NCERT Solutions for Class 12 Chemistry Chapter 8 - The d-and f-Block Elements
1. Write down the electronic configuration of: (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+ (v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.
Ans. (i) Cr3+ (Z = 24 – 3 = 21) = 1s2, 2s2, 2p6 3s2 3p6 3d3 or [Ar]3d3
(ii) Pm3+ (Z = 61 – 3 = 58) = 1s2, 2s2 2p6, 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 4f4 or [Xe] 4f4
(iii) Cu+ (Z = 29 – 1 = 28) = 1s2 2s2 2p6 3s2 3p6 3d10 or [Ar]3d10
(iv) Ce4+ (Z = 58 – 4 = 54 ) = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 = [Xe]
(v) Co2+ (Z = 27 – 2 = 25) = 1s2 2s2 2p6 3s2 3p6 3d7 or [Ar] 3d7
(vi) Lu2+ (Z = 71 – 2 = 69) = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 4f 14 5d1 or [Xe] 4f14 5d1
(vii) Mn2+ (Z = 25 – 2 = 23) = 1s2 2s2 2p6 3s2 3p6 3d5 or [Ar] 3d5
(viii) Th4+ (Z = 90 – 4 = 86) = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6p6 or [Rn]
2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?
Ans. Mn2+ (Z = 25 – 2 = 23) = 1s2, 2s2 2p6, 3s2 3p6 3d5
Fe2+ (Z = 26 – 2 = 24) = 1s2, 2s2 2p6, 3s2 3p6 3d6
Mn2+ is more stable due to half filled d-orbitals. Fe2+ is comparatively less stable as they have six elements in their 3 d-orbital. So, they tend to lose one electrons (From Fe3+) and get stable 3d5 configuration.
3. Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Ans. Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.
4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
Ans. When d-orbital are half filled or completely filled, the oxidation state is more stable.
Ex. 1. Mn (Z = 25) = [Ar] 3d54s2
Mn2+ (Z = 25 – 2 = 23) = [Ar] 3d5 (Most stable)
Mn3+ (Z = 25 – 3 = 22) = [Ar] 3d4
Mn4+ (Z = 25 – 4 = 21) = [Ar] 3d3
Ex. 2. Cu (Z = 29 – 1 = 28) = [Ar] 3d10 4s1
Cu+ (Z = 29 – 1 = 28) = [Ar] 3d10 (Most stable)
Cu2+ (Z = 29 – 2 = 27) = [Ar] 3d9
In manganiese, Mn2+ ion is more stable due to symmetry and half filled d-orbitals. In the same way in copper, Cu+ ion is more stable due to symmetry and completely filled d-orbitals.
5. What may be the stable oxidation state of the transition element with the following d-electron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?
Ans. Stable oxidation states :
3d3 : Vanadium (3d3 4s2) : Oxidation states +2, +3, +4 and +5
3d5 : Chromium (3d54s1) : Oxidation states +3, +4 and +6
3d8 : Cobalt (3d7 4s2) : Oxidation states +2, +3
3d4 : Such configuration does not exist in ground state.
6. 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.
Ans. Cr2O72– and CrO42– (Group number = Oxidation state of Cr = 6). MnO4– (Group number = Oxidation state of Mn = 7).
7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Ans. Lanthanoid Contraction: In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4f- electrons. However, the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is known as lanthanoid contraction.
Consequences of lanthanoid Contraction:
(i) Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.
(ii) Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.
(iii) Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.
(iv) Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.
$$\text{M}^{3+}(aq)+3e^{\normalsize-}\xrightarrow{}4 \text{M}(aq)$$
(v) Variation in physical properties like melting point, boiling point, hardness etc.
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. Characteristics of transition elements are as follows :
(i) Electronic configuration: (n – 1) d1–10 ns1–2
(ii) Metallic character: With the exceptions of Zn, Cd and Hg, they have typical metallic structures.
(iii) Atomic and ionic size: Elements of same charge in a given series show progressive decrease in radius with increasing atomic number.
(iv) Oxidation state: It varies from +2 to +7.
(v) Magnetic properties: These elements show diamagnetism and paramagnetism.
(vi) Ionisation enthalpies: Due to an increase in nuclear charge which accompanies the filling of inner d-orbitals, there is an increase in ionization enthalpies.
(vii) Formation of complex compounds: Due to small size and high charge density of metal ions.
(viii) They possess catalytic properties: Due to their ability to adopt multiple oxidation states.
(ix) Formation of interstitial compounds.
(xi) Alloy formation.
They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p-block elements. Zn, Cd and Hg have fully filled d-orbitals in their ground state hence may not be regarded as the transition elements.
9. In what way are the electronic configuration of the transition elements different from non-transition elements?
Ans. Electronic configuration of transition elements = (n – 1) d1–10 ns1 –2.
Electronic configuration of non-transition elements = ns1–2 or ns2 np1–6.
From the above electronic configuration, it is clear that transition elements have complete d–orbitals where as d–orbital is absent or completely filled in non–transition elements.
10. What are the different oxidation states exhibited by the lanthanoids?
Ans. Lanthanides exhibits +2, +3 and +4 oxidation states the most common oxidation state of lanthanoids is +3.
11. Explain giving reasons:
(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iii) The transition metals generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalyst
Ans. (i) Transition metals and many of their compounds show paramagnetic behaviour because they have unpaired electrons and each unpaired electron has a magnetic moment associated with its spin angular momentum and orbital angular momentum.
(ii) The enthalpies of atomisation of transition metals are high due to presence of large number of unpaired electrons in their atoms. These atoms have strong interatomic interaction and hence, stronger bonding between them.
(iii) The transition metals generally
form coloured compounds because they partially adsorbed of visible light and Jumps into next orbitals.
(iv) Transition metals are good catalyst because of following reasons:
(a) The ability of transition metal ion to pass easily from ane oxidation state to another and thus providing a new path to reaction with lower activation energy.
(b) The surface of transition metal acts as very good adsorbent and thus provides increased concentration of reactants on their surface causing reaction to occur.
12. What are interstitial compounds? Why are such compounds well known for transition metals?
Ans. Interstitial compounds: The compound in which small atoms like H, C, N, etc. occupies the interstital sites in the crystal lattice are called interstitial compounds.
These compounds are well known for transition metals because small atoms can easily occupy to positions in the voids present in the crystal lattic of transition metals.
13. How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Ans. The variability in oxidation states of transition elements is an important characteristic and it arises due to incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity. For example, vanadium, V show the oxidation.
States of +2, +3, +4 and +5 similarly, Cr shows oxidation states of +2, +3, +4, +5 and +6; Mn shows all oxidation states from +2 to +7.
This is contrasted with varibility of oxidation states of non-transition elements where oxidation states generally differ by units of two. For example, S shows oxidation states of –2, +2, +4, +6 while P shows +3 and +5 oxidation states. Halogenes like Cl, Br and I show oxidation states of –1, +1, +3, +5 and +7 states. In non-transition elements variability of oxidation states is caused due to unpairing of electrons in ns or np orbitals and their promotion to np or nd vacant orbitals.
14. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Ans. Preparation of potassium dichromate from iron chromite ore:
Step 1: Preparation of Sodium chromate : Potassium dichromate is prepared from chromate, which in turn is obtained by the fusion of chromite ore (FeCr2O4) with sodium or potassium carbonate in free excess of air. The reaction with sodium carbonate occurs as follows:
$$4\text{FeCr}_{2}\text{O}_{4}+8\text{Na}_{2}\text{CO}_{3}+7\text{O}_{2}\xrightarrow{}8\text{Na}_{2}\text{CrO}_{4}+2\text{Fe}_{2}\text{O}_{3}+8\text{CO}_{2}$$
Step 2: Conversion of Sodium chromate into sodium dichromate: The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7.2H2O can be crystallised.
$$2\text{Na}_{2}\text{CrO}_{4}+2\text{H}^{\normalsize+}\xrightarrow{}\text{Na}_{2}\text{Cr}_{2}\text{O}_{7}+2\text{Na}^{\normalsize+}+\text{H}_{2}\text{O}$$
Step 3: Conversion of sodium dichromate to potassium dichromate : Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.
$$\text{Na}_{2}\text{Cr}_{2}\text{O}_{7}+2\text{KCl}\xrightarrow{}\text{K}_{2}\text{Cr}_{2}\text{O}_{7}+2\text{NaCl}\\\text{Effect of increasing pH on K}_2\text{Cr}_2\text{O}_7 \text{Solution: On increasing pH, K}_2\text{Cr}_2\text{O}_7 \text{changes into K}_2\text{CrO}_4 \text{(orange to yellow)}\\\underset{\text{orange}}{\text{Cr}_{2}\text{O}_{7}^{2-}}+2\text{OH}^{\normalsize-}\xrightarrow{}\underset{\text{yellow}}{2\text{CrO}_{4}^{2-}+\text{H}_{2}\text{O}}$$
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. Potassium dichromate is a strong oxidising agent. In acidic medium the oxidation action can be represented as:
$$\text{Cr}_{2}\text{O}_{7}^{2-}+14\text{H}^{\normalsize+}+6e^{\normalsize-}\xrightarrow{}2\text{Cr}^{3+}+7\text{H}_{2}\text{O}$$
(i) Iodide: K2Cr2O7 oxidizes iodide to iodine.
$$\text{Cr}_{2}\text{O}_{7}^{2-}+14\text{H}^{\normalsize+}+6e^{\normalsize-}\xrightarrow{}2\text{Cr}^{3+}+7\text{H}_{2}\text{O}\\6\text{Fe}^{2+}\xrightarrow{}6\text{Fe}^{3+}+6e^{\normalsize-}\\\text{Cr}_{2}\text{O}_{7}^{2-}+6\text{Fe}^{2+}+14\text{H}^{\normalsize+}\xrightarrow{}2\text{Cr}^{3+}+6\text{Fe}^{3+}+7\text{H}_{2}\text{O}\\\text{(iii) H}_2\text{S}:\text{K}_{2}\text{Cr}_{2}\text{O}_{7}\space\text{oxides H}_{2}\text{S to S.}\\\text{Cr}_{2}\text{O}_{7}^{2-}+3\text{H}_{2}\text{S}+8\text{H}^{\normalsize+}\xrightarrow{}2\text{Cr}^{3+}+7\text{H}_{2}\text{O}+3\text{S} $$
16. 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.
Ans. Preparation of KMnO4 : Potassium permanganate (KMnO4) is prepared by the fusion of a mixture of pyrolusite (MnO2), potassium hydroxide and oxygen, first green coloured potassium manganate is formed. The potassium manganate is extracted by water, which then undergoes disproportionation in neutral or acidic solution to give potassium permanganate.
$$\text{(i) 2MnO}_{2}+ 4\text{KOH + O}_{2}\xrightarrow{} 2\text{K}_{2}\text{MnO}_{4}+ \text{2H}_{2}\text{O}\\\text{(ii) 3MnO}_4^{2-}+4\text{H}^{\normalsize+}\xrightarrow{}2\text{MnO}_{4}^{\normalsize-}+\text{MnO}_{2}+2\text{H}_{2}\text{O}\\\text{Reaction of KMnO}_4 \text{in acidic medium :}\\\text{MnO}_{4}^{\normalsize-}+ 8\text{H}^{\normalsize+}+5e^{\normalsize-}\xrightarrow{}\text{Mn}^{2+}+4\text{H}_{2}\text{O}\\\text{(i) Iron (II) solution: Ferrous (Fe}^{2+})\text{ion solution to ferric}(\text{Fe}^{3+})\space\text{ion solution}\\\text{MnO}_{4}^{\normalsize-}+8\text{H}^{\normalsize+}+5e^{\normalsize-}\xrightarrow{}\text{Mn}^{2+}+4\text{H}_{2}\text{O}\\5\text{Fe}^{2+}\xrightarrow{}5\text{Fe}^{3+}+5e^{\normalsize-}\\\text{MnO}_{4}^{\normalsize-}+5\text{Fe}^{2+}+8\text{H}^{\normalsize+}\xrightarrow{}\text{Mn}^{2+}+5\text{Fe}^{3+}+4\text{H}_{2}\text{O}\\\textbf{(ii)Sulphur dioxide (SO}_2)\\\lbrack \text{MnO}_{4}^{\normalsize-} + 8\text{H}^{+}+ 5 e^{\normalsize-}\xrightarrow{}\text{Mn}^{2+}+4\text{H}_{2}\text{O}\rbrack×2\\10\text{H}_{2}\text{O}+ 5\text{SO}_{2}\xrightarrow{}5\text{SO}_{4}^{2-}+2\text{Mn}^{2+}+10e^{\normalsize-}\\2\text{MnO}_{4}^{\normalsize-}+5\text{SO}_{2}+2\text{H}_{2}\text{O}\xrightarrow{}5\text{SO}_{4}^{2-}+2\text{Mn}^{2+}+4\text{H}^{\normalsize+}$$
(iii) Oxalic acid
$$\lbrack\text{MnO}_{4}^{\normalsize-}+8\text{H}^{\normalsize+}+5e^{\normalsize-}\xrightarrow{}\text{Mn}^{2+}+ 4 \text{H}_{2}\text{O}\rbrack×2\\5\text{(COO}^{\normalsize-})_{2}\xrightarrow{}10\text{CO}_{2}+10e^{\normalsize-}\\2\text{MnO}_{4}^{\normalsize-}+16\text{H}^{\normalsize+}+5\text{C}_{2}\text{O}_{4}^{2-}\xrightarrow{}2\text{Mn}^{2+}+10\text{CO}_{2}+8\text{H}_{2}\text{O}$$
17. For M2+/M and M3+/M2+ systems the E° values for some metals are as follows:
Cr2+/Cr → –0.9 V
Mn2+/Mn → –1.2V
Fe2+/Fe → –0.4 V
Cr3+/Cr2+ → –0.4 V
Mn3+/Mn2+ → + 1.5V
Fe3+/Fe2+ → + 0.8V
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. (i) Cr3+/Cr2+ has negative reduction potential. Hence, Cr3+ cannot be reduced to Cr2+. Mn3+/Mn2+ has a large positive reduction potential. Hence, Mn3+ can be easily reduced to Mn2+. Fe3+/Fe2+ has small positive reduction potential. Hence, Fe3+ is more stable than Mn3+ but less stable than Cr3+.
Mn3+ < Fe3+ < Cr3+
(ii) From the E° values, the order of oxidation of the metal to the divalent cation is : Mn > Cr > Fe.
18. Predict which of the following will be coloured in aqueous solution?
Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+, Co2+.
Ans. Only those ions will be coloured which have incomplete d-orbitals. The ions with either empty or filled d-orbitals are colourless.
Tr3+ = [Ar] 3d1—Coloured
V3+ = [Ar] 3d2 —Coloured
Cu+ = [Ar] 3d10—Colourless
Sc3+ = [Ar] —Colourless
Mn2+ = [Ar] 3d5 —Coloured
Fe3+ = [Ar] 3d5 — Coloured
Co2+ = [Ar] 3d7 —Coloured
19. Compare the stability of +2 oxidation state for the elements of the first transition series.
Ans. In general, the stability of +2 oxidation state in first transition series decreases from left to right due to increase in the sum of first and second ionisation energies. However Mn2+ is more stable due to half filled d-orbitals (3d5) and Zn2+ is more stable due to completely filled d-orbitals (3d10).
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: The general electronic configuration of lanthanoids is [Xe]54 4f 1–14 5d0–1 6s2 and that of actinoids is [Rn]86 5f 1–14 6d0–1 7s2.
(ii) Atomic and ionic sizes: Both lanthanoids and actinoids show decrease in size of their atoms or ions in +3 oxidation state as we go from left to right. In lanthanoids, the decrease is called lanthanoid contraction whereas in actinoids, it is called actinoid contraction. The contraction is greater from element to element in actinodes due to poorer shielding by 5f electrons.
(iii) Oxidation state: Lanthanoids show limited oxidation states (+2, +3, +4) out of which +3 is most common whereas actinoids show +3, +4, +5, +6, +7 oxidation states.This is because of large energy gap between 4f 5d and 6s orbitals. However, actinoids show a large number of oxidation states because of small energy gap between 5f 6d and is orbitals.
(iv) Chemical reactivity: The earlier members of the lanthanoids series are quite reactive similar to calcium but, with increase in atomic number, they behave more like aluminium. The metals combine with hydrogen when gently heated in the gas. Carbides, Ln3C, Ln2C3 and LnC2 are formed when the metals are heated with carbon. They liberate hydrogen from dilute acid and burn in halogens to form halides. They form oxides M2O3 and hydroxides M(OH)3.
Actinoids are highly reactive metals, especially when finely divided. The action of boiling water on them gives a mixture of oxide and hydride and combination with most non-metals take place at moderate temperatures. HCl attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers, alkalis have no action. Actinoids are more reactive than lanthanoids due to bigger atomic size and lower ionisation energy.
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 d1 configuration is very unstable in ions.
Ans. (i) E° value for Cr3+/Cr2+ is negative (–0.41 V) whereas E° values for Mn3+/Mn2+ is positive (+1.57 V). Hence, Cr2+ ion can easily undergo oxidation to give Cr3+ ion and, therefore, act as strong reducing agent whereas Mn3+ can easily undergo reduction to give Mn2+ and hence, act as an oxidising agent.
(ii) Co (III) has greater tendency to form coordination complexes
than Co (II). Hence, in the presence of ligands, Co (II) changes to Co (III), i.e., is easily oxidised.
(iii) The ions with dx configuration have the tendency to lose the only electron present in d-subshell to acquire stable d° configuration. Hence, they are unstable and undergo oxidation or disproportionation.
22. What is meant by disproportionation? Give two examples of disproportionate reaction in aqueous solution.
Ans. In disproportionate reactions are those in which the same substance undergoes oxidation as well as reduction, i.e., oxidation number of an element increases as well as decreases to form two different products.
$$\underset{(i)\space 3\text{MnO}_{4}^{2-}+4 \text{H}^{\normalsize+}}{\text{VI}}\xrightarrow{}\underset{2\text{MnO}_{4}^{\normalsize-}+\text{MnO}_{2}+2\text{H}_{2}\text{O}}{\text{VII}\qquad \text{IV}}\\\text{(ii) 2Cu}^{\normalsize+1}-\text{Cu}^{\normalsize+2}+\text{Cu}$$
23. Which metal in the first series of transition metal exhibits +1 oxidation state most frequently and why?
Ans. 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). Thus, in the first transition series, Cu exhibits +1 oxidation state very frequently as Cu ( +1) has an electronic configuration of [Ar] 3d10. The completely filled d-orbital makes it highly stable.
24. Calculate the number of unpaired electrons in the following gaseous ions : Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution.
Ans. Mn3+ (Z = 25) = [Ar] 3d4 = 4 unpaired electrons
Cr3+ (Z = 24) = [Ar] 3d3 = 3 unpaired electrons
V3+ (Z = 23) = [Ar] 3d2 = 2 unpaired electrons
Ti3+ (Z = 22) = [Ar] 3d1 = 1 unpaired electrons
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) The lower oxide of transition metal is basic because the metal atom has low oxidation state whereas higher once are acidic due to high oxidation state. For example, MnO is basic whereas Mn2O7 is acidic. Oxides in lower oxidation state are ionic hence basic. Oxides in higher oxidation state are covalent hence acidic
(ii) A transition metal exhibits higher oxidation states in oxides and fluorides because oxygen and fluorine are highly electronegative elements, small in size and strongest oxidising agents. For example, osmium shows an oxidation states of +6 in O5F6 and vanadium shows an oxidation states of +5 in V2O5.
(iii) Oxoanions of metal have highest oxidation state, e.g., Cr in Cr2O72– has an. oxidation state of + 6 whereas, Mn in MnO4– has an oxidation state of +7. This is again due to the combination of the metal with oxygen, which is highly electronegative and oxidizing agent.
26. Indicate the steps in the preparation of:
(i) K2Cr2O7 from chromite ore
(ii) KMnO4 from pyrolusite ore.
Ans. Preparation of potassium dichromate from iron chromite ore:
Step 1: Preparation of Sodium chromate: Potassium dichromate is prepared from chromate, which in turn is obtained by the fusion of chromite ore (FeCr2O4) with sodium or potassium carbonate in free excess of air. The reaction with sodium carbonate occurs as follows:
$$4\text{FeCr}_{2}\text{O}_{4}+8\text{Na}_{2}\text{CO}_{3}+7\text{O}_{2}\xrightarrow{}8\text{Na}_{2}\text{CrO}_{4}+2\text{Fe}_{2}\text{O}_{3}+8\text{CO}_{2}$$
Step 2: Conversion of Sodium chromate into sodium dichromate: The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7.2H2O can be crystallised.
$$2\text{Na}_{2}\text{CrO}_{4}+2\text{H}^{\normalsize+}\xrightarrow{}\text{Na}_{2}\text{Cr}_{2}\text{O}_{7}+2\text{Na}^{\normalsize+}+\text{H}_{2}\text{O}$$
Step 3: Conversion of sodium dichromate to potassium dichromate : Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride. Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.
$$\text{Na}_{2}\text{Cr}_{2}\text{O}_{7}+2\text{KCl}\xrightarrow{}\text{K}_{2}\text{Cr}_{2}\text{O}_{7}+2\text{NaCl}$$
Preparation of KMnO4: Potassium permanganate (KMnO4) is prepared by the fusion of a mixture of pyrolusite (MnO2),potassium hydroxide and oxygen, first green coloured potassium manganate is formed. The potassium manganate is extracted by water, which then undergoes disproportionate in neutral or acidic solution to give potassium permanganate.
$$\text{(i) 2MnO}_{2}+4\text{KOH + O}_{2}\xrightarrow{}2\text{K}_{2}\text{MnO}_{4}+2\text{H}_{2}\text{O}\\\text{(ii)\space 3\text{MnO}}_{4}^{2-}+4\text{H}^{\normalsize+}\xrightarrow{}2\text{MnO}_{4}^{\normalsize-}+\text{MnO}_{2}+2\text{H}_{2}\text{O}\\\text{Reaction of KMnO}_4\text{in acidic medium:}\\\text{MnO}_{4}^{\normalsize-}+8\text{H}^{\normalsize+}+5e^{\normalsize-}\xrightarrow{}\text{Mn}^{2+}+4\text{H}_{2}\text{O}$$
27. What are alloys? Name an alloy which contains some of the lanthanoid metals. Mention its uses.
Ans. Alloys : An alloy is a homogeneous mixture of two or more metals or a metal and a non-metal.
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, bullets, shell and lighter flint.
28. What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29,59,74,95,102,104.
Ans. The f-block elements in which the last electron enters into f-subshell are called inner-transition elements. These include lanthanoids (Z = 58 to 71) and actinoids (Z = 90 to 103).
Thus, the elements with atomic numbers 59,74,95 and 102 are the inner transition elements.
29. The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
Ans. Lanthanoids show limited number of oxidation state, viz, +2, +3 and +4 (out of which +3 is most common). This is because of large energy gap between 4f 5d and 6s subshells. The dominant oxidation state of actinoids is also +3 but they show a number of other oxidation states also. For example, uranium (Z = 92) and plutonium (Z = 94), show +3, +4, +5 and +6, neptunium (Z = 94) shows +3, +4, +5 and +7, etc. This is because of the small energy difference between. 5f, 6d and 7s orbitals of the actinoids.
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 (Lr, Z = 103) is the last element of actinoids.
Electronic configuration : Lr(Z = 103) = [Rn] 5f 14 6d1 7s2.
Its possible oxidation state = +3.
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.
Ans. Ce (Z = 58) = [Xe] 4f1 5d1 6s2
$$Ce^{3+} = [Xe] 4f^{1}\\\mu=\sqrt{n(n+2)}\\\text{Here, n = 1 i.e. only one unpaired electron.}\\\therefore\space \mu=\sqrt{1(1+2)}=\sqrt{3}=1.73\space\text{BM.}$$
32. 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 behaviour with the electronic configuration of these elements.
Ans. +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 4f 0 or 4f 7.
33. Compare the chemistry of actinoids with that of lanthanoids with reference to:
(i) Electronic configuration
(ii) Oxidation states
(iii) Chemical reactivity
Ans. (i) Electronic configuration: In lanthanoids 4f–orbitals are progressively filled whereas in actinoids 5f-orbitals are progressively filled.
(ii) Oxidation states: Lanthanoids shows +3 oxidation state. Some elements shows +2 and +4 oxidation state also. Actinoids shows +3, +4, +5 +6, +7 oxidation states. Although +3 and +4 are most common.
(iii) Chemical reactivity: Actinoids are more reactive than lanthanoids due to bigger atomic size and lower ionisation energy.
34. Write the electronic configurations of the elements with the atomic numbers 61,91,101 and 109.
Ans. Z = 61 (Promethium, Pm) ⇒ [Xe]4f5 5d0 6s2
Z = 91 (Protactinium, Pa) ⇒ [Rn] 5f2 6d1 7s2
Z = 101 (Mendelevium, Md) ⇒ [Rn] 5f 13 6d0 7s2
Z = 109 (Meitnerium, Mt) ⇒ [Rn] 5f 14 6d7 7s2
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: The elements in the same vertical column generally have similar electronic configuration. First transition series shows only two exceptions, (a) Cr = 3d5 4s1 (b) Cu = 3d10 4s1.
But second transition series shows more exceptions, (a), Y = 4d1 5s2, (b) Nb = 4d4 5s1, (c) Mo = 4d5 5s1, (d) Ru = 4d1 5s1, (e) Rh = 4d8 5s1, (f) Pd = 4d10 5s°, (g) Ag = 4d10 5s1.
In third transition, there are three exceptions, (a) Pt = 5d9 6s1 (b) Au = 5d10 6s1. (a) W = 5d4 6s1
Thus in the same vertical column, in a number of cases, the electronic configuration of the elements of three series are not similar.
(ii) Oxidation states: The elements in the same vertical column generally show similar oxidation states. The number of oxidation states shown by the elements in the middle of each series is maximum and minimum at the extreme ends.
(iii) Ionization enthalpies: The first ionization enthalpies in each series generally increases gradually as we more from left to right though some exceptions are observed in each series. The first ionization enthalpies of some elements in the second (4d) series are higher while some of them have lower value than the elements of 3d series in the same vertical column. However, the first ionization enthalpies of third (5d) series are higher than those of 3d and Ad series. This is because of weak shielding of nucleus by 4f-electrons in the 5d series.
(iv) Atomic sizes: In general, ions of the same charge or atoms in a given series show progressively decrease in radius with increasing atomic number though the decrease is quite small. But the size of the atoms of the Ad series is larger than the corresponding elements of the 3d series whereas size of elements of the 5d-series nearly the same as those of Ad series because of lanthanoid contraction.
36. 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)?
Ans.
| Sol. | Iron | Configuration | No. of 3d eclectrons | No. of unpaired electrons | Occupancy of 3d orbitals |
| (1) | Ti2+ | 3d2 | 2 | 2 |  |
| (2) | V2+ | 3d3 | 3 | 3 |  |
| (3) | Cr3+ | 3d3 | 3 | 3 |  |
| (4) | Mn2+ | 3d5 | 5 | 5 |  |
| (5) | Fe2+ | 3d6 | 6 | 4 |  |
| (6) | Fe3+ | 3d5 | 5 | 5 |  |
| (7) | Co2+ | 3d7 | 7 | 3 |  |
| (8) | Ni2+ | 3d6 | 8 | 2 |  |
| (9) | Cu2+ | 3d9 | 9 | 1 |  |
37. Comment on the statement that elements of first transition series possess many properties different from those of the heavier transition elements.
Ans. The heavier transition elements belong to fourth (4d) 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 4d and 5d series are more due to greater number of electron shells. However, the difference in 4d and 5d transition elements are comparatively less because of lanthanoid contraction.
(ii) Because of stronger inter atomic bonding, the melting point and boiling point of the elements of 4d 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.
38. What can be inferred from the magnetic moment values of the following complex species?
| Example | Magnetic moment (BM) |
| K4[Mn(CN)6] | 2.2 |
| [Fe(H2O)6]2+ | 5.3 |
| K2[MnCl4] | 5.9 |
$$\textbf{Ans.}\space\text{Magnetic moment,}\space \mu=\sqrt{1(1+2).}\text{B.M.}\\\text{For}\space n=1,\mu=\sqrt{1(1+2)}=\sqrt{3}=1.73\space\text{B.M.}\\n=2, \mu=\sqrt{(2(2+2))}=\sqrt{8}=2.83\space\text{B.M.}\\n=3, \mu=\sqrt{3(3+2)}=\sqrt{15}=3.87\space\text{B.M.}\\n=4,\mu=\sqrt{4(4+2)}=\sqrt{24}=4.89\space\text{B.M.}\\n=5,\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92\space\text{B.M.}$$
(i) K4[Mn(CN)6]: In this compound Mn is in +2 oxidation state and µ = 2.2
It means that it has only one unpaired electron. When CN– ligands approach Mn2+ ion, the electrons in 3d-orbital pair up.
Thus CN– is a strong ligand. The hybridisation is d2sp3 forming inner-orbital octahedral complex.
(ii) [Fe(H2O6)]2+: In this complex, Fe is in +2 oxidation state and m = 5.3. It means that there are four unpaired electrons in 3d. also, the 3d electrons do not pair up when the H2O molecules approach. Thus, H2O is a weak ligand.
The hybridisation involved is sp3 d2, forming an outer-orbital octahedral.
(iii) K2[MnCl4]: In this complex, Mn is in +2 oxidation state and m = 5.92. It means that there are five unpaired electrons. The hybridisation involved in sp3, forming a tetrahedral complex. Cl is a weak ligand.
NCERT Solutions for Class 12 Chemistry Chapter 8 Free PDF Download
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