NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements are part of NCERT Solutions for Class 12 Chemistry. Here we have given NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements.
|Chapter Name||d-and f-Block Elements|
|Number of Questions Solved||48|
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements
NCERT IN-TEXT QUESTIONS
Silver has completely filled d-orbitals (4d10) in its ground state. How can you say that it is a transition metal ?
Silver (Z = 47) belongs to group 11 of (7-block (Cu, Ag, Au) and its outer electronic configuration is 4d105s1. It shows + 1 oxidation state (4d10 configuration) in silver halides (e.g. AgCl). However, it can also exhibit + 2 oxidation state (4d9 configuration) in compounds like AgF2 and AgO. Due to the presence of half filled d-orbital, silver is a transition metal.
In the series Sc(Z = 21) to (Z = 30), the enthalpy of atomisation of zinc is the lowest i.e., 126 kJ mol-1. Why ?
The enthalpy of atomisation is directly linked with the stability of the crystal lattice and also the strength of the metallic bond. In case of zinc (3d104s2 configuration), no electrons from the 3d-orbitals are involved in the formation of metallic bonds since all the orbitals are filled. However, in all other elements belonging to 3d series one or more d-electrons are involved in the metallic bonds. This means that the metallic bonds are quite weak in zinc and it has therefore, lowest enthalpy of atomisation in the 3d series.
Which of the 3d series of transitional metals exhibits largest oxidation states and why ?
Mn (Z = 25) with electronic configuration [Ar]3d54s2 shows maximum oxidation state (+ 7) in its compounds since it has the maximum number of unpaired five i.e., seven. It shows largest variable oxidation state from + 2 to + 7 ( + 2, + 3, + 4, + 5, + 6, + 7) in its compounds.
The E°(M2+/M) value for copper is positive (+ 0·34 V). What is possibly the reason for this ? (C.B.S.E. Outside Delhi 2012, Sample Paper 2012)
E°(M2+/M) for any metal is based upon three factors which have been discussed in the text part.
M(s) + ∆aH → M(g) ; (∆aH = Enthalpy of atomisation)
M(g) + ∆fH → M2+(g) ; (∆fH = Ionisation enthalpy)
M2+(g) + aq → M2+(aq); (∆hydH = Hydration enthalpy)
Copper has very high enthalpy of atomisation (energy required) and low enthalpy of hydration (energy released). In nut shell, the ∆fH i.e. ionisation enthalpy needed is not compensated by the energy released. Therefore E°(Cu2+/Cu) is positive.
How would you account for the irregular variation in ionisation enthalpies (first and second) in first series of transition elements ?
The ionisation enthalpies of the transition metals are higher than those of s-block elements and less than the elements of p-block. Thus, these are less electropositive than the elements of s-block and at the same time more electropositive than the elements belonging to p-block present in the same period. In a transition series, the ionisation enthalpies increase from left to right. However, the gaps in the values of the two successive elements are not regular.
Why is the highest oxidation state of a metal exhibited by its fluoride and oxide only ? (C.B.S.E. Delhi 2010)
Both fluorine and oxygen have very high electronegativity values. They can oxidise the metals to the highest oxidation state. As a result, the highest oxidation states are shown by the fluorides and oxides of the metals; transition metals in particular.
Which is a stronger reducing agent Cr2+ or Fe2+ and why ? (SamplePaper 2011, C.B.S.E. Outside Delhi 2010, 2014)
Cr2+ is a stronger reducing agent than Fe2+. This is quite evident from the E° values ;
E°Cr3+/Cr2+ = – 0-41 V and E°Fe3+/Fe2+ = 0-77 V.
Reason : d4 → d3 occurs when Cr2+ changes to Cr3+ ion while d6 → d5 takes place when Fe2+ gets converted to Fe3+ ion.
Now, d4 → d3 transition is easier as compared to d6 → d5 transition because in the second case, an electron is removed from a paired orbital which is rather difficult. Therefore, Cr2+ is a stronger reducing agent than Fe2+.
Calculate the spin magnetic moment of M2+(aq) ion (Z = 27).
Electronic configuration of element M(Z = 27) : [Ar] 3d74s2
Electronic configuration M2+ (aq) ion : 3d7 or
Magnetic moment of M2+ (aq) ion with n = 3 ;
Explain why Cu+ ion is not stable in aqueous solution. (C.B.S.E. Delhi 2011)
In aqueous solution Cu+ (aq) undergoes disproportionation to form Cu2+ (aq) ion and Cu.
2Cu+(aq) → Cu2+(aq) + Cu(s)
The higher stability of Cu2+ (aq) in aqueous solution may be attributed to its greater negative ∆hydH⊝ than that of Cu+ (aq). It compensates the second ionisation enthalpy of Cu involved in the formation of Ci2+ ion. Thus, Cu+ (aq) ion changes to Cu2+ (aq) ion which is more stable.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why ? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)
The decrease or contraction in atomic radii as well as ionic radii in actinoid elements (actinoid contraction) is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.
Write down the electronic configuration of :
Why are Mn2+ compounds more stable than Fe2+ compounds towards oxidation to their +3 state ?
Electronic configuration of Mn2+ is 3d5 while that of Fe2+ is 3d6. This shows that Fe2+ ion has an urge to change to Fe3+ ion by losing an electron whereas Mn2+ ion has no such tendency. Thus, + 2 oxidation state of Mn is more stable as compared to + 2 oxidation state of Fe.
Explain briefly how + 2 oxidation state becomes more and more stable in the first half of the first row transition elements with increasing atomic number.
In all the elements listed, with the removal of valence 45 electrons (+2 oxidation state), the 3d-orbitals get gradually occupied. Since the number of the empty d-orbitals decreases or the number of unpaired electrons in 3d orbitals increases, the stability of the cations (M2+) increases from Sc2+to Mn2+.
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 an example.
The presence of half filled or completely filled orbitals imparts stability to a particular element/ion. Greater the number of such orbitals, more will be the relative stability. For example, let us write the different oxidation states of Mn (Z = 25) along with the electronic configurations. Mn: [Ar] 3d54s2 ; 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).
What must be the stable oxidation state of the transition elements with the following electronic configuration in the ground states of their atoms : 3d3, 3d5, 3d8, 3d4 ?
The maximum oxidation states of reasonable stability in the transition metals of 3d series correspond to the sum of s and d-electrons upto Mn. However, after Mn there is an abrupt decrease in oxidation states. In the light of this, most stable oxidation states of the elements are :
3d3 : 3d3s2 (+ 5);3 : 3d54s1 (+ 6) and 3d54s2 (+ 7)
3d8 : 3d84s2 (+ 2); 3d4 : 3d44s2 or 3d54s1 (+6)
Name the oxometal anions in the first transition series of transition metals in which the metal exhibits oxidation state equal to its group number.
Upto the element Mn, the oxidation state leading to the maximum stability corresponds to value equal to sum of s- and d-electrons i.e., the group number. For example,
[Sc(III)O2]–, [Ti(IV)O3]2-, [V(V)O3]–, [Cr(VI)O4]2-, [Mn(VII)O4]–
However, for the remaining elements, the oxidation states are not related to their group numbers.
What is lanthanoid contraction ? What are the consequences of lanthanoid contraction ? (C.B.S.E. Delhi 2013)
One common property associated with the elements in the periodic table is the variation in their atomic and ionic radii down the group and along a period. In general, these increase down the group due to the increase in the number of shells and decrease along a period considerably because of the increase in the magnitude of the effective nuclear charge.
Consequences of Lanthanoid Contraction
(a) Separation of Lanthanoids: Separation of lanthanoids is possible only due to lanthanoid contraction. All the lanthanoids have quite similar properties and due to this reason they are difficult to separate. However, because of lanthanoid contraction their properties (such as ionic size, ability to form complexes etc.,) vary slightly.
(b) Variation in basic strength of hydroxides: The basic strength of oxides and hydroxides decreases from La(OH)3 to Lu(OH)2. Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. The acidic strength which involves the cleavage of O—H bond follows the reverse trend i.e. it increases along the series.
(c) Similarly in the atomic sizes of the elements of second and third transition series present in the same group: We know that the atomic sizes of the elements generally increase appreciably down a group. Similar trend is also expected in the elements present in the different groups of d-block.
(d) Variation in standard reduction potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.
M3+ (aq) + 3e– → M (aq)
(e) Variation in physical properties like melting point, boiling point, hardness etc: Various physical properties like m.pt., b.pt., hardness etc., increase with the increase in atomic number. This is because the attraction forces between the atoms increase as the size decreases.
What are the characteristics of transition elements and why are they called transition elements ? Which of the d- block elements may not be regarded as the transition elements ?
General characteristics of transition elements.
- Electronic configuration – (n – 1) d1-10 ns1-2
- Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.
- Atomic and ionic size – Ions of same charge in a given series show progressive decrease in radius with increasing atomic number.
- Oxidation state – Variable ; ranging from + 2 to + 7.
- Paramagnetism – The ions with unpaired electrons are paramagnetic.
- Ionisation enthalpy – Increases due to increase in molecular charge.
- Formation of coloured ions – Due to unpaired electrons.
- Formation of complex compounds – Due to small size and high charge density of metal ions.
- They possess catalytic properties – Due to their ability to adopt multiple oxidation states.
- Formation of interstitial compounds.
- Alloy formation.
They are called transition elements due to their incompletely filled d-orbitals in ground state or any stable oxidation state and they are placed between s and p-block elements. Zn, Cd and Hg have fully filled d° configuration in their ground state hence may not be regarded as the transition elements.
In what way are the electronic configuration of the transition elements different from non-transition elements ?
Electronic configuration of transition elements : (n – 1)d1-10 ns1-2. Electronic configuration of non-transition elements : ns1-2 or ns2np1-6. From comparison, it is quite evident that the transition elements have incomplete d-orbitals (s- orbitals in some cases) while the non-transition elements have no d-orbitals present in the valence shells of their atoms. This is responsible for the difference in the characteristics of the elements belonging to these classess of elements.
What are the different oxidation states exhibited by lanthanoids ?
The common stable oxidation state of lanthanoids is + 3. However, some may also exhibit + 2 and + 4 oxidation states.
Explain giving reason :
(a) Transition metals and many of their compounds show paramagnetic behaviour. (H.P. Board 2014)
(b) The enthalpies of atomisation of transition metals are high. (Jharkhand Board 2010, C.B.S.E. Outside Delhi 2008, 2012, H.P. Board 2014)
(c) The transition metals generally form coloured compounds. (C.B.S.E. 2010. 2012)
(d) The transition metals and their compounds act as good catalysts. (C.B.S.E.Outside Delhi 2010) (C.B.S.E. Delhi 2008, Sample Paper 2010, H.P. Board 2017)
(a) Paramagnetism arises from the presence of unpaired electrons, each such electron has magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment,is determined by the number of unpaired electrons and is calculated by using the ‘spin’ only’ formula, i.e., µ = B.M
(b)Enthalpy of atomisation may be defined as the amount of heat energy needed to break the metal lattice of a crystalline metal into free atoms. Greater the magnitude of lattice energy, more will be the value of enthalpy of atomisation. The transition metals have high enthalpies of atomisation because the metallic bonds present are quite strong due to presence of large number of half-filled atomic orbitals. For more details, consult text-part.
(c) Due to presence of unpaired electrons and d-d transitions, the transition metals are generally coloured. When an electron from a lower energy d orbital is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed. The frequency of the light absorbed is determined by the nature of the ligand.
(d) The transition metals and their compounds are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes. Vanadium(V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in r Catalytic Hydrogenation) are some of the examples.Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst.
What are interstitial compounds ? Why are such compounds well known for transition metals ?
Interstitial compounds are the compounds formed as result of the trapping of atoms of small elements like H, N, C, B etc. in the crystal lattices of certain metals. These are non-stoichiometric in nature and are neither ionic nor covalent. In fact, no proper bonds exist in the atoms of metals and non-metals involved in these compounds. Transition metals havetendency to form such compounds. A few examples are: TiC,
Mn4N, Fe3H, VH0·56, Vse0·98 and Fe0·94O etc.
- These are generally non stoichiometric in nature. Therefore, they cannot be representived by a definite structure or formula.
- The compounds are neither covalent nor ionic and they donot represent the normal oxidation states of the metals.
- Since the strengths of the metallic bonds in these compounds increase due to greater electronic interactions, they show high melting points and high metallic conductivity. However, these compounds are chemically inert.
- The conductivity of the metals remains unaffected in the corresponding interstitial compounds.
How is the variability in oxidation states of transition metals different from that of the non-transition metals ? Illustrate with examples.
The transition metals show a number of variable oxidation states due to the participation of (n – 1) d electrons in addition to ns electrons in the bond formation. They therefore, exhibit a large number of variable oxidation states. On the other hand, the non-transition metals generally belonging to s-block do not show variable oxidation states because by the loss of valence s-electrons, they acquire the configuration of the nearest noble gas elements.
In the p-block the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of d-block. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO3 and WO3 are not.
Describe the preparation of potassium dichromate from chromite ore. What is the effect of increasing pH on a solution of potassium dichromate ?
Preparation from chromite:
Potassium dichromate is generally prepared from chromite ore (FeCr2O4). It is infact, a mixed oxide Fe0.Cr2O3 of iron and chrome also called ferro-chrome or chrome iron.
(i) Conversion of chromite ore into sodium chromate : Chromite ore is fused with sodium hydroxide or sodium carbonate in the presence of air.
(ii) Conversion of sodium chromate into sodium dichromate : The fused mass obtained above is extracted with water. Sodium chromate which is soluble in water goes into the solution leaving behind the insoluble ferric oxide (Fe2O3). The yellow solution of sodium chromate obtained above is treated with concentrated H2SO4 to form sodium dichromate which has an orange colour.
(iii) Conversion of sodium dichromate into potassium dichromate: Sodium dichromate is more soluble and less stable than potassium dichromate.
Effect of increasing pH : The solution of potassium dichromate (K2Cr2O7) in water is orange in colour. On increasing the pH i.e. on adding the base, the potassium dichromate changes to potassium chromate (K2CrO4) which is yellow in colour. Thus, on increasing the pH, the colour of the solution changes from orange to yellow.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with :
(b) iron (II) solution
Describe the preparation of potassium permanganate. How does acidified permanganate solution react with :
(a) iron (II) solution
(c) oxalic acid ?
Write the ionic equations for the reactions.
Potassium permanganate is prepared on a large scale from the mineral pyrolusite, MnO2.
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
For M2+/M and M3+/M2+ systems, the E° values of some metals are given :
Use this data to comment upon :
(a) The stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+.
(b) The ease with which iron can be oxidised as compared to the similar process for either chromium or manganese metal.
As is negative (- 0·4 V), this means that Cr3+ ions in solution cannot be reduced to Cr2+ ions or Cr3+ ions are very stable. As farther comparison of E° values shows that Mn3+ ions can be reduced to Mn2+ ion more readily than Fe3+ ions. Thus, in the light of this, the order of relative stabilities of different ions is :
Mn3+ < Fe3+ < Cr3+.
(b) From the E° values, the order of oxidation of the metal to the divalent cation is :
Mn > Cr > Fe.
Predict which of the following will be coloured in aqueous solution ?
Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+, Co2+.
Only those ions will be coloured which have incomplete d-orbitals. The ions with either empty or filled d-orbitals are colourless. Keeping this in view, the coloured ions among the given list are :
Ti3+(3d1), V3+(3d2), Mn2+(3d5), Fe3+(3d5), Co2+ (3d7)
Sc3+ (3d°) and Cu+ (3d10) ions are colourless.
Compare the stability of +2 oxidation state for the elements of the first transition series.
The common oxidation state of 3d series elements is + 2 which arises due to participation of only 4s electrons. The tendency to show highest oxidation state increases from Sc to Mn, then decreases due to pairing of electrons in 3d subshell. Thus in the series Sc(II) does not exist, Ti(II) is less stable than Ti(IV). At the other end of the series, oxidation state of Zn is +2 only.
Compare the chemistry of lanthanoids and actinoids with special reference to :
(a) electronic configuration
(b) oxidation state
(c) atomic and ionic sizes
(d) chemical reactivity.
(a) Electronic configuration: The actinoids involve the gradual filling of 5f-subshell in their atoms. Actinium has the outer electronic configuration of 6d17s2. From thorium (Z = 90) onward 5f-subshell gets progressively filled. Because of almost equal energies of 5fand 6d-subshells, there are some doubts regarding the filling of 5fand 6d-subshells.
(b) Oxidation states: Members of the actinoid family exhibit more variable oxidation states as compared to the elements belonging to lanthanoid family. This is due to the reason that there is less energy difference in 5d, 6fand 7s orbitals belonging to actinoid family than the energy difference in 5f, 4d and 65 orbitals in case of lanthanoid family.
(c) Atomic and ionic sizes : The atomic size of lanthanoids decreases from lanthanum to lutetium. Though the decrease is not regular, in case of atomic radii, the decrease in the ionic size (M3+) is regular. Decrease in size between two successive elements is higher in actinoids due to poor screening by 5f electrons.
(d) Chemical reactivity: Actinoids are highly reactive metals especially when these are in the finely divided form. Upon boiling with water, they form respective oxides and hydroxides. They combine with a variety of non-metals at moderate temperature. On reacting with hydrochloric acid, they form their respective chlorides.
How would you account for the following :
(a) Of the d4 species, Cr2+ is strongly reducing while Mn3+ is strongly oxidising in nature.
(b) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents, it is easily oxidised.
(c) d1 configuration is very unstable in ions.
(a) E° value of Cr3+/Cr2+ is negative (-0·41 V) while that of Mn3+/Mn2+ is positive (+ 1·57 V). This means Cr2+ ions can lose electrons to form Cr3+ ions and act as a reducing agent while Mn3+ ions can accept electrons and can act as oxidising agent.
(b) Cobalt (II) is stable in aqueous solution but in the presence of complexing agent, it undergoes change in oxidation state from +2 to +3 and is easily oxidised.
(c) The ion with d1 configuration is expected to be extremely unstable and has a great urge to acquire d° configuration (very stable) by losing the only electron present in the d-sub-shell.
What is meant by ‘disproportionation’ ? Give two examples of disproportionation reaction in aqueous solution.
In a disproportionation reaction, an element undergoes an increase as well as decrease in its oxidation state forming
two different compounds. In other words, we can say that it can act both as reducing agent as well as oxidising agent.
Which metal in the first transition metal series exhibits +1 oxidation state most frequently and why ?
Cu with configuration [Ar] 4s13d10 exhibits +1 oxidation state and forms Cu+ ion because by losing one electron, the cation or positive ion acquires a stable configuration of d-orbitals (3d10).
Calculate the number of unpaired electrons in the following gaseous ions :
Mn3+, Cr3+, V3+.
Give example and suggest reasons for the following features of transition metal chemistry.
(a) The lowest oxide of the transition metal is basic while the highest is acidic.
(b) A transition metal exhibits higher oxidation states in oxides and fluorides. (C.B.S.E. Sample Question Paper 2012)
(c) The highest oxidation state is exhibited in oxoanions of a metal.
(a) It may be noted that in the different oxides of the same element, the acidic strength increases with the increase in oxidation state of the element. MnO (Mn2+) is basic while Mn2O7 (Mn7+) is acidic in nature.
(b) Both oxygen and fluorine being highly electronegative can increase the oxidation state of a particular transition metal. In certain oxides, the element oxygen is involved in multiple bonding with the metal and this is responsible for the higher oxidation state of the metal. For detail, consult text part.
(c) This is also because of the high electronegativity of oxygen. For example, chromium exhibits oxidation state of (VI) in oxoanion [Cr(VI)O4]2- and manganese shows oxidation state of (VII) in oxoanion [Mn(VII)O4]– .
Give the steps in the preparation of : (C.B.S.E. Delhi 2009 comptt.)
(a) K2Cr2O7 from chromite ore
(b) KMnO4 from pyrolusite ore.
What are alloys ? Name an alloy which contains some lanthanoid metals. Mention its uses.
An alloy is a homogeneous mixture of different metals or metals and non-metals.
Misch metal is an alloy of cerium (Ce). lanthanum (La), neodymium (Nd), iron (Fe) and traces of carbon, sulphur, aluminium etc. It is used in making parts of jet engines.
What are inner transition elements ? Decide which of the following atomic numbers belong to inner transition elements :
29, 59, 74, 95, 102, 104.
The inner transition elements also called/-block elements include the series of lanthanoids (Z = 58 to 71) and actinoids (Z = 90 to 103). This means that the elements with atomic numbers 59, 95 and 102 belong to inner transition elements.
The chemistry of actinoid elements is not so smooth as that of lanthanoids. Justify this statement by giving some examples from the oxidation states of these elements.
The lanthanoids mostly exhibit +3 oxidation states in their compounds. Actinoid elements also normally show +3 oxidation states. But in their cases, 5f, 6d and 7s energy levels are comparable and have very small energy difference in them. As a result, they can exhibit a number of oxidation states. For example,
Neptunium (Np) : +3, +4, +5, +6, +7
Plutonium (Pu) : +3, +4, +5, +6, +7
Americium (Am) : +3, +4, +5, +6.
Which is the last element in the series of actinoids ? Write the electronic configuration of the element. Comment upon its possible oxidation state.
Lawrencium (Lr = 103); [Rn] 5f146d17s2 oxidation state = +3.
Use Hund’s rule to derive the electronic configuration of Ce3+ ion and calculate its magnetic moment on the basis of spin-spin formula.
Ce(Z = 58) = [Xe]54 4f15d16s2 ; Ce3+ = [Xe]54 4f4 (only one unpaired electron).
Magnetic moment (p) = = 1·73 BM.
Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to co-relate this type of behaviour with the electronic configuration of these elements.
+4 oxidation state : 58Ce, 59Pr, 65Tb
+ 2 oxidation state : 60Nd, 62Sm, 63Eu, 69Tm, 70Yb.
In general +2 oxidation state is exhibited by the elements with configuration 5d06s2 so that two electrons may be easily lost. Similarly +4 oxidation state is shown by the elements which after losing four electrons acquire configuration either close to 4f0 or 4f7.
Compare the chemistry of the actinoids with that of lanthanoids with reference to :
(i) Electronic configuration
(ii) Oxidation states
(iii) Chemical reactivity.
(i) Electronic configuration: The actinoids involve the gradual filling of 5f-subshell in their atoms. Actinium has the outer electronic configuration of 6d17s2. From thorium (Z = 90) onward 5f-subshell gets progressively filled. Because of almost equal energies of 5fand 6d-subshells, there are some doubts regarding the filling of 5fand 6d-subshells.
(ii) Oxidation states: Members of the actinoid family exhibit more variable oxidation states as compared to the elements belonging to lanthanoid family. This is due to the reason that there is less energy difference in 5d, 6fand 7s orbitals belonging to actinoid family than the energy difference in 5f, 4d and 65 orbitals in case of lanthanoid family.
(iii) Chemical reactivity: Actinoids are highly reactive metals especially when these are in the finely divided form. Upon boiling with water, they form respective oxides and hydroxides. They combine with a variety of non-metals at moderate temperature. On reacting with hydrochloric acid, they form their respective chlorides.
Write the electronic configuration of the elements with atomic numbers 61, 91, 101, 109.
Promethium or Pm (Z = 61) [Xe]544f55d06s2
Protactinium or Pa (Z = 91) [Rn] 4f26d17s2
Mendelevium or Md (Z = 101) [Rn] 5f16d07s2
Meitnerium or Mt (Z = 109) [Rn] 5f146d77s2
Compare the general characteristics of the first transition series of transition metals with those of the second and third transition series metals in the respective vertical columns. Give special emphasis on the following points :
(i) electronic configuration
(ii) oxidation states
(iii) ionisation enthalpies
(iv) atomic sizes.
(i) Electronic configuration. There are some exceptions in the electronic configurations in all the three series.
(ii) Oxidation state. The elements belonging to the different series but present in the same group have similar electronic configuration and therefore, exhibit almost same variable oxidation states. In general, these are maximum in the middle of the series while minimum towards the end. Transition elements show variable oxidation states due to the participation of ns and (n – 1) d electrons in bonding because the energies of ns and (n – 1) d-subshells are quite close. The stability of a particular oxidation state depends upon the nature of the element with which the transition metal forms the compound.
(iii) Ionisation enthalpies. In general, the ionisation enthalpies in all the three transition series increase from left to the right. However, the gaps in the two successive elements in a particular series are small and are also not regular. The first three ionisation enthalpies of the elements present in the first transition series are given in the text part. The ∆iH1 [ values of the elements belonging to 5d series and higher as compared to those belonging to 3d and Ad series in the same group because of poor shielding by intervening 4f electrons present.
(iv) Atomic size. In all the three transition series, the atomic as well as ionic radii of the elements increase from left to the right. The values for 3d series are given in the text part. However, the increase in their values are not as much as expected since the shielding by (n – 1 )d electrons is not as much as expected. In a particular group, the atomic radius of the elements belonging to Ad series is more than the elements in the 3d series. However, the gaps in the elements belonging to Ad and 5d series are negligible on account of lanthanoid contraction which the elements of 5d experience.
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 be occupied for these hydrated ions (octahedral).
The number of 3d electrons in the ions are :
For the explanation of the involvement of 3d orbitals in the hydrated ions (octahedral in nature) consult next unit on co-ordination compounds.
Comment on the statement that elements of first transition series possess many properties different from those of the heavier transition elements.
The heavier transition elements belong to fourth (Ad) and fifth (5d) and sixth (6d) transition series. Their properties are expected to be different from the elements belonging to the first (3d) series due to the following reasons :
(i) The atomic radii of the elements belonging to Ad and 5d series are more due to greater number of electron shells. However, the difference in Ad and 5d transition elements are comparatively less because of lanthanoid contraction.
(ii) Because of stronger inter atomic bonding, the m.p. and b.p. of the elements of Ad and 5d series are higher.
(iii) Ionisation enthalpies are expected to decrease as we move from one series to the other. However, the values for the elements of 5d series are higher as compared to the elements belonging to the other two series due to lanthanoid contraction.
Actually atomic size decreases on account of it and effective nuclear charge increases. As a result, there is an increase in ionisation energy in case of 3d elements.
What can be inferred from the magnetic moment values of the following complex species ?
The magnetic moment of a compound is given by the relation (µ) = B.M, where n is the number of unpaired electrons.
For one unpaired electron (n = 1) ; µ =
For two unpaired electrons (n – 2) ; µ =
For three unpaired electrons (n = 3); µ =
For four unpaired electrons (n = 4) ; µ =
For five unpaired electrons (n = 5) ; µ =
*In the light of the above value, let us gather the desired information about the complex species that are mentioned
Oxidation state of Mn : [Mn(CN)6]4- , x + 6(-l) = -4 or x = -4 + 6 = + 2
The magnetic value of 1·73 B.M. indicates the presence of one unpaired electron in the complex. When six, CN– ions (or ligands) approach Mn2+ ion, electrons in 3d orbitals pair up to make available six vacant orbitals involving d2sp3 hybridisation.
The complex is octahedral and is paramagnetic due to one unpaired electron.
Oxidation state of Fe : [Fe(H2O)6]2+ ; r + 6 (0) = +2
The magnetic moment value of 5·3 B.M. indicates that there are four unpaired electrons in the complex. This means that the electrons in Fe2+ ion do not pair up when six H20 molecules (or ligands) approach it. Since the desired number of vacant orbitals (six) are available, die complex formed is sp3d2 hybridised.
The complex is octahedral and is paramagnetic due to four unpaired electrons. It is also called outer orbital complex because 4d (n = 4) orbitals are involved.
Oxidation state of Mn : [MnCl4]2-, x + 4(-1) = -2 or x = -2 + 4= + 2
The magnetic moment value of 5·9 B.M. indicates that there are five unpaired electrons in the complex. This means that all the five 3d orbitals in Mn2+ ion are involved in the bond formation. The complex is sp3 hybridised in which one vacant 4s and three vacant 4p orbitals participate.
The complex is therefore, tetrahedral in nature.
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