Compounds of Transitional elements Questions in English

(D) The presence of $Cr$ (chromium) in steel forms a thin,protective oxide layer on the surface,which makes the steel resistant to chemical action,particularly oxidation. This property makes stainless steel highly useful for manufacturing cutlery and various kitchen appliances.

(C) The addition of transition metals like $Cr$,$Mn$,$W$,and $Ni$ to iron (to form alloy steels) significantly improves the mechanical properties of the steel,such as hardness,corrosion resistance,and tensile strength.
Therefore,it makes the steel more useful and alters its original properties.

(C) Stainless steel is an alloy of iron with chromium and nickel. The addition of titanium $(Ti)$ to steel helps in preventing intergranular corrosion by forming stable titanium carbides,which prevents the depletion of chromium at the grain boundaries. Therefore,the non-corrosive character is more prominent in $Ti$ steel.

(C) Vanadium is a transition metal that acts as a grain refiner when added to steel.
It forms carbides and nitrides,which significantly increase the hardness of the steel.
Additionally,it improves the tensile strength and toughness of the alloy.
Therefore,both hardness and tensile properties are enhanced.

(A) Stainless steel is an alloy of iron,chromium,and nickel.
It typically contains approximately $12-20$ $\%$ of chromium $(Cr)$ to provide corrosion resistance.
Among the given options,$14$ $\%$ is the most appropriate value.

(A) Steel is an alloy of iron and carbon. In the context of industrial production in India,specific alloying elements like Manganese $(Mn)$ and Chromium $(Cr)$ are commonly added to enhance the properties of steel,such as hardness,corrosion resistance,and tensile strength.

(B) Paris green is a chemical compound with the formula $Cu(CH_3COO)_2 \cdot 3Cu(AsO_2)_2$.
It is chemically known as copper$(II)$ acetoarsenite.
It is a double salt consisting of copper acetate and copper arsinite.

(C) Brass is an alloy of copper $(Cu)$ and zinc $(Zn)$.
German silver is an alloy of copper $(Cu)$,zinc $(Zn)$,and nickel $(Ni)$.
Therefore,the correct composition is $(i)$ zinc and copper; and $(ii)$ zinc,copper and nickel.

(B) $CuSO_{4}$ reacts with $KI$ to form $Cu_{2}I_{2}$ and $I_{2}$.
$CuSO_{4}$ reacts with $KCl$ to form $CuCl_{2}$ and $K_{2}SO_{4}$,not $Cu_{2}Cl_{2}$.
$CuSO_{4}$ reacts with $NaOH$ and glucose (a reducing agent) to form red precipitate of $Cu_{2}O$.
On strong heating,$CuSO_{4}$ decomposes to form $CuO$ and $SO_{3}$.
Therefore,the statement that it reacts with $KCl$ to give $Cu_{2}Cl_{2}$ is incorrect.

(A) Silver nitrate $(AgNO_3)$ is known as lunar caustic when in the fused state.
It is called lunar caustic because silver was called $luna$ by the ancient alchemists,who believed that silver was associated with the moon.

(B) In $CrO_2Cl_2$,the oxidation state of $Cr$ is $x + 2(-2) + 2(-1) = 0$,so $x = +6$. This is the group oxidation state of $Cr$ $(3d^5 4s^1)$.
In $MnO_4^-$,the oxidation state of $Mn$ is $x + 4(-2) = -1$,so $x = +7$. This is the group oxidation state of $Mn$ $(3d^5 4s^2)$.
Both $Cr$ and $Mn$ are in their highest possible oxidation states in these compounds.

(A) The borax bead test is used to identify transition metal ions. When borax $(Na_2B_4O_7 \cdot 10H_2O)$ is heated,it loses water and forms a glassy bead 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 cobalt oxide $(CoO)$,it forms cobalt metaborate,which is blue in color.
$CoO + B_2O_3 \rightarrow Co(BO_2)_2$ (Cobalt metaborate,blue bead).

(B) The borax bead test is used to identify transition metal ions. When cobalt salts are heated with borax,they form cobalt metaborate,which is blue in color.
$CoO + B_2O_3 \xrightarrow{\text{Heat}} Co(BO_2)_2$ (Cobalt metaborate,blue bead).

(C) Chromic acid $(H_2CrO_4)$ is formed when potassium dichromate $(K_2Cr_2O_7)$ reacts with concentrated sulfuric acid $(H_2SO_4)$.
The chemical reaction is: $K_2Cr_2O_7 + 2H_2SO_4 ightarrow 2CrO_3 + 2KHSO_4 + H_2O$.
$CrO_3$ (chromium trioxide) reacts with water to form chromic acid $(H_2CrO_4)$.

(B) The borax bead test is used to identify transition metal ions.
When borax $(Na_2B_4O_7 \cdot 10H_2O)$ is heated,it forms a glassy bead of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
When this bead is heated with cobalt salts,it forms cobalt metaborate,which has a characteristic blue color.
The reaction is: $CoO + B_2O_3 \rightarrow Co(BO_2)_2$ (blue bead).

(C) In a basic medium,the dichromate ion $(Cr_2O_7^{2-})$ is converted into the chromate ion $(CrO_4^{2-})$.
The chemical reaction is as follows:
$K_2Cr_2O_7 + 2KOH \rightleftharpoons 2K_2CrO_4 + H_2O$
The formation of potassium chromate $(K_2CrO_4)$,which is lemon yellow in color,causes the color change.

(C) Lanthanoid oxides are used in the manufacture of high refractive index optical glass for cameras and lenses. $CeO_2$ is specifically used in the glass industry for polishing and as a decolorizing agent,but the general class of compounds used for high refractive index glass is lanthanoid oxides.

(A) Potassium dichromate $(K_2Cr_2O_7)$ is widely used in the leather industry for chrome tanning. It is also used as an oxidant in the preparation of many azo compounds and as a volumetric reagent in analytical chemistry.

(B) Shape memory alloys are materials that can return to their original shape after being deformed. $Nitinol$,an alloy of $Nickel$ and $Titanium$,is a well-known shape memory alloy used in various applications like medical stents and actuators.

(B) When $Mn^{II}$ salt is fused with $KNO_3$ and $NaOH$,it undergoes oxidation to form a manganate salt.
The reaction is: $Mn^{2+} + 2NO_3^- + 4OH^- \rightarrow MnO_4^{2-} + 2NO_2^- + 2H_2O$.
In the product $MnO_4^{2-}$ (manganate ion),let the oxidation state of $Mn$ be $x$.
$x + 4(-2) = -2 \implies x - 8 = -2 \implies x = +6$.
Thus,the oxidation state of $Mn$ changes from $+2$ to $+6$.

(C) When an acidic solution of chromate or dichromate (chromic acid) reacts with hydrogen peroxide $(H_2O_2)$,it forms chromium pentoxide $(CrO_5)$,which is a deep blue colored compound.
The chemical reaction is:
$Cr_2O_7^{2-} + 2H^+ + 4H_2O_2 \rightarrow 2CrO_5 + 5H_2O$
Thus,the correct product is $CrO_5$ and water.

(A) The fusion of manganese dioxide $(MnO_2)$ with potassium hydroxide $(KOH)$ in the presence of air or an oxidizing agent like potassium nitrate $(KNO_3)$ results in the formation of potassium manganate $(K_2MnO_4)$.
The chemical reaction is: $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$.
Potassium manganate $(K_2MnO_4)$ is dark green in color.

(D) The colors of the given transition metal compounds are as follows:
$(1) \ NiCl_2 \cdot 6H_2O$ is green in color.
$(2) \ Co(NO_3)_2 \cdot 6H_2O$ is pink in color.
$(3) \ FeCl_3$ is yellow in color.
$(4) \ CuSO_4 \cdot 5H_2O$ is blue in color.
Therefore, the correct matching is: $(1$ $\rightarrow d, 2$ $\rightarrow a, 3$ $\rightarrow e, 4$ $\rightarrow c)$.

(C) Potassium permanganate $(KMnO_4)$ is a strong oxidizing agent.
In the textile industry,it is widely used as a bleaching agent for cotton,wool,and silk fibers.
It is also used in the tanning of leather and as a disinfectant.

(A) The structure of the dichromate ion $(Cr_2O_7^{2-})$ consists of two $CrO_4$ tetrahedra sharing a common oxygen atom ($Cr-O-Cr$ bridge).
In this structure,there is no direct $Cr-Cr$ bond,and there are no $O-O$ bonds.
Therefore,the correct statement is that it does not contain $Cr-Cr$ and $O-O$ bonds.

(C) The industrial preparation of potassium permanganate $(KMnO_4)$ involves the following steps:
$1$. Fusion of pyrolusite ore $(MnO_2)$ with $KOH$ in the presence of air $(O_2)$ to form potassium manganate $(K_2MnO_4)$:
$2MnO_2 + 4KOH + O_2 \to 2K_2MnO_4 + 2H_2O$
Here,$X = MnO_2$ and $Y = K_2MnO_4$.
$2$. The green potassium manganate $(K_2MnO_4)$ is then subjected to acidic disproportionation to form potassium permanganate $(KMnO_4)$:
$3K_2MnO_4 + 2H_2SO_4 \to 2KMnO_4 + MnO_2 + 2K_2SO_4 + 2H_2O$
Here,$Z = KMnO_4$.
Thus,$X = MnO_2, Y = K_2MnO_4, Z = KMnO_4$.

(C) The reaction between potassium dichromate $(K_2Cr_2O_7)$,potassium chloride $(KCl)$,and concentrated sulfuric acid $(H_2SO_4)$ is known as the chromyl chloride test.
The balanced chemical equation is:
$K_2Cr_2O_7 + 4KCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 6KHSO_4 + 3H_2O$.
The compound $CrO_2Cl_2$ (chromyl chloride) is produced as deep red vapors.

(B) When potassium chromate $(K_2CrO_4)$ reacts with dilute nitric acid $(HNO_3)$,the chromate ion $(CrO_4^{2-})$ is converted into the dichromate ion $(Cr_2O_7^{2-})$.
The chemical equation is:
$2K_2CrO_4 + 2HNO_3 \rightarrow K_2Cr_2O_7 + 2KNO_3 + H_2O$
Thus,$Cr_2O_7^{2-}$ and $H_2O$ are formed.

(A) Potassium dichromate $(K_2Cr_2O_7)$ is a powerful oxidizing agent in acidic medium.
It is widely used in volumetric analysis to oxidize ferrous ions $(Fe^{2+})$ to ferric ions $(Fe^{3+})$ according to the reaction:
$Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$.

(D) Cerium dioxide $(CeO_2)$ is widely used in the glass industry and as an abrasive in polishing glass. It is also used in the production of pigments and as an opacifier in enamels.

(B) $K_2Cr_2O_7$ is a primary standard and is preferred over $Na_2Cr_2O_7$ in volumetric analysis because $Na_2Cr_2O_7$ is deliquescent (absorbs moisture from the air),making it difficult to weigh accurately.
Therefore,the statement that $Na_2Cr_2O_7$ is more useful than $K_2Cr_2O_7$ is incorrect.

(C) The thermal decomposition of potassium dichromate $(K_2Cr_2O_7)$ is given by the following balanced chemical equation:
$4K_2Cr_2O_7 \xrightarrow{\Delta} 4K_2CrO_4 + 3O_2 + 2Cr_2O_3$
Comparing this with the given reaction,the species $X$ corresponds to $2Cr_2O_3$. Therefore,$X$ is $Cr_2O_3$.

(D) The common oxidation states of $Ti$ are $+2, +3$,and $+4$.
In the compound $K_2TiO_4$,the oxidation state of $Ti$ is calculated as: $2(+1) + x + 4(-2) = 0$,which gives $x = +6$.
Since $Ti$ $(Z = 22)$ has the electronic configuration $[Ar] 3d^2 4s^2$,it can lose a maximum of $4$ electrons.
Therefore,an oxidation state of $+6$ is not possible for $Ti$,and $K_2TiO_4$ does not exist.

(A) The fusion of manganese dioxide $(MnO_2)$ with potassium hydroxide $(KOH)$ in the presence of an oxidizing agent like potassium nitrate $(KNO_3)$ or air leads to the formation of potassium manganate $(K_2MnO_4)$:
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$
Potassium manganate $(K_2MnO_4)$ is dark green in color.
Therefore,the correct option is $A$.

(B) Bronze is an alloy consisting primarily of copper,usually with tin as the main additive. It is widely used for making coins,medals,and statues due to its durability and resistance to corrosion.

(B) The nature of metal oxides depends on the oxidation state of the metal.
$MnO$ is basic,$Mn_2O_3$ is basic,$MnO_2$ is amphoteric,and $Mn_2O_7$ is acidic.
Therefore,$MnO_2$ acts as an amphoteric oxide.

(A) The acidic or basic nature of transition metal oxides depends on the oxidation state of the metal.
$(i)$ Oxides with lower oxidation states are generally basic.
(ii) Oxides with intermediate oxidation states are often amphoteric.
(iii) Oxides with higher oxidation states are acidic.
In $Cr_2O_3$,the oxidation state of $Cr$ is $+3$. $Cr_2O_3$ is a well-known amphoteric oxide,as it reacts with both acids and bases.

(A) Silver halides (like $AgBr$) are used in photography because they are photosensitive. When exposed to light,they undergo a photochemical reaction to form metallic silver,which creates the image.

(C) Potassium dichromate $(K_2Cr_2O_7)$ is prepared from chromite ore $(FeCr_2O_4)$.
The process involves the following steps:
$1$. Fusion of chromite ore with sodium carbonate in the presence of excess air: $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$.
$2$. The yellow solution of sodium chromate is filtered and acidified with sulfuric acid to give a solution from which orange sodium dichromate $(Na_2Cr_2O_7 \cdot 2H_2O)$ can be crystallized.
$3$. Sodium dichromate is more soluble than potassium dichromate. The latter is prepared by treating the solution of sodium dichromate with potassium chloride: $Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl$.

(A) The fusion of manganese dioxide $(MnO_2)$ with potassium hydroxide $(KOH)$ in the presence of air (oxidizing atmosphere) results in the formation of potassium manganate $(K_2MnO_4)$.
The chemical equation is: $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$.
Potassium manganate $(K_2MnO_4)$ is a dark green colored compound.

(C) The most stable oxidation state of chromium is $+3$,which forms the most stable compounds known as chromic compounds. This is due to the stable $t_{2g}^3$ electronic configuration in an octahedral field.

(A) The compositions of the given alloys are as follows:
$(1)$ Brass: $Cu$ $(70\%)$,$Zn$ $(30\%)$ $\rightarrow$ $(d)$
$(2)$ Bronze: $Cu$ $(90\%)$,$Sn$ $(10\%)$ $\rightarrow$ $(c)$
$(3)$ Cupronickel: $Cu$ $(75-85\%)$,$Ni$ $(15-25\%)$ $\rightarrow$ $(e)$
$(4)$ Nichrome: $Ni$ $(60\%)$,$Cr$ $(40\%)$ $\rightarrow$ $(a)$
Therefore,the correct matching is $1$ $\rightarrow d, 2$ $\rightarrow c, 3$ $\rightarrow e, 4$ $\rightarrow a$.

(B) The fusion of manganese dioxide $(MnO_2)$ with potassium hydroxide $(KOH)$ in the presence of air (or an oxidizing agent like $KNO_3$) results in the formation of potassium manganate $(K_2MnO_4)$.
The chemical equation is: $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$.
Potassium manganate $(K_2MnO_4)$ is known for its characteristic dark green color.

(D) The process of adiabatic demagnetization is used to achieve temperatures close to absolute zero. Gadolinium sulphate $(Gd_2(SO_4)_3 \cdot 8H_2O)$ is commonly used in this process due to its high magnetic susceptibility and paramagnetic properties at low temperatures.

(B) The formula for magnetic moment is $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons.
Given $\mu = 4.9 \ BM$,we have $\sqrt{n(n+2)} = 4.9$.
Squaring both sides,$n(n+2) \approx 24$,which gives $n = 4$.
In $Mn_2O_3$,the oxidation state of $Mn$ is $+3$.
The electronic configuration of $Mn^{3+}$ is $[Ar] \ 3d^4$.
Since there are $4$ unpaired electrons,the magnetic moment is $\sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \ BM$.
Therefore,the oxide is $Mn_2O_3$.

(C) The reactions represent the interconversion between sodium chromate and sodium dichromate.
$1$. In the first reaction: $2Na_2CrO_4 + 2H^{+} \rightarrow Na_2Cr_2O_7 + 2Na^{+} + H_2O$. Here,$X = Na_2CrO_4$ and $Y = Na_2Cr_2O_7$.
$2$. In the second reaction: $Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl$. This confirms $Y = Na_2Cr_2O_7$.
Therefore,$X = Na_2CrO_4$ and $Y = Na_2Cr_2O_7$.

(A) $KMnO_4$ is prepared from the mineral pyrolusite,which is $MnO_2$.
The process involves fusing $MnO_2$ with an alkali metal hydroxide (like $KOH$) in the presence of air or an oxidizing agent like $KNO_3$ to form potassium manganate $(K_2MnO_4)$,which is then oxidized to potassium permanganate $(KMnO_4)$.

(A) $CrO_3$ (Chromium trioxide) is an acidic oxide.
In higher oxidation states,transition metal oxides exhibit acidic character.
Since the oxidation state of $Cr$ in $CrO_3$ is $+6$,it behaves as an acidic oxide and dissolves in water to form chromic acid $(H_2CrO_4)$.

(B) Dental amalgam is a liquid alloy of mercury with various metals.
The standard composition of dental amalgam alloy is approximately $Hg(50\%)$,$Ag(35\%)$,$Sn(12\%)$,$Cu(3\%)$,and sometimes a small amount of $Zn(0.2\%)$.
Option $B$ represents the most accurate composition including the trace element $Zn$.

(A) The alloy known for its shape memory property is $Nitinol$, which is an alloy of $Nickel$ and $Titanium$.
Its composition is approximately $55\% \, Ni$ and $45\% \, Ti$ by weight.
Therefore, the correct option is $A$ (noting that the provided option $A$ in the input was corrected to reflect the standard composition of $Nitinol$).