Chemical properties Questions in English

(A) . The $M^{3+}/M^{2+}$ reduction potential for $Mn$ $(+1.57 \ V)$ is greater than $Fe$ $(+0.77 \ V)$. Thus,statement $A$ is incorrect.
$B$. Higher oxidation states of $d$-block elements are stabilized by electronegative elements like oxygen and fluorine. Thus,statement $B$ is correct.
$C$. $Cr^{2+}$ is a strong reducing agent $(E^0_{Cr^{3+}/Cr^{2+}} = -0.41 \ V)$,so it can reduce $H^{+}$ to $H_2$. Thus,statement $C$ is correct.
$D$. $V^{2+}$ has $3$ unpaired electrons ($d^3$ configuration). The magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \ BM$. Thus,statement $D$ is incorrect.
Therefore,statements $B$ and $C$ are correct.

(C) The ions $Ti^{2+}, V^{2+},$ and $Cr^{2+}$ are strong reducing agents because their standard electrode potentials $(E^{\circ}_{M^{3+}/M^{2+}})$ are negative.
These ions can reduce $H^{+}$ ions in a dilute acid to liberate hydrogen gas.
The general reaction is: $2 M^{2+}_{(aq)} + 2 H^{+}_{(aq)} \longrightarrow 2 M^{3+}_{(aq)} + H_{2(g)}$.
Since all three ions $(Ti^{2+}, V^{2+}, Cr^{2+})$ possess this property,the total number of such ions is $3$.

(C) In aqueous solution,$Cu(II)$ is more stable than $Cu(I)$.
This is because the high hydration energy of the $Cu^{2+}$ ion compensates for the second ionization energy $(IE_2)$ required to convert $Cu^+$ to $Cu^{2+}$.

(D) The assertion is true because $Pb^{4+}, Bi^{5+}, Fe^{3+},$ and $Cu^{2+}$ are strong oxidizing agents in the presence of iodide ions $(I^-)$.
These metal ions oxidize $I^-$ to $I_2$ and get reduced to lower stable oxidation states (e.g.,$Pb^{2+}, Bi^{3+}, Fe^{2+}, Cu^+$).
For example: $PbI_4 \rightarrow PbI_2 + I_2$.
Thus,the higher oxidation state cannot exist in the presence of the strong reducing agent $I^-$.
Therefore,both Assertion and Reason are true,and the Reason is the correct explanation of the Assertion.

(B) $Mn^{+3}$ is a strong oxidising agent because $Mn^{+2}$ is stable due to $d^5$ configuration. $Co^{+3}$ is also a strong oxidising agent in aqueous solution. This statement is correct.
$(b)$ $Ti^{+2}$ $(d^2)$,$V^{+2}$ $(d^3)$,and $Cr^{+2}$ $(d^4)$ are strong reducing agents because they tend to lose electrons to achieve stable configurations. This statement is correct.
$(c)$ $CrO_3$ is an acidic oxide because it reacts with water to form chromic acid $(H_2CrO_4)$. This statement is correct.
$(d)$ For $Fe$,$E^{\circ}_{Fe^{+3}/Fe^{+2}} = +0.77 \ V$. Since the value is positive,this statement is incorrect.
Therefore,only $1$ statement is incorrect.

(C) The reaction $2 \ CuX_2 \rightarrow Cu_2X_2 + X_2$ represents the reduction of $Cu^{2+}$ to $Cu^{+}$ by the halide or pseudohalide ion $X^{-}$.
This reaction occurs spontaneously at room temperature when $X^{-}$ is a strong reducing agent,such as $I^{-}$ or $CN^{-}$.
In the case of $Cl^{-}$ and $Br^{-}$,the reduction of $Cu^{2+}$ to $Cu^{+}$ is not strong enough to occur at room temperature,and $F^{-}$ is not a reducing agent at all.

(B) $Pb_3O_4$ is a mixed oxide (combination of $PbO$ and $PbO_2$).
$N_2O$ is a neutral oxide.
$Mn_2O_7$ is an acidic oxide (oxide of a transition metal in a high oxidation state).
$Bi_2O_3$ is a basic oxide.
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

(C) $1$. $Cu^{+2}$ is a strong enough oxidizing agent to oxidize $I^{-}$ to $I_2$ $(2Cu^{+2} + 4I^{-} \rightarrow 2CuI + I_2)$. This statement is correct.
$2$. $MnO_3F$ is a known compound where manganese is in the $+7$ oxidation state. This statement is correct.
$3$. $MnF_7$ does not exist because fluorine is not capable of stabilizing the $+7$ oxidation state of manganese in a heptahalide form due to steric hindrance and electronic factors. This statement is incorrect.
$4$. $MnF_4$ is a known compound of manganese. This statement is correct.

(D) Interstitial compounds are formed when small atoms like $H, C, N$ are trapped inside the crystal lattice of transition metals.
They possess the following properties:
$1$. They have high melting points,higher than those of pure metals.
$2$. They are very hard.
$3$. They retain metallic conductivity.
$4$. They are chemically inert.
Therefore,the statement that metallic carbides are chemically inert is correct.

(D) Lanthanoids react with sulphur upon heating to form sulphides.
The general reaction is $2Ln + 3S \rightarrow Ln_2S_3$.
Thus,the general molecular formula of the product is $Ln_2S_3$.

(D) The reaction of acidified potassium dichromate with $SO_{2}$ is as follows:
$K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 3SO_{2} \longrightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O$
In this reaction,$SO_{2}$ acts as a reducing agent and is oxidized to sulphate ions,while the orange-colored dichromate ion $(Cr_{2}O_{7}^{2-})$ is reduced to green-colored chromic ion $(Cr^{3+})$ in the form of chromic sulphate $(Cr_{2}(SO_{4})_{3})$.
Therefore,the solution turns green.

(C) Amphoteric oxides are those that react with both acids and bases.
$Cr_2O_3$ is a well-known amphoteric oxide.
$V_2O_5$ also exhibits amphoteric character,although it is predominantly acidic,it reacts with strong bases to form vanadates and with strong acids to form oxovanadium ions.
$Mn_2O_7$ and $CrO_3$ are acidic oxides.
$CrO$ is a basic oxide.
$V_2O_4$ is a basic oxide.
Therefore,the correct pair is $V_2O_5$ and $Cr_2O_3$.

(B) When a transition metal is in its highest oxidation state,it cannot be further oxidized.
Instead,it can easily gain electrons to reach a lower oxidation state.
Therefore,it acts as an oxidising agent.

(A) $Cu^{2+}$ is a strong oxidizing agent and $I^{-}$ is a strong reducing agent. $Cu^{2+}$ oxidizes $I^{-}$ to $I_2$ as follows:
$2Cu^{2+} + 4I^{-}$ $\longrightarrow 2CuI_2$ $\longrightarrow 2CuI + I_2$.
Thus,$CuI_2$ is unstable and decomposes to form $CuI$ and $I_2$.

(A) $CuI_{2}$ is not known because it is unstable.
Copper$(II)$ iodide is not stable upon formation; it spontaneously decomposes to copper$(I)$ iodide and iodine.
The reaction is: $2CuI_{2} \rightarrow 2CuI + I_{2}$.

(D) The standard electrode potential $E^{\ominus}(M^{3+} / M^{2+})$ represents the ease of reduction of $M^{3+}$ to $M^{2+}$.
For $Mn^{3+} / Mn^{2+}$,the value is $+1.57 \ V$ (high positive due to stable $d^5$ configuration of $Mn^{2+}$).
For $Co^{3+} / Co^{2+}$,the value is $+1.97 \ V$.
For $Fe^{3+} / Fe^{2+}$,the value is $+0.77 \ V$.
For $Cr^{3+} / Cr^{2+}$,the value is $-0.41 \ V$.
Since the value for $Cr$ is negative,the correct option is $D$.

(D) The acidic or basic nature of transition metal oxides depends on the oxidation state of the metal.
Generally,oxides of transition metals in lower oxidation states are basic,while those in higher oxidation states are acidic or amphoteric.
Let us calculate the oxidation states of the metals in the given oxides:
$Cr_2O_3$: $2x + 3(-2) = 0 \implies x = +3$ (Amphoteric)
$CrO_3$: $x + 3(-2) = 0 \implies x = +6$ (Acidic)
$V_2O_5$: $2x + 5(-2) = 0 \implies x = +5$ (Acidic)
$V_2O_3$: $2x + 3(-2) = 0 \implies x = +3$ (Basic)
Among the given options,$V_2O_3$ has the metal in a lower oxidation state compared to the others,making it the most basic oxide.

(D) $Sc$ has an electronic configuration of $[Ar] 3d^1 4s^2$.
To form an oxide of the type $MO$,$Sc$ would need to exhibit a $+2$ oxidation state.
However,$Sc$ readily loses three electrons to attain the stable noble gas configuration of $[Ar]$,thus exhibiting a $+3$ oxidation state.
Therefore,$Sc$ does not form an $MO$ type oxide,whereas $V$,$Cr$,and $Mn$ are known to form $VO$,$CrO$,and $MnO$ respectively.

(A) In $VO_2^{+}$,the oxidation state of $V$ is $+5$.
In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$.
In $MnO_4^{-}$,the oxidation state of $Mn$ is $+7$.
Oxidizing power is related to the tendency of the central metal atom to undergo reduction,which is generally higher for transition metals in higher oxidation states.
Comparing the oxidation states: $Mn (+7) > Cr (+6) > V (+5)$.
Therefore,the correct order of oxidizing power is $VO_2^{+} < Cr_2O_7^{2-} < MnO_4^{-}$.

(A) The standard reduction potential for $Cu^{2+} + e^{-} \rightarrow Cu^{+}$ is $+0.15 \ V$ and for $I_2 + 2e^{-} \rightarrow 2I^{-}$ is $+0.54 \ V$.
Since the reduction potential of $I_2$ is higher,$Cu^{2+}$ is a strong enough oxidizing agent to oxidize $I^{-}$ to $I_2$.
The reaction is: $2Cu^{2+} + 4I^{-} \rightarrow 2CuI + I_2$.
Because $Cu^{2+}$ oxidizes $I^{-}$ to $I_2$,$CuI_2$ is unstable and does not exist; instead,$CuI$ is formed.
Thus,both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(C) $Yb^{2+}$ has the electronic configuration $[Xe] 4f^{14}$.
Since it has a stable $f^{14}$ configuration,it prefers to lose one electron to form $Yb^{3+}$,thus acting as a reductant.
$Lu^{3+}$ has the electronic configuration $[Xe] 4f^{14}$,which is fully filled,making it diamagnetic.
$CrO$ is a basic oxide because chromium is in a low oxidation state of $+2$.
Brass is an alloy of $Cu$ and $Zn$,whereas Bronze is an alloy of $Cu$ and $Sn$.

(A) For an ion to reduce dilute acid to hydrogen gas $(H_2)$,the standard electrode potential $(E^0)$ of the metal ion must be more negative than the standard reduction potential of hydrogen $(E^0_{H^+/H_2} = 0.00 \ V)$.
This means the metal must be a strong enough reducing agent to donate electrons to $H^+$ ions.
Among the given options,$Ti^{2+}$ $(E^0 = -0.37 \ V)$ and $Cr^{2+}$ $(E^0 = -0.41 \ V)$ have negative standard reduction potentials,making them capable of reducing $H^+$ to $H_2$ gas.
$Mn^{3+}$ and $Co^{3+}$ have positive reduction potentials and act as oxidizing agents,not reducing agents.

(A) $Cr^{2+}$ has the electronic configuration $[Ar] 3d^4$. It acts as a reducing agent because it gets oxidized to $Cr^{3+}$ $([Ar] 3d^3)$,which is more stable due to the half-filled $t_{2g}$ orbital.
$Mn^{3+}$ has the electronic configuration $[Ar] 3d^4$. It acts as an oxidizing agent because it gets reduced to $Mn^{2+}$ $([Ar] 3d^5)$,which is more stable due to the half-filled $d$-orbital.
Therefore,$Cr^{2+}$ is a reducing agent,while $Mn^{3+}$ is an oxidizing agent.
Thus,the correct statement is that $Cr^{2+}$ is a reducing agent.

(C) Iodide ions $(I^-)$ are strong reducing agents. When $Cu^{2+}$ reacts with $I^-$,the $Cu^{2+}$ ions oxidize the iodide ions to molecular iodine $(I_2)$ and are themselves reduced to $Cu^+$ ions. Consequently,$CuI_2$ is unstable and does not exist in nature. The reaction is: $2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2$.

(D) $i$. $CrO_3$ and $Mn_2O_7$ are acidic oxides,not basic oxides. Thus,statement $i$ is incorrect.
$ii$. $V_2O_3$ is a basic oxide and $V_2O_4$ is an amphoteric oxide. Thus,statement $ii$ is incorrect.
$iii$. $W(VI)$ is more stable than $Cr(VI)$ because stability of higher oxidation states increases down the group. Thus,statement $iii$ is incorrect.
Therefore,all statements $i, ii,$ and $iii$ are incorrect.

(D) The catalytic activity of transition elements is primarily attributed to their ability to exhibit $Variable \ oxidation \ states$ and their capacity to form $complexes$. These properties allow them to provide a large surface area and form intermediate compounds with reactants,thereby lowering the activation energy of the reaction.

(A) The acidic strength of metal oxides increases with an increase in the oxidation state of the metal.
Calculating the oxidation states of $Cr$ in the given oxides:
$1$. In $CrO$,the oxidation state of $Cr$ is $+2$.
$2$. In $Cr_2O_3$,the oxidation state of $Cr$ is $+3$.
$3$. In $CrO_3$,the oxidation state of $Cr$ is $+6$.
Since the acidic character is directly proportional to the oxidation state,the order of acidic strength is:
$CrO_3 (+6) > Cr_2O_3 (+3) > CrO (+2)$.

(B) The reaction between $Cu^{2+}_{(aq)}$ and $I^{-}_{(aq)}$ leads to the reduction of $Cu^{2+}$ to $Cu^{+}$ and the oxidation of $I^{-}$ to $I_2$,forming $CuI$ precipitate instead of $CuI_2$.
The reaction is: $2Cu^{2+}_{(aq)} + 4I^{-}_{(aq)} \longrightarrow 2CuI_{(s)} + I_{2(s)}$.
This occurs because $Cu^{2+}$ is a strong enough oxidizing agent to oxidize $I^{-}$ to $I_2$.
Therefore,Assertion $(A)$ is correct.
The reason $(R)$ stating that aqueous $Cu^{2+}$ is blue is a true statement regarding the color of the hydrated ion $[Cu(H_2O)_6]^{2+}$,but it does not explain why $CuI_2$ cannot be prepared.
Thus,both $(A)$ and $(R)$ are correct,but $(R)$ is not the correct explanation of $(A)$.

(B) The standard electrode potential $E^{\circ}_{M^{3+} \mid M^{2+}}$ represents the ease of reduction from the $+3$ oxidation state to the $+2$ oxidation state.
For the $3d$ series elements,the values are:
$V^{3+} \mid V^{2+} = -0.26 \ V$
$Cr^{3+} \mid Cr^{2+} = -0.41 \ V$
$Mn^{3+} \mid Mn^{2+} = +1.57 \ V$
$Fe^{3+} \mid Fe^{2+} = +0.77 \ V$
Comparing these values,$Mn^{3+} \mid Mn^{2+}$ has the highest positive value because $Mn^{2+}$ $(3d^5)$ is a stable half-filled configuration,making the reduction of $Mn^{3+}$ to $Mn^{2+}$ highly favorable.
Therefore,the correct option is $B$.

(C) The colour of transition metal ions in aqueous solution is due to $d-d$ transitions.
$Fe^{2+}$ (aquated) is pale green.
$Cu^{2+}$ (aquated) is blue.
$V^{3+}$ (aquated) is green.
$Fe^{3+}$ (aquated) is yellow or pale violet,not pink.
Therefore,the incorrect match is $Fe^{3+} - \text{Pink}$.

(A) $Mn$ is stable in the $+2$ oxidation state. Thus,$Mn^{3+}$ acts as an oxidizing agent by getting reduced to $Mn^{2+}$ and attaining the stable $d^5$ configuration.
$Cr$ is stable in the $+3$ oxidation state. Thus,$Cr^{2+}$ acts as a reducing agent by getting oxidized to $Cr^{3+}$.

(A) Fluorine is a small and highly electronegative atom,which allows it to stabilize higher oxidation states of transition metals effectively.
$VF_5$ is stable because the small size of fluorine atoms allows five of them to be accommodated around the vanadium atom without significant steric hindrance.
In contrast,$VCl_5$ is unstable because the larger size of chlorine atoms leads to significant steric hindrance,making it difficult for five chlorine atoms to bond with the central vanadium atom.
Additionally,the $V-F$ bond has a higher bond enthalpy compared to the $V-Cl$ bond,contributing to the stability of $VF_5$.
Therefore,both the assertion and the reason are true,and the reason is the correct explanation for the assertion.

(B) The oxidation states of the transition elements in the given ions are $V$ in $VO_2^{+}$ is $+5$,$Cr$ in $Cr_2O_7^{2-}$ is $+6$,and $Mn$ in $MnO_4^{-}$ is $+7$.
Oxidizing power is directly related to the ease of reduction,which depends on the stability of the lower oxidation states formed.
For these transition metals,the stability of the reduced products follows the trend $Mn^{2+} > Cr^{3+} > V^{3+}$.
Consequently,the oxidizing power increases as the reduction potential increases,leading to the order $VO_2^{+} < Cr_2O_7^{2-} < MnO_4^{-}$.

(D) Oxygen is capable of forming $ppi-dpi$ multiple bonds with transition metals,which stabilizes the higher oxidation states of the metal. Fluorine can only form single bonds and is limited by steric hindrance around the central metal atom. Thus,$Mn$ can achieve a higher oxidation state $(+7)$ with oxygen than with fluorine $(+4)$.

(B) Statement $I$ is incorrect because transition metals use both $3d$ and $4s$ electrons to form bonds with reactant molecules,not just $3d$ electrons.
Statement $II$ is incorrect because the catalyst works by adsorption,which actually weakens the bonds in the reacting molecules to lower the activation energy,rather than strengthening them.
Therefore,both statements are incorrect.