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.

Transcription

1 1. 1. Write down the electronic configuration of: (i) Cr 3+ (iii) Cu + (v) Co 2+ (vii) Mn 2+ (ii) Pm 3+ (iv) Ce 4+ (vi) Lu 2+ (viii) Th 4+ (a) Cr 3+ 1S 2 2S 2 2P 6 3S 2 3P 6 3d 3 (b) Cu + 1S 2 2S 2 2P 6 3S 2 3P 6 3d 10 (c) Co 2+ 1S 2 2S 2 2P 6 3S 2 3P 6 3d 7 (d) Mn 2+ 1S 2 2S 2 2P 6 3S 2 3P 6 3d 5 (e) Pm 2+ (Xe) 4F 5 (f) Ce 4+ (Xe) 4F o (g) Lu 2+ (Xe) 4f 14 5d 1 (h) Th 4+ (Rh) 5F o 2. Why are Mn 2+ compounds more stable than Fe 2+ towards oxidation to their +3 state? In Mn 2+ compounds the ion possess stable half electronic configuration of 3d 5 whereas in Fe 2+ compounds the ion has configuration 3d 6. If it loses one electron it can attain the configuration 3d 5 which is very stable. Therefore Fe 2+ ions tend to be oxidised easily to Fe 3+ to attain half filled electronic 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. With Scandium, having 3d 1 configuration, d-orbitals are not stabilised. That is why it exhibits stable +3 o.s, loosing both two 4S electrons and one 3d electron. After addition of each electron in 3d, the 3d - orbitals are more and more stabilised and hence they exhibits more stable +2 o.s due to loss of two 4s electrons only. Also the sum of IE 1 and IE 2 decides the stability of the +2 state. Thus upto Chromium +2 o.s. is less stable and from Chromium onwards +2 o.s. is more stable. Thus Mn 2+ is very stable because of stability of d orbitals (half filled orbitals). 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. Zero, half filled and completely filled electronic configuration makes the oxidation state of the first series of transition elements more stable. For e.g. in Scandium the +3 (O) state is very stable as the 1

2 Sc 3+ ion possesses 3d 0 configuration. Similarly Manganese exhibits +2 and +7 oxidation states in most of its compounds as the Mn 2+ and Mn +7 ions possesses the electronic configuration of 3d 5 and 3d O respectively. For Iron the +3 oxidation state is more stable as the electronic configuration is 3d 5. For Zinc, Zn 2+ ion is very stable as it possess 3d 10 structure. 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 : 3d 3,3d 5, 3d 8 and 3d 4? Ground State Electronic configuration The stable (o) state 3d d d d 4 does not exist 6. Name the oxo metal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number. Dichromate ion, Cr 2 O 7 2- State of the metal +6 Permanganate ion, MnO Chromate ion, CrO Vanadium dioxide ion, VO 2 +5 Scandium oxide ion +3 Chromic oxide, CrO Vanadium tetroxide ion, VO What is lanthanoid contraction? What are the consequences of lanthanoid contraction? The overall decrease in atomic and Ionic radii from Lanthanum to Luticium is a unique feature in the chemistry of Lantanides. This is due to the imperfect shielding of one electron by another in the same sub-shell. However the shielding of one 4f electron by another is less than one d electron by another and with the increase in nuclear charge along the series, there is fairly regular decrease in the sizes with increasing atomic numbers. Consequences: The atomic radii of Zr (160pm) and Hf (159 p.m) are almost similar. In general the atomic radii of pre Lanthanide elements and post Lanthanide elements are almost the same. Lanthanides occur together due to their similarities in Atomic radii and the difficulty is faced in their separation. 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? 2

3 The transition elements are the elements whose d atomic orbitals are incomplete and possess the general electronic configuration (n-1) d 1-10 ns 1-2. As a result of progressive filling of the electrons in the d orbital they show properties slightly different from the normal elements. As there is transition in chemical properties on moving from left to right of the periodic table they are known as Transition Elements. They possess certain general characteristic properties (i) variation in atomic and ionic sizes (ii) variable oxidation states (iii) Magnetic properties (iv) complex formation (v) Alloy formation (vi) catalytic properties (vii) Formation Interstitial compounds. Zn, Cd, Hg possess completely filled d orbitals in their atoms or important ions and do not satisfy the general electronic configuration. They may be considered as non Transition Elements, but they are regarded as transition elements because of their common properties with others. 9. In what way is the electronic configuration of the transition elements different from that of the non transition elements? In non transition elements,the incoming electron enters the nth electronic level or outermost level. But in transition elements, the last electron enters the penultimate or (n-1) level. 10. What are the different oxidation states exhibited by the lanthanoids? In Lanthanides La 3+ and Lu 3+ compounds are predominant species. However occasionally +2 and + 4 ions in solution or in solid compounds are also obtained. This irregularity arises mainly from the extra stability of empty, half filled or filled sub shell. Thus the formation of Ce 4+ is favoured by its noble gas configuration. Pr, Nd, Tb and Dy also exhibit +4 state in the oxides or the type MO 2. Eu 2+ is formed by losing two s electrons and its f7 configuration accounts for the formation of this ion. Eu 2+ is a strong reducing agent changing to common +3 state. 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. (i) Para magnetism arises from the presence of un paired electrons, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the formula, taking spin into consideration. µ = 3

4 The magnetic moment increases with increasing number of unpaired electrons. Thus the observed magnetic moment gives a useful indication about the number of unpaired electrons present in the atom or molecule or ion. (ii) Enthalpies of Atomisation: Strong metallic bonds between the atoms of the Transition elements are responsible for the high melting and boiling points. This is clearly seen in their high enthalpies of atomisation. The strength of the metallic bond depends upon the number of unpaired 'd' electrons (half filled d orbitals). Greater is the number of unpaird electrons stronger is the metallic bonding. Hence they possess high enthalpies of atomisation. (iii) Formation of coloured compounds: Most of the transition metal compounds (ionic as well as covalent) are coloured both in the solid state and in aqueous solution in contrast to compounds of S and P block elements. This property is due to presence of unpaired electron(s) in the d-orbital and its transition within the d-orbtials. The energy of excitation corresponds to the frequency of light absorbed. This frequency generally belongs to the visible region, as less energy is involved in the transition of electron. 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 legand. (iv) 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 form complexes. Vanadium (v) oxide, finely divided Iron oxide and nickel are some examples of catalysts. The solid surface involve the formation of bonds between reactant molecule s and atoms of the surface of the catalyst. This has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules. Also because the transition metal ions can change their oxidation states, they become more effective as catalysts. 12. What are interstitial compounds? Why are such compounds well known for transition metals? Interstitial compounds are those which are formed when small atoms like H N or C are trapped inside the crystal lattices of metals. They are usually non stoichiometric and are neither typically ionic nor covalent. Many of the transition metals form interstitial compounds, particularly with small non-metal atoms such as hydrogen, boron, carbon and nitrogen. These small atoms enter into the void sites between the packed atoms of the crystalline metal e.g TiC, Mn 4 N, TiH 2, etc. The formulae corresponds to non-stoichiometric compounds. The characteristics of these compounds are: (i) They have high melting points, higher than those of pure metals. (ii) They are very hard as compared to diamond (iii) They retain metallic conductivity. (iv) They are chemically inert. 13. How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples. In transition metals there exists less energy gap between (n-1) d and ns atomic orbitals. Therefore, electrons are lost from ns atomic orbital as well as and (n-1)d orbitals in order to attain empty d orbitals, half filled d orbitals and completely filled d orbital configuration which are extra stable 4

5 where as in non-transition metals the electrons are lost only from the same energy level. Transition metals show variable oxidation states, involving ns electrons and (n-1) d electrons.(eg) Mn exhibits +2 and +7 states. Non transition elements also show more than one valency, using ns and np electrons only. They show o.s. either equal to the valance electrons or (8-valance electrons). They exhibit variable valencies with a difference of one. (Eg) Iodine exhibits valencies ,+1,+3,+5,+7.Also in p block as you go down, lower oxidation states become more stable due to inert pair effect. 14. 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 by fusing the chromite ore Fe Cr 2 O 4 with sodium or potassium carbonate in free excess of air. 4FeCr 2 O 4 +8Na 2 CO 3 +7O 2 8Na 2 CrO 4 +2Fe 2 O 3 +8CO 2 The fused mass is dissolved in water. The yellow solution of sodium Chromate is filtered and acidified with sulphuric acid when sodium dichromate Na 2 Cr 2 O 7 2H 2 O is formed. 2Na 2 CrO 4 + 2H + Na 2 Cr 2 O 7 +2Na + + H 2 O Sodium dichromate is more soluble than potassium dichromate. The latter is prepared by treating the solution Sodium dichromate with potassium chloride. Na 2 CrO 7 + 2Kcl K 2 Cr 2 O Nacl Orange crystals of potassium dichromate crystallise out on fractional crystallisation. Potassium dichromate solution changes to potassium chromate by increasing the PH of the solution. Cr 2 O + 2OH - 2CrO + H 2 O On decreasing the PH of the chromate solution it charges dichromate solution. 2CrO + 2H + CrO + H 2 O 15. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:(a) iodide (b) iron(ii) solution and (c) H 2 S In acidic solution the oxidising action of potassium dichromate can be represented as follows 5

6 Cr 2 O + 14H + + 6e - 2Cr H 2 O (a) It oxidises I - to I 2 Cr 2 O + 14H + + 6I - 2Cr H 2 O + 3I 2 (b) It oxidises Fe 2+ to Fe 3+ ion Cr 2 O + 14H + + 6Fe 2+ 2Cr 3+ +7H 2 O + 6Fe 3+ (c) H 2 S 2H + + S 2- Cr 2 O + 14H + + 6e - 2Cr H 2 O 3[S 2- S + 2e - ] Cr 2 O + 14H + + 3S 2-2Cr H 2 O + 3S 16. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (a) iron(ii) ions (b) SO 2 and (c) oxalic acid? Write the ionic equations for the reactions. Potassium permanganate is prepared by fusion of MnO 2 with an alkali metal hydroxide and an oxidising agent like KNO 3. This produces dark green K 2 MnO 4 which disproportionates in neutral or acidic solution to give permanganate. 2MnO 2 + 4KOH + O 2 2K 2 MnO 4 + 2H 2 O 3 MnO + 4H + 2MnO + MnO 2 + 2H 2 O Potassium permanganate is a strong oxidising agent in acidic medium. The relevant half reactions are: Acidic solution MnO + 8H + + 5e - Mn 2+ +4H 2 O E = 1.55V (a) MnO + 8H + + 5Fe 2+ 5Fe 3+ + Mn H 2 O (b) 4MnO + 12H SO 2 4Mn H 2 O + 10SO 3 6

7 (c) MnO + 8H + +5 Mn H 2 O + 10CO For M 2+ /M and M 3+ /M 2+ systems the EV values for some metals are as follows: Cr 2+ /Cr -0.9V Cr 3+ /Cr V Mn 2+ /Mn -1.2V Mn 3+ /Mn V Fe 2+ /Fe -0.4V Fe 3+ /Fe V Use this data to comment upon: (a) the stability of Fe 3+ in acid solution as compared to that of Cr 3+ or Mn 3+ and (b) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal. (a) Comparing E values, they are in the increasing order Cr 3+ /Cr 2+ < Fe 3+ /Fe 2+ <Mn 3+ /Mn 2+. Hence the tendency to get reduced increases. It means Cr 3+ is more stable than Fe 3+ which is more stable than Mn 3+. (b) The oxidation electrode potentials of the metals are for Fe (+0.4V),Cr(+0.9V) and for. Mn(+1.2V).The order in which the metal gets oxidized is Mn>Cr>Fe. 18. Predict which of the following will be coloured in aqueous solution? Ti 3+, V 3+, Cu +, Sc 3+, Mn 2+, Fe 3+ and Co 2+. Give reasons for each. The transition metal ion exhibits colour in aqueous solution due to the presence of an unpaired electron in the d orbital. Ti 3+, V 3+, Mn 2+, Fe 3+, Co 2+ and MnO ions exhibit colour as they possess unpaired electrons. 19. Compare the stability of +2 oxidation state for the elements of the first transition series. Mn 2+ and Zn 2+ ions are more stable as contain half filled and fulfilled d- electrons. The rest of the M 2+ ions are less stable, with variation. Fe 2+ ion is less stable and easily oxidised to Fe 3+ which contains half filled d-orbital electrons. 7

8 20. 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. Lanthanides Electronic Configuration In Lanthanides 4f level gets Progressively filled Oxidation States In Lanthanides the +3(o.s) State is more stable Occasionally they also exhibit +2 and +4 o.s Actinides In actinides 5f level gets progressively filled In Actinides the 5f level get progressively filled. In actinides +4 oxidation state is very common. The other Oxidation states which are commonly exhibited are +5, +6 and +7 respectively. Atomic and Ionic Sizes There is over all decrease in Atomic and ionic radii The atomic and ionic sizes for actinides also from Lanthanium to Lutitium. decreases progressively. Chemical Reactivity They are radioactive elements. Actinides are All lanthanides show similar chemical properties. They tarnish readily on exposure to air. They form Ln 2 O 3 except Ce which forms CeO 2. All form Ln(OH) 3 which are stronger bases than Al(OH) 3. They combine with non metals like O 2, S and hologens. They act as reducing agents. They have less tendency to from complexes. highly reactive metals, especially when finely divided. With boiling water they give a mixture of oxide and hydroxide they combine with nonmetals at moderate temperatures. They do not react with alkalies. They are slightly affected by Nitric acid owing to the formation of Protective oxide layer. They have greater tendency to form complexes. 21. How would you account for the following: (a) Of the d 4 species, Cr 2+ is strongly reducing while manganese(iii) is strongly oxidising. (b) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (c) The d 1 configuration is very unstable in ions. The E 0 value of Cr 3+ /Cr 2+ is negative while that of Mn 3+ /Mn 2+ is positive.so Cr 2+ can easily get oxidized to Cr 3+ while Mn has greater tendency to remain in Mn 2+ state. 22. What is meant by disproportionation? Give two examples of disproportionation reaction in aqueous solution. When a particular oxidation state becomes less stable relative to other oxidation states, one lower, 8

9 one higher, it is said to undergo disproportionation, for e.g. Manganese (vi) becomes unstable relative to Mn (vii) and manganese (iv) in acidic solution 3Mn VI + 4H + = 2Mn VII O + Mn IV O 2 +2H 2 O 23. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why? Cu exhibits +1 oxidation most frequently in first series of transition metals. This is due to d orbitals consisting of 10 electrons giving rise to a stable configuration. 24. Calculate the number of unpaired electrons in the following gaseous ions: Mn 3+,Cr 3+, V 3+ and Ti 3+. Which one of these is the most stable in aqueous solution? Ions Number of unpaired electrons Mn 3+ 4 Cr 3+ 3 V 3+ 2 Ti 3+ 1 Of these Cr 3+ is most stable as it has half filled t 2g set of d orbitals. 25. Give examples and suggest reasons for the following features of the transition metal chemistry: (a) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic. (b) A transition metal exhibits highest oxidation state in oxides and fluorides. (c) The highest oxidation state is exhibited in oxoanions of a metal. (a) The lowest oxide of transition metal is basic due to its large size of central metal atom. It readily furnish electrons and behave as Lewis base whereas in higher oxidation state the central metal atom is smaller in size and possess more polarisability and readily accept the electrons. Thus the acidic nature is observed in these oxides. (b) The oxygen and fluorine are the most electronegative atoms hence when they combine with transition metals they readily oxidise the transition metal to a higher oxidation state. (c) The oxo anions of transition metals are generally formed from the reaction of metal with oxygen at high temperatures. As oxygen is more electronegative the transition metal atom loses the electrons to exist in higher oxidation states. 9

10 26. Indicate the steps in the preparation of:(a) K 2 Cr 2 O 7 from chromite ore. (b) KMnO 4 from pyrolusite ore. (a) The chromite ore is fused with Sodium Carbonate in excess of air or in the presence of oxidising agent. 4Fe.Cr 2 O Na 2 CO 3 8Na 2 CrO 4 + 2Fe 2 O 3 + 8CO 2 It is filtered. The filterate is yellow solution of Sodium Chromate. The yellow solution is now acidified with sulphuric acid. 2Na 2 CrO 4 + 2H Na 2 Cr 2 O 7 +2Na+ + H 2 O The Sodium dichromate is converted to K 2 Cr 2 O 7 by addition of potassium chloride Na 2 Cr 2 O 7 + 2KCl K 2 Cr 2 O NaCl The K 2 Cr 2 O 7 is seperated from NaCl by fractional crystallisation. (b) Potassium permanganate is obtained by the fusion of MnO 2 with alkali metal hydroxide and an oxidizing agent like KNO 3. 2MnO 2 + 4KOH+O 2 2K 2 MnO 4 +2H 2 O Potassium manganate is treated with Chlorine when KmnO 4 is formed. 2K 2 MnO 4 + Cl 2 2KCl + 2KMnO What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses. An alloy is a blend of metals prepared by mixing components. Alloys are homogeneous solid solutions in which atoms of one metal are randomly distributed among the atoms of the other. Lanthanides are used for the production of alloy steels for plates and pipes. A well known alloy is Misch metal which consists of a lanthanides metal on 95% and iron 5% and tracer of S, C, Ca and Al. A good deal of Misch metal is used in Mg based alloy to produce bullets, shell and lighter flint. Mixed oxides of Lanthanides are used as catalysts in petroleum cracking. 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. Inner Transition elements are the elements in which f-orbitals are filled. They possess the general electronic configuration (n-1)d 0-1. The neutral atom of ion should possess the following electronic configuration. 10

11 Atomic No. e - Configuration 29 1S 2, 2S 2, 2P 6, 3S 2, 3P 6, 4S 1, 3d S 2 2S 2 2P 6 3S 2 3P 6 4S 2 3d 10 4P 6 5S 2 4d 10 Sp 6 6S 2 4 f 3 74 [Xe] 4f 14 Sd 5 6S 1 95 [Rn] 5f 7 7S [Rn] 5 f 14 7S [Rn] 5f 14 6 d 2 7S 2 Except element with atomic number 29,all are inner transitional 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. The lanthanoid elements show only +3,+2 and +4 oxidation states. But actinoids exhibit +6 and+7 also in addition to this. Most of the Actinides elements are Radioactive in nature which makes the study of these elements rather difficult. When compared with lanthanides there is much variation in chemical properties in the first half of actinide series. 5f electrons are more effectively shielded from the nuclear charge than all the 4f electrons of the corresponding Lanthanides. The oxidation states of Actinides increases from +4 to +5 to +6 and +7 respectively in Pa, U and Np but decreases in succeeding elements. The ion in +3 and +4 oxidation states however tend to hydrolyse to form oxo ions due to the uneven distribution of oxidation states. 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. Last element in the actinoid series Atomic number Outer electronic configuration Outer electronic configuration of M 3+ ion Outer electronic configuration of M 4+ ion Lawrencium (Lr) 103 5f 14 6d 1 7s 2 5f 14 5f 13 The actinoids show in general +3 oxidation state. The first few elements of the actinoid series exhibit higher oxidation states. But, Lawrencium exhibit +3 oxidation state more commonly than other oxidation states because by doing so it gets a completely filled configuration. 31. Use Hund s rule to derive the electronic configuration of Ce 3+ ion, and calculate its magnetic moment on the basis of spin-only formula. The electronic configuration of Ce 3+ ion [Xe] 4f' 11

12 Magnetic moment, µ = µ = (where n is number of unpaired electrons) µ = 1.7BM. 32. Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements. Cerium exhibits Ce 4+ o.s as it as f 0 configuration and stable noble gas structure. Pr,Nd,Tb and Dy also exhibits o.s. IV in their oxides. Eu 2+ is due to f 7. Yb 2+ is given f 14 configuration. The existence of ions in +2 and +4 o.s. in solution or in solid the state is due to extra stability of empty, half filled and full filled f-orbitals. 33. Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109. Atomic No. Electronic Configuration 61 [Xe] 4f 5 6s 2 91 [Rn] 5f 2 6d 1 7s [Rn] 5f 13 7S [Rn] 5f 14 7S 2 6d 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. (i) The elements of first transition series involve progressive filling of 3d orbitals whereas that of second and third series involve filling of 4d and 5d subshells respectively. (ii) For the elements if first transition series +2 and +3 oxidation states are common and these elements form many stable complexes in these oxidation states. For the elements of second and third series higher oxidation states are more important. These elements form more stable compounds in higher oxidation states. For example, [CrCl 6 ] 3- is very stable whereas no equivalent complexes of Mo and W are known. On the contrary OsO 4 and PtF 6 are quite stable and no corresponding compounds for first transition series are known. (iii) The metals of second and third transition series have higher Ionisation enthalpies than the elements of first series. The elements of second and third series form many compounds with M-M bond. 12

13 (iv) The atomic radii of elements of second and third transition series are larger than those of the elements of first series. Because of lanthanoid contraction, the radii of third series are almost equal to those of second row. 35. Write down the number of 3d electrons in each of the following ions: Ti 2+, V 2+,Cr 3+, Mn 2+, Fe 2+, Fe 3+, Co 2+, Ni 2+ and Cu 2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral). (i) Ti 2+ : [Ar] 3d 2 Under the influence of an octahedral legand field the d-orbitals split into two groups of different energies. These are t 2g orbitals (d xy, d xz and d yz ) and e g orbitals (d z 2 and d x 2 -y 2 ),t 2g orbitals are of lower energy than e g orbitals. Thus, the two d- electrons in Ti 2+ would be present in two of the three t 2g orbitals. Thus there would be two unpaired electrons. (ii) V 2+ : [Ar] 3d 3 The three d-electrons in V2+ would be present in three different t 2g orbitals. Thus, there would be three unpaired electrons. (iii) Cr 3+ : [Ar] 3d 3 - (t 2g ) Three unpaired electrons (iv) Mn 2+ : [Ar] 3d 5 - (t 2g ) 3 (e g ) Five unpaired electrons (v) Fe 2+ : [Ar] 3d 6 - (t 2g ) 4 (e g ) Four unpaired electrons (vi) Fe 3+ : [Ar] 3d 5 - (t 2g ) 3 (e g ) Five unpaired electrons (vii) Co 2+ : [Ar] 3d 7 - (t 2g ) 5 (e g ) Three unpaired electrons (viii) Ni 2+ : [Ar] 3d 8 - (t 2g ) 6 (e g ) Two unpaired electrons (ix) Cu 2+ : [Ar] 3d 9 - (t 2g ) 6 (e g ) One unpaired electron 36. Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements. Some points of difference between properties of elements of the first transition series and those of heavier transition elements are given below: (i) For elements of first series +2 and +3 oxidation states are more common while for the heavier transition elements higher oxidation states are more common. (ii) M-M bonding is rare in the elements of first series but is quite common in heavier transition elements. 1

14 (iii) The elements of first transition series do not form complexes with coordination no. 7 or 8 whereas heavier transition elements do so. (iv) The elements of first transition series form low spin or high spin complexes depending upon strength of the ligand field. On the other hand heavier transition elements form low spin complexes irrespective of the strength of the ligand field. 37. What can be inferred from the magnetic moment values of the following complex species? Example Magnetic Moment (BM) (i)k 4 [Mn(CN) 6 ] 2.2 (ii)[fe(h 2 O) 6 ] (iii)k 2 [MnCl 4 ] 5.9 Calculate the magnetic moment by applying formula BM The magnetic moment of 2.2BM corresponds to n=1. Thus, in K 4 [Mn(CN) 6 ] there is only one unpaired electron in 3d subshell. Thus, the distribution of five 3d electrons in Mn(II) is (t 2g ) 5. This indicates that there are four unpaired electrons in the complex. Thus, the six 3d electrons in Fe(II) are distributed as (t 2g ) 4 (e g ) 2. This indicates that there are five unpaired electrons in the complex. Hence, the five 3d-electrons in Mn(II) are distributed as (t 2g ) 3 (e g ) 2 in the given complex. 38. Silver atom has completely filled d orbitals (4d10) in its ground state.how can you say that it is a transition element? Silver (Z = 47) can exhibit +2 oxidation state wherein it will have incompletely filled d-orbitals (4d), hence a transition element. 39. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kj mol 1. Why? In the formation of metallic bonds, no eletrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d series, electrons from the d-orbitals are always involved in the formation of metallic bonds. 14

15 40. The E (-) (M 2+ /M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high a H (-) and low hyd H (-) ) The positive value of E (-) (M 2+ /M) for copper is due to its high enthalphy of atomization and high values of enthalpies of ionization ( i H I + i H II ). 41. How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements? Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d-configurations (e.g., d 0, d 5, d 10 are exceptionally stable). 42. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state. 43. Which is a stronger reducing agent Cr 2+ or Fe 2+ and why? Cr 2+ is stronger reducing agent than Fe 2+ Reason: d 4 d 3 occurs in case of Cr 2+ to Cr 3+ But d 6 d 5 occurs in case of Fe 2+ to Fe 3+ In a medium (like water) d 3 is more stable as compared to d Calculate the spin only magnetic moment of M 2+ (aq) ion (Z = 27). The electronic configuration of the M 2+ (aq) ion (Z = 27) would be [Ar] 3d 7 It would contain three unpaired electrons. The spin only magnetic moment is given by the relation, BM = BM = 3.87 BM 45. Explain why Cu+ ion is not stable in aqueous solutions? Cu+ in aqueous solution undergoes disproportionation, i.e., 2Cu + (aq) Cu 2+ (aq) + Cu (s) 15

16 The E 0 value for this is favourable. 46. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? This is due to poor shielding effect of 5f electrons in actinoids compared to poor shielding effect by 4f electrons in lanthanoids. 16