Compounds of Transitional elements Questions in English

(D) The thermal decomposition of potassium permanganate $(KMnO_4)$ at $513 \ K$ is given by:
$2KMnO_4 (X) \xrightarrow{513 \ K} K_2MnO_4 (Y) + MnO_2 + O_2 (g)$
Thus,$X = KMnO_4$ and $Y = K_2MnO_4$.
$MnO_2$ reacts with $NaCl$ and concentrated $H_2SO_4$ to produce chlorine gas,which is a pungent gas:
$MnO_2 + 4NaCl + 4H_2SO_4 \to MnCl_2 + 4NaHSO_4 + 2H_2O + Cl_2 (Z)$
Thus,$Z = Cl_2$.
Therefore,the correct sequence is $X = KMnO_4, Y = K_2MnO_4, Z = Cl_2$.

(C) In the lanthanoid series,as we move from $La$ to $Lu$,the ionic radius decreases due to lanthanoid contraction.
As the ionic radius decreases,the covalent character of the $M-OH$ bond increases,which leads to a decrease in the basic strength of the hydroxides.
Therefore,the basic strength follows the order: $La(OH)_3 > Pr(OH)_3 > Pm(OH)_3 > Eu(OH)_3 > Lu(OH)_3$.
Among the given options,$Pr(OH)_3$ has the largest ionic radius and is therefore the most basic.

(D) Step $I$: Acidification of potassium dichromate: $K_2Cr_2O_7 + H_2SO_4 \rightarrow K_2SO_4 + H_2Cr_2O_7$
Step $II$: Reaction with hydrogen peroxide: $H_2Cr_2O_7 + 4H_2O_2 \rightarrow 2CrO_5 + 5H_2O$
Step $III$: The overall balanced chemical equation is $K_2Cr_2O_7 + H_2SO_4 + 4H_2O_2 \rightarrow 2CrO_5 + K_2SO_4 + 5H_2O$
The blue colour is due to the formation of chromium$(VI)$ peroxide,which is $CrO_5$. This compound is unstable and decomposes rapidly in aqueous solution,but it exhibits a characteristic deep blue colour.
Hence,the correct option is $D$.

(C) The intense color in $MnO^{-}_4$ arises from a ligand-to-metal charge transfer $(LMCT)$ transition.
In $MnO^{-}_4$,$Mn$ is in the $(+7)$ oxidation state,meaning it has a $d^0$ electronic configuration.
Since there are no $d$-electrons,a $d-d$ transition is impossible.
The color arises because an electron is transferred from the oxygen $2p$ orbitals to the empty $3d$ orbitals of the $Mn(+7)$ ion upon the absorption of light.
This electronic transition corresponds to the energy of yellow light,which is absorbed,resulting in the observation of the complementary color,which is intense pink or purple.

(C) The correct option is $(C)$ metal metaborates.
Explanation:
When borax $(Na_2B_4O_7 \cdot 10H_2O)$ is heated,it loses its water of crystallization and swells up. On further heating,it melts to form a transparent glassy bead consisting of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
$Na_2B_4O_7 \cdot 10H_2O \xrightarrow{\Delta} Na_2B_4O_7 + 10H_2O$
$Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3$
When this bead is heated with a transition metal salt in an oxidizing flame,the metal salt decomposes into the corresponding metal oxide,which then reacts with boric anhydride to form coloured metal metaborates,which are responsible for the characteristic colours observed in the test.
$MO + B_2O_3 \rightarrow M(BO_2)_2$ (coloured bead)

(C) When $CuCl_2$ is boiled with $Cu$ metal in the presence of concentrated $HCl$,a comproportionation reaction occurs.
The reaction is: $CuCl_2 + Cu + 2HCl \rightarrow 2H[CuCl_2]$.
The product formed is chlorocupric$(I)$ acid,which contains the complex ion $[CuCl_2]^-$.
Therefore,the correct option is $C$.

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

(D) The oxidation state of scandium $(Sc)$ is $+3$ and the oxidation state of oxygen $(O)$ is $-2$.
To form a neutral compound,the charges must be balanced.
Using the criss-cross method,the formula becomes $Sc_2O_3$.

(A) The mercurous ion does not exist as a monomeric $Hg^{+}$ ion because it would have an unpaired electron in the $6s$ orbital,making it paramagnetic.
Instead,it exists as a dimer $Hg_2^{2+}$,where two $Hg^{+}$ ions are linked by a metal-metal bond.
This pairing of electrons makes the $Hg_2^{2+}$ ion diamagnetic and colourless.
Therefore,the statement that the mercurous ion exists as $Hg^{+}$ is incorrect.

(A) Chrome yellow is $lead(II)$ chromate $(PbCrO_4)$.
It occurs naturally as the mineral crocoite.
$PbCrO_4$ is widely used as a yellow pigment in paints and coatings.
Therefore,the correct answer is $lead$ chromate.

(C) To find the oxidation state of $V$ in $VO^{2+}$: Let the oxidation state be $x$. Since the charge on oxygen is $-2$,we have $x + (-2) = +2$,which gives $x = +4$.
In $VO_2^+$,$x + 2(-2) = +1 \implies x = +5$.
In $V(OH)_4^+$,$x + 4(-1) = +1 \implies x = +5$.
In $[VO_3 \cdot OH]^{2-}$,$x + 3(-2) + (-1) = -2 \implies x - 7 = -2 \implies x = +5$.
Thus,$VO^{2+}$ is the only ion where vanadium is in the $+IV$ oxidation state.

(D) $1$. $Fe(OH)_3$ is a weaker base than $Fe(OH)_2$ because the oxidation state of $Fe$ in $Fe(OH)_3$ is $+3$,which makes it more acidic and less basic compared to $Fe(OH)_2$ where $Fe$ is in $+2$ oxidation state.
$2$. $Fe(OH)_3$ reacts with concentrated $KOH$ to form the complex $K_3[Fe(OH)_6]$,which is a known reaction.
$3$. $Fe(OH)_3$ is unstable and loses water upon heating or over time to form $Fe_2O_3 \cdot xH_2O$ and eventually $Fe_2O_3$.
$4$. Since all the statements $A$,$B$,and $C$ are chemically correct,the false statement is none of these.

(D) To find the oxidation state of the central atom in each oxoanion,we use the rule that the sum of oxidation states equals the charge on the ion.
$1$. For $SO_4^{2-}$: $x + 4(-2) = -2 \Rightarrow x = +6$. Sulfur is in Group $16$,so oxidation state $+6 \neq 16$.
$2$. For $VO_2^+$: $x + 2(-2) = +1 \Rightarrow x = +5$. Vanadium is in Group $5$,so oxidation state $+5 = 5$.
$3$. For $MnO_4^{2-}$: $x + 4(-2) = -2 \Rightarrow x = +6$. Manganese is in Group $7$,so oxidation state $+6 \neq 7$.
$4$. For $Cr_2O_7^{2-}$: $2x + 7(-2) = -2$ $\Rightarrow 2x = +12$ $\Rightarrow x = +6$. Chromium is in Group $6$,so oxidation state $+6 = 6$.
Note: Both $VO_2^+$ and $Cr_2O_7^{2-}$ satisfy the condition. However,in standard competitive chemistry contexts,$Cr_2O_7^{2-}$ is the classic example of an oxoanion where the oxidation state equals the group number $(6)$. Given the options,$D$ is the most appropriate answer.

(B) When iron reacts with carbon,it forms an interstitial compound known as cementite,which has the chemical formula $Fe_3C$.

(C) Gold $(Au)$ is a noble metal that is not affected by strong acids like concentrated $HNO_3$ or $H_2SO_4$,nor by concentrated solutions of strong alkalies like $NaOH$ or $KOH$.
$Au$ reacts to form gold$(III)$ chloride $(AuCl_3)$,which is widely used for toning in photography.
Silver $(Ag)$ forms $AgCl$,which is not $MCl_3$.
Therefore,the metal $M$ is $Au$.

(B) $1$. $CuSO_4$ reacts with $KI$ to form $Cu_2I_2$ and $I_2$ $(2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 + I_2)$. This is a correct statement.
$2$. $CuSO_4$ does not react with $KCl$ to produce $Cl_2$ because $Cl^-$ is not a strong enough reducing agent to reduce $Cu^{2+}$ to $Cu^+$. This is the wrong statement.
$3$. On strong heating,$CuSO_4 \cdot 5H_2O$ loses water and eventually decomposes to $CuO$ and $SO_3$. This is a correct statement.
$4$. The tartrate complex of copper$(II)$ (Fehling's solution) reacts with glucose (an aldehyde) in the presence of $NaOH$ to form a red precipitate of cuprous oxide $(Cu_2O)$. This is a correct statement.

(B) Commercially,sodium chromate $(Na_2CrO_4)$ is produced by roasting chromite ore $(FeCr_2O_4)$ with sodium carbonate $(Na_2CO_3)$ in the presence of excess air (oxygen).
The chemical reaction is:
$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$
This process converts the chromium in the ore into soluble sodium chromate.

(C) When $Fe(OH)_2$ is precipitated from $Fe(II)$ solutions,it initially appears as a white solid.
Upon exposure to air,it undergoes oxidation.
The dark green color is due to the presence of both $Fe(OH)_2$ and $Fe(OH)_3$.
Finally,it turns brown due to the complete oxidation of $Fe(II)$ to $Fe(III)$,forming hydrated ferric oxide,$Fe_2O_3 \cdot (H_2O)_n$ (also known as rust).

(D) $1$. $Hg$ does not form an amalgam with iron $(Fe)$,which is why iron containers are used to store mercury. Thus,statement $A$ is incorrect.
$2$. $Hg$ vapour is highly poisonous. Thus,statement $B$ is incorrect.
$3$. In mercurous compounds,mercury exists as the dimer $Hg_2^{2+}$,where each $Hg$ atom is monovalent,but the species is diatomic,not monoatomic. Thus,statement $C$ is incorrect.
$4$. Oxysalts of mercury (like $HgSO_4$,$Hg(NO_3)_2$) are thermally unstable and decompose upon heating to give $Hg$ metal,oxygen,and other products. Thus,statement $D$ is correct.

(C) $1$. Copper $(I)$ ions are unstable in aqueous solution and undergo disproportionation: $2Cu^+ (aq) \rightarrow Cu^{2+} (aq) + Cu (s)$. This statement is true.
$2$. Copper $(I)$ can be stabilized by forming insoluble complexes like $[CuCl_2]^-$ or $[Cu(CN)_2]^-$. This statement is true.
$3$. Copper $(II)$ oxide $(CuO)$ is a black powder,whereas Copper $(I)$ oxide $(Cu_2O)$ is a red powder. Therefore,the statement that Copper $(II)$ oxide is a red powder is false.

(B) Pig iron contains a high percentage of carbon $(3-4.5\%)$ and other impurities like $S, P, Si, Mn$.
To convert pig iron into steel,these impurities must be removed or reduced.
$(I)$ Oxygen or iron oxide is added to oxidize the impurities (like $C$ to $CO$,$Si$ to $SiO_2$,etc.).
$(II)$ Transition elements (like $Cr, Ni, Mn, V, W$) are added to impart specific properties such as hardness,corrosion resistance,and tensile strength to the steel.
$(III)$ Inner transition elements are generally not added for this purpose.
$(IV)$ Silica is an acidic impurity that is usually removed by adding a basic flux,not added to the steel.
Therefore,the correct additions are $(I)$ and $(II)$.

(C) In $CrO_4^{2-}$ (chromate ion),the central $Cr$ atom is $sp^3$ hybridized,and due to resonance,all four $Cr-O$ bonds are equivalent.
In $MnO_4^-$ (permanganate ion),the central $Mn$ atom is $d^3s$ hybridized,and due to resonance,all four $Mn-O$ bonds are equivalent.
In $Cr_2O_7^{2-}$ (dichromate ion),there is a $Cr-O-Cr$ bridge. The terminal $Cr-O$ bonds are not equivalent to the bridging $Cr-O$ bonds.
Therefore,only $CrO_4^{2-}$ and $MnO_4^-$ contain equivalent $M-O$ bonds.

(D) The reaction sequence is as follows:
$1$. Fusion of chromite ore $(Fe_2Cr_2O_4)$ with $Na_2CO_3$ and $O_2$ gives sodium chromate $[X]$ ($Na_2CrO_4$ where $Cr$ is $+6$).
$2$. Acidification of $[X]$ gives sodium dichromate $[Y]$ ($Na_2Cr_2O_7$ where $Cr$ is $+6$).
$3$. Reaction of $[Y]$ with $H_2O_2$ in acidic medium gives chromium pentoxide $[Z]$ ($CrO_5$ or $CrO(O_2)_2$ where $Cr$ is $+6$).
$4$. Statement $A$ is true as $Cr$ is in $+6$ oxidation state in $Na_2CrO_4$,$Na_2Cr_2O_7$,and $CrO_5$.
$5$. Statement $B$ is true as $CrO_5$ is a deep blue-violet compound that decomposes rapidly in aqueous solution into $Cr^{3+}$ and $O_2$.
$6$. Statement $C$ is true as saturated solution of dichromate $[Y]$ with conc. $H_2SO_4$ yields $CrO_3$ (chromic anhydride),which is a bright orange crystalline solid.

(D) In $CeO_2$,Cerium is in the $+4$ oxidation state $(Ce^{4+})$.
The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$. Therefore,$Ce^{4+}$ has the configuration $[Xe] 4f^0$.
$1$. Since there are no unpaired electrons in the $4f$ orbital,$CeO_2$ is diamagnetic. This statement is correct.
$2$. $Ce^{4+}$ has a strong tendency to gain an electron to return to the more stable $+3$ oxidation state $(Ce^{3+})$. Thus,it acts as a good oxidising agent. This statement is correct.
$3$. Due to the absence of $f-f$ transitions ($4f^0$ configuration),$CeO_2$ is colourless. This statement is correct.
Therefore,all three statements are correct.

(C) The reaction sequence is as follows:
$1$. $4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \xrightarrow{\Delta} 2CrO_2Cl_2(P) + 2KHSO_4 + 4NaHSO_4 + 3H_2O$. Here,$P$ is chromyl chloride $(CrO_2Cl_2)$,which appears as orange-red vapours.
$2$. $CrO_2Cl_2 + 4NaOH \to Na_2CrO_4(Q) + 2NaCl + 2H_2O$. Here,$Q$ is sodium chromate $(Na_2CrO_4)$,which forms a yellow solution.
$3$. $Na_2CrO_4 + Pb(CH_3COO)_2 \to PbCrO_4(R) + 2CH_3COONa$. Here,$R$ is lead chromate $(PbCrO_4)$,which is a yellow precipitate.
Therefore,the statement '$R$ is a yellow $ppt$' is correct.

(C) $Cr(OH)_3$ reacts with bases (hydroxide ions) to form the complex $[Cr(OH)_6]^{3-}$.
It also reacts with acids (hydrogen ions) to form the complex $[Cr(H_2O)_6]^{3+}$.
Since it reacts with both acids and bases,it is amphoteric in nature.
Thus,option $C$ is correct.

(C) The structure of the dichromate ion $(Cr_2O_7^{2-})$ consists of two $CrO_4$ tetrahedra sharing one oxygen atom at the corner.
Each chromium atom is bonded to $4$ oxygen atoms in total.
Specifically,each $Cr$ atom is linked to $3$ terminal oxygen atoms and $1$ bridging oxygen atom ($Cr-O-Cr$ linkage).

(C) To determine the color of the species,we look at the oxidation state of Vanadium $(V)$ and the presence of unpaired $d$-electrons.
$1$. In $VCl_3$,$V$ is in the $+3$ oxidation state ($3d^2$ configuration). It has unpaired electrons and is colored.
$2$. In $VOSO_4$,$V$ is in the $+4$ oxidation state ($3d^1$ configuration). It has one unpaired electron and is colored.
$3$. In $Na_3VO_4$,$V$ is in the $+5$ oxidation state ($3d^0$ configuration). There are no $d$-electrons available for $d-d$ transitions,making it colorless.
$4$. In $[V(H_2O)_6]SO_4 \cdot H_2O$,$V$ is in the $+2$ oxidation state ($3d^3$ configuration). It has unpaired electrons and is colored.
Therefore,the colorless species is $Na_3VO_4$.

(D) The aqueous solution of $FeSO_4$ is light green due to $[Fe(H_2O)_6]^{2+}$ ions.
The aqueous solution of $CrCl_3$ is green due to the formation of $[Cr(H_2O)_4Cl_2]^+$.
The aqueous solution of $NiCl_2$ is green due to $[Ni(H_2O)_6]^{2+}$ ions.
The aqueous solution of $MnCl_2$ is light pink or colourless due to $[Mn(H_2O)_6]^{2+}$ ions.
Therefore,$MnCl_2$ is the salt whose aqueous solution is not green.

(B) The thermal decomposition of $FeSO_4 \cdot 7H_2O$ occurs in steps:
$1$. Loss of water of crystallization: $2FeSO_4 \cdot 7H_2O \xrightarrow{\Delta} 2FeSO_4 + 14H_2O \uparrow$
$2$. Decomposition of anhydrous $FeSO_4$: $2FeSO_4 \xrightarrow{\Delta} Fe_2O_3 + SO_2 \uparrow + SO_3 \uparrow$
Comparing the products ($Fe_2O_3$,$SO_2$,$SO_3$,and $H_2O$ vapour) with the given options,$O_2$ is not produced.

(C) The color of acidified $K_2Cr_2O_7$ remains unchanged if the oxidation state of the transition metal,$Cr$,remains constant after the reaction.
The oxidation state of $Cr$ in $K_2Cr_2O_7$ is calculated as:
$2(+1) + 2x + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x = 12$
$x = +6$
Potassium dichromate reacts with hydrogen fluoride $(HF)$ to form potassium fluorotrioxochromate$(VI)$ and water:
$K_2Cr_2O_7 + 2HF \rightarrow 2K[CrO_3F] + H_2O$
In $K[CrO_3F]$,the oxidation state of $Cr$ is:
$(+1) + x + 3(-2) + (-1) = 0$
$1 + x - 6 - 1 = 0$
$x - 6 = 0$
$x = +6$
Since the oxidation state of $Cr$ remains $+6$ in both the reactant and the product,the color of the solution does not change. Other reagents like $H_2O_2$,$Sn^{2+}$,and $HBr$ act as reducing agents and reduce $Cr(VI)$ to $Cr(III)$,causing a color change.

(B) Step $I$: $K_2Cr_2O_7 + H_2SO_4 \rightarrow K_2SO_4 + H_2Cr_2O_7$
Step $II$: $H_2O_2 \rightarrow H_2O + [O]$
Step $III$: $H_2Cr_2O_7 + 4H_2O_2 \rightarrow 2CrO_5 + 5H_2O$
Overall reaction: $K_2Cr_2O_7 + H_2SO_4 + 4H_2O_2 \rightarrow 2CrO_5 + K_2SO_4 + 5H_2O$
The formation of $CrO_5$ (Chromium pentoxide) results in a deep blue color. $CrO_5$ has a butterfly structure with two peroxide linkages.

(A) The aqueous solution of chromate ions ($CrO_4^{2-}$,yellow) changes to dichromate ions ($Cr_2O_7^{2-}$,orange) in an acidic medium $(2CrO_4^{2-} + 2H^+ \rightarrow Cr_2O_7^{2-} + H_2O)$.
This orange dichromate solution can be reduced to green chromium$(III)$ ions $(Cr^{3+})$ using a reducing agent like $SO_2$ in an acidic medium $(Cr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O)$.
The green $Cr^{3+}$ ions can be oxidized back to yellow chromate ions $(CrO_4^{2-})$ in an alkaline medium using an oxidizing agent like $H_2O_2$ $(2Cr^{3+} + 3H_2O_2 + 10OH^- \rightarrow 2CrO_4^{2-} + 8H_2O)$.
Comparing these reactions with the given options,option $A$ correctly represents this cycle.

(C) In an acidic medium,the chromate ion ($CrO_4^{2-}$,yellow) is in equilibrium with the dichromate ion ($Cr_2O_7^{2-}$,orange).
The chemical equation for this reaction is:
$2CrO_4^{2-} (aq) + 2H^{+} (aq) \rightarrow Cr_2O_7^{2-} (aq) + H_2O (l)$

(D) The conversion of potassium manganate $(K_2MnO_4)$ to potassium permanganate $(KMnO_4)$ involves the oxidation of $MnO_4^{2-}$ to $MnO_4^-$.
$1$. Passing $CO_2$ gas: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$. This is a disproportionation reaction in acidic medium.
$2$. Passing $Cl_2$: $2MnO_4^{2-} + Cl_2 \rightarrow 2MnO_4^- + 2Cl^-$. This is an oxidation reaction.
$3$. Electrolytic oxidation: $2MnO_4^{2-} + H_2O + [O] \rightarrow 2MnO_4^- + 2OH^-$.
Since all three methods effectively convert $K_2MnO_4$ to $KMnO_4$,the correct option is $D$.

(B) The acidic nature of transition metal oxides is directly proportional to the oxidation state of the metal.
For $MnO$,the oxidation state of $Mn$ is $+2$ (basic).
For $Mn_2O_3$,the oxidation state of $Mn$ is $+3$ (basic).
For $MnO_2$,the oxidation state of $Mn$ is $+4$ (amphoteric).
For $Mn_2O_7$,the oxidation state of $Mn$ is $+7$.
Since $Mn_2O_7$ has the highest oxidation state,it is acidic in nature.

(C) Railway tracks are subjected to heavy loads and require high durability and resistance to wear and tear.
$Ni$ steel (Nickel steel) is used for making railway tracks because it possesses high tensile strength and toughness.

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

(B) $CrO_2$ is the transition metal oxide that exhibits electrical conductivity as high as that of metals.
It is used in making magnetic tapes.

(C) The compound $ReO_3$ (rhenium trioxide) exhibits metallic properties.
It has a conductivity and appearance similar to that of copper metal.
Therefore,the correct option is $C$.

(C) The borax bead test is used to identify colored transition metal ions. $Borax$ $(Na_2B_4O_7 \cdot 10H_2O)$ on heating forms a transparent glassy bead consisting of $NaBO_2$ and $B_2O_3$. Transition metal oxides react with this bead to form colored metaborates. Since $Sodium$ $(Na)$ is an alkali metal and does not form colored metaborates,it does not give the borax bead test. $Chromium$,$Ferrous$ salts,and $Cobalt$ are transition metals/ions that form characteristic colored beads.

(C) Nitric oxide $(NO)$ is a neutral oxide. It reacts with an aqueous solution of ferrous sulfate $(FeSO_4)$ to form a brown-colored complex known as nitrosoferrous sulfate. The reaction is: $Fe^{2+}(aq) + NO(g) + H_2O(l) \rightarrow [Fe(H_2O)_5(NO)]^{2+}(aq)$. This complex formation allows the aqueous solution of ferrous sulfate to absorb $NO$ gas in significant amounts.

(A) Borax $(Na_2B_4O_7 \cdot 10H_2O)$ when heated,loses water and swells up,then melts to form a transparent glassy bead consisting of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
$Na_2B_4O_7 \cdot 10H_2O$ $\xrightarrow{\Delta} Na_2B_4O_7$ $\xrightarrow{\Delta} 2NaBO_2 + B_2O_3$
When this glassy bead is heated with coloured transition metal salts,it forms coloured metaborates,which are characteristic of the metal cation. Thus,the borax bead test is used to identify coloured cations.
Since the formation of the transparent bead and its subsequent reaction with cations to form coloured beads explains why the test works,the Reason is the correct explanation of the Assertion.

(B) $Co(II)$ compounds,specifically cobalt oxide $(CoO)$,are widely used in the glass industry to impart a characteristic deep blue colour to the glass. This is a well-known property of cobalt ions in silicate matrices.

(A) In $KMnO_4$,Manganese $(Mn)$ is in the $+7$ oxidation state with a $d^0$ electronic configuration.
Since there are no $d$ electrons,$d-d$ transitions are not possible.
The intense purple color arises from Ligand-to-Metal Charge Transfer $(LMCT)$,where an electron is promoted from the $O^{2-}$ ligand to the empty $d$-orbitals of the $Mn^{7+}$ center.
Therefore,the transition is $L \to M$ charge transfer.

(B) The $Na_2CrO_4$ salt contains the chromate ion,$CrO_4^{2-}$.
In $CrO_4^{2-}$,the oxidation state of $Cr$ is $+VI$,which means its electronic configuration is $[Ar] 3d^0$.
Since there are no $d-$electrons,$d-d$ transitions are not possible.
The intense yellow colour of $Na_2CrO_4$ is due to charge transfer from oxygen to the metal $(Cr)$ atom.
Therefore,both the Assertion and the Reason are correct,but the Reason is not the correct explanation for the colour,as the colour arises from charge transfer,not the oxidation state itself.

(C) $K_3[Cu(CN)_4]$: $Cu$ is in $+1$ oxidation state $(3d^{10})$,so it has no unpaired electrons,making it diamagnetic and colourless.
$(NH_4)_2[TiCl_6]$: $Ti$ is in $+4$ oxidation state $(3d^0)$,so it has no unpaired electrons,making it diamagnetic and colourless.
$VOSO_4$: $V$ is in $+4$ oxidation state $(3d^1)$,so it has one unpaired electron,making it paramagnetic and coloured.
$K_2Cr_2O_7$: $Cr$ is in $+6$ oxidation state $(3d^0)$,so it has no unpaired electrons. It is diamagnetic,but appears coloured due to ligand-to-metal charge transfer $(LMCT)$.

(A) $MnO_{4}^{2-}$ (manganate ion) and $MnO_{4}^{-}$ (permanganate ion) both have a tetrahedral geometry.
In these ions,the manganese atom is in a high oxidation state and uses its $d-$orbitals to form $p\pi-d\pi$ bonds with the oxygen atoms.
Specifically,the $\pi-$bonding involves the overlap of filled $2p-$orbitals of oxygen with empty $3d-$orbitals of manganese.