NCERT Solutions for Class 12 Chemistry Part 1 Chapter 8
The d-and f-Block Elements Class 12
Chapter 8 The d-and f-Block Elements Exercise Solutions
In text : Solutions of Questions on Page Number : 212
Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?
Ag has a completely filled 4d orbital (4d10 5s1) in its ground state. Now, silver displays two oxidation states (+1 and
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol-1. Why?
Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
Mn (Z = 25) = 3d5 4s2
The Eθ(M2+/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high ΔaHθ and low ΔhydHθ)
The Eθ(M2+/M) value of a metal depends on the energy changes involved in the following:
1.
2. Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state.
3.
Now, copper has a high energy of atomization and low hydration energy. Hence, the Eθ(M2+/M) value for copper is positive.
How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?
Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as d0, d5, d10. Since these states are exceptionally stable, their ionization enthalpies are very high.
In case of first ionization energy, Cr has low ionization energy. This is because after losing one electron, it attains the stable configuration (3d5). On the other hand, Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals (3d10 4s2).
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Which is a stronger reducing agent Cr2+ or Fe2+ and why?
The following reactions are involved when Cr2+ and Fe2+ act as reducing agents. Cr2+ Cr3+ Fe2+ Fe3+
The value is - 0.41 V and is +0.77 V. This means that Cr2+ can be easily oxidized to Cr3+, but
(aq)
Calculate the 'spin only' magnetic
moment of M2+ ion (Z =
27).
Z = 27
[Ar] 3d7 4s2 M2+ = [Ar] 3d7
3d7 =
i.e., 3 unpaired electrons
μ ≈ 4 BM
Explain why Cu+ ion is not stable in aqueous solutions?
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Exercise : Solutions of Questions on Page Number : 234
Write down the electronic configuration of:
(ii) Pm3+ (iv) Ce4+ (vi) Lu2+ (viii) Th4+
(i) Cr3+: 1s22s22p63s23p63d3 Or, [Ar]183d3
(ii) Pm3+: 1s22s22p63s23p63d104s2 4p6 4d105s2 5p64f4 Or, [Xe]543d3
(iii) Cu+: 1s22s22p63s23p63d10 Or, [Ar]183d10
(iv) Ce4+: 1s22s22p63s23p63d104s2 4p6 4d105s2 5p6 Or, [Xe]54
(v) Co2+: 1s22s22p63s23p63d7 Or, [Ar]183d7
(vi) Lu2+: 1s22s22p63s23p63d104s2 4p6 4d105s2 5p6 4f14 5d1 Or, [Xe]542f143d3
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.
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
The oxidation states displayed by the first half of the first row of transition metals are given in the table below.
|
Sc |
Ti |
V |
Cr |
Mn |
|
+ 2 |
+ 2 |
+ 2 |
+ 2 |
|
+3 |
+ 3 |
+ 3 |
+ 3 |
+ 3 |
|
Oxidation state |
|
+ 4 |
+ 4 |
+ 4 |
+ 4 |
|
|
|
+ 5 |
+ 5 |
+ 6 |
|
|
|
|
+ 6 |
+ 7 |
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
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.
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 |
|
(i) |
3d3 (Vanadium) |
+2, +3, +4 and +5 |
||
(ii) |
3d5 (Chromium) |
+3, +4, +6 |
||
(iii) |
3d5 (Manganese) |
+2, +4, +6, +7 |
||
(iv) |
3d8 (Cobalt) |
+2, +3 |
||
(v) |
3d4 |
There is no3d4 configuration in ground state. |
||
|
|
|||
Q6 :
which the metal exhibits the oxidation state equal to its group number.
(i) Vanadate, Oxidation state of V is + 5.
(ii) Chromate, Oxidation state of Cr is + 6.
(iii)
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
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 4forbital also increases. The 4felectrons 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.
(i) There is similarity in the properties of second and third transition series.
ii. Separation of lanthanoids is possible due to lanthanide contraction.
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?
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.
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Transition metals have a partially filled d-orbital. Therefore, the electronic configuration of transition elements is (n - 1)d1-10ns0-2.
What are the different oxidation states exhibited by the lanthanoids?
Explain giving reasons:
(ii) The enthalpies of atomisation of the transition metals are high.
(iv) Transition metals and their many compounds act as good catalyst.
(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.
(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.
(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.
(iv) The catalytic activity of the transition elements can be explained by two basic facts.
(a) Owing to their ability to show variable oxidation states and form complexes, transition metals form unstable intermediate compounds. Thus, they provide a new path with lower activation energy, Ea, for the reaction.
(b)
What are interstitial compounds? Why are such compounds well known for transition metals?
How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Potassium dichromate is prepared from chromite ore in the following steps.
Step (2): Conversion of sodium chromate into sodium dichromate
Potassium dichromate being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filtration.
The dichromate ion exists in equilibrium with chromate ion at pH 4. However, by changing the pH, they can be interconverted.
Describe the oxidising action of potassium dichromate and write the ionicequations for its reaction with:
Answer :
takes up electrons to get reduced and acts as an oxidising agent. The reaction of K2Cr2O7with other iodide, iron (II) solution, and H2S are given below.
(ii) oxidizes iron (II) solution to iron (III) solution i.e., ferrous ions to ferric ions.
Describe the preparation of potassium permanganate. How does the acidifiedpermanganate solution react with (i) iron(II) ions (ii) SO2and (iii) oxalic acid?
Answer :
The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution.
At anode, manganate ions are oxidized to permanganate ions.
Oxidation by ozone
(i)
(ii) Acidified potassium permanganate oxidizes SO2 to sulphuric acid.
(iii)
For M2+/M and M3+/M2+ systems, the 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
(i) The stability of Fe3+in acid solution as compared to that of Cr3+or Mn3+ and
Answer :
(i) The 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 ofCr3+ 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+
(ii) The reduction potentials for the given pairs increase in the following order. Mn2+ / Mn < Cr2+ / Cr < Fe2+ /Fe
Predict which of the following will be coloured in aqueous solution? Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.
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 |
Ti |
22 |
T13+ |
[Ar] 3d1 |
V |
23 |
V3+ |
[Ar] 3d2 |
Cu |
29 |
Cu+ |
[Ar] 3d10 |
Sc |
21 |
Sc3+ |
[Ar] |
Mn |
25 |
Mn2+ |
[Ar] 3d5 |
Fe |
26 |
Fe3+ |
[Ar] 3d5 |
Co |
27 |
Co2+ |
[Ar] 3d7 |
Compare the stability of +2 oxidation state for the elements of the first transition series.
Sc |
|
|
+3 |
|
|
|
|
Ti |
+1 |
+2 |
+3 |
+4 |
|
|
|
V |
+1 |
+2 |
+3 |
+4 |
+5 |
|
|
Cr |
+1 |
+2 |
+3 |
+4 |
+5 |
+6 |
|
Mn |
+1 |
+2 |
+3 |
+4 |
+5 |
+6 |
+7 |
Fe |
+1 |
+2 |
+3 |
+4 |
+5 |
+6 |
|
Co |
+1 |
+2 |
+3 |
+4 |
+5 |
|
|
Ni |
+1 |
+2 |
+3 |
+4 |
|
|
|
Cu |
+1 |
+2 |
+3 |
|
|
|
|
Zn |
|
+2 |
|
|
|
|
|
From the above table, it is evident that the maximum number of oxidation states is shown by Mn, varying from +2 to
Compare the chemistry of actinoids with that of the lanthanoids with specialreference to:
(ii) atomic and ionic sizes and (iv) chemical reactivity.
(i) Electronic configuration
The general electronic configuration for lanthanoids is [Xe]544f0-145d0-16s2and that for actinoids is [Rn]865f1-146d0-17s2. Unlike 4forbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.
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 7slevels 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.
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 5forbitals.
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
How would you account for the following:
(ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
Answer :
(i) Cr2+is strongly reducing in nature. It has a d4configuration. While acting as a reducing agent, it gets oxidized to Cr3+(electronic configuration, d3). This d3configuration 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.
(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 3rdionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy.
(iii)
What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.
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.
Cr(V) is oxidized to Cr(VI) and reduced to Cr(III).
Which metal in the first series of transition metals exhibits +1 oxidationstate most frequently and why?
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?
|
Gaseous ions |
Number of unpaired electrons |
(i) |
|
4 |
(ii) |
|
3 |
(iii) |
|
2 |
(vi) |
|
1 |
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.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
(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, is basic and is acidic.
(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 OsF6and V2O5, the oxidation states of Os and V are +6 and +5 respectively.
(iii)
Indicate the steps in the preparation of:
(ii) KMnO4 from pyrolusite ore.
(i)
Potassium dichromate ( ) is prepared from chromite ore in the following steps.
Step (2):Conversion of sodium chromate into sodium dichromate
Potassium chloride being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filtration.
The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution.
At anode, manganate ions are oxidized to permanganate ions.
Oxidation by ozone
What are alloys? Name an important alloy which contains some of thelanthanoid metals. Mention its uses.
An alloy is a solid 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 possess different physical properties than those of the component elements.
An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94-95%), 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 tracer bullets and shells.
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.
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.
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.
The last element in the actinoid series is lawrencium, Lr. Its atomic number is 103 and its electronic configuration is
Use Hund's rule to derive the electronic configuration of Ce3+ ion and calculate its magnetic moment on the basis of 'spin-only' formula.
Where,
n = number of unpaired electrons
The electronic configuration of Ce3+ : 1s22s22p63s23p63d104s24p64d105s25p64f1 In Ce3+, n = 1
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.
The lanthanides that exhibit +2 and +4 states are shown in the given table. The atomic numbers of the elements are given in the parenthesis.
+2 |
+4 |
Nd (60) |
Ce (58) |
Sm (62) |
Pr (59) |
Eu (63) |
Nd (60) |
Tm (69) |
Tb (65) |
Yb (70) |
Dy (66) |
Ce after forming Ce4+ attains a stable electronic configuration of [Xe]. Tb after forming Tb4+ attains a stable electronic configuration of [Xe] 4f7. Eu after forming Eu2+ attains a stable electronic configuration of [Xe] 4f7. Yb after forming Yb2+ attains a stable electronic configuration of [Xe] 4f14.
Compare the chemistry of the actinoids with that of lanthanoids with reference to:
(ii) oxidation states and
Answer :
The general electronic configuration for lanthanoids is [Xe]544f0-145d0-16s2and that for actinoids is [Rn]865f1-146d0-17s2. Unlike 4forbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.
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 7slevels 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.
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.
Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.
|
Atomic number |
Electronic configuration |
|
61 |
|
||
91 |
|
||
101 |
|
||
109 |
|
||
|
|
||
Q35 :
(i) electronic configurations,
(iii) ionisation enthalpies, and
Answer :
(i) In the 1st, 2ndand 3rdtransition series, the 3d, 4d and 5dorbitals 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:
As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.
(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.
For example are stable complexes, but 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.
(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 4felectrons 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 2ndtransition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the 1sttransition series.
(iv)
Write down the number of 3d electrons in each of the following ions:
Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, CO2+, Ni2+ and Cu2+.
Answer :
|
Metal ion |
Number of d-electrons |
Filling of d-orbitals |
|
Ti2+ |
2 |
|
||
V2+ |
3 |
|
||
Cr3+ |
3 |
|
||
Mn2+ |
5 |
|
||
Fe2+ |
6 |
|
||
Fe3+ |
5 |
|
||
CO2+ |
7 |
|
||
Ni2+ |
8 |
|
||
Cu2+ |
9 |
|
||
|
|
|||
Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways.
(i) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2ndand 3rdtransition 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.
(ii) +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.
(iii) 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.
(iv) 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).
(v)
What can be inferred from the magnetic moment values of the following complex species? Example Magnetic Moment (BM)
[Fe(H2O)6]2+ 5.3
Answer :
Magnetic moment ( ) is given as . For value n = 1, .
For value n = 2, .
For value n = 3, .
For value n = 4, .
For value n = 5, .
We can see from the above calculation that the given value is closest to . Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.
Hence, we can say that CN - is a strong field ligand that causes the pairing of electrons.
(ii)
We can see from the above calculation that the given value is closest to . Also, in this complex, Fe is in the +2 oxidation state. This means that Fe has 6 electrons in the d-orbital.
Hence, we can say that H2O is a weak field ligand and does not cause the pairing of electrons.
(iii)
We can see from the above calculation that the given value is closest to . 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.