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|Chapter||8. The d and f Block Elements|
|Category||NCERT Solutions for Class 12|
NCERT Solutions for Class 12 Chemistry is an excellent tool to revise your syllabus quickly. These solutions are made by expert faculties and teachers of Chemistry. It provides you with complete information, making it easier for you to solve the problems. The NCERT textbook includes questions based on concepts and theories. They are important for testing students’ understanding of the topics. On this page, we have provided you with the NCERT Solutions for Class 12 Chemistry Chapter 8: The d and f Block Elements.
NCERT Solutions for Class 12 Chemistry Chapter 8
The d and f Block Elements Solutions
Q1) Write down the electronic configuration of:
(i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+ (v) Co2+ (vi) Lu2+ (vii) Mn2+ (viii) Th4+
Answer i) Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3
Or, [Ar]18 3d3
Answer ii) Pm3+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f4
Or, [Xe]54 3d3
Answer iii) Cu+: 1s2 2s2 2p6 3s2 3p6 3d10
Or, [Ar]18 3d10
Answer iv) Ce4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6
Answer v) Co2+: 1s2 2s2 2p6 3s2 3p6 3d7
Or, [Ar]18 3d7
Answer vi) Lu2+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d1
Or, [Xe]54 2f14 3d3
Answer vii) Mn2+: 1s2 2s2 2p6 3s2 3p6 3d5
Or, [Ar]18 3d5
Answer viii) Th4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6s6
Q2) Why are Mn2+compounds more stable than Fe2+ towards oxidation to their +3 state?
Electronic configuration of Mn2+ is [Ar]18 3d5.
Electronic configuration of Fe2+ is [Ar]18 3d6.
It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d5 configuration. This is the reason Mn2+ shows resistance to oxidation to Mn3+. Also, Fe2+ has 3d6 configuration and by losing one electron, its configuration changes to a more stable 3d5 configuration. Therefore, Fe2+ easily gets oxidized to Fe+3 oxidation state.
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?
Answer) The oxidation states displayed by the first half of the first row of transition metals are given in the table below.
It can be easily observed that except Sc, all others metals display +2 oxidation state. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also increases from 1 to 5.
Sc (+2) = d1
Ti (+2) = d2
V (+2) = d3
Cr (+2) = d4
Mn (+2) = d5
+2 oxidation state is attained by the loss of the two 4s electrons by these metals. Since the number of d electrons in (+2) state also increases from Ti(+2) to Mn(+ 2), the stability of +2 state increases (as d-orbital is becoming more and more half-filled). Mn (+2) has d5 electrons (that is half-filled d shell, which is highly stable).
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.
Answer) The elements in the first-half of the transition series exhibit many oxidation states with Mn exhibiting maximum number of oxidation states (+2 to +7). The stability of +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital. However, Sc does not show +2 oxidation state. Its electronic configuration is 4s2 3d1. It loses all the three electrons to form Sc3+. +3 oxidation state of Sc is very stable as by losing all three electrons, it attains stable noble gas configuration, [Ar]. Ti (+ 4) and V(+5) are very stable for the same reason. For Mn, +2 oxidation state is very stable as after losing two electrons, its d-orbital is exactly half-filled, [Ar] 3d5.
Q5) 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, 3d8and 3d4?
|Electronic configuration in ground state||Stable oxidation states|
|3d3 (Vanadium)||+2, +3, +4 and +5|
|3d5 (Chromium)||+3, +4, +6|
|3d5 (Manganese)||+2, +4, +6, +7|
|3d8 (Cobalt)||+2, +3|
|3d4||There is no3d4 configuration in ground state.|
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 number.
(i) Chromate, CrO2-4
Oxidation state of Cr is + 6.
(ii) Permanganate, MnO–4
Oxidation state of Mn is + 7.
(iii) Vanadate, VO–3
Oxidation state of V is + 5.
Q7) What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Answer) As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one. As electrons are being added to the same shell, the effective nuclear charge increases. This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron. Also, with the increase in atomic number, the number of electrons in the 4f orbital also increases. The 4f electrons have poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.
Consequences of lanthanoid contraction:
- There is similarity in the properties of second and third transition series.
- Separation of lanthanoids is possible due to lanthanide contraction.
- It is due to lanthanide contraction that there is variation in the basic strength of lanthanide hydroxides. (Basic strength decreases from La(OH)3 to Lu(OH)3.)
Q8) 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?
Answer) Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between s-block and p-block. Therefore, these are called transition elements.
Elements such as Zn, Cd, and Hg cannot be classified as transition elements because these have completely filled d-subshell.
Q9) In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Answer)Transition metals have a partially filled d−orbital. Therefore, the electronic configuration of transition elements is (n − 1)d1-10 ns0-2.
The non-transition elements either do not have a d−orbital or have a fully filled d−orbital. Therefore, the electronic configuration of non-transition elements is ns1-2 or ns2 np1-6.
Q10) What are the different oxidation states exhibited by the lanthanoids?
Answer) In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.
Q11) 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.
Answer 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.
Answer 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 is high.
Answer 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 toanother. 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.
Answer iv) The catalytic activity of the transition elements can be explained by two basic facts.
- 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.
- Transition metals also provide a suitable surface for the reactions to occur.
Q12) What are interstitial compounds? Why are such compounds well known for transition metals?
Answer) Transition metals are large in size and contain lots of interstitial sites. Transition elements can trap atoms of other elements (that have small atomic size), such as H, C, N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds.
Q13) How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Answer) In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation states differ by 1 (Fe2+ and Fe3+; Cu+ and Cu2+). In non-transition elements, the oxidation states differ by 2, for example, +2 and +4 or +3 and +5, etc.
Q14) Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Answer) From iron chromite ore, Potassium dichromate is prepared as in the following steps.
Step (1): Preparation of sodium chromate
4FeCr2O4 + 16NaOH + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8H2O
Step (2): Conversion of sodium chromate into sodium dichromate
2Na2CrO4 + conc.H2SO4 → Na2Cr2O7 + Na2SO4 + H2O
Step (3): Conversion of sodium dichromate to potassium dichromate
Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl
Potassium chloride being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filtration.
The dichromate ion (Cr2 O72-) exists in equilibrium with chromate (Cr2 O42-) ion at pH 4
However, by changing the pH, they can be inter converted.
2CrO2−4 → 2HCrO4− →2Cr2O72 −
Cr2O72− →2HCrO42−→ 2CrO42−
Q15) 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
Answer) k2Cr2O7 acts as a very strong oxidising agent in the acidic medium.
k2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O + 3 [O] k2Cr2O7 takes up electrons to get reduced and acts as an oxidizing agent. The reaction of k2Cr2O7 with H2S, iron (II) solution and other iodide are given below.
Answer i) k2Cr2O7 oxidizes iodide to iodine.
Answer ii) k2Cr2O7 oxidizes iron (II) solution to iron (III) solution i.e., ferrous ions to ferric ions.
Answer iii) k2Cr2O7 oxidizes H2S to sulphur.
Q16) 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.
Answer) Potassium permanganate can be prepared from pyrolusite (MnO2). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO3 or KClO4, to give K2MnO4.
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.
Q17) For M2+/M and M3+/M2+ systems the EV values for some metals are as follows:
Cr3/Cr2+ −0.4 V
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.
i) The reduction potentials for the given pairs increase in the following order.
Mn2+ / Mn < Cr2+ / Cr < Fe2+ /Fe
So, the oxidation of Fe to Fe2+ is not as easy as the oxidation of Cr to Cr2+ and the oxidation of Mn to Mn2+. Thus, these metals can be arranged in the increasing order of their ability to get oxidised as: Fe < Cr < Mn
The EΘ value for Fe3+/Fe2+ is higher than that for Cr3+/Cr2+ and lower than that for Mn3+/Mn2+. So, the reduction of Fe3+ to Fe2+ is easier than the reduction of Mn3+ to Mn2+, but not as easy as the reduction of Cr3+ to Cr2+. Hence, Fe3+ is more stable than Mn3+, but less stable than Cr3+. These metal ions can be arranged in the increasing order of their stability as: Mn3+ < Fe3+ < Cr3+
Q18) Predict which of the following will be coloured in aqueous solution? Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.
Answer) Only the ions that have electrons in d-orbital and in which d-d transition is possible will be coloured. The ions in which d-orbitals are empty or completely filled will be colourless as no d-d transition is possible in those configurations.
|Element||Atomic Number||Ionic State||Electronic configuration in ionic state|
All other ions, except Sc3+, will be coloured in aqueous solution because of d−d transitions. Not Sc3+ as it has an empty d-orbital.
Q19) Compare the stability of +2 oxidation state for the elements of the first transition series.
From the above table, it is evident that the maximum number of oxidation states is shown by Mn, varying from +2 to +7. The number of oxidation states increases on moving from Sc to Mn. On moving from Mn to Zn, the number of oxidation states decreases due to a decrease in the number of available unpaired electrons. The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the d-orbital.
Q20) Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic configuration (iii) oxidation state
(ii) atomic and ionic sizes and (iv) chemical reactivity.
Answer i) Electronic configuration: The general electronic configuration for lanthanoids is [Xe]54 4f0-14 5d0-1 6s2 and that for actinoids is [Rn]86 5f1-14 6d0-1 7s2. Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.
Answer ii) Oxidation states: The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of + 2 and + 4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.
Answer iii) Atomic and lonic sizes: Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater due to the poor shielding effect of 5f orbitals.
Answer iv) Chemical reactivity: In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).
Q21) 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 d1 configuration is very unstable in ions
Answer i) Cr2+ is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d3). This d3 configuration can be written as configuration, which is a more stable configuration. In the case of Mn3+ (d4), it acts as an oxidizing agent and gets reduced to Mn2+ (d5). This has an exactly half-filled d-orbital and is highly stable.
Answer 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.
Answer iii) The ions in d1 configuration tend to lose one more electron to get into stable d0 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.
Q22) What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.
Answer) 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.
Q23) Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
Answer) In the first transition series, Cu exhibits +1 oxidation state very frequently. It is because Cu ( +1) has an electronic configuration of [Ar] 3d10. The completely filled d-orbital makes it highly stable.
Q24) 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?
Cr3+ is the most stable in aqueous solutions owing to a t32g configuration.
Q25) 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.
Answer i) In the case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result, it can donate electrons and behave as a base. On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding and so, they are unavailable. There is also a high effective nuclear charge. As a result, it can accept electrons and behave as an acid.
For example, MnIIO is basic and MnVII2O7 is acidic.
Answer ii) Oxygen and fluorine act as strong oxidising agents because of their high electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits higher oxidation states in oxides and fluorides. For example, in OsF6 and V2O5, the oxidation states of Os and V are +6 and +5 respectively.
Answer iii) Oxygen is a strong oxidising agent due to its high electronegativity and small size. So, oxo-anions of a metal have the highest oxidation state. For example, in , the oxidation state of Mn is +7.
Q26) Indicate the steps in the preparation of:
(i) K2Cr2O7 from chromite ore.
(ii) KMnO4 from pyrolusite ore.
Answer) Potassium dichromate (K2Cr2O7) is prepared from chromite ore FeCr2O4 in the following steps:
Answer ii) Potassium permanganate ( KMnO4 ) can be prepared from pyrolusite (MnO2). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO3 or KClO4, to give K2MnO4.
Q27) What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Answer)An alloy is a solid solution of two or more elements in a metallic matrix. It can be a partial solid solution or a complete solid solution. Alloys generally are created as they have varied properties compared to their constituent elements. An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94−95%), iron (5%), and traces of Al, Ca, Si, C, S.
- It is used in tracer bullets and shells.
- Mischmetal is used in cigarettes and gas lighters.
- It is used in flame-throwing tanks.
Q28) 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.
Answer) 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. Among the given atomic numbers, the atomic numbers of the inner transition elements are 59, 95, and 102.
Q29) 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.
Answer) Lanthanoids project 3 oxidation states (+2, +3, +4). From these, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbital’s is quite large. While, the energy difference between 5f, 6d, and 7s orbital’s is very less. Hence, actinoids display a large number of oxidation states.
For example, neptunium displays +3, +4, +5, and +7 while uranium and plutonium display +3, +4, +5, and +6 oxidation states. The most common oxidation state in case of actinoids is also +3.
Q30) 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.
In the actinoid series, the last element is lawrencium, Lr. The atomic number of the element is 103 and its electronic configuration is [Rn] 5f14 6d1 7s2 . The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains stable f14 configuration.
Q31) Use Hund’s rule to derive the electronic configuration of Ce3+ ion and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Q32) Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to correlate this type of behavior with the electronic configurations of these elements.
In the parenthesis are the atomic numbers of the elements are given:
Tb after forming Tb4+ attains a stable electronic configuration of [Xe] 4f7.
Yb after forming Yb2+ attains a stable electronic configuration of [Xe] 4f14.
Eu after forming Eu2+ attains a stable electronic configuration of [Xe] 4f7.
Ce after forming Ce4+ attains a stable electronic configuration of [Xe].
Q33) Write the electronic configurations of the elements with the atomic numbers 61, 91, 101 and 109.
|Atomic number||Electronic configuration|
Q34) 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
Answer i) In the 1st, 2nd and 3rd transition series, the 3d, 4d, and 5d orbitals are respectively filled. We know that elements in the same vertical column generally have similar electronic configurations.
In the first transition series, two elements show unusual electronic configurations:
Cu (29) = 3d104s1
Cr (24) = 3d54s1
Similarly, there are exceptions in the second transition series. These are:
Ag(47) = 4d105s1
Pd(46) = 4d10 5s0
Rh(45) = 4d85s1
Ru (44) = 4d75s1
Tc (43) = 4d65s1
Mo (42) = 4d55s1
There are some exceptions in the third transition series as well. These are:
Au (79 ) = 5d10 6s1
Pt (78) = 5d9 6s1
W (74) = 5d4 6s2
As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.
Answer ii) In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends. However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states. The stability of the +2 and +3 oxidation states decreases in the second and the third transition series, wherein higher oxidation states are more important.
Some stable complexes are :
The issue is that no such complexes are known for the second and third transition series such as Mo, W, Rh, In. They form complexes in which their oxidation states are high.
For example: WCl6, ReF7, RuO4, etc
Answer iii) In each of the three transition series, the first ionisation enthalpy increases from left to right. However, there are some exceptions. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4f electrons in the third transition series. Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series. There are also elements in the 2nd transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the 1st transition series.
Answer iv) Atomic size generally decreases from left to right across a period. Now, among the three transition series, atomic sizes of the elements in the second transition series are greater than those of the elements corresponding to the same vertical column in the first transition series. However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.
Q35) 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).
Q36) Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
Answer) The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways:
- The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2nd and 3rd transition series).
However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.
- +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.
- The enthalpies of atomisation of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series.
- The melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of stronger metallic bonding (M−M bonding).
- The elements of the first transition series form low-spin or high-spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.
Q37) What can be inferred from the magnetic moment values of the following complex species?
Example Magnetic Moment (BM)
For in transition metals, the magnetic moment is calculated from the spin-only formula. Therefore,
√n(n + 2) = 2.2
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.
√n(n + 2) = 5.3
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.
√n(n + 2) = 5.9
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.
That’s it. These were the solutions of NCERT Class 12 Chemistry Chapter 8 – The d and f Block Elements. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.