D & F Block Elements
General electronic configuration
s block elements: - ns2
p block elements: - ns2np1-6
d block elements: - (n-1)d1-10 ns0-2
f block elements: - (n-2)f0-14(n-1)d0-1ns2
Example 1 On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?
Ans. On the basis of incompletely filled 3d orbitals in case of scandium atom in its ground state (3d 1 ), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d 10) in its ground state as well as in its oxidised state, hence it is not regarded as a transition element.
Intext 8.1 Silver atom has completely filled d orbitals (4d 10) in its ground state. How can you say that it is a transition element?
Ans. Silver (Ag) belongs to group 11 of d-block and its ground state electronic configuration is 4d10 5s1. It shows an oxidation state of +2 in its compounds like AgO and AgF2 in which its electronic configuration is d9 so it is a transition element.
Example 2 : Why do the transition elements exhibit higher enthalpies of atomisation?
Ans. Because of large number of unpaired electrons in their atoms they have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.
Intext Question 8.2 In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol–1. Why?
Ans. Zinc (4d10 5s2) has completely filled d-orbital and has no unpaired electron to take part in the formation of metallic bonds.
Example 8.3 - Name a transition element which does not exhibit variable oxidation states.
Ans. Scandium (Z = 21) does not exhibit variable oxidation states
Intext Question 8.3 Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
Ans. Manganese exhibit the largest number of oxidation states. It shows the oxidation states +2, +3, +4, +5 ,+6, and + 7. The reason for that is the maximum number of unpaired electrons present in its outermost shell i.e. 3d54s2
Example 8.4 Why is Cr2+ reducing and Mn3+ oxidising when both have d 4 configuration?
Ans. Cr2+ is reducing as its configuration changes from d 4 to d 3, the latter having a half-filled t2g level. On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d 5 ) configuration which has extra stability.
Intext Question 8.4 The E V (M2+/M) value for copper is positive (+0.34V). What is possible reason for this? (Hint: consider its high ∆aH V and low ∆hydH)
Ans. The Eθ(M2+/M) value of a metal depends on the energy changes involved in the following:
a. Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state.
b. Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state.
Example 8.5 How would you account for the increasing oxidising power in the series VO2 + < Cr2O7 2– < MnO4 – ?
Solution: - This is due to the increasing stability of the lower species to which they are reduced.
Intext Question 8.5 How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
Ans. Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner−d orbitals. The irregular variation of ionization enthalpies can be attributed to the extra stability of configuration such as d0, d5, d10. Since these states are exceptionally stable, their ionization enthalpy are very high
Example8.6: - For the first row transition metals the Eo values are:
Eo............... V..... Cr..... Mn..... Fe..... Co..... Ni..... Cu
(M2+/M) –1.18 – 0.91 –1.18 – 0.44 – 0.28 – 0.25 +0.34
Explain the irregularity in the above values.
Ans. The E(M2+/M) values are not regular which can be explained from the irregular variation of ionisation enthalpies (∆iH1 + ∆iH2) and also the sublimation enthalpies which are relatively much less for manganese and vanadium.
Example 8.7: - Why is the E V value for the Mn3+/Mn2+ couple much more positive than that for Cr3+/Cr2+ or Fe3+/Fe2+? Explain.
Ans. Much larger third ionisation energy of Mn (where the required change is d5 to d4 ) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance.
Intext Questions 8.6 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Ans. The highest oxidation state of a metal exhibited in its oxide or fluoride only since fluorine and oxygen are the most electronegative elements. The highest oxidation state shown by any transition element is +8.
Intext Questions 8.7 Which is a stronger reducing agent Cr2+ or Fe2+ and why ?
Ans. Cr+2 = 3d4 and Fe+2 = 3d6
reducing agent means it will reduce the other & itself gets oxidized. Means, one which will have a greater tendency to lose electron, would be better reducing agent.
Fe+2 having 3d6
The configuration will be more reducing because after losing an electron it will have d5 configuration which is stable as it is half-filled configuration. Hence, Fe+2 is stronger reducing agent than Cr+2
Example 8.8 Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.
Solution: - With atomic number 25, the divalent ion in aqueous solution will have d5 configuration (five unpaired electrons).
The magnetic moment, µ is µ = √ 5(5+2) = 5.92BM
Example 8.8 Calculate the ‘spin only’ magnetic moment of M2+ (aq) ion (Z = 27).
Ans.
Example 8 What is meant by ‘disproportionation’ of an oxidation state? Give an example.
Ans. When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation. For example, manganese (VI) becomes unstable relative to manganese(VII) and manganese (IV) in acidic solution.
3 MnO4 2– + 4 H+ → 2 MnO4 - + MnO2 + 2H2O
Intext Question 8.9 Explain why Cu+ ion is not stable in aqueous solutions?
Ans. In aqueous medium, Cu+2 is more stable than Cu+. This is because although energy is required to remove one electron from Cu+ to Cu2+, high hydration energy of Cu2+ compensates for it. Therefore, Cu+ ion in an aqueous solution is unstable.
Example 8.10 Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
Ans. Cerium (Z = 58)
Intext Question 8.10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why
Ans. From element to element, the actinoid contraction is greater than the lanthanoid contraction due to poor shielding by 5f-electrons in actinoids than that by 4f-electrons in the lanthanoids.
Q1 Write down the electronic configuration of: (i) Cr3+ (iii) Cu+ (v) Co2+ (vii) Mn2+ (ii) Pm3+ (iv) Ce4+ (vi) Lu2+ (viii) Th4+
Q2 Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?
Ans. E.C. of Mn2+ [18Ar] 3d5 (Stable configuration) E.C. of Fe2+; [18Ar] 3d6. Since Mn2+ has stable half filled electronic configuration, therefore Mn2+ compounds are more stable than Fe2+ towards oxidation to their +3 state. Fe2+(3d6) can lose one electron easily to give Fe3+(3d5, stable configuration).
Q3 Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Ans. +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number as 3d orbitals acquire only one electron in each of five 3d orbitals (remain half-filled) and the electronic repulsion is least and the nuclear charge increases. In the second half of the first row transition series, electrons pair up in 3d orbitals. This increases the electronic repulsion.
Q4 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.
Ans. The presence of half filled or completely filled d-orbitals imparts stability to a particular element/ion. For example, let us write the different oxidation states of Mn (Z = 25) along with the electronic configurations. The configuration of the element is [Ar] 3d5 4s2, Mn2+ ; [Ar] 3d5, Mn3+; [Ar] 3d4, Mn4+; [Ar] 3d3 +2 oxidation state of the element is likely to be the most stable because the corresponding electronic configuration of Mn2+ is highly symmetrical (all the five 3d-orbitals are half filled).
Q5 What may be the stable oxidation state of the transition element with the following d electron configuration in the ground state of their atoms?
3d3, 3d5, 3d8 and 3d4
Q6 Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group no. ?
Ans.
QNo. 8.7 What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Ans. Regular decrease (contraction) in the atomic and ionic radii with increasing atomic no. is known as lanthanoid contraction.
Consequences of lanthanoid contraction:
a. Difficulty in separation of lanthanoids: Change in ionic radii is small their chemical properties are similar, so, this makes the separation of the elements in the pure state difficult. Separation of individual lanthanoid elements by ion exchange methods.
b. Similarity in size of elements belonging to same group of second and third transition series: size of element in second transition series is larger than same group of the first transition series. Size of atom of third transition series is nearly same as that of the atom of the element lying in the same group of the second transition series.
c. Effect on the basic strength of hydroxides: As the size of the lanthanoid ions decreases from La3+ to Lu3+, the covalent character of the hydroxides increases and hence the basic strength decreases. Thus La(OH)3 is most basic whereas Lu(OH)3 is least basic.
Q.No.8.8 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?
Ans. Element whose atom in the ground state or ion in one of the common oxidation states, has incomplete d-subshell (partly filled), i.e. having electrons between 1 to 9.
This definition excludes zinc, cadmium and mercury from the transition elements because they do not have partly filled (or incomplete) d subshell in their atomic state or their common oxidation state (i.e. Zn2+, Cd2+, Hg2+). The d-subshell configurations of these atoms or their dipositive ions are 3d10, 4d10 and 5d10. They do not show properties of transition elements to any appreciable extent except for their ability to form complexes. However, being the end members of the three transition series, they are generally studied with d block elements.
Q8.9 In what way is the electronic configuration of the transition elements different from that of the non transition elements?
Ans .In transition elements, d orbitals are filled and in representative elements, s and p orbitals are filled.
Transition elements have general electronic configuration (n−1)d 1−10 ns1−2. Representative elements have general electronic configuration ns1−2 or ns2 np1−6.
In representative elements, the last shell is incompletely filled and in transition elements, the last two shells are incompletely filled.
Q8.10 What are the different oxidation states exhibited by the lanthanoids?
Ans. Apart from the +3 oxidation state, lanthanoids also show +2 and +4 oxidation states.
Q.No.8.11 Explain giving reason:
(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.
Ans. (i) Transition metals and many of their compounds show paramagnetic behaviour. This is due to presence of one or more unpaired electrons in d subshell.
(ii) The enthalpies of atomisation of the transition metals are high. This is due to high effective nuclear charge and large number of valence electrons. This results in formation of strong metallic bonds.
(iii) The transition metals generally form coloured compounds. This is due to d-d transition of unpaired electrons. In presence of ligands, the d orbitals split into two sets. Transition metal ions absorb radiation of a particular wavelength and reflect the remaining. This imparts colour.
(iv) Transition metals and their many compounds act as good catalyst. Transition metals show variable oxidation states and forms complexes. They form unstable intermediate compounds. They provide a new path with lower activation energy of reaction. They also provide a suitable surface for the reaction to occur.
Q.No.8.12 What are interstitial compounds? Why are such compounds well known for transition metals?
Ans. Interstitial compounds are those which are formed when small atoms like, H,N or C are trapped inside the crystal lattices of metals. They are usually non-stoichiometric and are neither typically ionic or covalent.
Interstitial compounds are well known for transition compounds due to its closed crystalline structure with voids in them. The atomic size of transition metals are very large hence have large voids to occupy these small atoms.
Q.No.8.13 How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Ans. In case of transition elements, the variability in the oxidation state is due to participation of (n-1) d orbitals and ns orbitals. The oxidation states differ by unity. Thus, V shows +2,+3,+4 and +5 oxidation states and Mn shows +2,+3,+4,+5,+6 and +7 oxidation states. On the other hand, the variable oxidation states shown by some p block elements differ by two units. Thus, tin shows +2 and +4 oxidation state and indium shows +1 and +3 oxidation states.
Q.No.8.14 Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Q 8.15 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
Ans.
Q. 8.16 Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (a) iron (II) ions (b) SO2 (c) oxalic acid? Write the ionic equations for the reaction.
Ans.
Q8.17 For M2+/M and M3+/M2+ systems the EV values for some metals are as follows:
Cr2+/Cr -0.9V
Cr3/Cr2+ -0.4 V
Mn2+/Mn -1.2V
Mn3+/Mn2+ +1.5 V
Fe2+/Fe -0.4V
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.
Ans. (a) The reduction potential of Cr3+/Cr2+ is negative and hence, Cr(III) is most stable as it cannot be reduced to Cr(II).
Mn3+/Mn2+ has large positive value and Mn(III) is least stable as it can be easily reduced to Mn(II).
Fe3+/Fe2+ has small positive value and Fe(III) is more stable than Mn(III) but less stable than Cr(III).
(b) Mn2+/Mn has most negative reduction potential and it is most easily oxidised. The ease of oxidation is Mn > Cr > Fe.
Q8.18 Predict which of the following will be coloured in aqueous solution? Ti 3+, V3+ , Cu+ , Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.
Ans. Transition metal ions with incompletely filled d orbitals are colored and those with completely filled d orbitals (or vacant d orbitals) are colorless.
Ti3+,V3+, Mn2+ and Co2+ are colored due to d-d transition in incompletely filled d orbitals.
Cu+,Sc 3+ are colorless due to completely filled and empty d orbitals respectively.
Q8.19 Compare the stability of +2 oxidation state for the elements of the first transition series.
Ans. The relative stability of +2 oxidation state increases on moving from Sc to Zn. This is because on moving from left to right, it becomes more and more difficult to remove the third electron from the d-orbital because of the increasing nuclear charge.
Q8.20 Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic configuration
(ii) atomic and ionic sizes and
(iii) oxidation state
(iv) chemical reactivity.
Ans. (i) Electronic configuration: -
lanthanoids have general electronic configuration [Xe]4f1−14 5d0−1 6s2. Actinoids have general electronic configuration [Rn]5f1−14 6d0−17s2. In lanthanoids, 4f orbital is filled and in actinoids, 5f orbital is filled.
(ii) Atomic and ionic sizes: -
The atomic and ionic sizes of lanthanoids and actinoids decrease with increase in the atomic number due to lanthanoid and actinoid contraction.
(iii) Oxidation state: -
Apart from +3 oxidation state, lanthanoids show +2 and +4 oxidation states due to large energy gap between 4f and 5d subshells. Whereas actinoids show large number of oxidation states due to small energy gap between 5f and 6d subshells.
(iv) Chemical reactivity: -
Highly electropositive lanthanoids have almost similar chemical reactivity. Actinoids (electropostive and highly reactive) are more reactive (specially in finely divided state) than lanthanoids.
Q8.21 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 reagents it is easily oxidised.
(iii) The d 1 configuration is very unstable in ions.
Ans. (i) Cr3+ /Cr2+ has negative value of standard electrode potential (−0.41 V) and Mn3+/Mn2+ have positive value (+1.57 V). Cr2+ can undergo oxidation and acts as reducing agent. Mn3+ can undergo reduction and is oxidizing agent.
(ii) Co (III) being more stable than Co(II), the change in the oxidation state of cobalt from +2 to +3 is easy in presence of complexing agent.
(iii) Once ns electrons are lost, removal of d1 electron will result in a stable electronic configuration. Such elements either acts as reducing agents or undergo disproportionation.
Q8.22 What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.
Ans. In disproportionation reactions, the same substance is oxidized as well as reduced.
For example,
(a) 2H2O2 → 2H2O + O2
(b) 3MnO42- + 4H+ → 2MnO4- + MnO2 + 2H2O
Q8.23 Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
Ans. Copper metal (with electronic configuration [Ar]3d10 4s1) in the first series of transition metals exhibits +1 oxidation state most frequently as it readily loses one electron (present in 4s orbital) to give stable 3d10 electronic configration.
Q8.24 Calculate the number of unpaired electrons in the following gaseous ions: Mn3+ , Cr3+, V3+ and Ti 3+. Which one of these is the most stable in aqueous solution?
Ans. The electronic configuration of various ions are given below:
Mn3+ :3d4, Number of unpaired electrons is 4.
Cr3+ :3d3, Number of unpaired electrons is 3.
V3+ :3d2, Number of unpaired electrons is 2.
Ti3+:3d4, Number of unpaired electrons is 1.
Cr3+ is most stable in aqueous solution.
Q8.25 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.
Ans. (i) Mn(II)O is basic and Mn2(VII)O7 is acidic.
Cr(II)O is basic and Cr2(III)O3 is amphoteric.
The ions in lower oxidation state can readily donate electrons and acts as bases and the ions in high oxidation states have a tendency to accept electrons and act as an acid.
(ii) Small sized (and highly electronegative) oxygen and fluorine can readily oxidize the metals. For example, Os (VI)F6 and V2(VI)O5.
(iii) Due to high electronegativity and high oxidizing power of oxygen, metal oxoanions have highest oxidation state. For example, In dichromate ion, Cr has +6 oxidation state and in permanganate ion, Mn has +7 oxidation state.
Q 8.26 Indicate the steps in the preparation of
i. K2Cr2O7 from chromite ore
ii. KMnO4 from pyrolusite ore.
Ans.
Q8.27 What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Ans. An alloy is a solid or solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to posses different physical properties than those of the component elements.
An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (4-95%), iron (5%) and traces of S,C,Si,Ca and Al
Uses
(1) Mischmetal is used in cigarettes and gas lighters
(2) It is used in flame throwing tanks
(3) It is used in flame bullets and shills.
Q8.28 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.
Ans. Inner transition metals are those elements in which the last electron enters the f-orbital. The elements in which the 4f and the 5f orbitals are progressively filled are called f-block elements. These includes lanthanoids (Z=58−71) and actinoids (Z=90−103).
Elements with atomic numbers 59, 95 and 102 are inner transition elements.
Q8.29 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.
Ans. Apart from +3 oxidation state, lanthanoids show +2 and +4 oxidation states due to large energy gap between 4f and 5d subshells. Whereas actinoids show large number of oxidation states due to small energy gap between 5f and 6d subshells.
For example, uranium (Z = 92) shows +3, +4, +5 and +6 oxidation states and neptunium (Z=94) shows +3,+4,+5,+6 and +7 oxidation states due to small energy gaps between 5f and 6d orbitals.
Q8.30 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.
Ans. Lawrencium (Z=103) is the last element in the actinoid series.
Its electronic configuration is [Rn]86 5f14 6d17s2.
The most common oxidation state displayed by it is +3 because after losing 3 electrons it attains stable f14 configuration.
Q8.31 Use Hund's rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.
Q8.32 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.
Ans.
Ce (Z=58), Pr (Z=59), Nd (Z=60), Tb (Z=65) and Dy (Z=66) shows +4 oxidation state. They acquire electronic configuration closer to 4f0 and 4f7.
Nd (Z=60), Sm (Z=62), Eu (Z=63), Tm (Z=69) and Yb (Z=70) shows +2 oxidation state. They acquire half filled (4f7) and completely filled (4f14) electronic configurations.
Q 8. 33 Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) electronic configuration (ii) atomic and ionic sizes and (iii) oxidation state (iv) chemical reactivity.
Ans.
Q8.34 Write the electronic configurations of the elements with the atomic numbers 61,91,101, and 109.
Q8.35 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
Ans.
(i) Electronic configuration
In the first transition series, 3d orbitals are progressively filled whereas, in second and third transition series, 4d and 5d orbitals are filled.
(ii) Oxidation states
+2 and +3 oxidation states of first transition series elements are more stable than +2 and +3 oxidation states of second and third transition series elements. Cobalt forms several complexes with +2 and +3 oxidation states but no such complexes are known for Rh,Ir which belong to the same group to which Co belongs. The maximum oxidation state in a group increases from first transition series to third transition series. In group 8, iron shows +2 and +3 oxidation states and ruthenium and osmium show +4, +6 and +8 oxidation states.
(iii) Ionization enthalpies
5d series elements have higher ionization enthalpies than 3d and 4d series elements. In 5d series, 4f orbitals are filled which have poor shielding effect and valence electrons experience higher effective nuclear charge. This results in higher ionization enthalpies for 5d series elements.
(iv) Atomic sizes
4d series elements have higher atomic size than 3d series elements.
4d and 5d series elements have nearly same atomic size due to lanthanoid contraction.
Q8.36
Q8.37 Comment on the statement that elements of the first transition series posses many properties different from those of heavier transition elements.
Ans. Elements of the first transition series possess many properties different from those of heavier transition elements in the following ways-
a. The atomic size of the 1st transition series is smaller than those of 2nd and 3rd series elements. But due to lanthanoid contraction, atomic size of the 2nd series elements are nearly the same as 3rd series element of the corresponding same vertical group.
b. In 1st transition series +2 and +3 oxidation states are more common but in the 2nd and 3rd series higher oxidation states are more common.
c. The enthalpy of atomisation of first series elements is lower than 2nd and 3rd series elements.
d. The melting and boiling point of the 1st transition series is less than that of heavier metals. This is because of strong metallic bonding in heavier metals.
Q8.38
