The d and f Block Elements

1) Write down the electronic configuration of :

i) Cr3+ ii) Pm3+ iii) Cu+ iv) Ce4+ v) Co2+ vi) Lu2+ vii) Mn2+ viii)Th4+

Solution

Transit

ion

metal

ion

Electronic configuration

Cr3+ 1s22s22p63s23p63d3

Pm3+ 1s22s22p63s23p63d104s24p64d105s25p64f4 Cu+ 1s22s22p63s23p63d10 Ce4+ 1s22s22p63s23p63d104s24p64d105s25p6

Co2+ 1s22s22p63s23p63d7 Lu2+ 1s22s22p63s23p63d104s24p64d105s25p64f14

5d1

Mn2+ 1s22s22p63s23p63d5 Th4+ 1s22s22p63s23p63d104s24p64d105s25p64f14

5s25p65d106s26p6

2) Why are Mn2+ compounds more stable than Fe2+ compounds towards oxidation to the +3 state?

Solution

Mn2+ compounds are more stable than Fe2+ compounds towards oxidation to their + 3 state, due to half filled d-orbitals in Mn2+ , while Fe2+ gains half filled d-orbitals by losing one electron by getting oxidised to Fe3+ . Hence , Mn2+ is not oxidised easily to Mn3+ .

Fe2+ → Fe3+ + e−

3d6 → 3d5

(Unstable) (Stable)

Mn2+ → Mn3+ + e−

3d5 → 3d4

(Stable) (Unstable)

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

Solution

The common oxidation sate of the first row transition metals except for scandium is +2, which arises due to the loss of 4s2 electrons. As in the first half elements the d- orbitals are filled with only one electron, due to which there is minimum repulsion and maximum stability, but as the pairing of electrons start in the second half, then due to repulsion between the electrons the stability of +2 state decreases.

4) 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 example.

Solution

The electronic configurations decide the stability of oxidation states in the first series of the transition elements . In general , the oxidations in which the element acquires a stable configuration, such as noble gas configuration , half-filled or fully filled d-subshell , are stable.

For example , Sc3+ , Mn2+ and Zn2+ are respectively the most stable of Sc, Mn and Zn.

Sl.No. Transition metal atom

Electronic configuration

Metal ion

Electronic configuration

Oxidation state

Type of stability

1. Sc [Ar]3d14s2 Sc3+ [Ar] +3 Inert gas configuration

2. Mn [Ar]3d54s2 Mn2+ [Ar]3d5 +2 Half-filled 3d5 subshell

3. Zn [Ar]3d104s2 Zn2+ [Ar]3d10 +2 Completely filled 3d10 subhell

5) What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d33d5, 3d8 and 3d4?

Solution

D electron configuration

Most stable oxidation state

3d3 +5 3d5 +2 and +7 3d8 +2 3d4 +6, +3

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

Solution

Oxometal ion Oxidation state

Group number

Chromate ion (CrO4-) +6 VI B

Dichromate ion (Cr2O72-) +6 VI B

Permangante ion (MnO4-) +7 VII B

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

Solution

The steady decreases in the size of lanthanides atoms and their trivalent ions with an increase in the atomic number due to poor screening of nuclear charge by 4f orbitals is called “ lanthanide contraction”.

In lanthanides , the differentiating electrons enters the ‘4f’ orbital. Due to its peculiar shape, the 4f electrons, which constitute the inner shells, do not shield the valence electrons from nuclear attraction. Consequently , the electron cloud of the nucleus shrinks, leading to a gradual decreases in the size of the lanthanoids with an increases in the atomic number.

Consequences:

i) A similarity in the size of the second and third lanthanoid series is observed.

Example:

Group III B IV B VB 3d series Sc Ti V Atomic size 144pm 132 pm 122 pm 4d series Y Zr Nb Atomic size 180 pm 160 pm 146 pm 5d series La Hf Ta Atomic size 187 pm 159 pm 146 pm

ii) The separation of lanthanoids is difficult due to their similar size, which, in turn, leads to similar properties . Due to their similar size, they differ slightly in their properties , like complex formation. Therefore, lanthanoids are separated by the ion exchange method.

iii) The basic strength of the hydroxides of lanthnoids decreases fro La(OH)3 to Lu(OH)3 due to the lanthanoids contraction. The decrease in the size of M3+ ion leads to an increase in the covalent character of M-OH bond, resulting in a decreases in the basic strength of the lanthanoid hydroxides.

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

Solution

The characteristics of the transition elements ar e:

i) They are metals.

ii) They exhibit magnetic character.

iii) They exhibit variable oxidation states.

iv) They form coloured compounds.

v) They form complexes with negative , positive and neutral ligands.

vi) They are their compounds act s good catalysts.

vii) They form alloys with other metals and non-metals.

viii) They form interstitial compounds.

he d- block elements that contain unpaired electrons in d-orbitals either in their elemental state or as ions are called transition elements . The transition elements are called so due to their position in the periodic table, which is between s-block elements(highly reactive metals) and p- block elements(less reactive metals).

Zinc , cadmium and mercury cannot be regarded as transition elements, as their atoms and ions do not have partially filled d-orbitals in their ground state or ionic state.

Element Atom Electronic configuration of atom

Ion EC of ion

Zinc Zn [Ar] 4s23d10 Zn2+ [Ar] 3d10 Cadmium Cd [Ar] 4s2 3d10 Cd2+ [Ar] 3d10 Mercury Hg [Ar] 4s2 3d10 Hg2+ [Ar] 3d10

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

Solution

The transition elements have partially filled d-subshell in their penultimate shell, that is, the d-subshell is incomplete , whereas non-transitoon do not have partially filled d –subshell. In non-transition elements, the last electron enters the s- or p – subhsell of the valence shell, whereas in transition elements , the last electron enters the d-subshell of the penultimate shell.

10) What are the different oxidation states exhibited by the lanthanoids?

Solution

The main oxidation state of lanthanoids is +3. The other oxidation sates are +2 and +4 . Eu (europium) exhibits an oxidation state of +2 , while Ce(cerium) exhibits an oxidation state of +4.

11) Explain giving reasons:

i) Transition metals and many of their compounds show paramagnetic behaviour.

ii) The enthalpies of atomisation 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.

Solution

i) The transition elements and many of their compounds show paramagnetic behaviour due to the presence of unpaired electrons. The paramagnetic behaviour is expressed in terms of magnetic moment.

Magnetic moment, µ = √[n(n+2)] BM

Where n= number of unpaired electrons in the atom or ion.

ii) Due to the presence of a large number of unpaired electrons in the atoms of the transition elements and due to the participation of the ns and (n-1)d electrons in bind formation. the transition elements have stronger inter-atomic interactions(metallic bonds). Hence, the enthalpy of atomisation of the transition elements is high.

iii) The transition elements exhibit colour due to d-d transition of electrons. The five d-orbitals split two sets, t2g(dxy, dxz, dyz) and eg (dx2-y2, dz2) set of orbitals . The energy difference between the types of d-orbitals is very less. When an electron from a lower energy d-orbitals (t2g) absorbs energy from visible light, it is excited to a higher energy d-orbitals (eg) . when the electron gets de-excited , it emits light of the colour that is complimentary to the colour absorbed . The frequency of light absorbed or emitted depends on the ligand in the compound. The colour of the transition compound is the emitted colour.

iv) Due to their variable valencies, the transition elements and their compounds can form unstable intermediate compounds and increase the speed of a reaction by lowering its

activation energy. This results in the reaction taking place at a relatively low temperature. Hence, they show catalytic activity.

Example

1. Vanadium pentoxide(V2O5) is used as a catalyst in the contact process, in the manufacture of sulphuric acid.

2. Finely divided iron is used as a catalyst in Haber’s process in the manufacture of ammonia.

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

Solution

Interstitial compounds are non-stoichiometric compounds formed when small atoms like C, N or He are trapped inside the crystal lattice of metals.

Example: TiC, Mn4N , Fe3H, VH0.56, TiH1.7

Transition metals form such type of compounds due to presence of interstitial spaces between atoms in their crystal lattice, in which small atoms like C, N, or H can easily be accommodated.

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

Solution

The transition elements show variable oxidation state due to the participation of ns and (n-1) d electrons in bond formation , due to less energy difference between ns and (n-1) orbitals.

Non-transition elements, s- and p-block elements , exhibit a limited number of oxidation states due to the participation of only the valence shell s- and p-orbitals in bond formation.

The variability in the oxidation states of the transition metals is by unity; as the number of unpaired electrons increases , the maximum oxidation state is exhibited.

Example: Manganese exhibits variable oxidation numbers of +2, +3,+4,+5, +6 and +7.

In non-transition elements, the variability in oxidation states is usually by two.

Example:

1. Gallium (group 13) exhibits +1 and +3 oxidation states.

2. Tin(group 14) , exhibits +2 and +4 oxidation states.

14) Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Solution

Potassium dichromate is prepared from chromite ore (FeCr2O4).

Step I

Preparation of sodium chromate:

Sodium chromate is prepared by the fusion of chromite ore with sodium carbonate in free supply of air.

4FeCr2O4+8Na2CO3 + 7O2 →8Na2CrO4 + 2Fe2O3 + 8CO2

Step II

Conversion of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate can be crystallised .

(Na2Cr2O7.2H2O)

2Na2Cr2O4 + 2H2SO4 → Na2Cr2O7 + 2Na2SO4 +H2O

On cooling the resultant solution, sodium sulphate separates out, and the solution containing sodium dichromate is crystallised to get orange crystals.

Step 3

Conversion of sodium chromate into potassium chromate:

Sodium dichromate is treated with potassium chloride to obtain potassium dichromate.

Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl

When the pH is increased , potassium dichromate gets converted into potassium chromate in basic solution.

K2Cr2O7 + 2OH- →2KCrO4 + H2O

15) 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

Solution

Potassium dichromate is used as an oxidising agent in acidic solution , and its oxidising action can be represented as follows:

Cr2O72- + 14 H+ + 6 e−

→ 2Cr3+ + 7H2O

i) It oxidises iodide to iodine.

Cr2O72- + 14 H+ + 6I- → 2Cr3+ + 7H2O + 3I2

ii) It oxidises iron (II) salt solution to iron (III) salt solution.

Cr2O72- + 14 H+ + 6Fe2+

→ 2Cr3+ + 6Fe3+ + 7H2O

iii) It oxidises H2S to S.

Cr2O72- + 8H+ + 3H2S → 2Cr3+ + 7H2O + 3S

16) Describe the preparation of potassium permanganate . How does acidified permanganate solution react with (i) iron ii)SO2 and iii) oxalic acid?

Write the equations for the reactions.

Solution

Preparation of potassium permanganate:

1. Potassium permanganate is prepared from pyrolusite (MnO2).

2. MnO2 is fused with an alkali metal hydroxide like KOH and an oxidising agent like KNO3 to form K2MnO4.

MnO2 + KOH + O2→ 2 K2MnO4 + 2H2O

3. K2MnO4 on disproportionation in neutral or acidic solution forms permanganate.

3MnO42- +4H+ → 2MnO4

- + MnO2 + 2H2O

I) Acidified potassium permanganate oxidises green iron (II) salkts to yellow iron(III) salts.

5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O

ii) Acidified potassium permanganate oxidises sulphur dioxide to sulphuric acid. (Highlight equation)

2KMnO4 (aq) + 2H2O (l) + 5SO2(g) →K2SO4(aq) + 2MnSO4 (aq) + 2H2SO4(aq)

Potassium Water Sulphur dioxide Potassium Manganese sulphate Sulphuric acid

permanganate sulphate

iii) Acidified potassium permanganate oxidises oxalic acid to carbon dioxide.

2KMnO4 + 3H2SO4 + 5H2C2O4 → K2SO4 + 2MnSO4 + 8H2O + 10CO2

17) For M2+ /M and M3+/M2+ systems, the Eθ 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+ and

ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Solution

The higher the value of Eθ , the greater is the tendency of its ion to undergo reduction. Therefore, Mn3+ . with the highest reduction potential , is the least stable as it easily undergoes reduction to Mn2+.

From the reduction potential values , it is evident that the stability of Fe3+ in an acid solution is more than that of Mn3+ , but less than that of Cr3+.

The lower the reduction potential value , the greater is tendency of its ions to get easily oxidised. Therefore , Mn , with the least reduction potential , is more readily oxidised to Mn2+ , and Fe is the least readily oxidised to Fe2+ , among to Fe2+ , among Mn, Fe and Cr.

Order of tendency to undergo oxidation : Mn > Cr > Fe.

18) Predict which of the following will be coloured in an aqueous solution: Ti3+, V3+ , Cu+

, Sc3+ , Mn2+ , Fe3+ and Co2+ . Give reasons for each.

Solution

Ions that contain unpaired electrons in their d-subshell are coloured . Ti3+ (d1), V3+ (d2) , Mn2+ (d5), Fe3+ (d5) and Co2+ (d7) contain unpaired electrons in their valence shell. Therefore, they are coloured . Cu+ (d10) and Sc3+ (d0) are colourless due to the absence of unpaired electrons in their d-subshell.

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

Solution

In the first row of elements of the transition series, the stability of the +2 oxidation state increases as we move from left to right. However , Mn2+ and Zn2+ are exceptionally stable in the +2 oxidation state due to the extra stability gained by the half filled d-orbital in Mn2+, and the completely filled d-orbitals i Zn2+.

20) Compare the chemistry of actinoids with that of lanthanoids with special reference to:

i) Electronic configuration iii) Oxidation state

ii) atomic and ionic sizes and iv) Chemical reactivity

Solution

Property Lanthanoids Actinoids i) Electronic configuration

The general outer electronic configuration is 6s2 , with variable occupancy of 4f level. The electronic configurations of all the lanthanoids are of the form 6s24fn , where n=1 to 14.

The general electronic configuration is 7s2 , with variable occupancy of 5f and 6 d subshells. The irregularities in the configurations are due to the stability of the 5f0 and 5f7 and 5f14 configurations.

ii) Atomic and ionic sizes

There is decrease in atomic and ionic radii from lanthanum to leutetium, which is known as lanthanoid contraction. This is due to the poor shielding of nuclear charge by 4f electrons.

A decrease in the radii of atoms and M3+ ions is observed in actinoids. The size of atoms also decreases. This is referred to as actonoid contraction. This is more in actinoids due to the poor shielding by 5f electrons.

iii) Oxidation states The +3 oxidation state is stable and the most common. The +2 and +4 states are also observed

+3 is the common oxidation state. The maximum oxidation state increases from +4 in thorium to +5, +6 and +7 on Pa, U and Np, but decreases in succeeding elements. The +2 state is rare in actinoids.

iv) Chemical reactivity

• They combine with hydrogen and carbon to form salt like hydrides and carbides.

• They form oxides and sulphides with oxygen and sulphur.

• The oxides react with water to form

• They are highly reactive metals and tarnish when exposed to air due to the formation of an oxide layer.

• They react with oxygen, hydrogen and halogens at moderate temperature.

insoluble hydroxides. The hydroxides react with CO2 to form carbonates.

• They are less basic than actinoids. They from ionic compounds.

• They have less tendency to form complexes.

• They react with boiling water. They are attacked by HCl, but not by nitric acid, due to the formation of the protective oxide layer. • They do not react with alkalis as they basic. They are more basic. • They have a strong tendency to form complexes.

21) How would you account for the following?

i) Of the d4 species, Cr2+ is strongly reducing while manganese (III) is strongly oxidising.

ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagent it is easily oxidised.

iii) The d1 configuration is very unstable in ions.

Solution

I) Chromium +3 state is more stable as compared to +2 state. therefore , chromium +2 readily changes to chromium +3 state, gets oxidised and behaves as a strong reducing agent. Manganese +2 state is more stable than +3 state. Cr2+ is reducing as its configuration changes from d4 to d3. The d3 configuration has a half-filled t2g level. When Mn3+ changes to Mn2+ . it results in d5 configuration, which has extra stability of a half-filled d –orbital. Hence, manganese +3 gains an electron and changes to manganese +2 . It acts as an oxidising agent.

II) In the presence of ligands , Co(II) is oxidised to Co(III) with a d6 configuration. Ligands that are strong cause spin pairing . Due to this, diamagnetic octahedral complexes are formed , which are highly stable due to their large crystal field stabilisation energy.

III) Ions that have d1 configuration tend to lose electrons and attain stable d0 configuration . Thus , the ions act as strong reducing agents.

22) What is meant by ‘disproportionation’ ? Give two examples of disproportionation reaction in aqueous solution.

Solution

Disproportionation:

Disproportionation is the conversion of an unstable metal ion two species , one with a lower oxidation state and the other with a higher oxidation state, by undergoing simultaneous reduction and oxidation.

Example:

1. Cr(+V) undergo disproportionation to Cr(+VI) and Cr(+III).

3CrO43- + 8H+ → 2CrO4

2- + Cr3+ + 4H2O

2. Mn(+VI) undergoes disproportionation to Mn(+VII) and Mn (+IV).

3MnO42- + 4H+ →2 MnO4 + MnO2 +2H2O

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

Solution

In the first series of transition elements, copper mostly exhibits the +1 oxidation state, as it is stable in this state, due to completely filled 3d10 orbitals.

Electronic configuration of Cu2+ = [Ar]3d10

24) 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?

Solution

Mn3+ :

Electronic configuration : [Ar] 3d4

Number of unpaired electrons= 4

Cr3+:

Electronic configuration = [Ar] 3d3

Number of unpaired electrons= 3

V3+

Electronic configuration = [Ar] 3d2

Number of unpaired electrons=2

Ti3+ :

Electronic configuration = [Ar] 3d1

Number of unpaired electrons=1

Cr3+ is most stable in aqueous solution.

25) Give examples and suggest reasons for the following features of the transition metal chemistry:

a) The lowest oxide of a transition metal is basic , while the highest is amphoteric/ acidic.

b) A transition metal exhibits the highest oxidation state in oxides and fluorides.

c) The highest oxidation state is exhibited in the oxoanions of metal.

Solution

a) Transition metal oxides are basic in their lowest oxidation state, and acidic in their higher oxidation states.

Example:

MnO and Mn2O3 are oxides of manganese , in which manganese is in +2 and + 3 oxidation states. These are basic.

Mn3O4 and MnO2 are oxides of manganese that are amphoteric . In these oxides, manganese is in +8/3 and +4 oxidation states.

In Mn2O7 , manganese is in the +7 oxidation state. In is acidic in nature.

In lower oxidation state, all the electrons are not involved in bond formation. Hence, its oxide can donate electrons, and thus , behave as a Lewis base. Its m=nuclear charge is not very high. When the metal is in high oxidation state, its nuclear charge is high and its oxide has a tendency to gain electrons. Therefore, it behaves as an acid.

b) Fluroine and oxygen are highly electronegative, and hence, a transition metal exhibits higher oxidation state in fluorides and oxides.

c) Oxoanions are formed when the oxides are dissolved in acids and bases . Since the metal is in a higher oxidation state in the oxides even in oxoanions the metal has a higher oxidation state.

Example:

1. The oxidation state of manganese in permanganate ion is +7, MnO4-1.

2. The oxidation state of chromium is +6 in dichromate anion , Cr2O72-

26) Indicate the steps in the preparation of:

i) K2Cr2O7 from chromite ore ii) KMnO4 from pyrolusite ore.

Solution

i) Preparation of K2Cr2O7 from chromite ore:

1. Potassium dichromate is obtained from chromite ore (FeCr2O4) . Chromite ore is powdered and fused with sodium carbonate in free supply of air.

4FeCr2O4 + 8Na2CrO4 + 2Fe2O3 + 8CO2

2. The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate can be crystallised .

(Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O

3. Sodium dichromate is treated with potassium chloride to obtain potassium dichromate.

Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl

ii) Preparation of KMnO4 from pyrolusite ore:

1. Potassium manganate is prepared by the fusion of pyrolusite (MnO2) with an alkali metal hydroxide and an oxidising agent like KNO3

2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O

2. Dark green K2MnO4 is produced , which disproportionate in a neutral or acidic solution to give permanganate.

3MnO42- + 4H+ → 2MnO4

- +MnO2 +2H2O

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

Solution

An alloy is a homogenous mixture of a metal with another metal or a non-metal.

Example: Mischmetal

Composition: It consists of a lanthnoid metal (95%) and iron (5%) and traces of S,C,Ca and Al.

Uses:

Mischmetal is used in magnesium –based alloys to produce bullets, shells and lighter flint.

28) 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.

Solution

The element in which the differentiating electron enters into the f-subshell of the anti-penultimate shell are called inner transition elements.

Lanthanoids(4f series) : These are 14 elements from lanthanum to lutetium (atomic numbers 58-71).

Actinoids(5f series) : These are 14 elements starting from actinium to lawrencium (atomic numbers 90-103).

In the given atomic numbers, 59, 95 and 102 belong to inner transition elements.

29) 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.

Solution

In lanthanoids , the +3 oxidation state is the most common . They also show the +2 and +4 states due to the stability of empty, half-filled and fully filled orbitals.

Example: Ce exhibits the +4 oxidation state due to its noble gas configuration. Other lanthanoids exhibit the +4 oxidation in their oxides. Eu2+ is formed by the loss of two s electrons due to its stable f7 configuration.

However, in actinoids , though the common oxidation state is +3, the elements in the first half of the series exhibit variable oxidation states of +4 in Th , to +5 , +6 and +7 in Pa, U and Np, which decreases in successive elements. Due to the uneven distribution of oxidation states in actinoids , which is due to the less energy difference between the 6d and 5f orbitals, the chemistry of the actinoid elements is not so smooth as that of the lanthanoids.

30) 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.

Solution

The last element in the series of actinoids is lawrencium with atomic number 103. Electronic configuration of Lr: [Rn] 5f14 6d17s2

The possible oxidation state of lawrencium is +3, as it gains extra stability of the completely filled f-subshell by the loss of 3 electrons Lr([Rn]5f14).

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

Solution

The electronic configuration of cerium is

Ce(Z=58) = [Xe]4f15d16s2

According to Hund’s rule , the electronic configuration of Ce3+ is [Xe] 4f1.

The spin only formula to calculate magnetic moment is µ = √(n(n+2) BM,

where n is the number of unpaired electrons.

For Ce3+ , n= 1

µ = √(1(1+2)

= √3 BM

= 1.732 BM

µ = 1.732 BM

32) 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.

Solution

Lanthanides that exhibit +4 oxidation state: Cerium , praseodymium, neodymium, terbium, dysprosium, holmium.

Lanthanides that exhibit +2 oxidation state: Neodymium, samarium, europium, thulium, ytterbium.

The +4 and +2 oxidation states arise due to the extra stability of fully filled, half-filled and empty f-subshells.

Example: Ce(+4) is 4f0 and has noble gas configuration. Tb(+4) and Eu(+2) have f7 sub shell.

Yb(+2) has completely filled f14 configuration.

33) Compare the chemistry of the actinoids with that of lanthanoids with reference to:

i) Electronic configuration ii) oxidation states and iii) chemical reactivity.

Solution

34) The electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

Solution

Atomic number

Element Electronic configuration

61 Promethium (Pm) [Xe] 4f56s2 91 Protactinium (Pa) [Rn] 5f26d17s2 101 Mendelevium (Md) [Rn] 5f137s2

109 Meitnerium (Mt) [Rn] 5f146d77s2

35) 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.

Solution

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

Solution

37) Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

Solution

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

Solution

From the formula , µ = n(n+2) BM

Number of unpaired electrons (n) can be calculated.

µ (BM) n

2.2 1

5.3 4

5.9 5

From the given data and the number of unpaired electrons we can infer:

1. K4[Mn(CN)6], there is only one unpaired electron in Mn2+ . There are five electrons in Mn2+, distributed as (t2g)

5.

2. In Fe(H2O)6]2+ , there are four unpaired electrons in Fe2+ . There are six electrons in Fe2+

distributed as (t2g)4 and (eg)2.

3. In K2[MnCl4] there are five unpaired electrons in Mn2+, distributed as (t2g)3 and (eg)2.