//M1//QN1//SUB//DL0

Write down the electronic configuration of: (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+ (v) Co2+ (vi) Lu2+ (vii) Mn2+ (viii) Th4+

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//M2//QN2//SUB//DL0

Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?

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//M2//QN3//SUB//DL0

Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

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//M2//QN4//SUB//DL0

To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

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//M1//QN5//SUB//DL0

What may be the stable oxidation state of the transition element with the following

d electron configurations in the ground state of their atoms: 3d3, 3d5, 3d8 and 3d4?

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//M0//QN6//SUB//DL0

Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

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//M0//QN7//SUB//DL0

What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

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//M3//QN8//SUB//DL0

What are the characteristics of the transition elements and why are they called transition elements? Which d-block elements may not be regarded as the transition elements?

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//M2//QN9//SUB//DL0

In what way is the electronic configuration of the transition elements different from that of the non-transition elements?

[Topic 4.0] [2 Marks]

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//M0//QN10//SUB//DL0

What are the different oxidation states exhibited by the lanthanoids?

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//M2//QN11//SUB//DL0//EQ

Explain giving reasons: (i) Transition metals and many of their compounds show paramagnetic behaviour. (ii) The enthalpies of atomization of the transition metals are high. (iii) The transition metals generally form coloured compounds. (iv) Transition metals and their many compounds act as good catalyst.

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(i) Transition metals show paramagnetic behaviour. Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum.

However, in the first transition series, the orbital angular momentum is quenched. Therefore, the resulting paramagnetism is only because of the unpaired electron.

Magnetic moment is calculated by following formula.

µ = B.M.

where, n = No. of unpaired electron

(ii) Transition elements have high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals in high.

(iii) Most of the complexes of transition metals are coloured. This is because of the absorption of radiation from visible light region to promote an electron from one of the d-orbitals to another.

In the presence of ligands, the d-orbitals split up into two sets of orbitals having different energies. Therefore, the transition of electrons can take place from one set to another.

The energy required for these transitions is quite small and falls in the visible region of radiation. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.

(iv) The catalytic activity of the transition elements can be explained by two basic facts.

(a) Owing to their ability to show variable oxidation states and form complexes, transition metals form unstable intermediate compounds. Thus, they provide a new path with lower activation energy, Ea, for the reaction.

(b) Transition metals also provide a suitable surface for the reactions to occur.

Example, V2O5 in contact process, Fe in Haber's process.

//M0//QN12//SUB//DL0

What are interstitial compounds? Why are such compounds well known for transition metals?

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//M0//QN13//SUB//DL0

How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.

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//M4//QN14//SUB//DL0

Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate? [Topic 4.4]

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//M0//QN15//SUB//DL0//EQ

Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with ; (i) iodide, (ii) iron(II) solution and (iii) H2S.

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Potassium dichromates are strong oxidising agents

In acidic solution, its oxidising action can be represented as follows

Cr2O2–7 + 14H+ + 6e– → 2Cr3+ + 7H2O

(1) Acidified potassium dichromate will oxidise iodides to iodine,

(2) Acidified potassium dichromate will oxidise iron (II) solution to iron (III) solution.

(3) Acidified potassium dichromate will oxidise H2S to sulphur.

//M3//QN16//SUB//DL0//EQ

Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO2 and (iii) oxalic acid? Write the ionic equations for the reactions.

OR State the balance reaction equations of acidic permanganate ion with (i) Fe2+ (ii) C2 O2–4 (iii) SO2–3 [June 2024] [3 Marks]

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Preparation:

Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate.

2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O

3MnO2–4 + 4H+ → 2MnO–4 + MnO2 + 2H2O

Commercially it is prepared by the alkaline oxidative fusion of MnO2 followed by the electrolytic oxidation of manganate (VI).

In the laboratory, a manganese (II) ion salt is oxidised by peroxodisulphate to permanganate.

2Mn2+ + 5S2O2–8 + 8H2O → 2MnO–4 + 10SO2–4 + 16H+

In Acidic Medium oxidising action of KMnO4

+ 8H+ + 5e– → Mn2+ + 4H2O

(i) Acidified KMnO4 solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous ions to ferric ions.

(ii) Acidified potassium permanganate oxidizes SO2 to sulphuric acid.

(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.

//M0//QN17//SUB//DL0

For M2+/M and M3+/M2+ systems the Eo values for some metals are as follows: Cr2+/Cr –0.9 V Cr3/Cr2+: –0.4 V Mn2+/Mn –1.2 V Mn3+/Mn2+ +1.5 V Fe2+/Fe –0.4 V Fe3+/Fe2 +0.8 V Use this data to comment upon: (i) The stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+. (ii) The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

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//M2//QN18//SUB//DL0

Predict which of the following will be coloured in aqueous solution? Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.

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//M0//QN19//SUB//DL0

Compare the stability of +2 oxidation state for the elements of the first transition series.

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//M0//QN20//SUB//DL0

Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) Electronic configuration, (ii) Oxidation state, (iii) Atomic and ionic sizes and (iv) Chemical reactivity.

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//M0//QN21//SUB//DL0//EQ

How would you account for the following: (i) Cr2+ is strongly reducing while manganese (III )is strongly oxidising. (ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (iii) The d 1 configuration is very unstable in ions.

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(i) Cr2+ is strongly reducing in nature. It has a d 4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d 3). This d 3 configuration can be written as configuration, which is a more stable configuration. In the case of Mn3+ (d 4), it acts as an oxidizing agent and gets reduced to Mn2+ (d 5). This has an exactly half-filled d-orbital and is highly stable.

(ii) Co (II) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co (III). Although the 3rd ionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy.

(iii) The ions in d 1 configuration tend to lose one more electron to get into stable d 0 configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.

//M2//QN22//SUB//DL0//EQ

What is meant by 'disproportionation'? Give an example of disproportionation reaction in aqueous solution.

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It is found that sometimes a relatively less stable oxidation state undergoes an oxidation – reduction reaction in which it is simultaneously oxidised and reduced. This is called disproportionation.

e.g. 3 + 4H+ → 2 + MnO2 + 2H2O

//M0//QN23//SUB//DL0

Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?

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//M0//QN24//SUB//DL0//EQ

Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution?

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Cr3+ is the most stable in aqueous solutions owing to a configuration.

//M3//QN25//SUB//DL0

Give examples and suggest reasons for the following features of the transition metal chemistry: (i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic. (ii) A transition metal exhibits highest oxidation state in oxides and fluorides. (iii) The highest oxidation state is exhibited in oxoanions of a metal.

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//M3//QN26//SUB//DL0

Indicate the steps in the preparation of: (i) K2Cr2O7 from chromite ore. (ii) KMnO4 from pyrolusite ore.

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//M0//QN27//SUB//DL0

What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

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//M0//QN28//SUB//DL0

What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.

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//M2//QN29//SUB//DL0

The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.

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//M1//QN30//SUB//DL0

Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

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//M0//QN31//SUB//DL0//EQ

Use Hund's rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.

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Electronic configuration of Ce: [xe] 4f 1 5d1 6s2

Ce3+: [xe] 4ƒ 1 (n = 1)

µ = B.M.

=

=

= 1.73 B.M.

//M0//QN32//SUB//DL0

Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.

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//M0//QN33//SUB//DL0

Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) Electronic configuration (ii) Oxidation states and (iii) Chemical reactivity.

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//M0//QN34//SUB//DL0

Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

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//M0//QN35//SUB//DL0

Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points: (i) Electronic configurations, (ii) Oxidation states, (iii) Ionisation enthalpies and (iv) Atomic sizes.

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//M3//QN36//SUB//DL0

Write down the number of 3d electrons in each of the following ions: Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to the occupied for these hydrated ions (octahedral).

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//M0//QN37//SUB//DL0

Elements of the first transition series possess many properties different from those of heavier transition elements.

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//M0//QN38//SUB//DL0//EQ

What can be inferred from the magnetic moment values of the following complex species?

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(i) K4[Mn(CN)6]

For in transition metals, the magnetic moment is calculated from the spin-only formula.

Therefore,

µ = = 2.2 B.M.

We can see from the above calculation that the given value is closest to n = 1. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.

Hence, we can say that CN– is a strong field ligand that causes the pairing of electrons.

(ii) [Fe(H2O)6]2+

µ = = 5.3 B.M.

• We can see from the above calculation that the given value is closest to n = 4. Also, in this complex, Fe is in the +2 oxidation state. This means that Fe has 6 electrons in the d-orbital.
• Hence, we can say that H2O is a weak field ligand and does not cause the pairing of electrons.

(iii) K2[MnCl4]

µ = = 5.9 B.M.

We can see from the above calculation that the given value is closest to n = 5. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.

Hence, we can say that Cl– is a weak field ligand and does not cause the pairing of electrons.