Chapter 8 The d and f Block Elements

X is in +3 oxidation state which means the X³⁺ has formed by the loss of 3 electrons. So the ground state electronic configuration is [Ar]3d⁶4s². The atomic number is equal to the number of electrons in a neutral atom, so it is 18+6+2=26.

Cu(II) is more stable than Cu(I) because it has a 2+ charge and is smaller than Cu(I). The electron density and effective nuclear charge is greater for Cu(II) and it forms stronger bonds(high hydration energy) and releases more energy and is more stable.

Fe,Co,Ni and Cu belong to the same period. Along the period, mass of the atom increases and the radius decreases. Density is defined as mass per unit volume. When mass increases and volume decreases, it leads to an increase in density. So, Cu which is to the extreme right of the period will have the highest density.

Electronic configuration of

Ag⁺ - [Kr] 4d¹⁰

Cu²⁺ - [Ar] 3d⁹

Zn^{2+ -} [Ar] 3d¹⁰

Cu⁺ - [Ar] 3d¹⁰ 4s¹

Among the four ions, only Cu²⁺ has an unpaired electron in the d orbital which will give rise to coloured compounds.

The following reaction happens when KMnO₄ reacts with concentrated H₂SO₄. The product formed Mn₂O₇ is a green oily compound.

2KMnO₄ + 2H₂SO₄ → Mn₂O₇ + 2KHSO₄ + H₂O

3d⁵ has the maximum number of unpaired electrons and will therefore have the highest magnetic moment.

Magnetic moment:

Where n is the number of unpaired electrons.

Putting the values in the above formula, we get

= = 5.91 BM,

Lanthanoids show +3 oxidation state more commonly. They have a general electronic configuration of [Xe] 4fⁿ 6s², where it is easier to remove the 6s electrons than 4f. Removing more than one 4f electron becomes difficult and therefore the oxidation state of +3 is more common.

A disproportionation reaction takes place when a molecule is transformed into two products, one of higher oxidation state and another of lower oxidation state than the reactant molecule. It is a typical reduction reaction.

Cu⁺→ Cu²⁺ + Cu

In this reaction, Cu⁺ with oxidation state +1 forms Cu²⁺(+2 higher oxidation state) and Cu ( 0 lower oxidation state)

Similarly in this reaction,

3MnO₄^2– + 4H⁺→ 2MnO₄^– + MnO₂ + 2H₂O

MnO₄^2– with oxidation state +6 forms two products MnO₄^– with +7(higher oxidation state) and MnO₂ with +4(lower oxidation state)

The following reaction takes place:

5C₂O₄^2– + 2MnO₄^– + 16H⁺→ 2Mn²⁺ + 8H₂O + 10CO₂

Here, KMnO₄ is a reactant in the reaction which forms Mn²⁺ which acts as an auto-catalyst. The reaction is slow initially because MnO₄^- is getting converted to Mn²⁺ but once Mn²⁺ forms, the reaction becomes faster due to the catalytic action

Thulium(Tm) belongs to the lanthanoid series and not actinoid series.

The following reaction takes place when KMnO₄ reacts with sulphide ions in acidic medium. 2 moles of KMnO₄ reacts with 5 moles of sulphide ions, so 2/5 moles of KMnO₄ reacts with 1 mole of sulphide ions.

5S^2– + 2MnO^-₄ + 16H⁺→ 2Mn²⁺ + 8H₂O + 5S

Amphoteric oxides react with both bases and acids. V₂O₅ and Cr₂O₃ are both amphoteric.

V₂O₅ reacts with alkalies as well as acids to give VO₄³⁻ and VO₄⁺ respectively.

In higher oxides, acidic character predominates and therefore V₂O₅ is slightly acidic.

Here, according to Aufbau principle electron should enter 6s, 4f and then 5d,

n+l rule for

4f= 4+3 =7

5d= 5+2=7

6s= 6+0=6

Atomic number of xenon is 54, the remaining 10 electrons are distributed into 4f, 5d and 6s. Only 7 electrons go into 4f because half filled orbital leads to stability. Now, the remaining 3 electrons, two of them go into 6s and one goes into 5d.

Compounds formed when small atoms like H,C and N get trapped inside the crystal lattice of transition metals are called interstitial compounds.

Some properties of interstitial compounds are called:

1) Compounds are very hard

2) They are chemically inert

3) Have high melting points, higher than pure metals

4) They retain metallic conductivity.

So, interstitial compounds are actually chemically inert.

Electronic configuration of Cr³⁺ is [Ar]3d³. The number of electrons that contribute towards spin only magnetic moment are 3. Spin only magnetic moment can be calculated using this formula

Where n is the number of unpaired electrons. So for Cr³⁺ , the number of unpaired electrons equals 3.

= 3.87 B.M

The following reaction takes place

When potassium permanganate reacts with potassium iodide, the iodide ion gets oxidized to iodate.

2MnO₄^– + H₂O + I^–→2MnO₂ + 2OH^– + IO₃^–

Copper is below hydrogen in the reactive series. Only a more reactive metal can displace a less reactive metal but Copper is less reactive than hydrogen and hence cannot liberate hydrogen from acids.

Cu + 2H₂SO₄→ CuSO₄ + SO₂ + 2H₂O

3Cu + 8HNO₃→ 3Cu(NO₃)₂ + 2NO + 4H₂O

The following reaction takes place when K₂Cr₂O₇ solution is added to Sn²⁺ salts

3 Sn²⁺→ 3Sn⁴⁺ + 6 e^–

Sn²⁺ gets converted to Sn⁴⁺

Fluorine is highly electronegative and form single bonds only. So, the oxidation state for fluorine is lower, but in the case of oxygen, double bond formation is possible and hence the presence of a higher oxidation state.

Due to lanthanide contraction, the f electrons shield very poorly which leads to an increase in effective nuclear charge and hence a contraction in size. So, the atomic radii of Zirconium and Hafnium are very similar and therefore they have similar physical and chemical properties.

KMnO₄ oxidises HCl to Cl₂ which is also an oxidizing agent. So,some amount of KMnO₄ gets used up in oxidizing Cl^- to Cl₂, thereby rendering the whole titration useless.

KMnO₄ + HCl → KCl + MnCl₂ + Cl₂ + H₂O

KMnO₄ in +7 oxidation state has no d electrons but it is deep purple in colour. The colour arises due to ligand to metal charge transfer where the ligand O^2- transfers charge to the metal centre.Ce (SO₄)₂ is also coloured due to charge transfer and is of intense yellow colour

Co²⁺ has electronic configuration of 3d⁷ and Cr³⁺ has electronic configuration of 3d³. Both have the same number of unpaired electrons, i.e 3.

Magnetic moment =

Where n is the number of unpaired electrons. So for Cr³⁺ and Co²⁺ , the number of unpaired electrons equals 3.

Magnetic moment = = 3.87 B.M

Higher oxidation states of heavier members of group 6 of transition series are more stable. So, Mo(VI) and W(VI) are more stable than Cr(VI). Because they are more stable, they will not get reduced(or act as oxidising agent). Cr(VI) on the other hand is not that stable and can easily get reduced to a lower oxidation state.

Pu and Np both show +7 oxidation state. U shows oxidation states +4 and +6, and Am shows oxidation states from +2 to +6.

Electronic configurations of all the elements are as follows:

U – [Rn] 5f³ 6d¹ 7s²

Np- [Rn] 5f⁴ 6d¹ 7s²

Pu-[Rn] 5f⁶ 6d⁰ 7s²

Am-[Rn] 5f⁷ 6d⁰ 7s²

So, only Uranium and Neptunium have 1 electron in 6d orbital

Electronic configurations of Eu and Yb are

Eu- [Xe] 4f⁷ 5d⁰ 6s²

Yb- [Xe] 4f¹⁴ 5d⁰ 6s²

Both show oxidation state of +2 besides the +3 oxidation state

Mn²⁺ (3d⁵) and Fe²⁺(3d⁶) have more number of unpaired electrons 5 and 4 than Ti³⁺(3d¹) and Co³⁺(3d⁷) which have only 1 and 3 unpaired electrons.

Magnetic moment =

Where n is the number of unpaired electrons.

Magnetic moment for all elements:

Mn²⁺ = =5.91

Fe²⁺= =2.89

Ti³⁺==1.73

Co³⁺==3.87

Greater the value of n, greater the magnetic moment.

So, Mn²⁺ has the highest magnetic moment

Cr and Co form compounds of the form MF₃. Fluorine stabilizes CoF₃ due to high lattice energy and CrF₃ due to high bond enthalpy.

Higher oxidation states of heavier members of group 6 of transition series are more stable. So, Mo(VI) and W(VI) are more stable than Cr(VI). Because they are more stable, they will not get reduced(or act as oxidising agent).

Electronic configuration of Ce – [Xe] 4f¹ 5d¹ 6s². Cerium losing 4 electrons makes it attain noble gas configuration which is very stable and it also attains f⁰ configuration.

A positive reduction potential means the reduced form of Cu is more stable than hydrogen. Thus, Cu is less reactive than hydrogen and cannot displace it from acids.

A negative E value means that the oxidized species is more stable than the reduced species, the metals will easily lose their electrons and get oxidized. Here, Mn²⁺(3d⁵) and Zn²⁺(3d¹⁰) have half-filled and fully filled d orbitals which gives them stability and therefore prefer to stay that way and not get reduced. As for Ni²⁺(3d⁸), it has very high negative hydration enthalpy which gets balanced by first and second ionization enthalpy.

The electronic configuration of Cr and Zn is given below

Cr – [Ar] 3d⁵ 4s¹

Zn – [Ar] 3d¹⁰ 4s²

Removing an electron from a half-filled 4s orbital requires lesser energy than removing an electron from a fully filled stable 4s orbital.

Transition elements have electrons in the d orbital which participate in metallic bonding. Metallic bonding arises from electrostatic force between conduction electrons and positively charged metal ions. Strong metallic bonding brings the atoms closer together and holds them tightly. Melting these metals involve breaking strong metallic bonds which becomes harder. So, melting points are higher for transition elements.

The following reaction takes place,

2Cu²⁺+ 4I^- → Cu₂I₂ + I₂

Cu²⁺ gets reduced to Cu⁺, and I^- gets oxidized to I₂.

CuCl₂ is more stable because Cu²⁺ has a higher electron density than Cu⁺. Cu²⁺ is smaller in size, has higher effective nuclear charge and therefore a higher hydration enthalpy ∆_hyd of Cu²⁺, which makes it more stable.

When brown co pound of manganese (A) is treated with HCl it gives chlorine gas.

MnO₂ + 4HCl → MnCl₂ + Cl₂ +2H₂O

^{(A) (B)}

The chlorine gas reacts with NH₃ to given NCl₃

3Cl₂ + NH₃ →NCl₃ + 3HCl

^{(B) (C)}

The brown compound A = MnO₂

Gas B = Cl₂

Explosive compound C= NCl₃

Electronic configuration of

F – 1s² 2s² 2p⁵

O – 1s² 2s² 2p⁴

Fluorine has one unpaired electron and forms a single bond, while Oxygen has two unpaired electrons and can form multiple bonds thereby stabilizing higher oxidation states.

The electronic configuration of

Cr³⁺ - [Ar] 3d³

Co²⁺- [Ar] 3d⁷

Cr³⁺ has a symmetrical electron distribution and will only have spin magnetic moment contribution whereas Co²⁺ has no symmetrical distribution of electrons so it will have orbital magnetic moment and spin magnetic moment contribution. Therefore, the total magnetic moment for Co²⁺ will be higher than Cr³⁺.

For Th, Pa and U, 5f electrons start filling and they have lower penetration than 4f electrons for Ce, Pr and Nd. Removal of weakly bound 5f electrons is easier than removing 4f electrons, so ionization enthalpy for Th, Pa and U is lower than Ce, Pr and Nd.

The 5f electrons will therefore, be more effectively shielded from the nuclear charge than 4f electrons of the corresponding lanthanoids. Therefore, outer electrons are less firmly held and they are available for bonding in the actinoids.

Zirconium and Hafnium do belong to the same group Lanthanide contraction due to lanthanoids contraction poor f orbital shielding leads to an increase in effective nuclear charge, which reduces the size of Hf. So the atomic radii of both Zr and Hf are similar which means they have similar physical and chemical properties and hence separation becomes difficult.

Ce – [Xe] 4f¹ 5d¹ 6s². Usually lanthanoids lose the 5d and 6s electrons and show +3 oxidation state, but Cerium loses the one 4f electron also so as to attain Xenon’s noble gas configuration which will make Ce⁴⁺ very stable.

The following reaction takes place when potassium permanganate reacts with oxalic acid. The deep purple colour of KMnO₄ disappears as MnSO₄ forms. This concept is used in redox titration

5H₂C₂O₄ + 2KMnO₄ +3H₂SO₄ ——>2MnSO₄ + 8H₂O + K₂SO₄ +10CO₂^{(deep purple) (Colourless)}

The following reaction takes place when Cr₂O₇^2–is treated with an alkali:

(orange) Cr₂O₇^2–+ OH^-→ 2CrO₄^2-(yellow)

When the yellow solution is treated with an acid, we get back the orange solution:

(yellow)2CrO₄^2- +2H⁺→Cr₂O₇^2–(orange)+ H₂O

This reaction is reversible under proper conditions.

In acidic medium, permanganate changes to manganous ion which is colourless.

MnO₄^-+8H⁺ + 5e^-→ Mn²⁺ + 4H₂O

^(colourless)

In alkaline medium, permanganate changes to manganate, which is a green solution

MnO₄^-+ e^- → MnO₄^2-

^(green)

In neutral medium, permanganate changes to manganese dioxide which is a brown precipitate,

MnO₄^-+ 2H₂O + 3e^-→ MnO₂ + 4OH^-

^(brown)

Due to poor f orbital shielding, effective nuclear charge increases and there is a contraction in size of the third row elements.This contraction in size is called Lanthanide contraction. The atomic radii of second and third rows are almost similar and therefore they resemble each other more.

Cu has electronic configuration [Ar] 3d¹⁰4s¹. Formation of Cu²⁺ involves disturbing a stable 3d¹⁰ configuration. So, the reduced form of Cu²⁺ is more stable than the oxidized form of Cu. Therefore, the value of E° is positive for Cu.

For Zinc, the electronic configuration is [Ar]3d¹⁰4s². Removing two electrons gives a stable configuration [Ar]3d¹⁰ with completely filled orbitals. So, the oxidized form is more stable than the reduced form. Therefore, the value of E° is negative for Zn

As the oxidation state increases, the charge on the atom increases and size of the ion of transition element decreases. Fajan’s rule’s states that greater the charge on an atom, greater the covalent character. So, halides become more covalent with increasing oxidation state.

Electrons are filled according to the n+l rule. If an orbital has lower n+l value, then electron will enter that orbital.

For 3d, n+l= 3+2=5

4s, n+l= 4+0=4

So, electron will first enter 4s and then 3d while filling.But, 4s electrons are loosely held by the nucleus and are outside 3d, so removing a 4s electron(ionization of the atom) becomes easier than removing a 3d electron.

When we move along a period from left to right, there is an increase in effective nuclear charge and the size reduces. The electrons are held more tightly and ionization enthalpy increases because removing the outermost electron becomes difficult. Therefore, reactivity also decreases. Sc is more reactive than Cu.

(i)-(c), (ii)-(d), (iii)-(b), (iv)-(e), (v)-(a)

(i)-(c) Hydrogenation of vegetable oil to ghee needs a catalyst and Ni in the presence of hydrogen acts as the catalyst

(ii)-(d) Copper salts are used as reagents and catalysts in Sandmeyer reaction, where diazonium salts are converted to aryl halides.

(iii)-(b) Contact process is the method used to produce sulphuric acid and V₂O₅ is used as a catalyst for the reaction.

(iv)-(e) Haber’s process is used in the manufacturing of ammonia. Finely divided iron is used as a catalyst for the reaction.

(v)-(a) Ziegler Natta catalyst is used in the synthesis of polymers from 1-alkenes. It usually consists of TiCl₄ with an aluminium based co-catalyst.

(i)-(b), (ii)-(a), (iii)-(d), (iv)-(e), (v)-(c)

(i)-(b) Lanthanoid oxides are used in television screens.

(ii)-(a) Lanthanoids are mixed with iron to form an alloy

(iii)-(d) Misch metal is an alloy of lanthanoids(95%) and iron(5%) and traces of S,Si, Al,Cetc

(iv)-(e) Magnesium alloy is used to make bullets

(v)-(c) Petroleum cracking involves breaking down complex organic molecules into simpler hydrocarbons. Mixed oxides of lanthanoids are used as catalyst for petroleum cracking.

(i)-(c), (ii)- (a), (iii)-(b)

(i)-(c) Osmium has the highest oxidation state of +8 because it can expand it’s octet and use all its 8 electrons.

Electronic configuration of Os- [Xe] 4f¹⁴ 5d⁶ 6s²

5d⁶ and 6s² can be excited to get an oxidation state of +8

(ii)-(a) Manganese can show upto +7 oxidation state

(iii)-(b) 3d block element with highest is Cr.

(i)-(c),(ii)-(a),(iii)-(e),(iv)-(b)

(i)-(c)- Oxidation state of Manganese in MnO₂ is

x-4=0

x=+4

(ii)-(a) Lower oxidation state of Mn is very stable so +2 is stable

(iii)-(e)In KMnO₄, the oxidation state of Mn is +7

(iv)-(b) Lanthanoid have 5d¹ and 6s² outermost configuration. They can lose these three electrons easily and thereby show an oxidation state of +3

(i)-(d), (ii)-(a), (iii)-(b), (iv)-(e), (v)-(f)

(i)-(d)FeSO₄.7H₂O, Fe²⁺ has 3d⁶ electronic configuration. The complementary colour is pale green

(ii)-(a) NiCl₂.4H₂O ,Ni²⁺ has 3d⁸ electronic configuration. The complementary colour is green

(iii)-(b)MnCl₂.4H₂O , Mn²⁺ has 3d⁵ electronic configuration. The complementary colour is light pink

(iv)-(e)CoCl₂.6H₂O, Co²⁺ has 3d⁷ electronic configuration. The complementary colour is pink

(v)-(f) Cu₂Cl₂is colourless, Cu⁺ has 3d¹⁰ electronic configuration. It has a fully filled d orbital without no unpaired electrons to get excited and absorb visible light. So, Cu₂Cl₂ is colourless.

(i)-(b), (ii)-(d),(iii)-(a),(iv)-(e),(v)- (c)

(i)-(b) Ce- [Xe]4f¹ 5d¹ 6s². Cerium can lose all 4 electrons so as to attain noble gas configuration.So, oxidation state of Ce is +4

(ii)-(d) Eu- [Xe]4f⁷ 5d⁰ 6s².Europium can show +2 and +3 oxidation states.

(iii)-(a) Promethium is a radioactive lanthanoid.

(iv)-(e)Gd-[Xe]4f⁷ 5d¹ 6s². Gadolinium shows oxidation state of +3.

(v) –(c) Lu-[Xe]4f¹⁴ 5d¹ 6s². Lutetium has a 4f¹⁴ electronic configuration and a +3 oxidation state.

(i)-(c), (ii)-(d), (iii)-(b), (iv)-(e)

(i)-(c) Cu⁺ has fully filled 3d orbital with 10 electrons. Removal of one more electron is not easy and will require very high ionization energy because removing that electron disturbs the stable configuration.

(ii)-(d) For Zn²⁺, the electronic configuration has fully filled d orbital with 10 electrons. So, removal of one more electron is very difficult. Therefore, third ionization enthalpy is higher.

(iii)-(b) Cr forms Cr(CO)₆

(iv)-(e) Nickel has the highest heat of atomization.

The following reaction takes place

2Cu²⁺+ 4I^- → Cu₂I₂ + I₂

Where iodide gets converted to iodine and cupric iodide becomes cuprous iodide

Zirconium and Hafnium do belong to the same group but that is not the reason why separating them is difficult. Lanthanide contraction due to poor f orbital shielding leads to an increase in effective nuclear charge, which reduces the size of Hf. So the atomic radii of both Zr and Hf are similar which means they have similar physical and chemical properties and hence separation becomes difficult.

Actinoids form more stable complexes than lanthanoids because their 5f orbitals extend outwards and can participate in bonding while the 4f orbitals of lanthanoids are well shielded by 5d and 6s orbitals and do not take part in bonding.

A positive reduction potential means the reduced form of Cu is more stable than hydrogen. Also, Cu is less reactive than hydrogen and cannot displace it from acids.

Osmium has the highest oxidation state of +8 because it can expand it’s octet and use all its 8 electrons.

Electronic configuration of Os- [Xe] 4f¹⁴ 5d⁶ 6s²

5d⁶ and 6s² can be excited to get an oxidation state of +8

A= Cu, B= Cu(NO₃)₂, C=[Cu(NH₃)₄](NO₃)₂, D=CO₂,

E=CaCO₃

Explanation:

CuCO₃→CuO + CO₂

^(D)

CO₂ + Ca(OH)₂→ CaCO₃+ H₂O

^(E)

2CuO+CuS → 3Cu +SO₂

^(A)

Cu+4HNO₃(conc) →Cu(NO₃)₂+ 2NO₂ +2H₂O

^(B)

Cu(NO₃)₂ + 4NH₃→ [Cu(NH₃)₄](NO₃)₂

^(C)

A=FeCr₂O₄, B=Na₂CrO₄, C=Na₂Cr₂O₇.2H₂O, D= K₂Cr₂O₇

Explanation:

When a chromite ore is fused with sodium carbonate, a yellow solution of sodium chromate is formed

4FeCr₂O₄(A) 8Na₂CO₃+ 7O₂→ 8Na₂CrO₄(B) + 2Fe₂O₃+8CO₂

When this yellow solution of sodium chromate is treated with acid, it changes to sodium dichromate

2NaCrO₄ +2H⁺→ Na₂Cr₂O₇ (C)+2Na⁺ +H₂O

Sodium dichromate on treatment with KCl gives orange crystals of potassium dichromate

Na₂Cr₂O₇ +2KCl → K₂Cr₂O₇ (D)+2NaCl

A=MnO₂, B=K₂MnO₄, C= KMnO₄, D= KIO₃

When Manganese dioxide(A) is fused with KOH, it gives a green solution of potassium manganate(B).

2MnO₂ (A) + 4KOH +O₂→ 2K₂MnO₄(B) +2H₂O

Potassium manganate disproportionates to give purple potassium permanganate(C).

3MnO₄^2- +H⁺→2MnO₄^-(C)+MnO₂ +2H₂O

Potassium permanganate on reacting with KI gives potassium iodate(D) and manganese dioxide again

2MnO₄^- +H₂O +KI → 2MnO₂ + 2OH^- + KIO₃(D)

(i) Due to lanthanide contraction, size reduces. With reduction in size, covalent character increases. Therefore, Lu₂O₃ is more covalent than La₂O₃

(ii) Oxosalts contain oxygen as anion. As the size of cation reduces from La to Lu, according to Fajan’s rules, the polarizing power of the cation will increase and it will distort the cloud of oxygen(anion) greatly, thus the bond weakens and the stability also reduces.

(iii)As the size of central atom reduces, stability of complex increases. A small metal ion with greater charge attracts the ligands better.

(iv) In 5d block elements, due to poor shielding of f orbitals, the effective nuclear charge increases, thereby reducing the size. This is called lanthanide contraction. So, the radii of 4d and 5d block elements end up being very similar.

(v) From La to Lu, the acidic character increases. As the size reduces from La to Lu, the ability to lose electrons(Lewis base character) reduces, so acidity will increase.

(i) Copper has electronic configuration [Ar] 3d¹⁰ 4s¹. Removing the loosely bound 4s electron is easier but removing a second electron from a completely filled d orbital is harder. So, the second ionization enthalpy will be the highest for Cu.

(ii) Zinc has electronic configuration [Ar]3d¹⁰ 4s². Removing the two 4s electrons will be easier but removing the third electron from completely filled d orbital will be harder. Therefore, zinc has the highest third ionization enthalpy

(iii) Zinc will have the lowest enthalpy of atomization because it has no unpaired electrons for metallic bonding.

(b)

(i) The metal M is Fe

(ii) The oxidation state of M is +7 and only Mn shows +7 oxidation state. D-electrons are not involved in bonding.

Compounds formed when small atoms like H,C and N get trapped inside the crystal lattice of transition metals are called interstitial compounds.

Some properties of interstitial compounds are called:

5) Compounds are very hard

6) They are chemically inert

7) Have high melting points, higher than pure metals

8) They retain metallic conductivity.

(a) Fe(III) acting as catalyst for the reaction between iodide and persulphate ions.

2Fe³⁺ + 2I^-→ 2Fe²⁺ + I₂

2Fe²⁺ + S₂O₈^2-→ 2Fe³⁺ + 2SO₄^2-

(b) Three processes where transition metals act as catalysts.

(i) Finely divided Nickel as catalyst for hydrogenation of vegetable oil

(ii) V₂O₅ as catalyst for contact process used in the manufacture of sulphuric acid

(iii) Finely divided iron as catalyst for Haber’s process in the manufacture of ammonia

A=KMnO₄, B=K₂MnO₄, C= MnO₂, D=MnCl₂

Explanation:

A violet compound of manganese which is potassium permanganate (KMnO₄), decomposes to liberate potassium manganate(K₂MnO₄) and manganese dioxide(MnO₂) along with oxygen.

KMnO₄(A) → K₂MnO₄(B) +MnO₂(C) + O₂

Manganese dioxide (MnO₂)reacts with KOH to give potassium manganate (K₂MnO₄)

MnO₂(C) + KOH + O₂→2K₂MnO₄(B) +2H₂O

On heating Manganese dioxide (MnO₂) with NaCl and H₂SO₄, we get Manganese(II) chloride(MnCl₂), chlorine gas and other products,

MnO₂(C) + 4NaCl + 4H₂SO₄→ MnCl₂(D) +4NaHSO₄+2H₂O +Cl₂