Q: Give Reasons(i) Copper formed by auto reduction of Cu2S is blister copper.(ii) Anodes should be repeatedly replaced during aluminium extraction.(iii) Gold is soluble in aqua regia, but silver is not.

(i) Copper obtained by self reduction of Cu2S contains SO2 gas. The copper obtained is called blister copper due to the blistered appearance produced by SO2.2Cu2 S+3O2⟶2Cu2O+2SO22Cu2O+Cu2 S⟶6Cu+SO2(ii) Aluminum is extracted by electrolytic reduction of Al2O3.Graphite lined steel tank is the cathode and graphite rods are the anode.During the process, Oxygen is liberated at the anode which reacts with the carbon rod giving CO2 gas. Hence the anode rods have to be replaced regularly.(iii) Gold dissolves in aqua regia due to formation of chloroauric acid.3HNO3+11HCl+2Au⟶2HAuCl4+6H2O+ 3NOCl−Whereas, silver remains insoluble due to the formation of insoluble AgCl on reaction with aqua regia.

Q: KMnO4 can be prepared by fusing together equimolar amounts of MnO2, KOH and KNO3, collecting the dark green solidified melt, which is K2MnO4, crushing it and treating it with dil.H2SO4. On adding dil.H2SO4 a solution of KMnO4 and a precipitate of MnO2 are formed. Write the relevant equations for the process.

The reaction taking place on fusion is3MnO2+4KOH+2KNO3→3 K2MnO4+2NO+2H2OK2MnO4 undergoes disproportionation in dil. H2SO43MnO42−+4H+→2MnO4−+MnO2+2H2O

Q: Give the electronic configuration of the metal in(i) FeO42−(ii) Cr(OH)3(iii) VCl5(iv) [Ni(H2O)6]2+Give their calculated magnetic moment.

(i) FeO42−Fe is in the +6 state[Ar]3d2μ=2(2+2)=8=2.84 BM(ii) Cr(OH)3Cr is in +3 state[Ar]3d3μ=3(3+2)=15=3.87 BM(iii) VCl5Vanadium is in +5 state[Ar]3d0μ=0(iv) [Ni(H2O)6]2+Nickel is in +2 state[Ar]3d8μ=2(2+2)=8=2.84 BM

Question 1

With example and reason, give the set of metals that exist in combined state and native state comment on their standard electrode potentials.

Accepted Answer

Metals in their free or native state have little or no affinity for oxygen and are resistant to attack by water and other chemical reagents. These are called noble metals.
e.g., gold, silver, platinum, etc. They occur as lumps of pure metal called nuggets. They have a positive standard reduction potential, i.e., their ions are easily reduced to the metal.
Metals that have a proclivity to be converted to ions and thus form compounds exist in nature as oxides, sulfides, and so on. These metals are called active metals and occur as compounds called minerals, e.g., Na, Fe, Cu, etc.
They have a negative standard-reduction potential.  They are easily oxidised to their ions or their ions do not have any tendency to be reduced to the metallic state.

Question 2

What is the principle behind Leaching? Explain with example.

Accepted Answer

Leaching is a method of concentrating the ore. The principle behind the process is to treat powdered ore with a suitable reagent such that only the ore particles completely dissolve in the reagent leaving behind the impurities. The impurities can be filtered off.Example, Bauxite ore is treated with NaOH solution. Only Al2​O3​ dissolves in NaOH forming soluble sodium meta aluminate while the impurities remain and can be filtered off.

Question 3

Give reasons for using zinc as a suitable solvent for extracting silver from lead.(Parke’s Process).

Accepted Answer

Zinc is a suitable solvent for extracting silver from lead because(a) molten zinc and lead are immiscible.(b) silver does not react with lead.(c) alloy of zinc and silver melts at a higher temperature than lead.(d) Molten zinc floats over molten lead.(e) Due to the low boiling point of zinc, it is easy to distil out.

Question 4

Give Reasons(i) Copper formed by auto reduction of Cu2​S is blister copper.(ii) Anodes should be repeatedly replaced during aluminium extraction.(iii) Gold is soluble in aqua regia, but silver is not.

Accepted Answer

(i) Copper obtained by self reduction of Cu2​S contains SO2​ gas. The copper obtained is called blister copper due to the blistered appearance produced by SO2​.2Cu2​ S+3O2​⟶2Cu2​O+2SO2​2Cu2​O+Cu2​ S⟶6Cu+SO2​(ii) Aluminum is extracted by electrolytic reduction of Al2​O3​.Graphite lined steel tank is the cathode and graphite rods are the anode.During the process, Oxygen is liberated at the anode which reacts with the carbon rod giving CO2​ gas. Hence the anode rods have to be replaced regularly.(iii) Gold dissolves in aqua regia due to formation of chloroauric acid.3HNO3​+11HCl+2Au⟶2HAuCl4​+6H2​O+ 3NOCl−Whereas, silver remains insoluble due to the formation of insoluble AgCl on reaction with aqua regia.

Question 5

6.87 kg of Mn3​O4​ was heated with 2.16 kg of Aluminium. Comment on this reaction and calculate the amount of oxide formed as product.

Accepted Answer

The reaction is alumino thermite process and it is a highly exothermic process. Aluminum being highly electropositive, it reduces the metal oxide to the metal when heated with itThe oxide formed is Al2​O3​ and the amount of Al2​O3​ formed is 4.08 kg

Question 6

All transition metals of 3d series except Zn exhibit a minimum of two oxidation states. Explain why Zn exhibits only +2 state. Does Zn form complexes?

Accepted Answer

Zn has 3d104s2 configuration. By losing two electrons, it acquires d10 configuration, which is very stable. Hence it does not exhibit other oxidation states. Zn(II) complexes are typically either tetrahedral (Example [ZnCl4​]2– ) or octahedral (Example [Zn(H2​O)6​]2+ )

Question 7

Give reason.(i) When Cu2+ is treated with H2​S, insoluble CuS is precipitated. But this method is not used to estimate Cu gravimetrically.(ii) K2​Cr2​O7​ is used as primary standard but not Na2​Cr2​O7​.(iii) NaI is not coloured but AgI is coloured.(iv) ZnSO4​ is white but KMnO4​ is pink.

Accepted Answer

(i) In addition to CuS, oxide of Cu is also produced.Heating of CuS will lead to its decomposition. Hence quantitative estimation by CuS is not possible.(ii) Na2​Cr2​O7​ is hygroscopic. Na2​Cr2​O7​ is hygroscopic. So it cannot be used as a primary standard.(iii) AgI is coloured because of polarisation.AgI is coloured because the small Ag+polarizes the large iodide ion.(iv) In ZnSO4​ there is no d−d transition. Colour of KMnO4​ is due to charge transfer.ZnSO4​ is a Zn2+ salt and has 3d10 configuration. There is no possibility of d−d transition. Since KMnO4​ contains Mn7+ which has a d0 configuration, no d−d transition is possible. The colour is due to charge transfer by which electrons are excited from oxygen to Mn7+.

Question 8

KMnO4​ can be prepared by fusing together equimolar amounts of MnO2​, KOH and KNO3​, collecting the dark green solidified melt, which is K2​MnO4​, crushing it and treating it with dil.H2​SO4​. On adding dil.H2​SO4​ a solution of KMnO4​ and a precipitate of MnO2​ are formed. Write the relevant equations for the process.

Accepted Answer

The reaction taking place on fusion is3MnO2​+4KOH+2KNO3​→3 K2​MnO4​+2NO+2H2​OK2​MnO4​ undergoes disproportionation in dil.  H2​SO4​3MnO42−​+4H+→2MnO4−​+MnO2​+2H2​O

Question 9

Give the electronic configuration of the metal in(i) FeO42−​(ii) Cr(OH)3​(iii) VCl5​(iv) [Ni(H2​O)6​]2+Give their calculated magnetic moment.

Accepted Answer

(i) FeO42−​Fe is in the +6 state[Ar]3d2μ=2(2+2)​=8​=2.84 BM(ii) Cr(OH)3​Cr is in +3 state[Ar]3d3μ=3(3+2)​=15​=3.87 BM(iii) VCl5​Vanadium is in +5 state[Ar]3d0μ=0(iv) [Ni(H2​O)6​]2+Nickel is in +2 state[Ar]3d8μ=2(2+2)​=8​=2.84 BM

Question 10

With reference to atomic size of transition metals, explain the observation that covalent radii decreases from left to right across a row in the transition series, but near the end size increases.

Accepted Answer

The increase in size towards the end of series can be explained by considering the increased force of repulsion on addition of electrons. The repulsive forces among the electrons exceeds the attractive forces from the nucleus. So there is an increase in the covalent radius towards the end of the series.

Question 11

The process of hydraulic washing for the concentration of ores is based on the:

Accepted Answer

Due to the difference in density between the ore particles and impurity particles, this separation is possible.

Question 12

Siderite ore is decomposed in excess of air, the solid product formed is

Accepted Answer

Fe2​O3​Siderite ore is Fe2​O3​4FeCO3​+O2​⟶2Fe2​O3​+4CO2​On roasting ores are converted to their oxides.

Question 13

The process that produces 99.99% pure aluminium is

Accepted Answer

Hoops process is electrolytic refining of aluminium.

Question 14

The most acidic oxide among the following is

Accepted Answer

The more the number of oxygen atoms bonded to the metal, greater is the acidity of the oxide.

Question 15

Ti3+ has same spin only magnetic moment as that of

Accepted Answer

Ti3+ has configuration d1 .Oxidation state of V in VBr4​ is +4. It is also a d1 ion.

Question 16

Statement 1:Roasting of sphalerite and cinnabar give the metal oxide and metal respectively.andStatement 2:During roasting, the ore is heated strongly in the presence of air above its melting point.

Accepted Answer

Ore is heated below its melting point during roasting.Sphalerite or Zinc blende is ZnS.Cinnabar is HgS.On roasting ZnS gives ZnO while HgS gives Hg. This is because the electrode reduction potential of Hg is less than zero.So HgO formed gets converted to Hg on heating.

Question 17

Statement 1:Impure titanium, when subjected to Van – Arkel method for purification, produces a stable volatile compound.andStatement 2:The compound is formed at a lower temperature than its decomposition temperature.

Accepted Answer

Van Arkel method of refining Titanium involves formation of stable, volatile TiI4​ which decomposes on heating.T1​+2I2​150∘ C​TiI4​1800∘ C​pureTi​+2I2(g)​

Question 18

Statement 1:Decrease in atomic radius from sodium to chlorine is greater than that from scandium to copper.andStatement 2:The increase in number of electrons on moving from sodium to chlorine is greater than from Sc to Cu.

Accepted Answer

Number of electrons added is less for Na1123​ to Cl1735​. The decrease in size is greater because of the low shielding power of the incoming d electrons into the penultimate 3d shell on moving from scandium to copper.

Question 19

Statement 1:In a transition element, the total magnetic moment of a substance is the resultant of magnetic moment of all the individual electrons.andStatement 2:Each electron behaves as a micromagnet having a certain value of magnetic moment.

Accepted Answer

Just like motion of electric charges creates a magnetic field, the motion of an electron, due to the presence of a negative charge on it creates a magnetic field. So it behaves as a micro magnet and the magnetic moment is dependent on the number of individual electrons.

Question 20

Statement 1:It is ordinarily not possible for Lanthanum to exhibit oxidation state greater than +3.andStatement 2:There are no electrons in the 4f orbital of Lanthanum.

Accepted Answer

Electronic configuration of La is [Xe]5d1 6s2 there are no electrons in 4f orbital.La3+[Xe]La in the 3+ state attains noble gas Xenon configuration so it is very difficult to remove an electron from La3+, so it is not possible to attain an oxidation state greater than 3+.V2​O5​ from V⟶(A) VX5​ will have V in d∘ configuration.So no d-d transition is possible.V+4HF⟶VF4​+2H2​2VCl4​⟶2VCl3​+Cl2​,2VF4​⟶VF5​+VF3​2VF3​⟶VF4​+VF2​VF5​+F−⟶[VF6​]−

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Solutions for IIT JEE Super Course in Chemistry: Inorganic Chemistry by Trishna Knowledge Systems

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