Compounds of Transitional elements Questions in English

(B) $ZnO$ (Zinc oxide) is white at room temperature. When heated,it loses oxygen to form non-stoichiometric $Zn_{1+x}O$ due to the presence of excess $Zn^{2+}$ ions in interstitial sites,which causes it to turn yellow.

(B) Calomel is the common name for the chemical compound mercury$(I)$ chloride,which has the formula $Hg_2Cl_2$.

(D) Hematite is $Fe_2O_3$,where iron exists in the $+III$ oxidation state.
Magnetite is $Fe_3O_4$,which is a mixed oxide $FeO \cdot Fe_2O_3$. In this mineral,iron exists in both $+II$ and $+III$ oxidation states.
Therefore,the correct answer is $D$.

(C) Gold chloride $(AuCl_3)$ is thermally unstable. When it is heated,it decomposes into metallic gold $(Au)$ and chlorine gas $(Cl_2)$. The chemical reaction is: $2AuCl_3 \rightarrow 2Au + 3Cl_2$.

(A) To find the oxidation state of $Cr$ in $K_2Cr_2O_7$:
Let the oxidation state of $Cr$ be $x$.
$2(+1) + 2(x) + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x = 12$
$x = +6$.
Therefore,in $K_2Cr_2O_7$,chromium has a $+6$ oxidation state.

(B) Heating $K_2Cr_2O_7$ gives $O_2$ gas: $4K_2Cr_2O_7 \rightarrow 4K_2CrO_4 + 2Cr_2O_3 + 3O_2$.
Heating $(NH_4)_2Cr_2O_7$ gives $N_2$ gas: $(NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 + 4H_2O$.
Heating $KClO_3$ gives $O_2$ gas: $2KClO_3 \rightarrow 2KCl + 3O_2$.
Heating $Zn(ClO_3)_2$ gives $O_2$ gas: $Zn(ClO_3)_2 \rightarrow ZnCl_2 + 3O_2$.
Therefore,$(NH_4)_2Cr_2O_7$ does not produce $O_2$ on heating.

(B) When $SO_2$ gas is passed through an acidified solution of $K_2Cr_2O_7$,it acts as a reducing agent and reduces $Cr(VI)$ to $Cr(III)$.
The balanced chemical equation is: $K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \longrightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.
The formation of $Cr_2(SO_4)_3$ results in a green-colored solution.

(B) . Passing $H_2S$ through acidified $K_2Cr_2O_7$ produces $S$ (sulfur) which causes a milky appearance. This is true.
$B$. $Na_2Cr_2O_7$ is more soluble than $K_2Cr_2O_7$,but it is hygroscopic,making it unsuitable for primary standards in volumetric analysis. Thus,$K_2Cr_2O_7$ is preferred. This statement is false.
$C$. $K_2Cr_2O_7$ exists as dichromate ions in acidic medium,which are orange. This is true.
$D$. In basic medium $(pH > 7)$,dichromate ions $(Cr_2O_7^{2-})$ convert to chromate ions $(CrO_4^{2-})$,which are yellow. This is true.

(A) $KMnO_4 \to K^{+} + MnO_4^-$.
In $MnO_4^-$,$Mn$ is in the $+7$ oxidation state,which means it has a $d^0$ electronic configuration (no electrons in $d$-orbitals).
The intense purple color of $KMnO_4$ is not due to $d-d$ transitions because there are no $d$-electrons.
Instead,it arises from $L \to M$ (Ligand to Metal) charge transfer,where an electron from the oxygen $2p$ orbital is excited to the vacant $Mn$ $3d$ orbital.
Since this charge transfer is both Laporte-allowed and spin-allowed,it results in an intense color.

(B) $Mn_2O_7$ is the acid anhydride of permanganic acid $(HMnO_4)$.
The reaction is: $Mn_2O_7 + H_2O \rightarrow 2HMnO_4$.
$KMnO_4$ is the potassium salt of this acid,formed by the reaction: $HMnO_4 + KOH \rightarrow KMnO_4 + H_2O$.
Therefore,$KMnO_4$ is the oxo-salt of the acid derived from $Mn_2O_7$.

(C) When ferric sulphate $(Fe_2(SO_4)_3)$ is heated,it undergoes thermal decomposition to form ferric oxide $(Fe_2O_3)$ and sulphur trioxide $(SO_3)$.
The balanced chemical equation is:
$Fe_2(SO_4)_3 \xrightarrow{\Delta} Fe_2O_3 + 3SO_3$

(D) In photography,gold toning is a process used to improve the appearance and stability of photographic prints.
$AuCl_3$ (Gold$(III)$ chloride) is the reagent commonly used for this purpose.
It replaces some of the silver in the image with gold,which provides a warmer tone and increases the archival permanence of the print.

(D) The thermal decomposition of $KMnO_4$ at $200^{\circ}C$ is given by: $2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$. Thus,'$X$' is $K_2MnO_4$ (potassium manganate).
$1$. In $K_2MnO_4$,the oxidation state of $Mn$ is $+6$ and the $MnO_4^{2-}$ ion has a tetrahedral geometry. So,statement $C$ is correct.
$2$. $K_2MnO_4$ undergoes disproportionation in acidic or neutral medium to form $KMnO_4$ (purple colour). $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$. Thus,reaction with $H_2O$ (neutral/acidic) gives a purple colour. So,statement $B$ is correct.
$3$. $K_2MnO_4$ reacts with $Cl_2$ to form $KMnO_4$ (purple colour): $2K_2MnO_4 + Cl_2 \rightarrow 2KMnO_4 + 2KCl$. So,statement $A$ is correct.
Since all statements are correct,the answer is $D$.

(B) The reaction of $Cr_2O_7^{2-}$ with $H_2O_2$ in acidic medium produces chromium pentoxide $(CrO_5)$.
$Cr_2O_7^{2-} + 2H^{\oplus} + 4H_2O_2 \rightarrow 2CrO_5 + 5H_2O$.
$CrO_5$ is unstable and has a deep blue color ('$X$').
On standing,$CrO_5$ decomposes to form $Cr^{3+}$ ions in aqueous solution ('$Y$'),which are green in color.
Therefore,'$X$' is dark blue and '$Y$' is green.

(A) The extraction of chromium from chromite ore $(FeCr_2O_4)$ involves the following steps:
Step $I$: The chromite ore is fused with sodium hydroxide in the presence of air to form sodium chromate $(Na_2CrO_4)$.
$4FeCr_2O_4 + 16NaOH + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8H_2O$
Here,$(A) = Na_2CrO_4$.
Step $II$: The yellow solution of sodium chromate is acidified with concentrated sulfuric acid $(H_2SO_4)$ to convert it into orange sodium dichromate $(Na_2Cr_2O_7)$.
$2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$
Here,$(B) = H_2SO_4$.
Thus,the compounds $(A)$ and $(B)$ are $Na_2CrO_4$ and $H_2SO_4$ respectively.

(D) Vanadium exhibits common oxidation states of $+2, +3, +4,$ and $+5$ in aqueous solutions.
In $VO_2^+$,the oxidation state of $V$ is $+5$.
In $VO^{2+}$,the oxidation state of $V$ is $+4$.
In $[V(H_2O)_6]^{3+}$,the oxidation state of $V$ is $+3$.
In $VO_2^{2+}$,the oxidation state of $V$ would be $+6$. Since the maximum oxidation state of vanadium is $+5$,the species $VO_2^{2+}$ is not stable or formed in aqueous solution.

(A) The interconversion between dichromate $(Cr_2O_7^{2-})$ and chromate $(CrO_4^{2-})$ ions is pH dependent.
In an alkaline medium $(OH^{-})$,the dichromate ion is converted into the chromate ion: $Cr_2O_7^{2-} + 2OH^{-} \rightarrow 2CrO_4^{2-} + H_2O$.
In an acidic medium $(H^{+})$,the chromate ion is converted into the dichromate ion: $2CrO_4^{2-} + 2H^{+} \rightarrow Cr_2O_7^{2-} + H_2O$.
Therefore,for the reaction $Cr_2O_7^{2-} \xrightarrow{X} 2CrO_4^{2-}$,$X$ must be $OH^{-}$.
For the reverse reaction $2CrO_4^{2-} \xrightarrow{Y} Cr_2O_7^{2-}$,$Y$ must be $H^{+}$.
Thus,$X = OH^{-}$ and $Y = H^{+}$.

(B) The correct answer is $(B)$.
The $MnO_4^-$ ion exhibits a deep purple colour due to ligand-to-metal charge transfer $(LMCT)$.
In $MnO_4^-$,the oxidation state of $Mn$ is $+7$,which corresponds to a $d^0$ electronic configuration.
Since there are no $d$-electrons available for $d-d$ transitions,the intense colour arises from the transfer of electrons from the oxygen $2p$ orbitals to the empty $3d$ orbitals of the $Mn^{7+}$ ion.

(D) $1$. In an acidic medium,$K_2Cr_2O_7$ acts as a strong oxidizing agent and oxidizes $I^-$ to $I_2$: $Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 7H_2O + 3I_2$. Thus,statement $A$ is correct.
$2$. $K_2Cr_2O_7$ is a primary standard used in redox titrations to estimate $Fe^{2+}$ ions: $Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$. Thus,statement $B$ is correct.
$3$. The n-factor for $K_2Cr_2O_7$ in an acidic medium is $6$ (change in oxidation state of $Cr$ from $+6$ to $+3$ for two $Cr$ atoms). Since $N = M \times n\text{-factor}$,we have $N = M \times 6$,which implies $M = N/6$. Thus,statement $C$ is correct.
$4$. Since all statements are correct,the answer is $D$.

(A) When acidified potassium dichromate or chromic acid reacts with $H_2O_2$,it forms chromium pentoxide $(CrO_5)$ which is blue in color.
The reaction is: $H_2CrO_4 + 2H_2O_2 \rightarrow CrO_5 + 3H_2O$.
Here,$X$ is $CrO_5$ and $Y$ is $H_2O$.

(B) The reaction of $CuSO_4$ with $KI$ produces $Cu_2I_2$ and $I_2$ gas: $2Cu^{2 } 4I^- \rightarrow Cu_2I_2 I_2(g)$. Thus,$Y = I_2$.
When $CuSO_4$ is dissolved in dilute $H_2SO_4$,it forms the hydrated copper$(II)$ ion,which is blue in color: $CuSO_4 4H_2O \rightarrow [Cu(H_2O)_4]^{2 } SO_4^{2-}$. Thus,$X = [Cu(H_2O)_4]^{2 }$.

(D) $1$. $MnO_2$ reacts with conc. $HCl$ to form $MnCl_2$,$Cl_2$ gas,and $H_2O$. It does not form $Mn^{4+}$ ions because $Mn^{4+}$ is unstable in the presence of $Cl^-$ ions. This statement is correct.
$2$. $MnO_2$ reacts with hot conc. $H_2SO_4$ to produce $MnSO_4$,$O_2$ gas,and $H_2O$. This statement is correct.
$3$. When $MnO_2$ is fused with $KOH$ in the presence of an oxidizing agent like air,$KNO_3$,$PbO_2$,or $NaBiO_3$,it forms green potassium manganate $(K_2MnO_4)$. This statement is correct.
$4$. The decomposition of $KMnO_4$ is indeed catalyzed by light (sunlight). Therefore,the statement that it is 'not catalysed by sunlight' is incorrect.

(A) Invar is an alloy of $Fe$ $(64\%)$ and $Ni$ $(36\%)$.
It has a very low coefficient of thermal expansion.
Due to this property,it does not change its length significantly with temperature variations,making it suitable for manufacturing precision instruments like metre scales.

(D) Amphoteric oxides are oxides that behave as both acidic and basic oxides.
$MnO_2$ acts as an amphoteric oxide.
$Mn_3O_4$ is a mixed oxide,which can be represented as $MnO \cdot Mn_2O_3$. It also exhibits amphoteric character.
$Mn_2O_7$ is an acidic oxide.
Therefore,both $MnO_2$ and $Mn_3O_4$ are amphoteric.

(C) The statement in option $C$ is incorrect because $KMnO_4$ acts as a strong oxidizing agent in an alkaline medium,where it is reduced to $MnO_4^{2-}$ (manganate ion).
Option $A$ is correct as $MnO_4^{2-}$ is stable only in strongly alkaline solutions and disproportionates in neutral or acidic media.
Option $B$ is correct as $KMnO_4$ is a powerful oxidizing agent in acidic media,reducing to $Mn^{2+}$.
Option $D$ is correct as it describes the standard industrial preparation of $KMnO_4$ from pyrolusite $(MnO_2)$.

(A) Railway tracks are made of manganese steel,which is an alloy of iron containing approximately $10-15\%$ manganese.
This specific composition makes the steel extremely hard,tough,and resistant to wear and abrasion,which is essential for the heavy loads and constant friction experienced by railway tracks.
The addition of $Mn$ increases the hardness and strength of the steel while reducing its brittleness.
Therefore,the correct answer is that it is hard due to the high percentage of $Mn$.
Thus,the correct option is $A$.

(A) The reaction of potassium dichromate $(K_2Cr_2O_7)$ with hydrogen peroxide $(H_2O_2)$ in an acidic medium produces chromium pentoxide $(CrO_5)$,which is blue in color.
$K_2Cr_2O_7 + 4H_2O_2 + 2H^+ \to 2CrO_5 + 5H_2O + 2K^+$
In $CrO_5$,the chromium atom is bonded to four oxygen atoms via peroxide linkages $(-O-O-)$ and one double-bonded oxygen atom.
The structure is a butterfly structure.
Calculating the oxidation state of $Cr$ in $CrO_5$:
Let the oxidation state be $x$.
$x + 4(-1) + 1(-2) = 0$
$x - 4 - 2 = 0$
$x = +6$
Thus,the oxidation state of the central atom $Cr$ in $CrO_5$ is $6$.

(A) The colour of metal salts is generally due to $d-d$ transitions,which require the presence of unpaired electrons in the $d$-orbitals.
In $TiCl_4$,the oxidation state of $Ti$ is $+4$,which corresponds to a $3d^0$ configuration (no unpaired electrons).
In $Cu_2Cl_2$,the oxidation state of $Cu$ is $+1$,which corresponds to a $3d^{10}$ configuration (no unpaired electrons).
In $Ce(SO_4)_2$,the oxidation state of $Ce$ is $+4$. Although $Ce^{4+}$ has a $4f^0$ configuration,$Ce(SO_4)_2$ is known to be yellow in colour. This colour arises due to charge transfer transitions rather than $d-d$ transitions.
Therefore,$Ce(SO_4)_2$ is the coloured compound among the given options.

(D) $1.$ Thermal decomposition of $KMnO_4$: $2 KMnO_4 \xrightarrow{\Delta } K_2MnO_4 + MnO_2 + O_2$. Here,$K_2MnO_4$ is formed.
$2.$ Reaction of $KMnO_4$ with concentrated $KOH$: $4 KMnO_4 + 4 KOH \xrightarrow{\Delta } 4 K_2MnO_4 + O_2 + 2 H_2O$. Here,$K_2MnO_4$ is formed.
$3.$ Fusion of $MnO_2$ with $KOH$ in the presence of air: $2 MnO_2 + 4 KOH + O_2 \xrightarrow{\Delta } 2 K_2MnO_4 + 2 H_2O$. Here,$K_2MnO_4$ is formed.
$4.$ Reaction of $KMnO_4$ with concentrated $H_2SO_4$: $KMnO_4 + H_2SO_4 \text{ (conc.)} \xrightarrow{\Delta } K_2SO_4 + Mn_2O_7 + H_2O$. In this reaction,manganese heptoxide $(Mn_2O_7)$ is formed,not potassium manganate $(K_2MnO_4)$.

(A) The interconversion between chromate $(CrO_4^{2-})$ and dichromate $(Cr_2O_7^{2-})$ ions is dependent on the $pH$ of the solution.
In an acidic medium $(pH < 7)$,the chromate ion ($CrO_4^{2-}$,yellow) is converted into the dichromate ion ($Cr_2O_7^{2-}$,orange) according to the reaction: $2CrO_4^{2-} + 2H^+ \rightarrow Cr_2O_7^{2-} + H_2O$.
In a basic medium $(pH > 7)$,the dichromate ion $(Cr_2O_7^{2-})$ is converted back into the chromate ion $(CrO_4^{2-})$ according to the reaction: $Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O$.
Given the reaction $CrO _4^{2-} \underset{ pH = y }{\stackrel{ pH = x }{\longleftrightarrow}} Cr _2 O _7^{2-}$,$x$ represents the acidic condition $(pH < 7)$ and $y$ represents the basic condition $(pH > 7)$.
Comparing the given options,option $A$ $(6, 8)$ correctly represents an acidic $pH$ $(6)$ for the forward reaction and a basic $pH$ $(8)$ for the reverse reaction.

(A) The highest oxidation state of $Mn$ is $+7$.
In fluorides,$Mn$ forms $MnF_4$ as its highest fluoride because the high electronegativity of fluorine and the size of the $Mn$ atom limit the coordination number.
In oxides,$Mn$ forms $Mn_2O_7$ (manganese heptoxide) where the oxidation state of $Mn$ is $+7$.
Therefore,the highest fluoride is $MnF_4$ and the highest oxide is $Mn_2O_7$.

(B) Misch metal is a well-known alloy of lanthanoid metals $(Ln)$.
It typically consists of approximately $95\%$ lanthanoid metal and $5\%$ iron $(Fe)$,along with traces of sulfur $(S)$,carbon $(C)$,silicon $(Si)$,and aluminum $(Al)$.
Among the given options,the most accurate representation of its composition is $Ln + Fe$ (often represented as $Ce + Fe$ in simplified contexts,but $Ln + Fe$ is the standard definition).
Since the provided options are limited,$Ln + Fe$ is the intended answer.

(D) In aqueous solution,$Cu^{2+}$ is more stable than $Cu^{+}$ due to its much more negative hydration enthalpy,which compensates for the second ionization enthalpy of copper.
$Cu^{2+}$ has a $3d^9$ configuration (one unpaired electron),making it paramagnetic,while $Cu^{+}$ has a $3d^{10}$ configuration (no unpaired electrons),making it diamagnetic.
$Cu^{2+}$ compounds are generally blue due to $d-d$ transitions,while $Cu^{+}$ compounds are colourless due to the filled $d$-subshell.
$Cu^{+}$ is unstable in aqueous solution and undergoes disproportionation into $Cu$ and $Cu^{2+}$.
Therefore,the statement that both $Cu^{+}$ and $Cu^{2+}$ are stable with $F^{-}$ is incorrect,as $Cu^{+}$ is generally unstable in aqueous environments.

(B) $K_2Cr_2O_7$ is produced from chromite ore,which has the chemical formula $FeO \cdot Cr_2O_3$ (also written as $FeCr_2O_4$).

(D) The extraction of chromium from chromite ore $(FeCr_2O_4)$ involves the following steps:
$1$. Fusion of chromite ore with sodium carbonate in the presence of air: $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \to 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$. Here,$A$ is $Na_2CrO_4$ (sodium chromate).
$2$. Conversion of sodium chromate to sodium dichromate by acidification with sulfuric acid: $2Na_2CrO_4 + H_2SO_4 \to Na_2Cr_2O_7 + Na_2SO_4 + H_2O$. Here,$B$ is $Na_2Cr_2O_7$ (sodium dichromate).
$3$. Sodium dichromate can be converted back to sodium chromate by adding $NaOH$: $Na_2Cr_2O_7 + 2NaOH \to 2Na_2CrO_4 + H_2O$.
Thus,both $A = Na_2CrO_4$ and $B = Na_2Cr_2O_7$ are correct.

(D) The thermal decomposition of potassium permanganate $(KMnO_4)$ occurs according to the following balanced chemical equation:
$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$
From the reaction,it is evident that $K_2MnO_4$,$MnO_2$,and $O_2$ are formed.
Therefore,$MnO$ is not formed during this process.

(C) An amphoteric oxide is a substance that can react with both acids and bases to form salt and water.
$Cr_2O_3$ is a well-known amphoteric oxide.
It reacts with acids to form $Cr^{3+}$ salts and with bases to form chromites,such as $NaCrO_2$.

(B) The thermal decomposition of potassium dichromate is given by the reaction:
$4K_2Cr_2O_7 \xrightarrow{\Delta} 4K_2CrO_4 + 2Cr_2O_3 + 3O_2$
In this reaction,$K_2CrO_4$ is yellow,$Cr_2O_3$ is green,and $O_2$ is a gas.
Since $B$ is the green-colored compound,it is $Cr_2O_3$.

(B) The stability of higher oxidation states of transition metals is determined by the nature of the ligand.
Oxygen,being a small and highly electronegative atom,has the ability to form multiple bonds (specifically $p\pi-d\pi$ back-bonding) with the transition metal.
This multiple bonding stabilizes the metal in its highest oxidation state,such as $Mn$ in $Mn_2O_7$ ($+7$ oxidation state).
In contrast,fluorine can only form single bonds with the metal,which is insufficient to stabilize the $+7$ oxidation state of $Mn$,limiting it to $MnF_4$.

(B) Vermilion is a bright red pigment,which is chemically mercury$(II)$ sulfide,$HgS$.
It has been used historically in cosmetics,as well as in traditional Ayurvedic and Yunani medicines.

(A) The starting material for the manufacture of $KMnO_4$ is pyrolusite.
Pyrolusite is a mineral consisting essentially of manganese dioxide $(MnO_2)$ and is the primary ore of manganese used in this industrial process.
Hence,option $A$ is correct.

(C) The aqueous solution of sodium chromate $(Na_2CrO_4)$ is yellow due to the presence of chromate ions $(CrO_4^{2-})$.
When $CO_2$ gas is passed through this solution,it reacts with water to form carbonic acid $(H_2CO_3)$,which provides $H^+$ ions.
The reaction is: $2Na_2CrO_4 + 2CO_2 + H_2O \rightarrow Na_2Cr_2O_7 + 2NaHCO_3$.
The formation of sodium dichromate $(Na_2Cr_2O_7)$ results in the orange-red color of the solution due to the presence of dichromate ions $(Cr_2O_7^{2-})$.

(B) In $7B$ $(OsO_4)$,the oxidation state of $Os$ is calculated as: $x + 4(-2) = 0$,so $x = +8$.
In $7C$,the osmium atom is bonded to two oxygen atoms via single bonds (from the cyclic ester linkage) and two oxygen atoms via double bonds. The oxidation state is calculated as: $x + 2(-1) + 2(-2) = 0$,which gives $x - 2 - 4 = 0$,so $x = +6$.

(A) The reaction of $MnO_2$ with $KOH$ and $KNO_3$ is: $MnO_2 + 2KOH + KNO_3 \to K_2MnO_4 + KNO_2 + H_2O$.
The product $K_2MnO_4$ (potassium manganate) is dark green in color.
In an acidic medium,the manganate ion $(MnO_4^{2-})$ undergoes disproportionation: $3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$.
The $MnO_4^-$ (permanganate) ion is dark purple in color.
Thus,$X$ is $Mn$.

(B) The stability of species in aqueous solution depends on their tendency to undergo disproportionation or oxidation-reduction reactions.
$Cr^{2+}$ is a strong reducing agent and is easily oxidized to $Cr^{3+}$ in the presence of air or water.
$Cu^{+}$ undergoes disproportionation in aqueous solution: $2Cu^{+} \to Cu^{2+} + Cu$.
$MnO_4^{3-}$ is highly unstable and disproportionates rapidly.
$MnO_4^{2-}$ (manganate ion) is stable in strongly alkaline solutions,whereas it disproportionates in neutral or acidic media. Among the given options,$MnO_4^{2-}$ is the most stable species in specific aqueous conditions (alkaline medium).

(C) ruby is a crystal of alumina,aluminum oxide $(Al_2O_3)$,containing a trace of chromium $(III)$ ions replacing some of the aluminum ions.
In ruby,each $Al^{3+}$ ion and $Cr^{3+}$ ion is surrounded by six oxide ions in an octahedral arrangement.
The origin of the color of emeralds is similar to that of the color of rubies.
However,the bulk of an emerald crystal is composed of beryl,beryllium aluminum silicate $(Be_3Al_2(SiO_3)_6)$,instead of the alumina which forms rubies.
The color is produced by chromium $(III)$ ions,which replace some of the aluminum ions in the crystal.
In emeralds,the $Cr^{3+}$ is surrounded by six silicate ions,rather than the six oxide ions in ruby.
Therefore,both ruby and emerald derive their color from the presence of $Cr^{3+}$ ions.

(A, B) $Na_2Cr_2O_7$ is deliquescent and is $NOT$ used as a primary standard in volumetry,whereas $K_2Cr_2O_7$ is used as a primary standard.
$Na_2Cr_2O_7$ is more soluble than $K_2Cr_2O_7$ due to the higher hydration energy of $Na^+$ ions compared to $K^+$ ions.
$CrO_4^{2-}$ has a tetrahedral structure.
$Cr_2O_7^{2-}$ contains a $Cr-O-Cr$ bridge bond.

(C) When potassium dichromate $(K_2Cr_2O_7)$ is heated with concentrated sulphuric acid $(H_2SO_4)$ and a soluble chloride (like $NaCl$),it produces reddish-brown vapours of chromyl chloride $(CrO_2Cl_2)$.
This is known as the chromyl chloride test.
The reaction is: $K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \rightarrow 2KHSO_4 + 4NaHSO_4 + 2CrO_2Cl_2 + 3H_2O$.

(A) $Cu_2O$ (cuprous oxide) is red in colour,not colourless. Therefore,the statement '$Cu_2O$ is colourless' is incorrect.

(B) The given reactions represent the chemistry of manganese compounds:
$1$. $2 MnO_2 + 4 KOH + O_2 \rightarrow 2 K_2MnO_4 (Green) + 2 H_2O$. Thus,$A = MnO_2$ and $B = K_2MnO_4$.
$2$. $3 K_2MnO_4 + 4 HCl \rightarrow 2 KMnO_4 (Purple) + MnO_2 + 2 H_2O$. Thus,$C = KMnO_4$.
$3$. $2 KMnO_4 + H_2O + KI \rightarrow 2 MnO_2 + 2 KOH + KIO_3$. Thus,$D = KIO_3$.
Comparing these,$A = MnO_2$ and $D = KIO_3$.