Chemistry Study Material – Chemistry is fun.



CHAPTER – 8d-AND f-BLOCK ELEMENTSTransition elements - Those elements which have partially filled d- shell in their elementary form or in their commonly occurring oxidation states The d-block elements are called Transition element because of their position between s-block and p-block element of the periodic table. Zn, Cd and Hg are not regarded as Transition element because the d- orbital in these element are completely filled in the ground state and in their common oxidation states. The electronic configurations of Zn, Cd and Hg are represented by the general formula (n-1)d10ns2. The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements Sc is the lightest Transition element and Os is the heaviest one.The Transition element exhibit the following characteristics:They exhibit several oxidation statesThey form coloured ions (due to presence of unpaired electrons and d-d transition of electrons).The metal and their compound act as catalyst (due to their ability to exhibit multiple oxidation state).The element and many of their compounds are paramagnetic. They have the ability to form complexes. (Due to highly charged ion and contain vacant d-orbital).Electronic configuration The general Electronic configuration of the d- block element is [Inert gas] (n-1) d1-10 ns1-2. Exceptions in electronic configuration: 1.First (3d) transition series (Sc-Zn) 24Cr [Ar] 3d5 4s1, 29Cu [Ar] 3d10 4s1. 47Ag [kr] 4d10 5s1 . Ag++ has unpaired d-electron ( d9 ) * Exceptions in electronic configuration are either because of symmetry or release of maximum exchange energy due to more number of exchange of electrons are possible amongst the degenerate d-orbitals in half-filled and fulfilled configurations # iron, cobalt and nickel are known as ferrous metalWith partly filled d orbitals these elements exhibit certain characteristic properties such as display of a variety of oxidation states, formation of coloured ions and entering into complex formation with a variety of ligands. The transition metals and their compounds also exhibit catalytic property and paramagnetic behaviour. There are greater horizontal similarities in the propertiesStrong interatomic interaction:- High Enthalpy of atomization The high melting points of these metals are attributed to the involvement of greater number of electrons from (n-1)din addition to the ns electrons in the inter-atomic metallic bonding. In any row the melting points of these metals rise to a maximum at d5 except for anomalous values of Mn and Tc and fall regularly as the atomic number increases. The maxima at about the middle of each series indicate that one unpaired electron per d orbital is Particularly favourable for strong interatomic interaction. In general, greater the number of valence electrons, stronger is the resultant bonding. Since the enthalpy of atomisation is an important factor in determining the standard electrode potential of a metal, metals with very high enthalpy of atomisation(i.e., very high boiling point) tend to be noble in their reactions (see for electrode potentials). Melting point of transition elements The m.p and b.p of transition element are very high due to stronger interparticle bonds. In the period from left to right the m.p of the metal first increases to maximum then gradually decreases to maximum then gradually decreases towards the end of the period. (Mn, Tc posses anomalous m.p). M.P of Tungsten is 34100C The strength of interparticle bond in the transition element is roughly related to the number of half filled-orbital.In the beginning he number of half filled d-orbital increases till the middle of the period then decreases .so strength of interparticle bonds initially increases and at the end it is minimum . So m.p and enthalpy of atomization increases from left to middle and then decreases at the right end in a period The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d5 (Mn2+) and d10 (Zn2+) ions superimposed upon the general increasing trend. In general, the third ionisation enthalpies are quite high and there is a marked break between the values for Mn2+ and Fe2+. Also the high values for copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements. The variability of oxidation states, a characteristic of transition elements, arises out of incomplete filling of d orbitals in such a way that their oxidation states differ from each other by unity,e.g., V2+, V3+, V4+, V5+. This is in contrast with the variability of oxidation states of p-block elements where oxidation states normally differ by a unit of two. Cause:-The tendency to take part ‘d’ as well as outermost ‘s’-electrons in bond formation –due to more or less similar energy i.e. (n-1)d and ns (both are available for bonding purposes). When ns electron participate in bonding—Low O.S and when (n-1)d and ns take part in bonding—Higher OS results. Relative Stabilities of different Oxidation States depend upon the following factors:-1) Electronic Configuration 2) lattice and Solvation Enthalpy3) Nature of the element with which the TM is combined—Oxygen, Fluorine etc. 4) Nature of bonding 5) Nature of solvent,ligand,complexing agents etc. 6) Stereochemistry(octahedral,tetrahedral etc.) Cr+2 Unstable in H2O w.r.t oxidation Cr+3 Stable in water (in acidic solution also) Fe+2 Unstable in aerated water w.r.t. oxidation Cr+6 stable in alkaline solution Group no of Mn=7,oxidation state of Mn =+7 in permanganate ion MnO4ˉ Ggroup no of Cr=6,oxidation state of Cr =+6 in Cr2O72-and CrO4ˉ. Magnetic properties:-Paramagnetism arises from the presence of unpaired electrons in an atom ,ion or molecule, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. Diamagnetic substances are repelled by the applied field while the paramagnetic substances are attracted by the applied field. Substances which are attracted very strongly are said to be ferromagnetic. In fact, ferromagnetism is an extreme form of paramagnetism. Many of the transition metal ions are paramagnetic.Q:-Transition metals generally from color compound ? Ans:- Due to d-d transition of unpaired d-electron . The unpaired electron excited from lower energy d-orbital to higher energy d-orbital(CFT), the energy is absorbed from visible region of radiation and the complementary part of absorbed light( i.e. reflected light) will decide the colour of the compound Q:-Transition metals and many of their compound act as a good catalyst . Ans-- catalytic activity is ascribed to their 1) ability to adopt multiple oxidation states and 2) to form complexes. (form different intermediate with any oxidation states which may lead through lower activation energy path). ** In some cases , the T.Metals provide s suitable large surface area with free valencies on which reactants are adsorbed . **(Due to their tendency to show variable oxidation states TM forms unstable intermediate compounds and provide a new path for the reaction with lower activation energy path) Q:-Transtion metals form complex compounds. Ans-- This is due to the high (charge / radius) ratio and empty d-orbital . 1) comparatively smaller sizes of the metal ions, 2) their high ionic charges and 3) the availability of empty d orbitals for bond formation.Redox Reaction :- 1# In Acid Solution-- Reduction half cell : MnO4ˉ (permanganate) Mn+2(manganous) Cr2O72ˉ (dichromate) 2Cr+3 Oxidation half cell : Iˉ(Iodide) I2 (Iodine) , S 2ˉ (Sulphide) S(Sulphur) , Sulphite(SO32ˉ) sulphate(SO42ˉ) , Nitrite (NO2ˉ ) Nitrate (NO3ˉ ) Fe+2(Ferrous) Fe+3 (Ferric) , C2O42 ˉ(oxalate) CO2 (Carbon Dioxide) Sn+2 (stannous) Stannic(Sn+4)2# In neutral or faintly alkaline solution : Reduction half cell : MnO4ˉ MnO2Oxidation half cell : Iˉ(Iodide) IO3ˉ (Iodate) , Sulphite(SO32ˉ) sulphate(SO42ˉ) LANTHANOIDS:-Lanthanoid Contraction:The overall decrease in atomic and ionic radii from lanthanum to lutetiumCause- due to the imperfect shielding of one f-electron by another in the same sub-shell. and the dominance of nuclear charge . [The extra orbital electrons incompletely shielded the extra nuclear charge. Thus effective nuclear charge increases.Thus all the electrons are pulled in closer ]Consequences:-Thus pairs of elements such as Zr/Hf ,Nb/Ta , Mo/W are almost identical in size.The close similarities of properties in such a pair makes chemical separation very difficult.[The almost identical radii of Zr (160 pm) and Hf (159 pm), a consequence of the lanthanoid contraction, account for their occurrence together in nature and for the difficulty faced in their separation. ]La (OH)3 are more basic than Lu(OH)3 [ since Lu+3 is the smallest ion ,it is most heavily hydrated, it forms strongest complexes among all ,it forms weakest base – Fajan’s Rule ] The first IE of 5d elements lie higher than those of 3d and 4d elements [This is due to the greater effective nuclear charge acting on outer valence electrons because of weak shielding of the nucleus by 4f electrons. ]The radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called Lanthanoid contraction which essentially compensates for the expected increase in atomic size with increasing atomic number. The net result of the lanthanoid contraction is that the second and the third d series exhibit similar radii (e.g., Zr 160 pm, Hf (159 pm) and have very similar physical and chemical properties much more than that expected on the basis of usual family relationship.# APPLI CATIONS:-The best single use of the lanthanoids is for the production of alloy steels for plates and pipes. A well known alloy is mischmetall which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. *A good deal of mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint.*Mixed oxides of lanthanoids are employed as catalysts in petroleum cracking.*Some individual Ln oxides are used as phosphors intelevision screens and similar fluorescing surfaces This is indicated from the abnormally low value of the third ionisation enthalpy of lanthanum, gadolinium and lutetium. the exchange enthalpy considerations appear to impart a certain degree of stability to empty, half-filled and completely filled orbitals f level. The first ionisation enthalpies of the lanthanoids are around 600 kJ mol–1, the second about 1200 kJ mol–1 comparable with those of calcium. Many trivalent lanthanoid ions are coloured both in the solid state and in aqueous solutions. Colour of these ions may be attributed to the presence of f electrons. Neither La3+ nor Lu3+ ion shows any colour but the rest do so.However, absorption bands are narrow, probably because of the excitation within f level.The lanthanoid ions other than the f 0type (La3+ and Ce4+) and the f 14 type (Yb2+ and Lu3+) are all paramagnetic. The paramagnetism rises to maximum in neodymium. All the lanthanoids are silvery white soft metals and tarnish rapidly in air. The hardness increases with increasing atomic number, samarium being steel hard. Their melting points range between 1000 to 1200 K but samarium melts at 1623 K. They have typical metallic structure and are good conductors of heat and electricity. Density and other properties change smoothly except for Eu and Yb and occasionally for Sm and Tm. Oxidation States :-the sum of the first three I.E for each elements are low . Thus the oxidation state(III) is ionic and Ln+3 dominates the chemistry of these elements . The Ln+2 and Ln+4 ions that do occur in solution or in solid compounds are also obtained This irregularity (as in ionization enthalpies) arises mainly from the extra stability of empty, half-filled or filled f subshell. Pr, Nd, Tb and Dy also exhibit +4 state but only in oxides, MO2 1.a noble gas(empty) configuration.e.g.Ce+4(f0). Thus, the formation of Ce(IV) is favoured by its noble gas configuration, but it is a strong oxidant reverting to the common +3 state. The Eo value for Ce4+/ Ce3+ is + 1.74 V which suggests that it can oxidize water. However, the reaction rate is very slow and hence Ce(IV) is a good analytical reagent.2. a half filled f –shelle.g. Eu+2& Tb+4 (f 7)3. a completely filled f-level e.gYb+2.(f 14) ----------------------- but always less stable than Ln+3 . The behaviour of samarium is very much like europium, exhibiting both +2 and +3 oxidation states. ** Sm+2(f 7) ,Eu+2(f 7),Yb+2(f 14) (tendency to change into +3 O.S.) Sm+3,Eu+3, Yb+3 ( Stable , common O.S)[ Acts as reducing agent)** Tb+4(f 7) , Ce+4(f0) (tendency to change into +3 O.S.) Tb+3 , Ce+3 ( Stable , common oxidation states) [ Acts as oxidizing agent]ACTINOIDS :-They show wide range of oxidation states due to the fact t hat the 5f,6d and 7s energy levels are of comparable energies. Therefore all these three subshells can participate. Common Oxidation States is +3 .Th(+4) , Pa(+5) , U (+6) ,Np(+7)Binding energy of 5f are lower.5f electrons have poor shielding effect. Therefore, contraction in their sizes is more .They have greater tendency to form complexes .They are more basic . , They are more radioactive .They form oxoions such as UO2+ ,NpO2+ , PuO2+,UO2+Most of the actinides are coloured.e.g.U3+ (red) , U4+ (green) ,UO2+ (yellow) .They are also paramagnetic like lanthanoids but their magnetic properties cannot be explained easilyThe ionization enthalpies of the early actinoids are lower than for the early lanthanoids. This is quite reasonable since it is to be expected that when 5f orbitals are beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Because the outer electrons are less firmly held, they are available for bonding in the actinoids. The actinoids are highly reactive metals, especially when finely divided. The action of boiling water on them, for example, gives a mixture of oxide and hydride and combination with most non metals takes place at moderate temperatures. Hydrochloric acid attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers; alkalies have no action. actinoid contraction :- There is a gradual decrease in the size of atoms or M3+ ions across the series. The actinoid contraction is, however, greater from element to element in this series resulting from poor shielding by 5f electrons.Although the 5f orbitals resemble the 4f orbitals in their angular part of the wave-function, they are not as buried as 4f orbitals and hence 5f electrons can participate in bonding to a far greater extent. Comparative study of Lanthanoids and Actinoids:-PropertyLanthanoidsActinoids1.Oxidation statesLanthanoids show mainly +3 Oxidation state except in a few cases where it is +2 and +3 Actinoids shows wide range of oxidation states due to the fact t hat the 5f,6d and 7s energy levels are of comparable energies.Therefore all these three subshells can participate. Common Oxidation States is +3 Th(+4),Pa(+5), U (+6) ,Np(+7)2.Binding energyBinding energies of 4f are higherBinding energies of 5f are lower3.shielding effect4f electrons have greater shielding effect as compared to 5f .Therefore , the contraction in ionic radii is less .5f electrons have poor shielding effect as compared to 4f .Therefore , the contraction in ionic radii is more .4. Tendency to form complexesTendency to form complexes is less .They have greater tendency to form complexes .5.Basic characterLanthanoid compounds are less basicActinoid compounds are more basic6.Tendency to form oxo ionsThey do not form oxo ions They form oxo ions such as UO2+ ,NpO2+ , PuO2+,UO2+7.RadioactivityExcept promethium , these are non-radioactiveAll the acinoids are radioactive8.ColoursMost of their ions are colourless. But exceptions are there .Most of the actinoidions are colourede.g. U3+(red) , U4+(green) ,UO2+(yellow)9.Paramagnetic characterThey are paramagnetic and their magnetic properties can be easily explainedThey are also paramagnetic but their magnetic properties cannot be easily explainedII# IDENTIFY THE FOLLOWING:1# A mixed oxide of iron and chromium FeO.Cr2O3 is fused with Sodium Carbonate in presence of air to form a yellow coloured compound (A). On acidification the compound (A) forms an orange coloured compound (B) which is a strong oxidizing agent. (i) Identify the compounds (A) and (B) (ii) Write balanced chemical equations for each step.2 # A green chromium compound (A) on fusion with alkali gives a yellow compound (B) which on acidification gives an orange coloured compound (C) . “C” on treatment with NH4Cl gives an orange coloured product (D) , which on heating decomposes to give back (A) . Identify A,B,C,D .Write equations for reactions .3# (a) A blackish brown coloured solid (A) when fused with alkali metal hydroxides in presence of air produces a dark green compound (B), which on electrolytic oxidation in alkaline medium gives a dark purple compound (C). Identify (A), (B) and (C) and write balanced chemical equations for the reactions involved. (b) What happens when an acidic solution of the green coloured compound (B) is allowed to stand for some time? Give the equation of the reaction involved. What is this type of reaction called? (Hint: MnO42- changes to MnO4-)4# (A) reacts with H2SO4 to form purple coloured solution (B) which reacts with KI to form colourless compound (C). The colour of (B) disappears with acidic solution of FeSO4. With concentrated H2SO4 (B) forms (D) which can decompose to give a black compound (E) andO2. Identify (A) to (E) and write equations for the reactions involved.5# When an orange coloured crystalline compound ‘A’ was heated with common salt and concentrated H2SO4, an orange red coloured gas ‘B’ was evolved. The gas ‘B’ on passing through NaOH solution gave an yellow solution C (i) Identify A,B and C. (ii) Write balanced chemical equation involved in the reactions.6# Two gases (A) and (B) turns acidified K2Cr2O7 solution green. Gas(A) turns lead acetate paper black and when passed through gas(B) in aq. Solution , yellowish turbidity appears . Identify A and B and explain.7# Explain why a green solution of K2MnO4 turns purple and a brown solid is precipitated when CO2 gas is bubbled into the solution .PREVIOUS YEAR QUESTION PAPERS – 2008 TO 2015 – SOLVEDQ.1# Why Zn ,Cd, and Hg are normally not considered as Transition metals.Ans—They do not have unpaired d-electrons in the ground state as well as in their common oxidation states.**The electronic configurations of Zn, Cd and Hg are represented by the general formula (n-1)d10ns2. The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements.Q.2# Transition elements show variable oxidation states.Write all the possible O.S. of an element (Z=25) Ans--due to very little energy difference of energy between (n-1)d and ns orbitals. The tendency to take part ‘(n-1)d’ as well as outermost ‘ns’-electrons in bond formation –**When ns electron participate in bonding—Low O.S and when (n-1)d and ns take part in bonding—Higher OS resultsQ.3# Transition metals have high boiling points and have high enthalpies of atomization . (AI-05)Ans-- Because of large number of unpaired d-electrons in their atoms they have stronger inter-atomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.(both metallic and d-d- overlapping of covalent bond ) **One unpaired electron per d orbital is particularly favourable for strong interatomic interactionQ.4# Transition metals and many of their compounds show paramagnetic behaviour . (AI-05)Ans—Due to spin and orbital motion of negatively charged unpaired d-electron(s) in an atom ,ion or molecule Q.5# Transition metals generally from color compound (AI-08)Ans--Due to d-d transition of unpaired d-electron .The unpaired electron excited from lower energy d-orbital to higher energy d-orbital(CFT), the energy is absorbed from visible region of radiation and the complementary part of absorbed light( i.e. reflected light) will decide the colour of the compoundQ.6# Transition metals and many of their compound act as a good catalyst . (AI-05)Ans-- catalytic activity is ascribed to their 1) ability to adopt multiple oxidation states and 2) to form complexes. (form different intermediate with any oxidation states which may lead through lower activation energy path).** In some cases , the T.Metals provide s suitable large surface area with free valencies on which reactants are adsorbed . **(Due to their tendency to show variable oxidation states TM forms unstable intermediate compounds and provide a new path for the reaction with lower activation energy path) Q.7# Transtion metals form complex compounds.Ans--This is due to the high (charge / radius) ratio and empty d-orbital . 1) comparatively smaller sizes of the metal ions, 2) their high ionic charges and 3) the availability of empty d orbitals for bond formation. Q.8 # Transition metal generally form alloys with other transition metals.Ans— Because of similar radii and other characteristics of transition metals, alloys are readily formed by these metals. Alloys are formed by atoms with metallic radii that are within about 15 percent of each other. Substitutional and interstitial alloy .Q.9 # Transition metals form a number of interstitial compounds Ans-- It because transition metals have voids in their crystal lattice , in which small atoms of H , C , N can fit forming interstitial compounds which do not have fixed composition . The trapped atoms get bonded to the atoms of TE . e.g. TiC , Fe3H , Mn4N Q. 10 # There is hardly any increase in atomic size with increasing at.no. in a series of transition metals . ANS- This is because with increase in atomic number in a series , the increases nuclear charge is partly cancelled by the increased shielding effect of electrons in the d-orbitals of penultimate shell . Q.11# What is lanthanide contraction ? State the cause and two consequences of lanthanide contraction . Ans-- The overall decrease in atomic and ionic radii from lanthanum to lutetiumCause- due to the imperfect shielding of one f-electron by another in the same sub-shell. and the dominance of nuclear charge .The extra orbital electrons incompletely shielded the extra nuclear charge. Thus effective nuclear charge increases.Thus all the electrons are pulled in closer Consequences—(1) Thus pairs of elements such as Zr/Hf (2)The first IE of 5d elements lie higher than those of 3d and 4d elements (3) The metallic radii of 5d and 4d are nearly same . (4) it is difficult to separate lanthanoid elements (5) La(OH)3 is more basic than Lu(OH)3Q.12 # (a)Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (b)There is a greater range of oxidation states among the Actinoids than that in LanthanoidsAns—(a) The 5f electrons themselves provide poor shielding from element to element in the series as compared to 4f –electrons .As because 5f orbitals are comparatively larger and more diffused . Imperfect screening is greater in case of 5f than in case of 4f .Ans—(b) Due to the fact that 5f, 6d and 7s levels are of comparable energiesi.e the energy differences are very less , so electron can participate from all three sub-shells . Q.13 # WhyZr and Hf show similar properties? Ans—It is due to similar ionic size which is due to lanthanoid contraction . (Due to the imperfect shielding of one f-electron by another in the same sub-shell.)Q.14# The 4d- and 5d-series of transition metals have more frequent Metal-Metal bonding in their compounds than do the 3d-metals . [(OR ,There occurs much more frequent metal-metal bonding in compounds of heavy transition elements (3rd series).]OR , The metallic radii of 5d and 4d series are nearly the same .OR ,There is a close similarity in physical and chemical properties of the 4 d and 5 d series of thetransition elements, much more than expected on the basis of usual family relationship.Ans – Due to lanthanoid contraction Due to the imperfect shielding of one f-electron by another in the same sub-shell. and the dominance of nuclear charge .The extra orbital electrons incompletely shielded the extra nuclear charge. Thus effective nuclear charge increases.Thus all the electrons are pulled in closer Q.15 # The E0(M2+/M) value for copper is positive (+0.34V). What is possibly the reason for this? Copper is the only metal in the first series of Transition elements showing this behavior ?Ans— The reduction potential value depends on relative value of three enthalpies viz ΔaH0 , ΔiH0 and - ΔhydH0 . The copper has high ΔaH0 and ΔiH0. Therefore high enthalpy required to transform Cu(s) to Cu2+(aq) is not balanced by its low ΔhydH0 . Q.16 # Why is Cr2+ strongly reducing and Mn3+ strongly oxidising when both have d4 configuration. Ans:- Cr2+ is reducing as its configuration changes from d4 to d3 (t2g3 ), Cr3+ (t2g3 )-- As the having a half-filled t2g level .It has higher CFSE than Mn2+ ( t2g3 eg2)On the other hand, the change from Mn3+( d4) to Mn2+ (d5) results in the half-filled (d5) configuration which has extra stability.***In aqueous solution Cr3+(t2g3) has more stability than Mn2+ ( t2g3 eg2) Q.17 # Explain why Cu+ ion is not stable in aqueous solutions? Ans--Cu+ in aqueous solution underoes disproportionation, i.e., 2Cu+(aq) → Cu2+(aq) + Cu(s) . Cu ++ is more stable .The E0 value for this is favourable. Cu++ has relatively higher hydration enthalpy than Cu+Q.18 # What is meant by ‘disproportionation’ of an oxidation state? Give an exampleWhen a particular oxidation state of an element becomes less stable relative to other oxidation state, one lower and one higher, it is said to undergo disproportionation. For example, in aqueous solution: For example, (i) manganese (VI) becomes unstable relative to manganese(VII) and manganese (IV) in acidic solution. (i) 3 MnO42- + 4H+ ——> 2MnO42- + MnO2 + 2H2O (ii) Cu+ disproportionate into Cu2+and Cu : 2Cu2+ Cu2+ +Cu (iii) Mn3+ Mn2+ + MnO2How would you account for the following:Q.19# Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.OR ,Co (II) is easily oxidised in the presence of strong ligands.Ans— Co(III) has greater tendency to form coordination complexes than Co(II) .Due to higher charge by radius ratio and also due to greater CFSE( Crystal Field Stabilisation Energy) .Co (III) has d6 configuration In case of of strong ligand , It is t2g 6 eg0 (stable configuration)So , Co(II) changes to Co(III) i.e. easily oxidized .Q.20# Thed1 configuration is very unstable in ions.Ans-- The ions with d1 configuration have the tendency to lose the only electron present in d-subshell to acquire stable d0 configuration (Argon configuration). Hence they are unstable and undergo oxidation and disproportionation .Q.21 # Mn+2 compounds are more stable than Fe+2 toward oxidation to their + 3 state .(AI-06)OR , Why E0 for Mn+3 / Mn+2 couple is more positive than for Fe+3 / Fe+2 or Cr+3 / Cr+2 .Ans-- Mn 3+ + e- ——>Mn 2+ (d4 system , less stable ) (d5 system , more stable ) extra stable due to half-filled confN Fe3+ + e- Fe 2+ (d5system , more stable )(d6 system , less stable ) Cr3+ + e- ——> Cr 2+ (t2g3 - half filled- more stable )(d4 system , less stable) Much larger third Ionisation enthalpy of Mn( where the required change is d5 to d4) is mainly responsible for this . this also explains why the +3 state of Mn is of little importanceCr3+ (d3), t2g 3 - half filled ) Higher Crystal Field stabilization energy . (CFSE)Q.22 # (a) There is no regular trend in E 0 values in the series Vanadium( At No –23) to Copper(At No 29) for M2+/M system..(ii) EoM2+ /M values are not regular for first row transition metals (3d series).Ans— The E0(M2+/M) values are not regular which can be explained from the irregular variationof ionization energies (IE1 + IE2) and also the sublimation energy which are relatively much less for Mn and Vanadium. (240kj/mol for Mn and 470 KJ/mol for Vanadium)Q.23 # How would you account for the increasing oxidising power in the series VO2+< Cr2O72–< MnO4– ?Ans--This is due to the increasing stability of the lower species to which they are reduced.Q.24 # How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?Ans-- Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d-configurations (e.g., d0, d5, d10 are exceptionally stable).Q.25 # Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? OR , Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF4, whereas the highest oxide is Mn2O7.—Explain OR Why the highest oxidation state is exhibited in oxo-anions of a transition metal.?Ans-- Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidationstate.**The ability of fluorine to stabilise the highest oxidation state isdue to either higher lattice energy as in the case of CoF3, or higher bond enthalpy terms for the higher covalent compounds, e.g., VF5 and CrF6.The ability of oxygen to stabilise these high oxidation states exceeds that of fluorine. The ability of oxygen to form multiple bonds with transition metal explains its superiority.Thus the highest Mn fluoride is MnF4 whereas the highest oxide is Mn2O7.In the covalent oxide Mn2O7, each Mn is tetrahedrally surrounded by O’s including a Mn–O–Mn bridge.Q.26# Which is a stronger reducing agent Cr2+ or Fe2+ and why ?Ans-- Cr2+ is stronger reducing agent than Fe2+Reason:d4 → d3 occurs in case of Cr2+ to Cr3+ ((t2g3 - half filled- more stable )But d6 → d5 occurs in case of Fe2+ to Fe3+ .In a medium (like water) d3is more stable (higher CFSE) as compared to d5 .Q.27 # Why is the E0 value for the Mn3+/Mn2+ couple much more positive than that for Cr3+/Cr2+ or Fe3+/Fe2+? Explain.Ans--Much larger third ionisation energy of Mn (where the required change is d5 to d4) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance.Q. 28 # Among lanthanoids , Ln(III) compounds are predominant . However , occasionally in solutions or in solid compounds , +2 & +4 ions are also obtained .ANS- lanthanoid metals show +2 and +4 O.S to attain extra stability of f0 f7 and f14 configuration . Eu2+ ,Tb4+ -(4f7)-half filled . Ce4+ ( 4f0) Q.29 # How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.ANS- In transition elements the oxidation states vary by oneFor example, for Mn it may vary as +2, +3, +4, +5, +6, +7. In the nontransition elements the variation is selective, always differing by 2, e.g. +2, +4, or +3, +5 or +4, +6 etc.Q. 30 # Why there is general increase in density from Ti to Cu .ANS- The decrease in metallic radius coupled with increase in atomic mass results in a general increase in density .Q. 31 # The higher O.S are usually exhibited by the members in the middle of the series of Transition Elements.ANS- Due to presence of maximum number of unpaired electrons in a TM which is present in the middle of the series . Q. 32 # Mn2+ is much more resistant than Fe2+ towards oxidation . ANS-Mn2+(d5 , half-filled ) is stabler than Mn3+(d4) , so Mn2+ not easily get oxidized Fe2+(d6) is less stable than Fe3+ (d5 , half-filled) . so Fe2+ is easily get oxidized to get stable configuration .Q.38 #.Why is HCl not used to acidify a permanganate solutions in volumetric estimation of Fe+2 or C2O42￣ .Ans-Since Clˉ of Hydrochloric acid is oxidized to chlorine by MnO4ˉ . so , actual amount of KMnO4 required to oxidize C2O42￣couldnot be determinedQ.33 # Give reason for it?. Ni+2 compounds are thermodynamically more stable than Pt+2compounds whilstPt(IV) are relatively more stable than Ni(IV) compounds . OR, K2PtCl6 is well known cpds. but corresponding Ni-compound is not known .Ans--The oxidation state of Pt in K2 Pt Cl6is +4 .The sum of the first four ionization energies (IE1+IE2 +IE 3+IE 4) of pt is less than those of Ni.Q.34 # E0 for Mn+3 / Mn+2 couple is more positive than for Fe+3 / Fe+2 or Cr+3 / Cr+2 .Ans-- Much larger third Ionisation enthalpy of Mn ( where the required change is d5 to d4) is mainly responsible for this . this also explains why the +3 state of Mn is of little importance .Q.35 # Why do transition elements have high enthalpy of hydration? Ans-- the high enthalpy of hydration of transition elements is due to small size of the cation andlarge positive charge e.g. Cu++ has higher enthalpy of hydration than Cu+Q.36 # For the first row transition metals the Eo values are: Eo V Cr Mn Fe Co Ni Cu M2+/M) -1.18 – 0.91 –1.18 – 0.44 – 0.28 – 0.25 +0.34 -------- Explain the irregularity in the above values. Ans--The E0 (M2+/M) values are not regular which can be explained from the irregular variation of ionisation enthalpies ( Δi H1+Δ iH2) and also the sublimation enthalpies which are relatively much less for Mn and V..Q. 37 # What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses. OR , Write two uses of mischmetallAns- Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other A well known alloy is mischmetallwhich consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. A good deal of mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint.Q.38 # What is the effect of increasing pH on a solution of K2Cr2O7Ans- It will change to yellow colour by forming Potassium chromate ( in alkaline medium) K2Cr2O7 ( orange) ------ ( OH-)K2CrO4 ( Yellow) ** K2CrO4 ( Yellow) ------ ( H+)-K2Cr2O7 ( orange) Q.39# W h a t chemical changes takes places w h e n – (i) Pyrolusite is fused with KOH in the presence of air(ii) Chromite ore is fused with molten NaOH in presence of air (iii)When a solution of chromate is acidified(iv)NH3 is added to a suspension of AgCl(v) Alkaline KMnO4 reacts with KI Ans (i) Potassium manganate (K2MnO4 –green , paramagnetic)(ii) Pottasium Chromate (K2CrO4 – Yellow , diamagnetic)K2Cr2O7 – Orange ) (iv) Soluble complex , [Ag(NH3)2]Cl(v) Iodate (IO3? ) and Mn2+Q.40# 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(103) 5f 146d17s2 +3Q.41 # On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?Ans--On the basis of incompletely filled 3d orbitals in case of scandium atom in its ground state (3d1), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d10) in its ground state as well as in its oxidisedstate, hence it is not regarded as a transition element.Q.42# Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?Ans--Silver (Z = 47) can exhibit +2 oxidation state wherein it will have incompletely filled d-orbitals (4d), hence a transition element.Q.43 # 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?Ans- 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.Q.44 # Name a transition element which does not exhibit variable oxidation states.Ans--Scandium (Z = 21) does not exhibit variable oxidation states. Only oxidation states i.e Sc3+45# Name the element showing maximum number of oxidation states among the first series of transition metals from Sc (Z = 21) to Zn (Z = 30).OR , Which of the 3-d series of the transition metals exhibits the largest number oxidation states and why?Ans :Manganese (Z = 25), as its atom has the maximum number of unpaired electrons. i.e five ( +2 , +3 , +4 , +5 , +6 , +7) Q.46 # Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.Ans-- With atomic number 25, the divalent ion in aqueous solution will have d5 configuration (five unpaired electrons). The magnetic moment, μ is μ = √5(5+2) = 5.92BMQ.47 # Calculate the ‘spin only’ magnetic moment of M2+ (aq) ion (Z = 27).Ans—three unpaired electron .μ = √3(3+2) = 3.87 BMQ.48 # Name a member of the lanthanoid series which is well known to (a) exhibit +4 oxidation state. (AI14)(b) exhibit +2 oxidation state. (D-14)Ans—(a) Cerium (Z = 58) Ce4+ : Xe] 4f0 accounts for its formation . (b) Eu . Eu2+ is formed by losing the two s- electrons and its Xe] 4f7 accounts for its formation . Q.49 # Among the species Sc3+,.Ce4+, and Eu2+ which one is good oxidising agent ? Sc3+ ——>[A r] 3d0 4s0 Ce4+ ——> [Xe] 4f0 5d0 6s0 Eu2+ ——>[Xe] 4f75d0 6s0Ans--In all the above ion the most stable oxidising state is +3 .therefore Sc3+ will remain unchanged. Ce4+has the tendency to accept one electron to get the +3 oxidation state. Hence Ce4+ is a good oxidising agent.Eu2+ has the tendency to lose one electron to get the +3 oxidation state. Hence Eu2+ is a good reducing agent.Q.50 # Mn (III) undergoes disproportionation reaction easily.Ans : Mn (III) unstable oxidation states , it undergoes disproportionation to Mn2+ and MnO2 ( +4 oxidation states ) 51 # Out of Mn3+ and Cr3+ , which is more paramagnetic and why ?Ans :- Mn3+ (3d4 4s0 , four unpaired d- electrons ) -- more paramagnetic - higher no. of unpaired d- electrons Cr3+( 3d3 4s0 , three unpaired electron) 52# Which transition metal of 3d series has positive E0(M2+/M) value and why Ans :- Cu . The sum of high enthalpy of atomization and ionization enthalpy to transform Cu(s) Cu2+(aq) is not balanced by hydration enthalpy .53 # Sc3+ is colourless in aqueous solution whereas Ti3+ is coloured (2014-F)Ans :-Sc3+ have no unpaired d- electron . No d—d transition of unpaired d-electron . BUT Ti3+ have one unpaired d- electron .d—d transition of unpaired d-electron in visible region . The complementary colour is seen .54# La3+ (Z = 57) and Lu3+ (Z = 71) do not show any colour in solutions.Ans: La3+ : (Xe] 4f0 and Lu3+ (Xe] 4f14 - absence of unpaired f- electrons for f—f transition . so no f—f transition in the visible region . 55# Compare actinoids and lanthanoids with special reference to their : (i) electronic configurations of atoms (ii) oxidation states of elements (iii) general chemical reactivity of elements. (iv) atomic and ionic sizes (i) In Lanthanoids electrons are filled in antepenultimate shell i.e 4f subshell BUT in Actinoids 5f subshell (ii) +3 , +2 and +4 ( do not show greater range of O .S ) BUT in Actinoids +3 , +4 , +5, +6 &7 (iiI) Lanthanoids are less reactive whereas Actinoids( low I.E) are more reactive chemically (iv) Lanthanoids are smaller than the actinoids .Q.56 # Calculate the number of unpaired electrons in the following gaseous state ions: Mn2+, Cr3+, V3+ and Fe2+ Which one of these is the most stable in aqueous solutions? (At. nos. V = 23, Cr = 24, Mn = 25, Fe = 26)Mn2+ = 5 , Cr3+ = 3 , V3+ = 2 , Fe2+ = 4 Cr3+ , due to half-filled configuration ( t2g3 eg0 ----- Higher crystal field stabilization energy . Q.57 # Which of the following cations are coloured in aqueous solutions and why? Sc3+, V3+, Ti4+, Mn2+ ,Ti3+, Cu+, Fe3+ and Co2+.Cations are coloured in aqueous solutions:- V3+ (d1- )1-unpaired electron, Mn2+(d5) - 5 unpaired electron ,Ti3+(d1) - 1 unpaired electron, Fe3+(d5- 5 unpaired electron) and Co2+(d7) - 3 unpaired electron. Due to d-d transition of unpaired d- electronsQ.58# Which metal in the first transition series (3d series) exhibits +1 oxidation state most frequently and why? Cu as Cu+Q.59# Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements.Atomic size decreases from Ti to Cu .REASON:-Along a period ( 3d – series) Relative increase in nuclear charge is more dominant than relative increase in screening effect in d- subshell( imperfect screening) Atomic mass increases but volume decreases that’s why density increases from Ti to Cu . Q.60# The higher oxidation states are usually exhibited by the members in the middle of a series of transition elements. Due to the presence of higher number of unpaired electrons. (Mn shows O.S from +2 to +7)—present in the middle of the series .Q.61# There is a greater horizontal similarity in the properties of the transition elements than of the main group elements.Atomic sizes are very close by along the d- series Q.62# The third ionization enthalpy of manganese (Z = 25) is exceptionally high. Due to half –filled configuration.( extra stable- ) Removing third electron from Mn2+ needs high I.E ( tightly held) Q.63# Discuss the relative stability in aqueous solutions of +2 oxidation state among the elements : Cr,Mn, Fe and Co. How would you justify this situation? (At. Nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)Mn-- 3d5 4s 2Due to half –filled configuration.( extra stable- )-- Mn2+ - -- 3d54s0Cr--3d 5 4s 1 ; Cr2+--3d 4 4s 0 (unstable)Cr3+ , Stable-due to half-filled configuration – Stable O.S. In aq. Solution ( t2g3 eg0 ----- Higher crystal field stabilization energy ) CFTCo--3d 7 4s 2 ; Co2+--3d 7 4s 0(unstable) Fe --3d 6 4s 2 ; Fe++- 3d 6 4s0 (unstable)Q.64 # What may be the possible oxidation states of the transition metals with the following d electronic configurations in the ground state of their atoms: 3d 3 4s 2 , 3d5 4s 2 and 3d 6 4s 2 . Indicate relative stability of oxidation states in each case.3d 3 4s 2--- (+2 ,+3 ,+5- stable)3d5 4s 2---( +2 , +3 , +4 , +5 , +6 , +7) 3d 6 4s 2 –(+2, +3)Q.65 # Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism.Ans : Due to highest number of unpaired d- electron ( i.e 5 ) [ 3d5 4s0 ]Q.66 # Mn3+ is a good oxidising agent Mn (II) stable – d5 configuration . (-half filled configuration- stable) . So Mn (III) being unstable get itself reduced to stable Mn(II) and act as oxidizing agent . Q.67 # Give reason (i) The +3 oxidation state of 57La , 64Gd , 71Lu are especially stable . (ii) Ce3+ can be easily oxidized to Ce4+(i) 57La—[Xe]54 4f15d06s2 ; 57La3+—[Xe]54 4f05d06s0 -- fulfilled configurtaion64Gd--- —[Xe]54 4f75d16s2 ; 64Gd3+--- —[Xe]54 4f75d06s0 -half –filled 71Lu----—[Xe]54 4f145d16s2 ; 71Lu3+----—[Xe]54 4f145d06s0- fulfilled configuration(ii)Due to fulfilled configuration of Ce4+ - (Noble gas configuration- Configuration of Xe) –stable . Ce3+ can be easily oxidized to Ce4+Q.68# What are the characteristics of the transition elements and why are they called transition elements? High enthalpy of atomization. 2. Show paramagnetism 3. Show colour compounds 4. Complex formation ability 5 . Show variable oxidation states .etc …( unique properties of Transitional metals) They show transitional characteristics between s- block and p- block elements .Q.69 # What are the different oxidation states exhibited by the lanthanoids? +3 ( commonOxidatyion States) +2 and +4 are also seen due to half –filled and fulfilled configuration of various ions.Q. 70 # Which of the d-block elements may not be regarded as the transition elements? Zn , Cd and Hg - Due to absence of Unpaired d- electrons either in the ground state or in their commonly found Oxidation states . Q.71 # 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. Permanganate ( MnO4 ? ) , +7-- Oxidation state , Mn belongs to 7th group . Q.72 # How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.Oxidation states in Transition metals differs by one BUT , in Non- transition metals O.S differs by 2 Phosphorous - +3 and +5 ( as in P2O3 and P2O5 ) Iron – Fe2+ and Fe3+Q.73 # What are interstitial compounds? Why are such compounds well known for transition metals?Ans-- It because transition metals have voids in their crystal lattice , in which small atoms of H , C , N can fit forming interstitial compounds which do not have fixed composition . The trapped atoms get bonded to the atoms of TE .e.g. TiC , Fe3H , Mn4N ,VH0.56 TiH1.7 .** Due to variable O.s of Transition metals , Metal deficient defects are seen in solid state compound like FeO and NiOQ.74 # 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.due to half –filled and fulfilled configuration of various ions.Mn-- 3d5 4s 2Mn2+ - -- 3d54s0(half –filled ) Zn --3d10 4s 2, Zn++ --3d10 4s 0 (fulfilled configuration)Q.75 # 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).Hydrated ions (octahedral)–water is the ligand ( weak field – high spin complex ) Δ0< PTi2+ ----(2 unpaired d- elctrons)3d24s0--- t2g2eg0 V2+--(3 unpaired d- elctrons)--3d34s0---- t2g3eg0Cr3+--(3 unpaired d- elctrons)-3d34s0------ t2g3eg0 Mn2+-(5 unpaired d- elctrons)3d54s0-- t2g3eg2Fe2+--(4 unpaired d- elctrons)-3d64s0- t2g4eg2Fe3+--(5 unpaired d- elctrons)-3d54s0-- t2g3eg2Co2+---(3 unpaired d- elctrons)-3d74s0-- t2g5eg2 Ni2+---(2 unpaired d- elctrons)-3d84s0-- t2g8eg2Cu2+---(1 unpaired d- elctrons)3d94s0-- t2g6eg3ExampleMagnetic Moment (BM)K4[Mn(CN)6]2.2[Fe(H2O)6]2+5.3K2[MnCl4]5.9Q.76 # What can be inferred from the magnetic moment values of the following complex species ?Magnetic moment depends on spin and orbital motion of negatively charged unpaired electron .by knowing magnetic value we can predict the number of unpaired d- electron(s).Spin –only formula , ? = [n(n+2)]1/2 = [1(1+2)]1/2 = 1.73 BM (example) ExampleMagnetic Moment(BM)No.of unpaired electronconfigurationK4[Mn(CN)6]2.21 t2g5eg0 (Strong field)Oh--CFT[Fe(H2O)6]2+5.34t2g4eg2 (weak field)K2[MnCl4]5.95eg2t2g3(weak field)Q.77 # Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109. Atomic numbers 61--[Xe]544f 5 5d0 6s 2Atomic numbers 91--[Rn]86 5f 26d17s 2Atomic numbers 101-[Rn]86 5f 136d07s 2Atomic numbers 109 ---[Rn]86 5f 146d77s 2Q.78# What are inner transition elements? Decide which of the following At.nos are of the inner transition elements :29, 59, 74, 95, 102, 104.Lanthanoids ( 4f – series from 6th period and 3rd group) and actinoids( 5f – series from 7th period and 3rd group) are known as Inner transition elements : Lanthanoids-- 59Actinoids-- 95, 102Q.79 # Sm2+ , Eu2+ and Yb+2 ions in solutions are good reducing agents but an aq. Solution of Ce4+ is a good oxidizing agent** Sm+2(f 7) ,Eu+2(f 7),Yb+2(f 14) Sm+3,Eu+3, Yb+3 ( Stable , common O.S)(Sm2+ , Eu2+ and Yb+2 ions tendency to change into +3 O.S.)-- [ Acts as reducing agent)** Tb+4(f 7) , Ce+4(f0) Tb+3 , Ce+3 ( Stable , common oxidation states) (tendency to change into +3 O.S.)-- [ Acts as oxidizing agent]Q.80 # The outer electronic configurations of two members of the lanthnoid series are as follows : 4f 1 5d 1 6s 2 and 4f 7 5d 0 6s 2 . What are their atomic numbers?Predict the Oxidation States exhibited by these elements in their compounds . (AI-05)Xe is the noble gas core whose atomic number is 54 [Xe]544f 1 5d 1 6s 2-- Atomic number is 58 ; stable O.S . +4 (empty conf.) [Xe]544f 7 5d 0 6s 2-- Atomic number is 63 ; stable O.S +2 (half-filled conf.)Q.81 # Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.58Ce—[Xe]54 4f15d06s0 Only one unpaired electron is present Spin –only formula , ? = [n(n+2)]1/2 = [1(1+2)]1/2 = 1.73 BMPreparation , properties and uses of Potassium dichromate K2Cr2O7 and Potassium permanganate KMnO4Describe the preparation of potassium dichromate from iron chromite ore. Describe the preparation of Potassium permanganate KMnO4Potassium dichromate K2Cr2O7 – Preparation , and usesSTEP-1: Fusion of chromite ore (FeCr2O4) with sodium potassium carbonate in free access of air ( Alkaline Oxidation)4 FeCr2O4 + 8 Na2CO3 + 7 O2 → 8 Na2CrO4 + 2 Fe2O3 + 8 CO2STEP-2: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.2Na2CrO4 + 2 H+ → Na2Cr2O7 + 2 Na+ + H2O***Sodium dichromate is more soluble than potassium dichromate.STEP-3:Potassium dichromate is therefore, prepared by treating the solution of sodium dichromate with potassium chloride. Na2Cr2O7 + 2 KCl → K2 Cr2O7 (Orange crystals crystallise out.) + 2 NaCl**The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. The oxidation state of chromium in chromate and dichromate is the same. 2 CrO42– + 2H+ → Cr2O72– + H2O Cr2O72– + 2 OH- → 2 CrO42-USES: K2Cr2O7 is used as (1) primary standard in volumetric analysis (2) good preparative Oxidant in organic chemistry (3) In leather industry for tanning purpose(4) preparation of many azo compoundsPotassium permanganate KMnO4- Preparation , and usesPotassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidicsolution to give permanganate.STEP-1 : 2MnO2 + 4KOH + O2 → 2K2MnO4 (dark green)+ 2H2OSTEP-2 :3MnO42– + 4H+ → 2MnO4– (dark purple or almost black crystals)+ MnO2 + 2H2OCommercially it is prepared by the alkaline oxidative fusion of MnO2 followed by the electrolytic oxidation of manganate (Vl).STEP-1 : MnO2 ( fused with KOH, oxidized with air or KNO3) MnO42– (manganate ion)STEP-2 :MnO42– (manganate ion) on Electrolytic oxidation in alkaline solution MnO4– ( permanganate ion)KMnO4 is used as (1) a strong oxidant use in labs and industry as volumetric reagent in alkaline and acidic media (2) Disinfectant for well water (3) as bleaching agent for wood ,cotton ,silk and other textiles.Question for you *****Q.1# Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with: (i) iodide (ii) iron(II) solution and (iii) H2SQ.2# 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.Redox Reaction :-1# In Acid Solution--Reduction half cell :MnO4ˉ (permanganate) +8H+ + 5e- Mn+2(manganous) + 4 H2OCr2O72ˉ (dichromate) + 14H+ + 6e- 2Cr+3 +7H2OOxidation half cell : Iˉ(Iodide) I2 (Iodine) , S 2ˉ (Sulphide) S(Sulphur) , Sulphite(SO32ˉ) sulphate(SO42ˉ) , Nitrite (NO2ˉ ) Nitrate (NO3ˉ ) Fe+2(Ferrous) Fe+3 (Ferric) , C2O42 ˉ(oxalate) CO2 (Carbon Dioxide) Sn+2 (stannous) Stannic(Sn+4)2# In neutral or faintly alkaline solution : Reduction half cell : MnO4ˉ + 3e- MnO2Oxidation half cell : Iˉ(Iodide) IO3ˉ (Iodate) , Sulphite(SO32ˉ) sulphate(SO42ˉ)TOP-20 --PREVIOUS YEAR QUESTIONS OF d-BLOCK REACTIONS( MUST DO …) Sl noComplete and write the balanced chemical equation YearSl noComplete and write the balanced chemical equation Year1 MnO4? +C2O42 ?+ H+11MnO2 +KOH+ O22KMnO4 +Heat 12FeCr2O4 + Na2CO3 +O2 3Cr2O7 2 ?+H2S + H+13MnO4? +SO32 ?+ H+4MnO42? + H+14Cr2O7 2 ?+I? + H+5Na2CrO4 + H+ →15Cr2O7 2 ?+14 H++ 6e-6Cr2O7 2 ?+ OH?16MnO4? +NO2? + H+7MnO4? + H+ +3e? 17MnO4? +S2 ?+ H+8Cu2+ + I? 18MnO4? +MnO2 +H2O9Cr2O7 2 ?+Fe2+ + H+19MnO4? +S2O32- +H2O10Mn2+ 5S2O82- + 8H2O 20MnO4? + I? + H2OVALUE BASED QUESTIONS- CHAPTER-8: d and f- block elementsQ.1#People residing near villages have a tendency to dispose waste in water . A person was disposing mercury cells in water. A student Raju, asked the person not to do so. (a)What are the harmful effects of mercury? (b)What values are associated with the above discussion?Answer:-: a) impaired neurological development, carcinogenic (bCare for environment and mankindQ.2. KMnO4 is commercially obtained from ore Pyrolusite. It exists in the form of dark purple crystals. It is moderately soluble in water at room temperature and its solubility increases with the rise in temperature. KMnO4 is useful oxidising agent and oxidises under neutral, acidic and basic conditions.(i) What happens when KMO4 is heated?(ii) What is Baeyer’s reagent?( iii) How does it act as a test for unsaturation?(iv)What is value associated with the use of KMnO4 in our daily life?Answer- i) 2KMnO4 ----? K2 MnO4 + MnO2 + O2 ii) Baeyer’s reagent is dil cold alkaline KMnO4(aq) solution. iii) It is decoloured (pink to colourless) by the reaction of any unsatured organic compound (alkenes or alkynes or unsaturated fat etc)iv) It useful to purify water .It destroys undesirable bacteria by oxidation. It is an antiseptic for washing wounds. It has bleaching action too for textile industries.Q.3# Nickel is the second most abundant element by weight in earth crust. The mond’s process provides us high purity of nickel. Most of the nickel produced is used to make ferrous and non ferrous alloys. Nickel provides both the strength of steel and its resistance to chemical attack. Nickel is used in making alloys for coins in USA. It is widely used as catalyst especially in hydrogenation of vegetable oils to get vegetable ghee. Nickel is also used in Ni-Cd cell? i) What is the use of Nickel steel and why? ii) How is Nickel of high purity, obtained by Mond’s process? iii) Why is Ni-Cd cell preferred over lead storage cell? iv) Why is vegetable oil better than vegetable ghee? What values are possessed by people, taking vegetable oils? v) Why is nickel used for alloys used in making coins?Answer- i) It is used for making clock pendulums because it has coefficient of thermal expansion. ii) Nickel when heated in steam of carbon monoxide forms Ni(CO)4 which decomposes to from pure Ni. iii) Lead creates pollution therefore; Ni-Cd cell is preferred.iv) Vegetable oils are unsaturated and they do not lead to the formation of cholesterol where as vegetable ghee is saturated and leads to the formation of high cholesterol. Nickel gets mixed up with vegetable oil which is carcinogenic. People taking vegetable oils are more conscious about their health. v) It is because it does not get corroded and has lustre like silver ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download