Compounds of Transitional elements Questions in English

(B) When lanthanoids $(Ln)$ are heated with carbon at elevated temperatures,they form carbides of the general formula $LnC_2$.
These carbides are typically ionic in nature and are known as acetylides,as they react with water to produce acetylene gas $(C_2H_2)$.

(D) The manufacturing process of $K_{2}Cr_{2}O_{7}$ from chromite ore $(FeCr_{2}O_{4})$ involves three main steps:
$1$. Conversion of chromite ore into sodium chromate:
$4FeCr_{2}O_{4} + 8Na_{2}CO_{3} + 7O_{2} \longrightarrow 2Fe_{2}O_{3} + 8CO_{2} + 8Na_{2}CrO_{4}$
$2$. Conversion of sodium chromate into sodium dichromate:
$2Na_{2}CrO_{4} + H_{2}SO_{4} \longrightarrow Na_{2}Cr_{2}O_{7} + Na_{2}SO_{4} + H_{2}O$
$3$. Conversion of sodium dichromate to potassium dichromate:
$Na_{2}Cr_{2}O_{7} + 2KCl \longrightarrow K_{2}Cr_{2}O_{7} + 2NaCl$
In the first step,$Na_{2}CrO_{4}$ (sodium chromate) is produced. This $Na_{2}CrO_{4}$ is then used as a reactant in the second step to produce $Na_{2}Cr_{2}O_{7}$.

(C) The correct answer is $C$ (Carnotite).
Carnotite is a mineral that contains vanadium. Its chemical formula is approximately $K_2(UO_2)_2(VO_4)_2 \cdot 3H_2O$.
It contains approximately $53 \% \text{ uranium}$,$12 \% \text{ vanadium}$,and trace quantities of radium.
It is a radioactive,bright-yellow,soft,and earthy mineral that serves as a significant source of uranium and vanadium.

(C) The oxidation state of chromium in $Cr_2O_3$ is $+3$.
$Cr_2O_3$ is an amphoteric oxide because it reacts with both acids and bases.
$CrO$ is basic,while $CrO_3$ is acidic in nature.

(D) $KMnO_4$ is a crystalline substance,not an amorphous one. It forms dark purple,almost black,orthorhombic crystals. Therefore,the statement that it is an amorphous substance is incorrect. The other statements are correct: it acts as an antiseptic,a strong oxidizing agent,and is used for bleaching in textile industries.

(B) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 4.9 \ BM$,we have $\sqrt{n(n+2)} = 4.9$,which implies $n(n+2) \approx 24$,so $n = 4$.
For $MnO$,$Mn$ is in $+2$ oxidation state $(3d^5)$,having $5$ unpaired electrons.
For $Mn_2O_3$,$Mn$ is in $+3$ oxidation state $(3d^4)$,having $4$ unpaired electrons.
For $Mn_2O_7$,$Mn$ is in $+7$ oxidation state $(3d^0)$,having $0$ unpaired electrons.
For $MnO_2$,$Mn$ is in $+4$ oxidation state $(3d^3)$,having $3$ unpaired electrons.
Since $Mn^{3+}$ has $4$ unpaired electrons,the oxide is $Mn_2O_3$.

(C) The reaction between potassium chromate $(K_2CrO_4)$ and dilute $H_2SO_4$ is as follows:
$2CrO_4^{2-} (aq) + 2H^+ (aq) \rightleftharpoons Cr_2O_7^{2-} (aq) + H_2O (l)$
In this reaction,the yellow chromate ion $(CrO_4^{2-})$ is converted into the orange dichromate ion $(Cr_2O_7^{2-})$.
The chromate ion is a monocentric species (containing one $Cr$ atom),while the dichromate ion is a dicentric species (containing two $Cr$ atoms).
Therefore,the color change indicates that the monocentric complex is converted into a dicentric complex.

(D) $CeO_2$ (Cerium dioxide) is widely used as a pigment in the glass and ceramic industries. It is also used as an opacifier and in the polishing of glass.

(A) The ion $MnO_4^{2-}$ (manganate ion) is green in colour.
$MnO_4^{-}$ (permanganate ion) is purple.
$Cr_2O_7^{2-}$ (dichromate ion) is orange.
$CrO_4^{2-}$ (chromate ion) is yellow.
Therefore,the correct option is $A$.

(A) The compound $K_2MnO_4$ is potassium manganate.
In this compound,manganese is in the $+6$ oxidation state.
Potassium manganate $(K_2MnO_4)$ is known for its characteristic green color.

(A) The $Cu^{2+}$ ion in aqueous solution exists as the hydrated complex $[Cu(H_2O)_6]^{2+}$.
Due to the presence of one unpaired electron in the $d$-orbital,$d-d$ transitions occur when light is absorbed.
The absorption of energy from the visible region results in the transmission of the complementary colour,which is blue.

(A) In an aqueous medium,$Cu^{2+}$ is more stable than $Cu^{+}$ because the higher negative enthalpy of hydration of $Cu^{2+}_{(aq)}$ more than compensates for the energy required to remove the second electron from $Cu^{+}$.
Therefore,$CuCl_{2}$ (containing $Cu^{2+}$) is more stable than $Cu_{2}Cl_{2}$ (containing $Cu^{+}$) in aqueous solution.

(A) Amphoteric oxides are those which react with both acids as well as bases.
$V_{2}O_{5}$ and $Cr_{2}O_{3}$ are amphoteric oxides.
$V_{2}O_{5}$ reacts with both acids and bases to form salts.
$Cr_{2}O_{3}$ also exhibits amphoteric character.

(B) The thermal decomposition of potassium permanganate $(KMnO_4)$ at $513 \ K$ is given by the following reaction:
$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$
From the reaction,it is clear that $K_2MnO_4$,$MnO_2$,and $O_2$ are produced.
Therefore,$MnO$ is not obtained upon heating potassium permanganate.

(C) The production of $ATP$ during the oxidation of glucose occurs primarily through the electron transport chain $(ETC)$ in the mitochondria.
$Fe^{3+}$ ions are essential components of cytochromes,which are heme-containing proteins that act as electron carriers in the $ETC$.
These cytochromes undergo reversible oxidation and reduction,specifically involving the transition between $Fe^{2+}$ and $Fe^{3+}$ states,to facilitate the transfer of electrons and the subsequent generation of the proton gradient required for $ATP$ synthesis.

(D) The colors of the aquated ions are as follows:
$1$. $[Ni(H_2O)_6]^{2+}$ is green.
$2$. $[Fe(H_2O)_6]^{3+}$ is yellow.
$3$. $[Mn(H_2O)_6]^{3+}$ is violet.
$4$. $[V(H_2O)_6]^{4+}$ is blue.
Thus,the correct matching is $A-V, B-III, C-I, D-II$.

(D) The alloy containing copper $(Cu)$ and zinc $(Zn)$ is known as Brass $(x)$.
The alloy containing copper $(Cu)$ and nickel $(Ni)$ is known as 'Silver' $UK$ coin $(y)$,which is a cupronickel alloy.
Therefore,$x$ is Brass and $y$ is 'Silver' $UK$ coin.

(C) The amphoteric oxide of Vanadium is $V_2O_5$,where the oxidation state of $V$ is $+5$.
$V_2O_5$ reacts with alkali $(NaOH)$ to form vanadate ion $(VO_4^{3-})$,which is the oxoion '$X$'. In $VO_4^{3-}$,the oxidation state of $V$ is $x + 4(-2) = -3$,so $x = +5$.
$V_2O_5$ reacts with acid $(HCl)$ to form the vanadyl ion $(VO_2^+)$,which is the oxoion '$Y$'. In $VO_2^+$,the oxidation state of $V$ is $x + 2(-2) = +1$,so $x = +5$.
Therefore,the oxidation states of $V$ in both '$X$' and '$Y$' are $+5$ and $+5$.

(C) $TiO_2$ is a widely used white pigment in the paint and coating industry.
$MnO_2$ is used as a depolarizer in dry battery cells (Leclanché cells).
$UK$ 'copper' coins are primarily made of copper-plated steel or cupronickel alloys,but the statement $(iii)$ as presented is often considered incorrect in the context of standard chemistry curriculum which identifies them as cupronickel or bronze alloys,not simply $Cu/Ni$ in the context of pure extraction examples.
Therefore,only $(i)$ and $(ii)$ are correctly matched.

(B) Ruby is $Al_2O_3$ in which the red colour is obtained when $Cr^{3+}$ ions replace $Al^{3+}$ ions in the octahedral sites.
Emerald is $Be_3Al_2(SiO_3)_6$ in which the green colour is obtained when $Cr^{3+}$ ions replace $Al^{3+}$ ions.

(C) The electrical conductivity of $Cu$ is approximately $1 \times 10^7 \ S \ m^{-1}$.
Among the given transition metal oxides,$ReO_3$ exhibits metallic properties and its electrical conductivity is comparable to that of copper.
Therefore,option $(C)$ is the correct answer.

(C) Statement $(3)$ is incorrect because interstitial compounds of $d$- or $f$-block metals show conductivity,which is similar to their parent metals.

(B) The acidification of chromate ions $(CrO_4^{2-})$ leads to the formation of dichromate ions $(Cr_2O_7^{2-})$,which is '$Z$'.
$2 CrO_4^{2-} + 2 H^{+} \longrightarrow Cr_2O_7^{2-} + H_2O$
In the dichromate ion $(Cr_2O_7^{2-})$,let the oxidation state of $Cr$ be $x$.
$2(x) + 7(-2) = -2$
$2x - 14 = -2$
$2x = 12$
$x = +6$
Therefore,the oxidation state of chromium in '$Z$' is $6$.

(C) The purple colored compound $(X)$ is $KMnO_4$.
Upon heating,$KMnO_4$ decomposes to liberate oxygen and forms $K_2MnO_4$ $(Y)$ and $MnO_2$ $(Z)$.
The reaction is: $2KMnO_4(X) \xrightarrow{\Delta} K_2MnO_4(Y) + MnO_2(Z) + O_2$.
Compound $MnO_2$ $(Z)$ reacts with $KOH$ in the presence of an oxidizing agent like potassium nitrate $(KNO_3)$ to form potassium manganate $(K_2MnO_4)$ $(Y)$.
The reaction is: $2MnO_2(Z) + 4KOH + O_2 \rightarrow 2K_2MnO_4(Y) + 2H_2O$.
Thus,the compounds are $X = KMnO_4, Y = K_2MnO_4, Z = MnO_2$.

(B) When potassium permanganate $(KMnO_4)$ is heated to $513 \ K$,it undergoes thermal decomposition to form potassium manganate $(K_2MnO_4)$,manganese dioxide $(MnO_2)$,and oxygen gas $(O_2)$.
The balanced chemical equation for this reaction is:
$2KMnO_4 \xrightarrow{513 \ K} K_2MnO_4 + MnO_2 + O_2$

(D) In the process of petroleum cracking,mixed oxides of lanthanoids are used as catalysts. These catalysts are particularly effective in the catalytic cracking of hydrocarbons to produce lighter,more valuable products.

(B) In reactions $I$ and $III$,dichromate acts as an oxidising reagent.
$(I)$ $Cr_2O_7^{2-} + 6Fe^{2+} + 14H^{+} \longrightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
In this reaction,$Cr^{6+}$ is reduced to $Cr^{3+}$ and $Fe^{2+}$ is oxidised to $Fe^{3+}$.
$(II)$ $Cr_2O_7^{2-} + 2OH^{-} \longrightarrow 2CrO_4^{2-} + H_2O$
In this reaction,there is no change in the oxidation state of $Cr$.
$(III)$ $Cr_2O_7^{2-} + 6I^{-} + 14H^{+} \longrightarrow 2Cr^{3+} + 3I_2 + 7H_2O$
In this reaction,$Cr^{6+}$ is reduced to $Cr^{3+}$ and $I^{-}$ is oxidised to $I_2$.
$(IV)$ $Na_2Cr_2O_7 + 2KCl \longrightarrow K_2Cr_2O_7 + 2NaCl$
In this reaction,no change in the oxidation state of $Cr$ occurs.

(C) For Manganate ion $(MnO_4^{2-})$:
$1$. It has a tetrahedral structure.
$2$. $p\pi-d\pi$ bonding is present between $Mn$ and $O$.
$3$. It is paramagnetic due to the presence of one unpaired $d$-electron in $Mn^{+6}$ $(3d^1)$.
For Permanganate ion $(MnO_4^{-})$:
$1$. It has a tetrahedral structure.
$2$. $p\pi-d\pi$ bonding is present between $Mn$ and $O$.
$3$. It is diamagnetic due to the absence of unpaired $d$-electrons in $Mn^{+7}$ $(3d^0)$.
$4$. It is a stronger oxidising agent than manganate.
Comparing these,statement $C$ is incorrect because manganate is paramagnetic while permanganate is diamagnetic.

(A) . $Co^{2+} \longrightarrow (V)$ Pink colour
$B$. $Fe^{2+} \longrightarrow (IV)$ Pale green colour
$C$. $Ni^{2+} \longrightarrow (II)$ Dark green colour
$D$. $Cu^{2+} \longrightarrow (III)$ Blue colour
Hence,the correct matching is $A-V, B-IV, C-II, D-III$.
Therefore,option $(A)$ is the correct answer.

(C) The thermal decomposition of potassium permanganate $(KMnO_4)$ at $513 \ K$ is given by the following reaction:
$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2 \uparrow$
In the products,the oxidation states of manganese are:
$K_2MnO_4$: $Mn$ is in $+6$ oxidation state.
$MnO_2$: $Mn$ is in $+4$ oxidation state.
The product with the higher oxidation state is $K_2MnO_4$ $(Mn^{+6})$.
The electronic configuration of $Mn^{+6}$ is $[Ar] 3d^1 4s^0$.
Since it has one unpaired electron,it is paramagnetic.
$K_2MnO_4$ is known to have a characteristic green colour.
Therefore,the product with the higher oxidation state $(Mn^{+6})$ is paramagnetic and green.

(B) The basicity of lanthanide hydroxides decreases as the ionic radius decreases with an increase in atomic number (lanthanide contraction).
Therefore,the basicity follows the order: $La(OH)_3 > Ce(OH)_3 > ... > Lu(OH)_3$.
Comparing $Ce(OH)_3$ and $Ce(OH)_4$,the hydroxide with the lower oxidation state of the metal $(Ce^{3+})$ is more basic than the one with the higher oxidation state $(Ce^{4+})$ because the higher charge density of $Ce^{4+}$ increases the covalent character of the $Ce-OH$ bond.
Thus,$Ce(OH)_3$ is the most basic among the given options.

(A) The lightest element of group $7$ is Manganese $(Mn)$.
The oxoanion of $Mn$ in the $+6$ oxidation state is the manganate ion,$MnO_4^{2-}$.
The potassium salt of this ion is potassium manganate,$K_2MnO_4$.
Potassium manganate $(K_2MnO_4)$ is known for its characteristic green color.

(B) The reaction between common salt $(NaCl)$,potassium dichromate $(K_{2}Cr_{2}O_{7})$,and concentrated sulfuric acid $(H_{2}SO_{4})$ is as follows:
$4 \ NaCl + K_{2}Cr_{2}O_{7} + 6 \ H_{2}SO_{4}$ $\longrightarrow 2 \ KHSO_{4} + 2 \ CrO_{2}Cl_{2} + 4 \ NaHSO_{4} + 3 \ H_{2}O$
The gas evolved is chromyl chloride $(CrO_{2}Cl_{2})$.
In $CrO_{2}Cl_{2}$,let the oxidation state of $Cr$ be $x$.
$x + 2(-2) + 2(-1) = 0$
$x - 4 - 2 = 0$
$x = +6$
Thus,the formula is $CrO_{2}Cl_{2}$ and the oxidation state is $+6$.

(B) . $Mn$ in $Mn_2O_7$ has an oxidation state of $+7$,which is its highest oxidation state. This statement is correct.
$B$. Oxygen is a small,highly electronegative atom that can form multiple bonds ($p\pi-d\pi$ bonding) with transition metals,stabilizing them in higher oxidation states. This statement is correct.
$C$. $Mn_2O_7$ is a covalent oxide,not an ionic one. This statement is incorrect.
$D$. The structure of $Mn_2O_7$ consists of two $MnO_3$ tetrahedra sharing a common oxygen atom (bridged oxygen). This statement is correct.
Therefore,statements $A, B$ and $D$ are correct.

(C) Mixed oxides are those which are formed by the combination of two simple oxides of the same metal in different oxidation states.
Analyzing the given oxides:
$1. Mn_{3}O_{4} = MnO \cdot Mn_{2}O_{3}$ (Mixed oxide)
$2. Fe_{3}O_{4} = FeO \cdot Fe_{2}O_{3}$ (Mixed oxide)
$3. Co_{3}O_{4} = CoO \cdot Co_{2}O_{3}$ (Mixed oxide)
The other oxides like $Ti_{2}O_{3}, V_{2}O_{4}, Cr_{2}O_{3},$ and $Fe_{2}O_{3}$ are simple oxides.
Therefore,there are $3$ mixed oxides in the given list.

(B) Statement $A$ is correct: Potassium dichromate $(K_2Cr_2O_7)$ is a strong oxidising agent in acidic medium and oxidises ferrous sulfate $(FeSO_4)$ to ferric sulfate $(Fe_2(SO_4)_3)$.
Statement $B$ is incorrect: Sodium dichromate $(Na_2Cr_2O_7)$ is hygroscopic in nature,meaning it absorbs moisture from the atmosphere. Therefore,it cannot be used as a primary standard. Potassium dichromate is preferred as a primary standard.
Statement $C$ is correct: Chromate $(CrO_4^{2-})$ and dichromate $(Cr_2O_7^{2-})$ ions exist in equilibrium in aqueous solution,and their interconversion depends on the $pH$ of the solution: $2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$.
Statement $D$ is correct: The structure of the dichromate ion involves two $CrO_4$ tetrahedra sharing a corner oxygen atom,and the $Cr-O-Cr$ bond angle is experimentally determined to be $126^\circ$.
Thus,statements $A$,$C$,and $D$ are correct.

(B) The correct matches are as follows:
$A$. $V_2O_5$ is used as a catalyst in the Contact Process for the industrial preparation of $H_2SO_4$ from $SO_2$. Thus,$A-III$.
$B$. $Fe$ is used as a catalyst in the Haber process for the preparation of ammonia from a $N_2/H_2$ mixture. Thus,$B-I$.
$C$. $PdCl_2$ is used as the Wacker catalyst for the oxidation of ethyne to ethanal. Thus,$C-IV$.
$D$. $Ni$ complexes are used to catalyze the polymerization of alkynes. Thus,$D-II$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.