V$.
For $Fe$: $E^{0}_{Fe^{2+}/Fe^{3+}} = -(+0.77 \ V) = -0.77 \ V$.
For $Co$: $E^{0}_{Co^{2+}/Co^{3+}} = -(+1.97 \ V) = -1.97 \ V$.
$A$ higher (more positive) oxidation potential indicates that the oxidation is easier.
Since $Cr$ has the highest positive oxidation potential $(+0.41 \ V)$,the change from $+2$ to $+3$ is easiest for $Cr$.

(B) The standard electrode potential $(E^o\,M^{2+} | M)$ values for the given transition metals are:
$Mn^{2+} | Mn = -1.18 \ V$
$Cr^{2+} | Cr = -0.91 \ V$
$Fe^{2+} | Fe = -0.44 \ V$
$Co^{2+} | Co = -0.28 \ V$
Comparing the magnitude of these negative values,the order is $Mn > Cr > Fe > Co$. Therefore,the correct option is $B$.

(A) The basic character of metal oxides depends on the oxidation state and the nature of the metal. For transition metal monoxides $(MO)$,the basic character decreases as the atomic number increases across the period because the electronegativity of the metal increases and the ionic character of the $M-O$ bond decreases.
Therefore,the order of basic character is $TiO > VO > CrO > FeO$.

(B) $Cu^{2+}$ is a strong oxidizing agent and $I^-$ is a strong reducing agent.
When $Cu^{2+}$ reacts with $I^-$,$Cu^{2+}$ oxidizes $I^-$ to $I_2$ and itself gets reduced to $Cu^+$.
The reaction is: $2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2$.
Therefore,$CuI_2$ is unstable and does not exist.

(C) $pH$ of $12$ indicates a basic medium.
In an alkaline (basic) solution,the dichromate ion $(Cr_2O_7^{2-})$ is converted into the chromate ion $(CrO_4^{2-})$.
The reaction is: $Cr_2O_7^{2-} + 2OH^- \to 2CrO_4^{2-} + H_2O$.
The color changes from orange to yellow.

(D) The acidic or basic nature of transition metal oxides depends on the oxidation state of the metal.
Generally,lower oxidation states of metals form basic oxides,while higher oxidation states form acidic or amphoteric oxides.
$1$. $Mn_2O_7$ ($Mn$ in $+7$): Acidic oxide.
$2$. $V_2O_3$ ($V$ in $+3$): Basic oxide.
$3$. $V_2O_5$ ($V$ in $+5$): Amphoteric (predominantly acidic) oxide.
$4$. $CrO$ ($Cr$ in $+2$): Basic oxide.
$5$. $Cr_2O_3$ ($Cr$ in $+3$): Amphoteric oxide.
Thus,$V_2O_3$ and $CrO$ are basic oxides.
The correct option is $D$.

(D) The basic character of metal oxides decreases as the oxidation state of the metal increases or as the electronegativity/ionization potential of the metal increases.
For the given monoxides $(TiO, VO, CrO, FeO)$,the metallic character decreases from $Ti$ to $Fe$ across the $3d$ series.
Therefore,the basic strength follows the order: $TiO > VO > CrO > FeO$.

(B) The standard electrode potential $E^o_{M^{3+}/M^{2+}}$ depends on the stability of the $M^{2+}$ and $M^{3+}$ oxidation states.
For $Mn \ (Z = 25)$,the electronic configuration is $[Ar] \ 3d^5 \ 4s^2$.
The $Mn^{2+}$ ion has a stable $d^5$ configuration (half-filled),which is very stable.
However,the $Mn^{3+}$ ion has a $d^4$ configuration.
Because $Mn^{2+}$ is exceptionally stable,it is very difficult to oxidize $Mn^{2+}$ to $Mn^{3+}$,which makes the reduction potential $E^o_{Mn^{3+}/Mn^{2+}}$ very high (positive).
Therefore,$Mn$ has the maximum $E^o_{M^{3+}/M^{2+}}$ value among the given elements.

(B) Iron $(Fe)$ reacts with steam $(H_2O(g))$ at high temperatures to produce iron oxide $(Fe_3O_4)$ and hydrogen gas $(H_2)$.
The balanced chemical equation is:
$3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)$

(A) The oxidizing power of transition metal oxoanions depends on the reduction potential of the metal in its highest oxidation state.
The standard reduction potentials $(E^\circ)$ for the given species are:
$VO_2^+ + 2H^+ + e^- \rightarrow VO^{2+} + H_2O$ $(E^\circ = 1.00 \ V)$
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ $(E^\circ = 1.33 \ V)$
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ $(E^\circ = 1.51 \ V)$
Since the oxidizing power increases with the increase in standard reduction potential,the order is $VO_2^+ < Cr_2O_7^{2-} < MnO_4^-$.

(D) The $Fe^{+3}$ ion has a higher charge density than the $Fe^{+2}$ ion,which makes $Fe^{+3}$ compounds more acidic and more prone to hydrolysis. Therefore,the statement that ferrous $(Fe^{+2})$ compounds are more easily hydrolysed than ferric $(Fe^{+3})$ compounds is incorrect.

(D) Transition elements with lower oxidation states act as reducing agents because they can easily lose electrons to reach a higher oxidation state.

(C) Transition metals and their compounds are excellent catalysts. For example,$CuCl_2$ is used in the Deacon process for the manufacture of chlorine,and finely divided transition metals are used in the hydrogenation of oils.

(A) The acidic character of metal oxides is directly proportional to the oxidation state of the metal atom.
As the oxidation state of the metal increases,the acidic character increases.
The oxidation states of $Mn$ in the given oxides are:
$MnO$: $+2$
$Mn_2O_3$: $+3$
$MnO_2$: $+4$
$MnO_3$: $+6$
$Mn_2O_7$: $+7$
Therefore,the correct order of acidic character is $MnO < Mn_2O_3 < MnO_2 < MnO_3 < Mn_2O_7$.

(C) Transition elements are characterized by high melting points due to strong metallic bonding involving $d$-electrons.
They exhibit variable oxidation states and many of their ions are paramagnetic due to the presence of unpaired $d$-electrons.
However,not all transition elements dissolve in acids. For example,noble metals like $Au$ (gold) and $Pt$ (platinum) are chemically inert and do not dissolve in common mineral acids like $HCl$ or $H_2SO_4$.

(C) The catalytic activity of transition metals and their compounds is primarily attributed to their ability to adopt variable oxidation states and form complexes.
For example,in the contact process,$V_{2}O_{5}$ acts as a catalyst:
$2SO_{2} + O_{2} \xrightarrow{V_{2}O_{5}} 2SO_{3}$
This occurs because vanadium can easily switch between oxidation states:
$SO_{2} + V_{2}O_{5} \longrightarrow SO_{3} + 2VO_{2}$
$2VO_{2} + \frac{1}{2}O_{2} \longrightarrow V_{2}O_{5}$

(A) The stability of the $+2$ oxidation state in transition metals is influenced by the electronic configuration and exchange energy.
$1$. $Mn^{2+}$ has a $d^5$ configuration,which is half-filled and highly stable due to maximum exchange energy.
$2$. $Fe^{2+}$ has a $d^6$ configuration. While it has one electron paired,it still possesses significant exchange energy,making it more stable than $Cr^{2+}$ $(d^4)$.
$3$. $Cr^{2+}$ has a $d^4$ configuration,which is less stable than $d^5$ and $d^6$ due to lower exchange energy.
$4$. $Co^{2+}$ has a $d^7$ configuration,which is the least stable among these due to increased inter-electronic repulsion.
Thus,the order of stability is $Mn^{2+} > Fe^{2+} > Cr^{2+} > Co^{2+}$.

(A) In reaction $I$,sodium dichromate $(Na_2Cr_2O_7)$ is reduced to chromium$(III)$ oxide $(Cr_2O_3)$ using carbon $(C)$ as a reducing agent:
$Na_2Cr_2O_7 + 2C \xrightarrow{\Delta} Na_2CO_3 + Cr_2O_3 + CO$.
In reaction $II$,chromium$(III)$ oxide $(Cr_2O_3)$ is reduced to metallic chromium $(Cr)$ using aluminum $(Al)$ in the thermite process:
$Cr_2O_3 + 2Al \xrightarrow{\Delta} Al_2O_3 + 2Cr$.
Therefore,$X = C$ and $Y = Al$.

(C) Manganese is added to steel to produce manganese steel,which is extremely hard and resistant to wear.
In the metallurgical process,manganese acts as a deoxidizer and desulfurizer.
It reacts with oxygen and sulfur impurities present in the molten steel to form manganese oxide $(MnO)$ and manganese sulfide $(MnS)$,which are then removed as slag.
This purification process significantly improves the mechanical properties of the steel,making it ideal for heavy-duty applications like railroad rails.

(B) Iron $(Fe)$ becomes passive when treated with concentrated nitric acid $(HNO_3)$.
This passivity is due to the formation of a thin,non-reactive,and protective oxide layer on the surface of the metal.
The oxide formed is iron($II$,$III$) oxide,which is $Fe_3O_4$ (or sometimes represented as $Fe_2O_3 \cdot FeO$).
Therefore,the correct answer is $80\% \text{ conc. } HNO_3$ and $Fe_3O_4$.

(B) $I^-$ ions possess strong reducing properties. They reduce $Fe^{3+}$ to $Fe^{2+}$ and $Cu^{2+}$ to $Cu^+$.
Consequently,$Fe^{3+}$ cannot coexist with $I^-$ to form $FeI_3$,and $Cu^{2+}$ cannot coexist with $I^-$ to form $CuI_2$.
Therefore,both $CuI_2$ and $FeI_3$ do not exist.

(D) The ability of oxygen to form multiple bonds with metals is the key factor.
In $Mn_2O_7$,each $Mn$ atom forms multiple bonds with oxygen,which stabilizes the $+7$ oxidation state.
In contrast,fluorine can only form single bonds.
To achieve a $+7$ oxidation state with fluorine,$7$ fluorine atoms would need to surround the $Mn$ atom,which causes significant steric crowding and makes the compound unstable.
Therefore,oxygen's ability to form $p\pi-d\pi$ multiple bonds allows it to stabilize higher oxidation states compared to fluorine.

(A) In the $d$-block elements,the stability of the higher oxidation state increases down the group.
$Cr$ is in the $3d$ series,$Mo$ is in the $4d$ series,and $W$ is in the $5d$ series.
As we move down the group,the stability of the $+6$ oxidation state increases,meaning the tendency to get reduced decreases.
Therefore,the oxidizing power decreases as we go down the group: $CrO_3 > MoO_3 > WO_3$.
Thus,$CrO_3$ acts as the best oxidizing agent among the given options.

(A) For an ion to liberate $H_2$ gas from dilute acids,it must act as a strong reducing agent such that its standard reduction potential is more negative than that of the hydrogen electrode $(E^{\circ}_{H^+/H_2} = 0.00 \ V)$.
Among the given species,only $Cr^{2+}$ has a standard reduction potential $(E^{\circ}_{Cr^{3+}/Cr^{2+}} = -0.41 \ V)$ which is lower than $0.00 \ V$,allowing it to reduce $H^+$ to $H_2$.
The reaction is: $Cr^{2+} + H^+ \to Cr^{3+} + \frac{1}{2}H_2 \uparrow$.
Therefore,only $1$ ion $(Cr^{2+})$ liberates $H_2$ gas.

(A) Amphoteric oxides are those that react with both acids and bases to form salt and water.
$Fe_2O_3$ (Ferric oxide) is a basic oxide.
$Pb_3O_4$ (Red lead) is a mixed oxide $(2PbO \cdot PbO_2)$ and exhibits amphoteric character.
$Mn_2O_3$ is a basic oxide,but in the context of common inorganic chemistry questions regarding amphoterism,$Fe_2O_3$ is strictly basic,while $Sb_2O_3$ is amphoteric.
However,among the given options,$Fe_2O_3$ is the most well-known basic oxide that does not show amphoteric properties.

(B) Amphoteric oxides are those that can react with both acids and bases to form salt and water.
$Cr_2O_3$ (Chromium$(III)$ oxide) and $ZnO$ (Zinc oxide) are both well-known amphoteric oxides.
$CO$ and $N_2O$ are neutral oxides.
$SiO_2$ is an acidic oxide.
$CaO$ is a basic oxide.