Compounds of Transitional elements Questions in English

(B) The structure of the dichromate dianion $(Cr_2O_7^{2-})$ consists of two $CrO_4$ tetrahedra sharing a common oxygen atom.
In this structure,there are $6$ terminal $Cr-O$ bonds (three on each chromium atom) and $2$ bridging $Cr-O-Cr$ bonds.
The $6$ terminal $Cr-O$ bonds are equivalent due to resonance,while the $2$ bridging bonds are different from the terminal ones.
Therefore,$6$ $Cr-O$ bonds are equivalent.

(A) When $MnO_2$ is fused with $KOH$ in the presence of air (or an oxidizing agent like $KNO_3$),it forms potassium manganate $(K_2MnO_4)$.
The balanced chemical equation is: $2MnO_2 + 4KOH + O_2 \xrightarrow{\Delta} 2K_2MnO_4 + 2H_2O$.
The compound $K_2MnO_4$ is dark green or purple-green in colour.

(B) The correct option is $(B)$.
When potassium iodide $(KI)$ reacts with copper sulfate $(CuSO_4)$,it undergoes a redox reaction:
$2CuSO_4 + 4KI \to 2CuI_2 + 2K_2SO_4$
The unstable $CuI_2$ immediately decomposes to form cuprous iodide $(CuI)$ and iodine $(I_2)$:
$2CuI_2 \to 2CuI + I_2$
Thus,the overall reaction is: $2CuSO_4 + 4KI \to 2CuI + I_2 + 2K_2SO_4$.

(C) The borax-bead test is primarily given by transition metals that form colored metaborates upon heating with borax $(Na_2B_4O_7 \cdot 10H_2O)$.
$Cr$,$Ni$,and $Mn$ are transition metals and form colored beads.
$Pb$ $(Lead)$ is a post-transition metal and does not form a colored bead in the borax-bead test.
Therefore,the correct option is $C$.

(C) The borax bead test is primarily given by transition metals that form colored metaborates upon heating with borax $(Na_2B_4O_7 \cdot 10H_2O)$.
$Na$ is an alkali metal and its salts are colorless,meaning it does not form colored metaborates.
Therefore,$Na$ does not give the borax bead test.

(B) The borax bead test is used to identify transition metal ions. When cobalt salts are heated with borax $(Na_2B_4O_7 \cdot 10H_2O)$,they form cobalt metaborate,which is blue in color. The reaction is: $CoO + B_2O_3 \xrightarrow{\text{Heat}} Co(BO_2)_2$ (Cobalt metaborate,blue).

(B) The colors of the sulphides are as follows:
$CdS$: Yellow color
$ZnS$: White color when pure and black if impure
$PbS$ and $NiS$: Black color.

(B) $CuSO_4$ reacts with $KI$ to form $Cu_2I_2$ and $I_2$.
$CuSO_4$ reacts with $NaOH$ and glucose to form a red precipitate of $Cu_2O$.
On strong heating,$CuSO_4$ decomposes to give $CuO$ and $SO_3$.
However,$CuSO_4$ reacts with $KCl$ to form $CuCl_2$,not $Cu_2Cl_2$.
Therefore,the statement in option $B$ is incorrect.

(C) When the mixture is heated with excess $Na_2O_2$ (a strong oxidizing agent),the following reactions occur:
$1$. $Fe^{2+}$ is oxidized to $Fe^{3+}$,which then precipitates as brown $Fe(OH)_3$.
$2$. $Al^{3+}$ forms soluble $[Al(OH)_4]^-$ (aluminate) in the filtrate.
$3$. $Cr^{3+}$ (from chrome alum) is oxidized to $CrO_4^{2-}$ (chromate),which is yellow and soluble in the filtrate.
Therefore,the filtrate contains $CrO_4^{2-}$ (yellow) and $[Al(OH)_4]^-$,while the residue contains $Fe(OH)_3$ (brown).

(D) $HgCl_2$ (mercuric chloride) is known as corrosive sublimate.
It has a highly corrosive action on tissues and is extremely poisonous.
It sublimes upon heating,which is why it is historically referred to as 'corrosive sublimate'.

(B) Corundum is a naturally occurring crystalline form of aluminum oxide.
It is best known for its gem varieties,Ruby and Sapphire.
Corundum is a very hard,tough,and stable mineral.
The chemical formula of corundum is $Al_2O_3$.

(C) The transition metal is $Cr$ (Chromium).
In the $+3$ oxidation state,$Cr^{3+}$ compounds (like $Cr_2(SO_4)_3$) are typically green.
In the $+6$ oxidation state,$Cr^{6+}$ compounds (like $K_2Cr_2O_7$) are typically yellow or orange.
Therefore,the correct option is $C$.

(A) The flint used in cigarette or gas lighters is made up of an alloy known as $Misch \ metal$.
$Misch \ metal$ consists of a lanthanoid metal $(\approx \ 95\%)$,iron $(\approx \ 5\%)$,and traces of $S$,$C$,$Ca$,and $Al$.
Due to its pyrophoric nature,it produces sparks when struck,making it ideal for lighters.

(C) The steel used for making gun-barrels is known as tungsten steel,which contains $Fe$ and $W$ (Tungsten).
This alloy is chosen for its high hardness,heat resistance,and durability,which are essential for withstanding the high temperatures and pressures generated during firing.

(D) In electric bulbs,$Mo$ $(Molybdenum)$ is used as a support for the tungsten filament because of its high melting point and excellent mechanical strength at high temperatures.

(C) Among the transition metals,$Mn$ forms the maximum number of stable oxides.
The oxides of $Mn$ include:
$MnO$ (basic),$Mn_3O_4$ (amphoteric),$Mn_2O_3$ (amphoteric),$MnO_2$ (amphoteric),and $Mn_2O_7$ (acidic).

(B) $Wood's$ metal is a low-melting alloy composed of $Bi$,$Pb$,$Sn$,and $Cd$.
Its melting point is approximately $70 \ ^\circ C$.
Since the boiling point of water is $100 \ ^\circ C$,$Wood's$ metal melts in boiling water.

(A) The tip of a fountain pen nib is made from an alloy of $Osmium$ $(Os)$ and $Iridium$ $(Ir)$.
These metals are members of the platinum group and are extremely hard,corrosion-resistant,and durable,making them ideal for the tip of a pen nib to prevent wear.

(A) Ferrous sulphate $(FeSO_4 \cdot 7H_2O)$ is commonly known as green vitriol. One of its primary industrial applications is in the manufacture of blue-black ink,where it reacts with tannic acid to form a complex that provides the dark color. Therefore,the correct option is $(A)$.

(D) As fertilizer. Copper sulphate is not used as a fertilizer because it is not a primary nutrient required by plants and can be toxic in high concentrations.

(C) The compound $CoO \cdot Al_2O_3$ is chemically known as cobalt aluminate.
It is also commonly referred to as Thenard's blue,named after the French chemist Louis Jacques Thénard who discovered it.
Therefore,both $(a)$ and $(b)$ are correct.

(C) $ZnO \cdot CoO$ is known as Cobalt zincate.
It is also commonly referred to as Rinman's green.
Therefore,both $(a)$ and $(b)$ are correct.

(B) The correct answer is $B$.
When copper turnings are heated with concentrated $HCl$ and copper sulphate $(CuSO_4)$,the reaction produces cuprous chloride $(CuCl)$.
The balanced chemical equation is:
$Cu_{(s)} + CuSO_{4(aq)} + 2HCl_{(conc)} \to 2CuCl_{(s)} + H_2SO_{4(aq)}$

(A) The reaction of $Cu_2Cl_2$ (cuprous chloride) with $HCl$ in the presence of an oxidizing agent (represented as $[O]$) leads to the oxidation of $Cu(I)$ to $Cu(II)$.
The balanced chemical equation is:
$2CuCl + 2HCl + [O] \to 2CuCl_2 + H_2O$
Thus,the product formed is $CuCl_2$.

(A) When potassium dichromate $(K_2Cr_2O_7)$ reacts with an aqueous solution of sodium hydroxide $(NaOH)$,it forms potassium chromate and sodium chromate.
The chemical equation is: $K_2Cr_2O_7 + 2NaOH \rightarrow K_2CrO_4 + Na_2CrO_4 + H_2O$.
In this reaction,the orange dichromate ion $(Cr_2O_7^{2-})$ is converted into the yellow chromate ion $(CrO_4^{2-})$.

(A) Corrosive sublimate is the common name for Mercury $(II)$ chloride,which has the chemical formula $HgCl_2$.

(C) Alums are double salts with the general formula $M_2(SO_4)_3 \cdot M'_2SO_4 \cdot 24H_2O$,where $M$ is a trivalent metal and $M'$ is a monovalent metal.
$Sc^{3+}$ is a $d^0$ ion,meaning it has no unpaired electrons in its $d$-orbitals.
Since $Sc^{3+}$ has a $d^0$ configuration,it does not undergo $d-d$ transitions.
Therefore,the compound is colorless or white.

(B) is the incorrect statement.
$CuSO_4$ does not react with $KCl$ to form $Cu_2Cl_2$.
$CuSO_4$ reacts with $KI$ to form $Cu_2I_2$ and $I_2$.
$CuSO_4$ reacts with $NaOH$ and glucose (a reducing sugar) to form a red precipitate of $Cu_2O$.
On strong heating,$CuSO_4$ decomposes to form $CuO$ and $SO_3$.

(D) $Cr_2O_3$ (Chromium$(III)$ oxide) imparts a dark green color to the glass.
$MnO_2$ is used as a decolorizing agent to remove the green tint caused by iron impurities and can impart a purple or amethyst tint.
$Cu_2O$ (Cuprous oxide) typically produces a red or turquoise color depending on the concentration and conditions.
$CdS$ (Cadmium sulfide) is used to produce yellow colored glass.

(A) $AgNO_3$ is known as lunar caustic.
Silver nitrate,especially when fused and molded into sticks or small cones for use as a caustic agent,is referred to as lunar caustic.

(A) To determine the oxidation state of vanadium $(V)$ in each compound,we assign the oxidation states of other elements:
$1$. In $VOCl_3$: Let the oxidation state of $V$ be $x$. Oxygen is $-2$ and Chlorine is $-1$. Thus,$x + (-2) + 3(-1) = 0$,which gives $x - 5 = 0$,so $x = +5$.
$2$. In $VCl_4$: $x + 4(-1) = 0$,so $x = +4$.
$3$. In $VCl_3$: $x + 3(-1) = 0$,so $x = +3$.
$4$. In $VCl_2$: $x + 2(-1) = 0$,so $x = +2$.
Comparing these values,the maximum oxidation state of vanadium is $+5$,which is exhibited by $VOCl_3$.

(C) The correct answer is $(C)$.
When $K_2Cr_2O_7$ reacts with an aqueous base like $KOH$,the dichromate ion $(Cr_2O_7^{2-})$ is converted into the chromate ion $(CrO_4^{2-})$.
The chemical reaction is: $K_2Cr_2O_7 + 2KOH \to 2K_2CrO_4 + H_2O$.
The $K_2CrO_4$ (potassium chromate) formed is lemon-yellow in colour.

(B) The reaction of pyrolusite $(MnO_2)$ with $KOH$ in the presence of air (oxygen) is a fusion reaction used in the industrial preparation of potassium manganate.
The balanced chemical equation is:
$2MnO_2 + 4KOH + O_2 \to 2K_2MnO_4 + 2H_2O$
Thus,the product obtained is potassium manganate $(K_2MnO_4)$.

(A) When $Mn(OH)_2$ is heated with $PbO_2$ and concentrated $HNO_3$,the manganese is oxidized to the permanganate ion $(MnO_4^-)$.
The reaction is: $2Mn(OH)_2 + 5PbO_2 + 6HNO_3 \rightarrow 2HMnO_4 + 5Pb(NO_3)_2 + 4H_2O$.
The pink colour is due to the formation of permanganic acid $(HMnO_4)$,which dissociates to give the characteristic pink-coloured $MnO_4^-$ ions in the solution.

(A) $TiO_2$ is used as a white pigment.
It is white in color because the $Ti^{4+}$ ion has a $3d^0$ electronic configuration,meaning there are no unpaired electrons to undergo $d-d$ transitions,which would otherwise result in color.

(B) The metal oxide used to impart a blue color to glass is $CoO$ (Cobalt$(II)$ oxide).
In this compound,cobalt exists as $Co^{2+}$ ions,which are responsible for the characteristic blue color in glass.

(D) Ferrosilicon is an alloy of iron and silicon. In the steel industry,it is primarily used as an alloying agent to improve the properties of steel,such as strength,hardness,and corrosion resistance. It also acts as a deoxidizer during the steel-making process.

(D) High-temperature superconductors are typically complex ceramic materials,such as $YBa_2Cu_3O_{7-x}$,which are derived from compounds of $d$-block transition elements.

(C) To find the oxidation state of $Mn$ in each compound:
$1$. In $MnO_3$: $x + 3(-2) = 0 \implies x = +6$.
$2$. In $Mn_3O_4$: $3x + 4(-2) = 0 \implies 3x = 8 \implies x = +8/3 \approx +2.67$.
$3$. In $KMnO_4$: $1 + x + 4(-2) = 0 \implies x - 7 = 0 \implies x = +7$.
$4$. In $K_2MnO_4$: $2(1) + x + 4(-2) = 0 \implies x - 6 = 0 \implies x = +6$.
The highest oxidation state of $Mn$ is $+7$,which is achieved in $KMnO_4$.

(A) $CrO_3$ and $Mn_2O_7$ are acidic oxides because they react with water to form acids.
$CrO_3 + H_2O \to H_2CrO_4$ (Chromic acid)
$Mn_2O_7 + H_2O \to 2HMnO_4$ (Permanganic acid)
In contrast,$ZnO$ and $Al_2O_3$ are amphoteric,$CaO$ is basic,and $Na_2O$ is basic.

(A) In photography as it is sensitive towards light.

(A) $CrO_3$ is an acidic oxide. When it reacts with an aqueous base like $NaOH$,it forms a chromate salt.
The balanced chemical equation is:
$CrO_3 + 2NaOH \to Na_2CrO_4 + H_2O$
In the aqueous solution,$Na_2CrO_4$ dissociates to give $CrO_4^{2-}$ ions,which are yellow in color.

(C) The thermal decomposition of ammonium dichromate is given by the reaction:
$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$
In this reaction,$(NH_4)_2Cr_2O_7$ decomposes to form nitrogen gas,water vapor,and chromium$(III)$ oxide $(Cr_2O_3)$.
$Cr_2O_3$ is a green-coloured powder.

(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,$X$ corresponds to $2Cr_2O_3$.

(A) Guignet's green is a hydrated chromium$(III)$ oxide,which is represented by the formula $Cr_2O_3 \cdot 2H_2O$.
It is a permanent green pigment used in paints and inks.

(B) The reaction between $KI$ and $CuSO_4$ is: $2CuSO_4 + 4KI \to Cu_2I_2 + I_2 + 2K_2SO_4$.
In this reaction,$Cu^{2+}$ is reduced to $Cu^+$ (forming $Cu_2I_2$) and $I^-$ is oxidized to $I_2$.
Then,$I_2$ reacts with $Na_2S_2O_3$: $I_2 + 2Na_2S_2O_3 \to Na_2S_4O_6 + 2NaI$.
Here,$Na_2S_2O_3$ is oxidized to $Na_2S_4O_6$ and $I_2$ is reduced to $I^-$.
$CuI_2$ is not formed because $Cu^{2+}$ oxidizes $I^-$ to $I_2$ immediately,resulting in the formation of $Cu_2I_2$ (cuprous iodide).

(B) Stainless steel is an alloy primarily composed of iron $(Fe)$,chromium $(Cr)$,and nickel $(Ni)$.
The typical composition is approximately $82\% \ Fe$,$11\% \ Cr$,and $7\% \ Ni$.
Therefore,the correct option is $(B)$.

(C) When iron reacts with carbon at high temperatures,it forms iron carbide,also known as cementite,which has the chemical formula $Fe_3C$.

(A) Stainless steel is resistant to corrosion by air,moisture,and minor changes in $pH$. It is primarily composed of $Fe$ $(82\%)$,$Ni$ $(7\%)$,and $Cr$ $(11\%)$,making it ideal for automobile parts and kitchen utensils.

(C) Manganese $(Mn)$ is added to steel to act as a deoxidizer and to improve its mechanical properties.
Specifically,the presence of $Mn$ increases the tensile strength and hardness of the steel,while also enhancing its elasticity and toughness.