Compounds of Transitional elements Questions in English

(D) The disproportionation reaction of manganate $(MnO_{4}^{2-})$ in acidic medium is given by:
$3 MnO_{4}^{2-} + 4 H^{+} \longrightarrow 2 MnO_{4}^{-} + MnO_{2} + 2 H_{2}O$
In this reaction,manganese is oxidized to $Mn^{+7}$ (in $MnO_{4}^{-}$) and reduced to $Mn^{+4}$ (in $MnO_{2}$).
The product with the higher oxidation state is $MnO_{4}^{-}$,where the oxidation state of $Mn$ is $+7$.
The electronic configuration of $Mn^{+7}$ is $[Ar] 3d^{0}$.
Since there are no unpaired electrons $(n = 0)$,the spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \, B.M.$

(B) The standard electrode potential $E^{\circ}$ for the reduction of $H^{+}$ to $H_{2}$ is $0.00 \ V$.
$A$ metal ion can liberate $H_{2}$ from a dilute mineral acid if its standard reduction potential $E^{\circ}(M^{n+}/M)$ is negative.
Among the given ions,$Co^{3+}$ has a high positive reduction potential $(E^{\circ}(Co^{3+}/Co^{2+}) = +1.82 \ V)$,meaning it is a strong oxidizing agent and will not liberate $H_{2}$ from acid; instead,it is reduced to $Co^{2+}$.
The electronic configuration of $Co^{3+}$ $(Z=27)$ is $[Ar] 3d^{6}$.
In the $3d^{6}$ configuration,the number of unpaired electrons $(n)$ is $4$.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.899 \ B.M.$
The nearest integer value is $5$.

(C) The structure of $Mn_2O_7$ consists of two $MnO_4$ tetrahedra sharing a common oxygen atom at one corner.
In this structure,each $Mn$ atom is bonded to three terminal oxygen atoms via double bonds $(Mn=O)$ and one bridging oxygen atom via a single bond $(Mn-O-Mn)$.
Therefore,there are $3$ $Mn=O$ bonds around each $Mn$ atom.
Total number of $Mn=O$ bonds = $3 + 3 = 6$.

(C) $X$ is $CrO_{2}Cl_{2}$ and $Y$ is $Na_{2}CrO_{4}$.
When a mixture of $NaCl$,$K_{2}Cr_{2}O_{7}$ and conc. $H_{2}SO_{4}$ is heated in a dry test tube,a red vapour $(X)$,$CrO_{2}Cl_{2}$ (chromyl chloride),is evolved.
This vapour $CrO_{2}Cl_{2}$ reacts with an aqueous solution of $NaOH$ to form $Na_{2}CrO_{4}$ (sodium chromate),which is yellow in color.
The chemical reactions are as follows:
$4Cl^{-} + Cr_{2}O_{7}^{2-} + 6H^{+} \longrightarrow 2CrO_{2}Cl_{2} (X) + 3H_{2}O$
$CrO_{2}Cl_{2} + 4OH^{-} \longrightarrow CrO_{4}^{2-} + 2Cl^{-} + 2H_{2}O$
The resulting $CrO_{4}^{2-}$ ions in the presence of $Na^{+}$ ions form $Na_{2}CrO_{4} (Y)$.

(A) The oxidising ability of an anion depends on the oxidation state of the central metal atom. Higher oxidation state of the central atom leads to higher oxidising power.
Calculating the oxidation states:
$Ti$ in $TiO_{4}^{4-}$: $x + 4(-2) = -4 \implies x = +4$
$V$ in $VO_{4}^{3-}$: $x + 4(-2) = -3 \implies x = +5$
$Cr$ in $CrO_{4}^{2-}$: $x + 4(-2) = -2 \implies x = +6$
$Mn$ in $MnO_{4}^{-}$: $x + 4(-2) = -1 \implies x = +7$
The order of oxidation states is $+4 < +5 < +6 < +7$.
Therefore,the order of oxidising ability is $TiO_{4}^{4-} < VO_{4}^{3-} < CrO_{4}^{2-} < MnO_{4}^{-}$.

(A) The fusion of chromite ore $(FeCr_{2}O_{4})$ with $Na_{2}CO_{3}$ in the presence of air produces sodium chromate $(Na_{2}CrO_{4})$,which forms a yellow-coloured aqueous solution.
$8Na_{2}CO_{3} + 4FeCr_{2}O_{4} + 7O_{2} \longrightarrow 8Na_{2}CrO_{4} + 2Fe_{2}O_{3} + 8CO_{2}$
Subsequent acidification of the yellow solution with $H_{2}SO_{4}$ converts the chromate ion into the dichromate ion,resulting in an orange-coloured solution.
$2Na_{2}CrO_{4} + H_{2}SO_{4} \longrightarrow Na_{2}Cr_{2}O_{7} + Na_{2}SO_{4} + H_{2}O$
Thus,the yellow colour is due to $Na_{2}CrO_{4}$ and the orange colour is due to $Na_{2}Cr_{2}O_{7}$.

(A) $MnO_2$,when fused with $KOH$ and oxidised in air,gives a dark green compound $X$,which is potassium manganate $(K_2MnO_4)$.
In acidic solution,$K_2MnO_4$ undergoes disproportionation to give an intense purple solution of potassium permanganate $Y$ $(KMnO_4)$ and $MnO_2$.
The chemical reactions are:
$2MnO_2 + 4KOH + O_2 \longrightarrow 2K_2MnO_4 + 2H_2O$
$3K_2MnO_4 + 4H^{+} \longrightarrow 2KMnO_4 + MnO_2 + 2H_2O + 4K^{+}$
Thus,$X$ is $K_2MnO_4$ and $Y$ is $KMnO_4$.

(B) When $Cu^{2+}$ ions react with $KI$,$CuI_2$ is initially formed,which is unstable and decomposes to form a white precipitate of $Cu_2I_2$ and iodine $(I_2)$:
$2Cu^{2+} + 4I^-$ $\rightarrow 2CuI_2$ $\rightarrow Cu_2I_2 \downarrow + I_2$
The liberated $I_2$ dissolves in excess $KI$ to form $KI_3$ (triiodide ion),which gives a brown color to the solution:
$I_2 + I^- \rightarrow I_3^-$
This solution is then titrated against sodium thiosulphate $(Na_2S_2O_3)$,where $I_2$ is reduced to $I^-$ and thiosulphate is oxidized to tetrathionate $(Na_2S_4O_6)$:
$I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$
Thus,$X$ is $Cu_2I_2$ and $Y$ is $Na_2S_4O_6$.

(A) The structure of $Mn_2O_7$ consists of two $MnO_4$ tetrahedra sharing a common oxygen atom at one corner.
Thus,each $Mn$ atom is tetrahedrally surrounded by four oxygen atoms (statement $A$ is correct).
The structure contains an $Mn-O-Mn$ bridge (statement $C$ is correct).
There is no direct $Mn-Mn$ bond in $Mn_2O_7$ (statement $D$ is incorrect).
Therefore,statements $A$ and $C$ are correct.

(D) In the lanthanoid series,$Eu^{2+}$ (europium$(II)$) is a strong reducing agent because it tends to lose an electron to achieve the stable $f^7$ configuration $(Eu^{3+})$.
$Ce^{4+}$ (cerium$(IV)$) is a strong oxidizing agent because it tends to gain an electron to achieve the stable $f^0$ configuration $(Ce^{3+})$.
Therefore,the correct pair is $Eu^{2+}$ and $Ce^{4+}$.

(D) Statement $A$ is correct as transition metals form ionic $MO$ oxides.
Statement $B$ is correct as the highest oxidation state equals the group number from $Sc$ $(Group \ 3)$ to $Mn$ $(Group \ 7)$.
Statement $C$ is incorrect because the acidic character increases as the oxidation state of the metal increases $(V_2O_3 < V_2O_4 < V_2O_5)$. Therefore,the basic character decreases.
Statement $D$ is incorrect because $V_2O_4$ dissolves in acids to form $VO^{2+}$ ions,not $VO_4^{3-}$ salts.
Statement $E$ is correct as $CrO$ is basic and $Cr_2O_3$ is amphoteric.
Thus,statements $C$ and $D$ are incorrect.

(A) $4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 \ [(B) \text{Reddish brown fumes}] + 2KHSO_4 + 4NaHSO_4 + 3H_2O$
$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 \ [(C) \text{Yellow solution}] + 2NaCl + 2H_2O$
Therefore,$(B)$ is $CrO_2Cl_2$ and $(C)$ is $Na_2Cr_4$.

(D) The thermal decomposition of potassium permanganate $(KMnO_4)$ at $513 \ K$ is given by the following reaction:
$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$
Thus,$KMnO_4$ decomposes to form potassium manganate $(K_2MnO_4)$,manganese dioxide $(MnO_2)$,and oxygen gas $(O_2)$.

(A) In $K_2Cr_2O_7$,the oxidation state of $Cr$ is $+6$,which corresponds to a $d^0$ configuration. Thus,no $d-d$ transition is possible.
In $KMnO_4$,the oxidation state of $Mn$ is $+7$,which corresponds to a $d^0$ configuration. Thus,no $d-d$ transition is possible.
The intense colours in both compounds arise due to charge transfer transitions from the oxygen ligands to the metal centre.

(B) The alkaline oxidative fusion of $MnO_2$ with $KOH$ in the presence of air (or $KNO_3$) produces potassium manganate $(K_2MnO_4)$,where the manganate ion is $MnO_4^{2-}$. Thus,$A = MnO_4^{2-}$.
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$
Electrolytic oxidation of the green manganate solution $(MnO_4^{2-})$ in an alkaline medium converts it into purple permanganate $(MnO_4^{-})$. Thus,$B = MnO_4^{-}$.
$2MnO_4^{2-} + H_2O + [O] \rightarrow 2MnO_4^{-} + 2OH^{-}$
Therefore,$A$ and $B$ are $MnO_4^{2-}$ and $MnO_4^{-}$ respectively.

(A) The reaction of chromyl chloride $(CrO_2Cl_2)$ with $NaOH$ produces sodium chromate $(Na_2CrO_4)$ as follows:
$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$
Therefore,$A = Na_2CrO_4$.
Sodium chromate reacts with hydrogen peroxide $(H_2O_2)$ in an acidic medium $(HCl)$ to form the deep blue colored chromium pentoxide $(CrO_5)$:
$Na_2CrO_4 + 2HCl + 2H_2O_2 \rightarrow CrO_5 + 2NaCl + 3H_2O$
Therefore,$B = CrO_5$.

(D) . The chromate ion $(CrO_4^{2-})$ is tetrahedral,not square planar.
$B$. Dichromates are prepared from chromates by adding acid $(2CrO_4^{2-} + 2H^{+} \rightarrow Cr_2O_7^{2-} + H_2O)$. This statement is correct.
$C$. The green manganate ion $(MnO_4^{2-})$ has $1$ unpaired electron ($d^1$ configuration),making it paramagnetic,not diamagnetic.
$D$. $K_2MnO_4$ disproportionates in neutral or acidic solution to give permanganate $(MnO_4^-)$ and manganese dioxide $(MnO_2)$. This statement is correct.
$E$. As the oxidation number of the transition metal increases,the ionic character of the oxides decreases (covalent character increases). This statement is correct.
Therefore,statements $B, D,$ and $E$ are correct.

(C) $Mn_2O_7$ is a green oil at room temperature. This statement is correct.
$(B)$ $V_2O_4$ dissolves in acids to give $VO^{2+}$ salts. This statement is correct.
$(C)$ $CrO$ is a basic oxide. This statement is correct.
$(D)$ $V_2O_5$ is amphoteric; it reacts with both acids and bases. Therefore,the statement that it does not react with acid is incorrect.
Thus,statements $(A), (B),$ and $(C)$ are correct.

(A) The compound $CuSO_4 \cdot 5 H_2 O$ exhibits colour due to $d-d$ transition.
In $CuSO_4 \cdot 5 H_2 O$,the copper ion is in the $Cu^{2+}$ oxidation state.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3 d^9$.
Since there is an unpaired electron in the $d$-orbital,the $d-d$ transition is possible,which results in the observed colour.
In contrast,$K_2 Cr_2 O_7$,$K_2 CrO_4$,and $KMnO_4$ exhibit colour primarily due to charge transfer transitions,as the metal ions ($Cr^{6+}$ and $Mn^{7+}$) have a $d^0$ configuration.

(B) The first row transition metal with the highest enthalpy of atomization is Vanadium $(V)$.
Vanadium reacts with oxygen to form oxides like $V_2O_3$,$V_2O_4$,and $V_2O_5$.
Among these,$V_2O_4$ $(VO_2)$ is amphoteric in nature.
In $V_2O_4$,the oxidation state of $V$ is $+4$.
The electronic configuration of $V^{+4}$ is $[Ar] 3d^1$.
It has $1$ unpaired electron $(n=1)$.
The 'spin-only' magnetic moment is calculated as $\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ BM$.
The nearest integer value is $2$ $BM$.

(B) In the borax bead test,point $A$ represents the oxidizing flame (outer part) and point $B$ represents the reducing flame (inner part).
Copper salts $(Cu^{2+})$ produce a green-coloured bead in the oxidizing flame (hot) and a red-coloured bead in the reducing flame.
However,iron salts $(Fe^{3+})$ produce a yellow-brown bead in the oxidizing flame and a bottle-green bead in the reducing flame (point $B$).
For $Fe^{3+}$ $([Ar] 3d^5)$,the number of unpaired electrons $(n)$ is $5$.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
The nearest integer value is $6$.

(B) The chemical reaction for the fusion of chromite ore $(FeCr_2O_4)$ with sodium carbonate $(Na_2CO_3)$ in the presence of air is:
$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$
Here,$A = Na_2CrO_4$ and $B = Fe_2O_3$.
$1$. For $Na_2CrO_4$ $(Cr^{6+})$: The electronic configuration is $[Ar]3d^0$. The number of unpaired electrons $(n)$ is $0$. Therefore,the spin-only magnetic moment $\mu_s = \sqrt{0(0+2)} = 0 \ B.M.$
$2$. For $Fe_2O_3$ $(Fe^{3+})$: The electronic configuration is $[Ar]3d^5$. The number of unpaired electrons $(n)$ is $5$. Therefore,the spin-only magnetic moment $\mu_s = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \ B.M.$
Sum of magnetic moments $= 0 + 5.91 = 5.91 \ B.M.$
The nearest integer value is $6$.

(A) $CrO$ is a basic oxide.
$Cr_2O_3$ is an amphoteric oxide.
In $CrO$,$Cr$ is in the $+2$ oxidation state $([Ar] 3d^4)$,so it has $4$ unpaired electrons. The spin-only magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
In $Cr_2O_3$,$Cr$ is in the $+3$ oxidation state $([Ar] 3d^3)$,so it has $3$ unpaired electrons. The spin-only magnetic moment $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
The sum of the spin-only magnetic moments is $4.90 + 3.87 = 8.77 \ BM$.
Expressed as $10^{-2} \ BM$,this is $877 \times 10^{-2} \ BM$.

(A) $MnO_2 + 2 KOH + \frac{1}{2} O_2 \rightarrow K_2MnO_4 + H_2O$ (Dark green color).
Statement $I$ is true because the fusion of $MnO_2$ with $KOH$ in the presence of an oxidizing agent like $KNO_3$ or atmospheric oxygen yields potassium manganate $(K_2MnO_4)$,which is dark green in color.
Statement $II$ is true because the manganate ion $(MnO_4^{2-})$ undergoes electrolytic oxidation in an alkaline medium to form the purple-colored permanganate ion $(MnO_4^-)$.
The reaction at the anode is: $MnO_4^{2-} \rightarrow MnO_4^- + e^-$.

(C) The metal halide is $MnCl_2$.
$MnCl_2 + 2NaOH \rightarrow Mn(OH)_2 \downarrow (P) + 2NaCl (Q)$
$P$ is $Mn(OH)_2$ (white precipitate).
$Mn(OH)_2$ reacts with $PbO_2$ in the presence of $H_2SO_4$ to form $MnO_4^-$ (purple coloured species,$X$):
$2Mn(OH)_2 + 5PbO_2 + 6H_2SO_4 \rightarrow 2HMnO_4 + 5PbSO_4 + 6H_2O$
$Q$ is $NaCl$. $Cl^-$ ions react with $MnO(OH)_2$ (or $MnO_2$) and conc. $H_2SO_4$ to liberate $Cl_2$ gas $(Y)$:
$MnO_2 + 2Cl^- + 4H^+ \rightarrow Mn^{2+} + Cl_2 + 2H_2O$
$Cl_2$ reacts with $KI$ to liberate $I_2$,which gives a blue color with starch paper.
Thus,$X = MnO_4^-$ and $Y = Cl_2$.

(A) The reaction for the alkaline oxidative fusion of manganese dioxide is:
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$
The product formed is potassium manganate,$K_2MnO_4$,which contains the manganate ion,${MnO_4}^{2-}$.
To find the oxidation state of $Mn$ in ${MnO_4}^{2-}$:
Let the oxidation state of $Mn$ be $x$.
$x + 4 \times (-2) = -2$
$x - 8 = -2$
$x = +6$
Thus,the oxidation number of $Mn$ in the product is $6$.

(D) Haematite is $Fe_2O_3$,where the oxidation state of iron is $+3$.
Magnetite is $Fe_3O_4$,which is a mixed oxide represented as $FeO \cdot Fe_2O_3$.
In $FeO$,the oxidation state of iron is $+2$,and in $Fe_2O_3$,it is $+3$.
Therefore,the oxidation states of iron in haematite and magnetite are $III$ and $II, III$ respectively.

(A) In the borax bead test,the chromium $(III)$ salt decomposes upon heating to form chromium $(III)$ oxide $(Cr_2O_3)$.
Borax $(Na_2B_4O_7 \cdot 10H_2O)$ on heating loses water and forms sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
The reaction is: $Cr_2O_3 + 3B_2O_3 \longrightarrow 2Cr(BO_2)_3$.
The green colour is due to the formation of chromium metaborate,$Cr(BO_2)_3$.

(B) $MnO_2 + 2 KOH + \frac{1}{2} O_2 \xrightarrow{\Delta} K_2MnO_4 + H_2O$ $(W)$
$W = K_2MnO_4 \rightleftharpoons 2K^{\oplus} + MnO_4^{2-} (Y)$
$K_2MnO_4 + H_2O \xrightarrow{\text{Electrolytic}} H_2 + KOH + KMnO_4 (X)$
Anion of $X = MnO_4^{-} (Z)$
$1$. $Y$ $(MnO_4^{2-})$ is $d^1$ (paramagnetic) and $Z$ $(MnO_4^{-})$ is $d^0$ (diamagnetic). Statement $(1)$ is incorrect.
$2$. Both $MnO_4^{2-}$ and $MnO_4^{-}$ are coloured due to charge transfer and have tetrahedral geometry. Statement $(2)$ is correct.
$3$. In both,$\pi$-bonding occurs between $p$-orbitals of oxygen and $d$-orbitals of manganese ($p\pi-d\pi$ bonding). Statement $(3)$ is correct.
$4$. In acidic solution,$MnO_4^{2-}$ undergoes disproportionation: $3MnO_4^{2-} + 4H^{\oplus} \longrightarrow 2MnO_4^{-} + MnO_2 + 2H_2O$. Statement $(4)$ is correct.
Thus,statements $(2, 3, 4)$ are correct.

(C) The reaction between acidified potassium chromate and $H_2O_2$ produces chromium pentoxide $(CrO_5)$,which is a blue-colored compound.
The chemical reaction is: $K_2CrO_4 + 2H_2O_2 + 2H^+ \rightarrow CrO_5 + 2K^+ + 2H_2O$.
$CrO_5$ has a butterfly-like structure.
In this structure,there is one double-bonded oxygen atom $(Cr=O)$ and four single-bonded oxygen atoms $(Cr-O)$ which are part of two peroxo groups $(-O-O-)$.
Therefore,the number of oxygen atoms bonded to chromium through only single bonds is $4$.

(C) The reaction of potassium dichromate $(K_2Cr_2O_7)$ with potassium hydroxide $(KOH)$ yields potassium chromate $(K_2CrO_4)$ and water:
$K_2Cr_2O_7 + 2KOH \rightarrow 2K_2CrO_4 + H_2O$
Thus,$[A] = K_2CrO_4$.
Next,the reaction of potassium chromate $(K_2CrO_4)$ with sulfuric acid $(H_2SO_4)$ yields potassium dichromate $(K_2Cr_2O_7)$,potassium sulfate $(K_2SO_4)$,and water:
$2K_2CrO_4 + H_2SO_4 \rightarrow K_2Cr_2O_7 + K_2SO_4 + H_2O$
Thus,$[B] = K_2Cr_2O_7$.

(D) The preparation of potassium permanganate $(KMnO_4)$ from pyrolusite ore $(MnO_2)$ occurs in two steps.
In the $1^{st}$ step,$MnO_2$ is fused with potassium hydroxide $(KOH)$ in the presence of an oxidizing agent like potassium nitrate $(KNO_3)$ or air to form potassium manganate $(K_2MnO_4)$.
The reaction is: $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$ or $MnO_2 + 2KOH + KNO_3 \rightarrow K_2MnO_4 + KNO_2 + H_2O$.
Thus,the product of the $1^{st}$ step is $K_2MnO_4$.

(C) Among the given oxides,$V_2O_5$ is amphoteric in nature and reacts with alkali to form vanadate ions.
The reaction is: $V_2O_5 + 6OH^- \rightarrow 2VO_4^{3-} + 3H_2O$.
In the $VO_4^{3-}$ ion,let the oxidation state of $V$ be $x$.
$x + 4(-2) = -3$
$x - 8 = -3$
$x = +5$.
Therefore,the oxidation state of $V$ in the oxide anion is $+5$.

(A) The strongest oxidising agent among the given compounds is $Mn_2O_3$ due to the high reduction potential of the $Mn^{3+}/Mn^{2+}$ couple $(E^{\circ} = +1.57 \ V)$.
In $Mn_2O_3$,the oxidation state of $Mn$ is $+3$.
The electronic configuration of $Mn^{3+}$ is $[Ar] 3d^4$.
The number of unpaired electrons $(n)$ is $4$.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$
$\mu = \sqrt{4(4+2)} \ B.M. = \sqrt{24} \ B.M. \approx 4.89 \ B.M.$
The nearest integer value is $5$.

(A) The reactions are as follows:
$1$. $K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2KHSO_4 + 2NaHSO_4 + 3H_2O$ (Chromyl chloride test).
$2$. $CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$ (Compound $B$ is $Na_2CrO_4$).
$3$. $2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O$ (Compound $C$ is $Na_2Cr_2O_7$).
In the dichromate ion $(Cr_2O_7^{2-})$,there are two $Cr$ atoms linked by an oxygen bridge $(Cr-O-Cr)$.
Each $Cr$ atom is bonded to three terminal oxygen atoms (two double-bonded and one single-bonded with a negative charge).
Total terminal oxygen atoms = $3 + 3 = 6$.

(C) In the $3d$ transition series,Vanadium $(V)$ has the highest enthalpy of atomisation.
The aquated ion $V^{3+}$ exists in a green colour.
The oxide formed by $V^{3+}$ is $V_2O_3$.
$V_2O_3$ is a basic oxide.

(D) The structure of the dichromate ion,$Cr_2O_7^{2-}$,consists of two $CrO_4$ tetrahedra sharing a common oxygen atom.
In this structure,there are six terminal $Cr-O$ bonds which are equivalent due to resonance.
Additionally,there is one $Cr-O-Cr$ bridge bond.
Therefore,the structure contains six equivalent $Cr-O$ bonds and one $Cr-O-Cr$ bond.

(D) In $KMnO_4$,Manganese is in the $+7$ oxidation state.
The electronic configuration of $Mn^{+7}$ is $[Ar] 3d^0 4s^0$.
Since there are no $d$-electrons,the purple colour of $KMnO_4$ cannot be due to $d-d$ transitions.
Instead,the colour arises due to ligand-to-metal charge transfer $(LMCT)$ from oxygen to manganese.
Thus,both the Assertion and the Reason are true,and the Reason correctly explains the Assertion.

(D) $1$. Tungsten $(W)$ has the highest melting point among all metals due to strong metallic bonding,so statement $A$ is correct.
$2$. Chromyl chloride $(CrO_2Cl_2)$ has a tetrahedral geometry around the central $Cr$ atom,so statement $B$ is correct.
$3$. The yellow colour of $K_2CrO_4$ is due to the ligand-to-metal charge transfer $(LMCT)$ from $O^{2-}$ to $Cr^{6+}$,so statement $C$ is correct.
$4$. $KMnO_4$ acts as a strong oxidising agent in acidic,neutral,and alkaline media. In an alkaline medium,it is reduced to manganate $(MnO_4^{2-})$,so statement $D$ is incorrect.

(C) The aqueous solution of sodium chromate $(Na_2CrO_4)$ is yellow due to the presence of the chromate ion $(CrO_4^{2-})$.
When $CO_2$ gas is passed through this solution,it reacts with water to form carbonic acid $(H_2CO_3)$,which increases the acidity of the solution.
In an acidic medium,the chromate ions $(CrO_4^{2-})$ are converted into dichromate ions $(Cr_2O_7^{2-})$,which are orange-red in color.
The chemical reaction is as follows:
$2Na_2CrO_4 + 2CO_2 + H_2O \rightarrow Na_2Cr_2O_7 + 2NaHCO_3$
Thus,the formation of sodium dichromate $(Na_2Cr_2O_7)$ is responsible for the color change.

(C) The ability of $F$ and $O$ to stabilize higher oxidation states is due to their small size and high electronegativity.
For $F$,the higher lattice energy (in solids) or higher bond enthalpy (in covalent compounds) is responsible.
For $O$,the ability to form multiple bonds ($p\pi-d\pi$ bonding) is responsible.
Fluorine is more electronegative than Oxygen,so it has a greater ability to stabilize higher oxidation states.
Therefore,the statement that 'Oxygen has less ability to stabilize higher oxidation state than Fluorine' is actually a correct statement.
However,the question asks for the incorrect statement. Looking at the options,all statements $A$,$B$,and $C$ are scientifically correct. Option $D$ is also a fact,but in the context of standard chemistry questions of this type,the comparison in $C$ is often debated; however,based on electronegativity,$F$ is superior. Upon re-evaluation,all statements provided are technically correct in standard textbooks. Given the constraints,if one must be chosen as 'incorrect' in a typical exam context,it is often a trick question,but here,all are factually true. Assuming the question implies a comparison,$C$ is the most common target for 'incorrect' in some contexts,but strictly speaking,all are correct.

(B) The given reaction sequence is as follows:
$1$. Pyrolusite is $MnO_2$,which is $(A)$.
$2$. When $MnO_2$ is fused with $KOH$ in the presence of air,it forms potassium manganate $(K_2MnO_4)$,which is a green compound $(B)$.
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$
$3$. Potassium manganate $(K_2MnO_4)$ on reaction with $H_2SO_4$ (acidic medium) disproportionates to form potassium permanganate $(KMnO_4)$,which is a purple crystal $(C)$.
$3K_2MnO_4 + 2H_2SO_4 \rightarrow 2KMnO_4 + MnO_2 + 2K_2SO_4 + 2H_2O$
Therefore,$(A) = MnO_2$,$(B) = K_2MnO_4$,and $(C) = KMnO_4$.

(D) Potassium permanganate $(KMnO_4)$ is a strong oxidizing agent.
It is widely used for the bleaching of wool,cotton,silk,and other textile fibres.
It is also used for the decolourisation of oils.
In analytical chemistry,it is extensively used in redox titrations (permanganometry) due to its strong oxidizing properties.
Therefore,all the given options are correct uses of $KMnO_4$.

(A) The oxidation state of $Mn$ in $MnO_4^{2-}$ is $+6$.
The electronic configuration of $Mn^{6+}$ is $[Ar] 3d^1$.
Due to the presence of one unpaired electron in the $3d$ orbital,the $MnO_4^{2-}$ ion is paramagnetic.
The manganate ion $(MnO_4^{2-})$ exhibits a green colour.

(C) Trophies are typically made of bronze,which is an alloy of copper $(Cu)$ and tin $(Sn)$.
Therefore,the correct pair of elements is $Cu$ and $Sn$.

(C) Stainless steels are used in the construction of the outer fuselage of ultra-high-speed aircraft due to their high strength and resistance to corrosion at elevated temperatures.

(D) Nichrome is an alloy of $Ni$ and $Cr$. Due to its high resistance to oxidation and heat,it is used in heating elements and gas turbine engines.

(C) When a mixture of manganese dioxide $(MnO_2)$,potassium hydroxide $(KOH)$,and potassium chlorate $(KClO_3)$ is fused,the potassium chlorate acts as an oxidizing agent and decomposes to provide oxygen.
The overall reaction is:
$2KClO_{3(s)} \xrightarrow{\Delta} 2KCl_{(s)} + 3O_{2(g)}$
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$
The final product obtained is potassium manganate $(K_2MnO_4)$.

(A) The chemical formula for the chromate ion is $CrO_4^{2-}$,which carries a charge of $-2$.
The chemical formula for the dichromate ion is $Cr_2O_7^{2-}$,which also carries a charge of $-2$.
Therefore,the ionic charges are $-2$ and $-2$ respectively.

(A) In the industrial production of potassium dichromate,the yellow solution of sodium chromate $(Na_2CrO_4)$ is acidified with concentrated sulphuric acid $(H_2SO_4)$ to convert it into orange sodium dichromate $(Na_2Cr_2O_7)$.
The reaction is: $2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O$.