d AND f- Block Elements Notes PDF

d- AND f- BLOCK ELEMENTS. d-Block Elements. The elements of groups 3, 4, 5, 6, 7, 8, 9,10, 11 and 12 in which the differentiating electron enters into the d- subshell of the penultimate shell [(n-1) d subshell] are called d-block elements or transition e...

d- AND f- BLOCK ELEMENTS. d-Block Elements. The elements of groups 3, 4, 5, 6, 7, 8, 9,10, 11 and 12 in which the differentiating electron enters into the d- subshell of the penultimate shell [(n-1) d subshell] are called d-block elements or transition elements. The d-block elements show a transitional behaviour between the highly electropositive s-block elements and weakly electropositive (highly electro negative) p-block elements. This is why the elements of d-block are referred to as transition elements. Classification of Transition Elements The transition elements or d-block elements are classified into four series on the basis of (n-1) d-subshell which gets filled up. (i) First transition series or 3d-series: This series consists of ten elements of fourth period from Scandium (Z = 21) to Zinc (At. No. = 30) which involves the filling of 3d-subshell. (ii) Second transition series or 4d-series This series consists of ten elements of fifth period from Yttrium (Z = 39) to Cadmium (Z= 48) which involves the filling of 4d-subshell. The element Technetium is a synthetic metal which was made artificially. (iii) Third transition series or 5d-series: This series consists of ten elements of sixth period from Lanthanum (Z= 57) and those from Hafnium to Mercury (Z= 80) which involves the filling of 5d-subshell. (iv) Fourth transition series or 6d-series: This series consists of ten elements of seventh period from actinium (Z= 89) and those from Rutherfordium to Copernicium (Z= 112) which involves the filling of 6d-subshell. All elements of this series are radioactive and except actinium, all are synthetic elements. Electronic Configuration of Transition Elements. The general outer electronic configuration of transition elements is (n-1)d1-10ns1-2 Most of the transition elements have 2 electrons in their valence shell. Some of the elements like chromium, copper, silver, platinum gold etc. have only one electron in their outermost ns shell. Pd has no electron in its ns- subshell. ([ Kr] 4d10 5s0) Exceptional configuration of Chromium and copper. Chromium and copper have anomalous electronic configurations. We know that half-filled and completely-filled electronic configurations (i.e., d5 and d10) have extra stability. Thus, to acquire increased ability, one of the 4s- electrons goes to nearby 3d-orbitals so that 3d-orbitals become half-filled in case of chromium and completely- filled in case of copper. Therefore, the electronic configuration of Cr is [Ar] 3d5 4s1 rather than [Ar] 3d10 4s2 while that of Cu is [Ar] 3d10 4s¹ instead of [Ar] 3d9 4s2. Definition of Transition elements. Elements which have incompletely filled (partially filled) d-orbitals (subshell)in their ground state or in any one of their commonly occurring oxidation states, are called transition elements. Example 1. Electronic configuration of Cu, Cu+, and Cu2+ are given below. Cu – 1s2 2s22p6 3s2 3p6 3d10 4s1 Cu+ - 1s2 2s22p6 3s2 3p6 3d10 Cu2+ - 1s2 2s22p6 3s2 3p6 3d9 Both Cu and Cu+ ion possess completely filled d-subshell. But Cu2+ ion possesses incompletely filled d-subshell. Therefore, copper is regarded as a transition element. Zinc, cadmium and mercury are not regarded as transition elements. Why? According to definition, transition elements are those which have partially filled d-subshell in their ground state or in their most common the oxidation states. Zinc, cadmium and mercury have completely filled d-orbitals and they don’t have partially filled orbitals in their ground state or in their commonly occurring oxidation state. Thus, they are not regarded as transition elements. (They are also known as pseudo transition elements) For example, electronic configuration of Zn and Zn2+ are Zn – 1s2 2s22p6 3s2 3p6 3d10 4s2 Zn2+ - 1s2 2s22p6 3s2 3p6 3d10 Neither Zn nor Zn2+ ion possess partially filled d-orbitals and therefore zinc cannot be regarded as a transition element. Thus, all transition elements are d-block elements but all d-block elements are not transition elements. GENERAL CHARACTERISTICS OF TRANSITION ELEMENTS. 1. All the transition elements are metals which have typical metallic properties such as high melting and boiling points, high tensile strength, malleability, ductility, metallic lustre, high thermal and electrical conductivity etc. (Exception: Mercury is a liquid transition metal at room temperature) 2. The first ionisation energies of d-block elements are higher than those of s-block elements but are lesser than those of p-block elements. They are electropositive in nature. 3. Most of them form coloured compounds. 4. hey have good tendency to form complexes. 5. They exhibit several oxidation states. 6. Transition metals and their compounds are generally paramagnetic in nature. 7. They form several alloys with other metals. 8. They form interstitial compounds with elements such as hydrogen, boron, carbon, nitrogen, etc. 9. Most of the transition metals such as Mn, Ni, Co, Cr, V, Pt, etc. and their compounds are used as good catalysts. GENERAL PROPERTIES OF TRANSITION ELEMENTS 1. ATOMIC RADII Atomic radii of transition elements are intermediate between those of s-block and p-block elements. The following trends in atomic radii of transition elements are observed. i) The atomic radii of elements of a particular transition series decrease with increase in atomic number but the decrease in atomic radii becomes small after midway. Explanation: In the beginning, the decrease in atomic radii with increase in the atomic number is due to an increase in the nuclear charge. As the atomic number increases, the added electrons enter into (n - 1) d-subshell and shield the outermost electrons. The shielding effect increases with increase in the number of d-electrons, i.e., with increase in the atomic number. Thus, the effect of the increased nuclear charge due to increase in atomic number is counterbalanced by the increased shielding effect of the (n - 1) d-electrons. This is why, the atomic radii remain almost constant after mid-way in each series. ii) Near the end of each series, there is a slight increase in the atomic radii. Explanation. Towards the end of a series, the electron repulsions between the added electrons in the similar orbitals predominate over the attractive forces due to the increase in nuclear charge. Therefore, electron cloud expands and atomic size increases. iii) The atomic radii of transition elements increase on moving down the group. However, the atomic radii of the elements of second and third transition series are nearly the same. Explanation. The increase in atomic radii on moving down the group is due to an increase in the number of electronic shells. This is why the atomic radii of the elements of second transition series are larger than those of the first transition series. The almost similar values of atomic radii of the elements of second and third transition series are due to lanthanoid contraction. 2. IONIC RADII Ionic radii follow the same trend as the atomic radii. For the same oxidation state, the ionic radii generally decrease with increase in nuclear charge. Ionic radii decrease with increase in oxidation state. 3. METALLIC CHARACTER All transition elements are metals. Explanation. The metallic character of transition metals is due to their low ionisation energies, presence of unpaired electrons and the presence of several vacant orbitals in their outermost shell. This favours the formation of strong metallic bond in them. This is why transition elements exhibit typical metallic properties. Higher the number of unpaired electrons in d- subshell stronger is the metallic bond and greater is the hardness of the metal. Chromium, molybdenum and tungsten have maximum number of unpaired electrons and therefore they are very hard and have maximum enthalpy of atomisation. Zinc, cadmium and mercury do not have any unpaired electrons; therefore, these elements are not very hard. 4. DENSITY All transition metals have high density. i) Within a particular transition series, the density of transition elements increases from left to right. Explanation.. On moving from left to right in a series, atomic radii decrease due to increase in nuclear charge. Therefore, atomic volume decreases, but at the same time atomic mass increases. Hence density (atomic mass/atomic volume) increases. ii) The density of transition metals increases on moving down a group. Explanation. The atomic radii of transition elements increase on moving down the group. The atomic radii of elements of 2nd and 3rd transition series are nearly same. But their atomic masses increase nearly two times. Therefore, decrease in atomic radii coupled with increase in atomic mass results in increase in density on moving down a group. Osmium is the densest or heaviest metal (22.59 gcm-3) Iridium is the second heaviest metal (22.56 gcm-3)) 5. MELTING AND BOILING POINTS. i) Transition metals have very high melting and boiling points. Explanation: The high melting and boiling points of transition metals are due to their close-packed structures. In these structures, the transition metal atoms are held together by strong metallic bonds. Transition metals contain large number of unpaired electrons in their d-subshells. Higher the number of unpaired electrons in d- subshell stronger is the metallic bond and greater is the melting and boiling points of the metal. Therefore, it requires considerable amount of energy to break the metallic bonds in order to melt the metal. Consequently, these metals have very high melting and boiling points. ii) In a particular transition series, the melting points first increase, attain a maximum value and then steadily decrease as the atomic number increases. Explanation. The strength of metallic bonds depends upon the number of unpaired electrons. Greater the number of unpaired electrons, stronger is the metallic bonding. In a particular transition series, the number of unpaired electrons in (n - 1) d-subshell increases up to the middle, ie., up to d5 configuration. Therefore, the strength of the metallic bonds and hence the melting and boiling points increase up to d5 configuration, i.e., up to the middle of a series. Beyond d5 configuration, the electrons start pairing up and the number of unpaired electrons decrease steadily on moving further in the given series. This progressively decreases the strength of metallic bond and hence the melting points and boiling points decrease progressively after the middle of the series. For example, in first transition series, as the atomic number increases, the number of unpaired electrons in the 3d-subshell increases up to chromium (Sc has 1, Ti has 2, V has 3 and Cr has 5 unpaired electrons). Therefore, the strength of metallic bonds and hence the melting and boiling points increase from Sc to Cr. Chromium has the maximum melting point in the first transition series. After Cr, the  number of unpaired electrons goes on decreasing (Fe has 4, Co has 3, Ni has 2 unpaired electrons). Therefore, the strength of metallic bonds and hence the melting and boiling points decrease in going from Cr to Cu. Zn, Cd and Hg have no unpaired electrons and therefore, metallic bond is very weak. This is why these metals are soft and have low melting and boiling points Hg is a liquid at room temperature and has melting point of 234 K. The unexpectedly lower melting points of Mn and Te are probably due to the complicated lattice structures. 6. IONISATION ENTHALPIES/ENERGIES i) The first ionisation enthalpy increases with increase in atomic number across a given transition series, although the increase is not very regular. Explanation: The increase in ionisation enthalpy with increase in atomic number across a given series is due to an increase in the nuclear charge with increase than the atomic number. The added electrons enter into(n-1)d-subshell and shield the valence electrons from the nucleus. Thus, the effect of the increased nuclear charge is opposed by the screening effect of (n-1)d-electrons. This is why the increase in ionisation enthalpy with increase in atomic number is rather slow and not very regular. ii) Chromium and copper have exceptionally high ionisation enthalpy values than those of their neighbours. Explanation This is because of the extra stability of half-filled d- subshell in chromium (3d5) and fully filled d-sub shell in copper (3d10). iii) The first ionisation enthalpies of the elements of third transition series are higher than those of the elements of first and second transition series. Explanation. In the atoms of third transition series, there are filled 4f orbitals. The 4f-orbitals have very poor shielding effect. As a result, the outer electrons experience greater nuclear attraction. Consequently, greater amount of energy is required to ionise elements of third transition series. Therefore, their ionisation energies are higher. iv) Thermodynamic stability of a transition metal compound depends on the sum of all the ionisation enthalpies needed for a transition metal to attain a particular oxidation state. Smaller the sum of all the ionisation enthalpies, greater is the stability of the complex. Example 1.: Ni(II) compounds are thermodynamically more stable than Pt(II) compounds. Explanation: The sum of first and second ionisation enthalpies for Ni (2.49x103) is lesser than that for Pt (2.66x103). This means that, ionisation of Ni to Ni2+ requires lesser energy than that for Pt to Pt2+. This is why, Ni(II) compounds are thermodynamically more stable than Pt(II) compounds. Example 2.: Pt(IV) compounds are thermodynamically more stable than Ni(IV) compounds. Explanation. The sum of first four ionisation enthalpies (IE1+IE2+ IE3 +IE4) for Pt (9.36x103) is lesser than that for Ni (11.29x103). This means that, ionisation of Pt to Pt4+ requires lesser energy than that for Ni to Ni4+. This is why, Pt(IV) compounds are thermodynamically more stable than Ni(IV) compounds. Question: Third ionisation enthalpy of Mn is greater than that of Fe. Why? OR Which is more stable, Fe2+ or Mn2+? Why? Answer: Electronic configuration of: Fe2+ - 1s2 2s22p6 3s2 3p6 3d6 Mn2+ - 1s2 2s22p6 3s2 3p6 3d5. Mn2+ possesses half-filled d-subshell (d5) which is more stable than d6 configuration of Fe2+. Consequently, greater amount of energy is required to remove an electron from Mn2+. Therefore, third ionisation energy of Mn is higher than that of Fe. 7. *** STANDARD ELECTRODE POTENTIALS. (Standard Reduction Potentials) [E0] 1. Trends in E0 (M2+‫׀‬M). E0 (M2+‫׀‬M) represents the standard reduction potential for the reaction: M2+(aq.) + 2e̅ → M There is no regular trend in these values. It depends on ionisation enthalpies (IE1+ IE2), sublimation enthalpy and enthalpy of hydration. If E0 (M2+‫׀‬M) value is high (more positive or less negative), it means: i) M2+ ions can be readily reduced to metal. ii) M2+ ions are less stable than M. iii) Metal cannot readily lose electrons to form M2+ ions. ie. Metal is not readily oxidised. 0 2+ If E (M ‫׀‬M) value is small (less positive or more negative), it means: i) M2+ ions cannot be readily reduced to metal. ii) M2+ ions are more stable than M. iii) Metal can readily lose electrons to form M2+ ions. ie. Metal is readily oxidised and can act as good reducing agents. Except copper, all other elements of first transition series possess negative value of standard reduction potential. This is why they act as good reducing agents. The values of E0 for Mn(-1.18V) and Zn(-0.76V) are more negative than expected. This due to the greater stability of half filled d-subshell(d5) in Mn2+ and completely filled d-subshell (d10) in Zn2+. 2. Trends in E0 (M3+‫׀‬M2+). E0 (M3+‫׀‬M2+) represents the standard reduction potential for the reaction: M3+(aq.) + e̅ → M2+ If E0 (M3+‫׀‬M2+) value is high (more positive or less negative), it means: i) M3+ ions can be readily reduced to M2+ ions. ii) M3+ ions are less stable than M2+ ions in aqueous solutions. iii) M2+ ions are more stable than M3+ ions and cannot readily lose electrons to form M3+ ions. ie. M2+ ion is not readily oxidised. If E0 (M3+‫׀‬M2+) value is small (less positive or more negative), it means: i) M3+ ions cannot be readily reduced to M2+ ions. ii) M3+ ions are more stable than M2+ ions in aqueous solutions. iii) M3+ ions are more stable than M2+ ions and M2+ ions can readily lose electrons to form M3+ ions. ie. M2+ ion is readily oxidised. The comparatively high value of E0 (Mn3+‫׀‬Mn2+) shows that Mn2+ion is more stable than Mn3+ ion on account of stable d5 configuration of Mn2+. The comparatively low value of E0 (Fe3+‫׀‬Fe2+) shows that Fe2+ion is less stable than Fe3+ ion on account of stable d5 configuration of Fe3+.  8. OXIDATION STATES Most of the transition elements exhibit variable oxidation states (more than one oxidation states) in their compounds. Explanation. This is due to the participation of inner (n-1)d electrons in addition to outer ns electrons because, the energies of the ns and (n-1)d orbitals are almost equal. The +1 and +2 oxidation states involve the participation of ns electrons but higher oxidation states such as +3, +4, +5, +6 etc. involve the use of both ns and (n-1) d electrons for bond formation. Most common oxidation state of elements of first transition series is +2 (except Sc. [+3]) The bonds formed by transition metals in +2 and +3 oxidation states are mostly ionic. On the other hand, the bonds formed in the higher oxidation states are essentially covalent. The covalent compounds of these elements are usually formed with fluorine or oxygen. Example. In MnO4-, all the bonds between Mn and O are covalent. ***Within a group, the maximum oxidation state increases with atomic number. For example, in group 8, common oxidation states of Fe are +2, +3 whereas ruthenium and osmium show +4, +6 and +8 also. The highest oxidation states of transition metals are found in their fluorides and oxides. This is because fluorine and oxygen are the most electronegative elements. The highest oxidation state shown by any transition element is eight. [Ru and Os] Transition metals also form compounds in low oxidation states such as +1 and 0. For example, the oxidation state of nickel in nickel tetracarbonyl, Ni (CO)4, is zero. ****The relative stability of a particular oxidation state in aqueous medium can be known on the basis of the standard electrode potential value (E0) for the electrode system. Greater the negative (or lesser the positive) value of E0 greater is the stability of the element in the particular oxidation 0 0 state. For example, 𝐸𝐶𝑢 + /𝐶𝑢 is +0.52V and 𝐸𝐶𝑢 2+ /𝐶𝑢 is +0.34V. Standard reduction potential of Cu2+/Cu is less positive than that of Cu+/Cu, Cu2+ion is more stable in aqueous solution than Cu+ ion. Crystal Field splitting. A d-subshell consists of five orbitals namely dxy, dxz ,dyz, dx2-y2 and dz2. The five d-orbitals are of the same energy and are said to be degenerate. However, under the influence of the combining anions or neutral molecules, the degeneracy of the five -orbitals is destroyed and they split into two groups of orbitals called t2g and eg groups. This splitting d-orbitals into t2g and eg group of orbitals under the influence of combining groups is called crystal Field splitting. The t2ag group of orbitals consists of dxy dyz and dxz orbitals, whereas eg group consists of dx2-y2 and dz2 orbital d- d Transition The electronic transition from lower energy d-orbitals to higher energy d-orbitals by the absorption of light energy of suitable wavelength is known as d-d transition. 9. FORMATION OF COLOURED IONS. Most of the compounds of transition metals are coloured both in the solid state as well as in the aqueous solutions. Explanation: The colour of transition metal ions is due to d-d transitions taking place between the split d-orbitals. All transition metals usually possess one or more unpaired electrons in the (n- 1)d subshell. When visible light falls on a transition metal compound or hydrated ion, the electrons present in the lower energy d-orbitals  get promoted to the higher energy d-orbitals (d-d transitions) due to the absorption of a specific wave length. The remaining wavelengths present in the visible light get transmitted. Therefore, transmitted light shows a complementary colour corresponding to the absorbed colour (wavelength) and the compound or the hydrated ion shows the same colour. For example, hydrated Cu2+ ions absorb red radiation and transmit the complementary greenish-blue radiation. Thus, cupric compounds have greenish-blue colour. Transition metal ions having partially filled d-subshell (d1 to d9) are coloured, because when d- subshell is partially filled the transition or promotion of an electron from a lower energy state to a higher energy state (d-d transition) is possible. Transition metal ions having completely filled d-subshell(d10) are colourless, because, when the d-subshell is fully filled (d10), there is no available space in the higher energy level for d-d transition to take place. Thus, Cu+, Zn2+, Cd2+, Hg2+ etc. having d10 configuration are colourless. Transition metal ions having completely empty d-subshell(d0) are colourless, because, when there are no electrons (d0), d-d transition is not possible. Thus, Sc3+, Ti4+, etc having d0 configuration are colourless. Compounds of s and p block elements are colourless. `This is because, in these elements d- orbitals are either missing or fully filled. Therefore they are unable to undergo d-d transition. Question. Ti3+ is coloured(purple) whereas Ti4+ is colourless. Why? Ans. Electronic configuration of: Ti - 1s2 2s22p6 3s2 3p6 3d2 4s2 Ti3+- 1s2 2s22p6 3s2 3p6 3d1 Ti4+ - 1s2 2s22p6 3s2 3p6 3d0 Ti3+ has incompletely filled d- orbital and hence, d-d transition takes place when white light falls on it. Therefore, Ti3+ is coloured. In Ti4+, 3d subshell is completely empty and hence d-d transition is not possible. Therefore, Ti4+ is colourless. Question. Which ion is coloured, Cu+ or Cu2+? Explain 10. Magnetic Properties. Most of the transition elements and their compounds are paramagnetic in nature. Explanation. Paramagnetism is due to the presence of unpaired electrons in (n-1) d orbitals. Higher the number of unpaired electrons greater is the paramagnetic behaviour. The magnetic character is expressed in terms of magnetic moment. The larger the number of unpaired electrons in a substance, the greater is paramagnetic character and larger is the magnetic moment. The magnetic moment is expressed in Bohr magnetons abbreviated as B.M. For the first transition series elements, the magnetic moments arise only from the spin of the electrons. This can be calculated from the relation: μ =√𝑛(𝑛 + 2) B.M. Where n is the number of unpaired electrons and μ is magnetic moment in Bohr magneton (BM) units.  The paramagnetic character in first transition series (3d series) elements increases up to Mn and then decreases. Why? As we move from scandium (at.no=21) to Manganese (at.no. 25) the number of unpaired electrons increases and hence paramagnetic character increases. After Mn, pairing of electrons in the d-subshell starts and the number of unpaired electrons decreases and hence paramagnetic character decreases. Question. Cu2+ is paramagnetic but Cu+ is diamagnetic. Explain Ans. The electronic configuration of : Cu - 1s2 2s22p6 3s2 3p6 3d10 4s1 Cu+ - 1s2 2s22p6 3s2 3p6 3d10 Cu2+ - 1s2 2s22p6 3s2 3p6 3d9 Cu+ has completely filled d-subshell and it does not contain unpaired electron. Hence, it is diamagnetic. Cu2+ contains an unpaired electron and hence it is paramagnetic. Question. Which of the two, ferrous or ferric ion has larger magnetic moment? Why? Ans: Fe2+ - 1s2 2s22p6 3s2 3p6 3d6. It contains 4 unpaired electrons. Fe3+ - 1s2 2s22p6 3s2 3p6 3d5. It contains 5 unpaired electrons. Fe3+ has larger magnetic moment because it has a greater number of unpaired electrons than Fe2+. 11. Catalytic Properties. Many of the transition metals and their compounds act as catalysts for a number of chemical reactions. Explanation. The catalytic activity of transition metals and their compounds may be attributed to the following factors. (i) Presence of vacant d-orbitals (ii) Ability to exhibit variable oxidation states (variable valencies) (iii) Tendency to form complexes. Due to their ability to exhibit variable valency and the tendency to form complexes, these elements can form unstable reaction intermediates with reactants. The unstable reaction intermediates provide an alternate path of lower activation energy. This causes an increase in the rate of the reaction. 12. Complex formation. Transition metal ions form a large number of complex compounds or coordination compounds. Explanation. The high tendency of transition metal ions to form complexes is due to:  i) Small size of transition metal ions ii) High nuclear charge iii) Availability of vacant d-orbitals to accommodate lone pair of electrons donated by the ligands. 13. Alloy Formation. Transition metals form a number of alloys among themselves. Explanation. Transition metals are quite similar in atomic size and therefore the atoms of one metal can substitute the atoms of other metal in its crystal lattice to form solid solution called alloy. 14. Formation of Interstitial Compounds. Transition metals form interstitial compounds with elements such as H, B, C, and N. The small atoms of these elements occupy the vacant spaces (interstitial sites) in the lattice of transition metals to form interstitial compounds. As vacant spaces of the transition metals are filled up by small atoms, these compounds are hard and rigid. The chemical properties of the parent transition metals are not altered during the formation of interstitial compounds. However, there are various changes in the physical properties such as density, rigidity, hardness, malleability, ductility, electrical conductivity etc. Steel and cast iron are the interstitial compounds of iron which are formed with carbon. In the formation of these compounds, the malleability and ductility of iron are lost to a great extent, but the tenacity of the metal increases. These compounds have very high melting points, higher than that of the parent transition metals. The conductivity exhibited by them is similar to their parent metal. These compounds are chemically inert in nature. Important compounds of transition metals. Potassium dichromate, K2 Cr2O7 Preparation. It is prepared from chromite ore, FeO.Cr2O3 or (FeCr2O4) The process involves the following steps: (i) Conversion of chromite ore into sodium chromate. The chromite ore is mixed with sodium carbonate (soda ash) or sodium hydroxide and quick lime. The mixture is roasted in the presence of excess of air. 4FeCr2O4 + 8Na2CO3 + 7O2 → 2Fe2O3 + 8Na2CrO4 + 8CO2 OR 4FeCr2O4 + 16NaOH + 7O2 → 2Fe2O3 + 8Na2CrO4 + 8H2O (ii) Conversion of sodium chromate into sodium dichromate. Sodium chromate in the above step is acidified with conc.sulphuric acid to form sodium dichromate. 2Na2CrO4 + H₂SO4 → Na2Cr2O7 +Na2SO4 +H2O On cooling, sodium sulphate crystallizes out as Na2SO4.10H2O The resulting solution contains sodium dichromate.  (iii) Conversion of sodium dichromate into potassium dichromate. Potassium dichromate is prepared by mixing a hot concentrated solutions of sodium dichromate and potassium chloride in equimolar proportions. Na2Cr2O7 +2KCl → K2Cr2O7 + 2NaCl The above mixture on cooling gives orange crystals of potassium dichromate. Chemical properties. 1. Oxidising character. In acidic medium (in the presence of dilute sulphuric acid) nascent oxygen is liberated and potassium dichromate acts as a strong oxidising agent. K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O +3[O] Since one mole of potassium dichromate takes 6 moles of electrons (produces 6 grameq. of oxygen) 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝟐𝟗𝟒 Eq.Wt of K2Cr2O7 = 𝟔 = 𝟔 = 49 Examples. i) It oxidises KI to iodine. K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O +3[O] [ 2KI + H2SO4 + [O] → K2SO4 + H2O + I2 ] x 3 ----------------------------------------------------------------------------------- K2Cr2O7 + 7H2SO4 + 6KI → 4K2SO4 + Cr2(SO4)3 + 7H2O + 3I2 The above reaction is used in volumetric analysis (redox titration) for the estimation of iodide ions. ii) It oxidises ferrous sulphate to ferric sulphat e. K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O +3[O] [2FeSO4 + H2SO4 + [O] → Fe2 (SO4)3 + H2O + ] x 3 ---------------------------------------------------------------------------------- K2Cr2O7 + 7H2SO4 +6FeSO4 → K2SO4 + Cr2(SO4)3 + 3Fe2 (SO4)3 + 7H2O The above reaction is used in volumetric analysis (redox titration) for the estimation of ferrous ions(Fe2+ ions) iii) It oxidises Hydrogen sulphide to sulphur. K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O +3[O] [ H2S + [O] → H2O + S ] x 3 ----------------------------------------------------------------------------------- K2Cr2O7 + 4H2SO4 +3H2S → K2SO4 + Cr2(SO4)3 + 7H2O + 3S iv) It oxidises Sulphur dioxide to sulphuric acid K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O +3[O] [ SO2 + H2O + [O] → H2SO4 ] x 3 -------------------------------------------------------------------------------- K2Cr2O7 + H2SO4 + 3SO2 → K2SO4 + Cr2(SO4)3 + H2O This reaction is used as a test to differentiate between sulphur dioxide and carbon dioxide. When passed through acidified potassium dichromate solution, SO2 changes the colour of potassium dichromate from orange to green. CO2 does not give this reaction. v) It oxidises conc. HCl to Chlorine K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O +3[O] [ 2HCl + [O] → H2O + Cl2 ] x 3 K2Cr2O7 + 4H2SO4 +6HCl → K2SO4 + Cr2(SO4)3 + 7H2O + 3Cl2  Similarly, it oxidises HBr to Br2 and HI to I2 vi) It oxidises sodium sulphite to sodium sulphate. K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O +3[O] [Na2SO3 + [O] → Na2SO4 ] x 3 -------------------------------------------------------------------------------- K2Cr2O7 + 4H2SO4 + 3Na2SO3 → K2SO4 + Cr2(SO4)3 +4H2O + 3Na2SO4 2. Action of alkali ( inter conversion of chromate and dichromate ion) or effect of pH On heating with alkali (KOH), the orange colour of potassium dichromate changes to yellow due to the formation of potassium chromate (chromate ions) K2Cr2O7 + 2KOH → 2K2CrO4 + H2O (orange) (yellow) On acidifying the solution, the above yellow coloured solution changes to orange due to the formation of potassium dichromate (dichromate ions) 2K2CrO4 + H2SO4 → K2Cr2O7 + K2SO4 + H2O (yellow) (orange) Thus, chromate ion and dichromate ion exist in equilibrium and are interconvertible by changing the pH of the solution. 2CrO42- + 2H+ ⇋ Cr2O72- + H2O chromate ion dichromate ion (yellow) (orange) When alkali is added ie pH of the solution is increased, the OH- ions combine with H+ ions and the equilibrium shifts towards left. This increases the concentration of chromate ions and the solution becomes yellow. When an acid is added ie pH of the solution is decreased, the concentration of H+ ions increases and the equilibrium shifts towards right. This increases the concentration of dichromate ions and the colour of the solution becomes orange. Uses Of K2Cr2O7 1. Used as an oxidising agent. 2. Used for the estimation of Fe2+ and I- ions in volumetric analysis (Redox titration). 3. Used in dyeing and tanning Structure of dichromate ion.  Potassium permanganate (KMnO4) Preparation. Potassium permanganate is prepared from mineral pyrolusite (MnO2). The preparation of KMnO4 involves the following steps: (i) Conversion of pyrolusite ore to potassium manganate. The pyrolusite (MnO2) is fused with caustic potash (KOH) or potassium carbonate in the presence of air or oxidising agent such as potassium nitrate or potassium chlorate to give a green mass due to the formation of potassium manganate. 𝒉𝒆𝒂𝒕 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H₂O Potassium manganate (Green mass) OR 2MnO2 + 2K2CO3 + O2 → 2K2MnO4 + 2CO₂ (ii) Oxidation of potassium manganate to potassium permanganate. The green mass is extracted with water resulting in green solution potassium manganate. The solution is then, treated with a current of chlorine or ozone or carbon dioxide to oxidise potassium manganate to potassium permanganate. The solution is evaporated to dark purple/violet crystals of potassium permanganate. 2K2MnO4 + Cl2 → 2KCl + 2KMnO4 OR 2K2MnO4+O3 + H₂O → 2KMnO4 + 2KOH + O2 OR 3K2MnO4 + 2CO2 →2K2CO3 + MnO2 + 2KMnO4 Chemical Properties Oxidising Properties. Potassium permanganate is a powerful oxidising agent in acidic, alkaline and neutral medium due to the liberation of nascent oxygen. 1. In acidic medium. In acidic medium it liberates oxygen as follows. 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5[O] In acidic medium KMnO4 is reduced to MnSO4. [MnO4- is reduced to Mn2+ ions.] Since one mole of potassium permanganate takes 5 moles of electrons (produces 5 grameq. of oxygen) 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝟏𝟓𝟖 Eq.Wt of KMnO4 in acidic medium = 𝟓 = 𝟓 = 31.6 Examples. a) It oxidises potassium iodide to iodine. 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5[O] [ 2KI + H2SO4 + [O] → K2SO4 + H2O + I2 ] x 5 2KMnO4 + 8H2SO4 + 10KI → 6K2SO4 + 2MnSO4 + 8H2O + 5I2  b) It oxidises ferrous sulphate to ferric sulphate 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5[O] [2FeSO4 + H2SO4 + [O] → Fe2 (SO4)3 + H2O ] x 5 ----------------------------------------------------------------------------------- 2KMnO4 + 8H2SO4 + 10FeSO4 → K2SO4 + 2MnSO4 + 5Fe2 (SO4)3 + 8H2O This reaction is used in volumetric analysis for the estimation of ferrous ions. c) It oxidises oxalic acid to carbon dioxide and water 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5[O] [ H2C2O4 + [O] → H2O + 2CO2 ] x 5 --------------------------------------------------------------------------------------- 2KMnO4 + 3H2SO4 + 5H2C2O4 → K2SO4 + 2MnSO4 + 10 CO2 + 8H2O This reaction is used in volumetric analysis for the estimation of oxalic acid and oxalate ions. d) **** It oxidises sulphur dioxide to sulphuric acid [Test for SO2 – disappearance of pink/purple colour of KMnO4 solution] 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5[O] [ SO2 + H2O + [O] → H2SO4 ] x 5 2KMnO4 + 5SO2 + 2H2O + → K2SO4 + 2MnSO4 + 2H2SO4 e) **** It oxidises conc. HCl to chlorine 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5[O] [ 2HCl + [O] → H2O + Cl2 ] x 5 2KMnO4 + 3H2SO4 + 10HCl → K2SO4 + 2MnSO4 + 5Cl2 + 8H2O f) It oxidises hydrogen sulphide to sulphur. 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5[O] [H2S + [O] → H2O + S ] x 5 2KMnO4 + 3H2SO4 + 5 H2S → K2SO4 + 2MnSO4 + 8H2O + 5S 2). In alkaline/basic medium. In alkaline medium it liberates oxygen as follows.  2KMnO4 + 2KOH → 2K2MnO4 + H2O + [O] [K2MnO4 + H2O → 2KOH + MnO2 + [O] ] x2 2KMnO4 + H2O → 2MnO2 + 2KOH + 3[O] In alkaline medium KMnO4 is reduced to MnO2. [MnO4- is reduced to MnO2.] Since one mole of potassium permanganate takes 3 moles of electrons (produces 3 grameq. of oxygen) 𝟏𝟓𝟖 Eq.Wt of KMnO4 in alkaline medium = = = 52.67 𝟑 Example. It oxidises potassium iodide to potassium iodate. 2KMnO4 + H2O → 2MnO2 + 2KOH + 3[O] KI + 3[O] → KIO3 2KMnO4 + H2O + KI → 2MnO2 + 2KOH + KIO3 3). In neutral medium. In neutral medium it liberates oxygen as follows. 2KMnO4 + H2O → 2MnO2 + 2KOH + 3[O] In neutral medium KMnO4 is reduced to MnO2. [MnO4- is reduced to MnO2.] Since one mole of potassium permanganate takes 3 moles of electrons (produces 3 grameq. of oxygen) 𝟏𝟓𝟖 Eq.Wt of KMnO4 in neutral medium = = 𝟑 = 52.67 Example. It oxidises hydrogen sulphide to sulphur. 2KMnO4 + H2O → 2MnO2 + 2KOH + 3[O] [H2S + [O] → H2O + S ] x 3 2KMnO4 + 3H2S → 2MnO2 + 2KOH + + 2H2O + 3S Uses of KMnO4 i) Used as oxidising agent ii) Used as disinfectant and germicide (for sterilising well water, for cleaning wounds etc.} iii) Used in volumetric analysis for the estimation of ferrous ion, oxalic acid, oxalate ion etc.  iv) Dilute alkaline KMnO4 solution (Baeyer’s reagent) is used to identify unsaturated compounds in organic chemistry. Unsaturated compounds discharge the pink/purple colour of Baeyer’s reagent. This is known as Baeyer’s test for unsaturation. Structure of Manganate ion. Note. The purple Colour of potassium permanganate is not due to d-d transition but due to charge transfer from O to Mn. f- BLOCK ELEMENTS (INNER TRANSITION ELEMENTS) (RARE EARTH ELEMENTS) The elements in which the last electron enters the f - orbital of their pre-penultimate or anti- penultimate shell are called f-block elements. Their general electronic configuration is: (n-2)f1-14 (n-1)d⁰-Ins2 These elements are also called inner transition elements. They consist of two series of elements placed at the bottom of the periodic table. These two series are generated by the filling of characteristic electrons in the 4f- and 5f-orbitals. (i) Lanthanoids/first inner transition series The 14 elements of 6th period from cerium (58) to lutetium (71) in which 4f orbitals are progressively filled Lanthanoids. Lanthanoids follow lanthanum in the periodic table and the physical and chemical properties of lanthanoids are similar to those of lanthanum. This is why they are known as lanthanoids. Electronic Configuration. General outer electronic configuration may be written as 4f1-14 5d⁰-I6s2. The electronic configuration of lanthanoids is not known with certainty.  Since the energy of 4f subshell is very close to that of 5d subshell, it is very difficult to predict whether the differentiating electron is in 4f subshell or 5d subshell. This is why, the observed electronic configuration of most of the transition elements differ from the predicted or expected electronic configuration. The electronic configuration of La (d-block) is: [Xe] 4f0 5d1 6s2. In La the differentiating electron enters in 5d subshell because the energy 5d subshell is lower than that of 4f subshell. After La the next element is cerium, which is the first member of lanthanoids. The predicted electronic configuration of cerium is : 4f1 5dI 6s2. After filling electron in 4f orbital, the energy of 4f orbital becomes much lower than that of 5d subshell. So, the 4f sub shell is occupied by two electrons and 5d subshell remains vacant. Thus, the observed electronic configuration of Cerium is: 4f2 5d⁰ 6s2. Similarly, the observed electronic configuration of Praseodymium is: 4f3 5d⁰ 6s2. Europium (Eu-63). The observed electronic configuration of Europium is: 4f7 5d⁰ 6s2. After Eu the next element is Gadolinium (Ga-64). The observed electronic configuration of Gadolinium is: 4f7 5d1 6s2. At this stage 4f subshell is half filled and stable and therefore one electron occupies 5d subshell. After Gd the next element is Terbium. The observed electronic configuration of terbium is: 4f9 5d0 6s2. This is because, with increase of electrons in 4f subshell, its energy decreases and the 5d electrons start occupying in the 4f subshell. This continues up to Ytterbium (Z=70). The observed electronic configuration of Ytterbium is: 4f14 5d⁰ 6s2. After Yb the next element is Lutetium. The observed electronic configuration of lutetium is: 4f14 5d1 6s2. At this stage 4f subshell is completely filled and stable and therefore one electron occupies 5d subshell. Oxidation States. The most common and stable oxidation state of lanthanoids is +3. In addition to +3, some lanthanides show less common oxidation states like +2 and +4. Lanthanoid metal ions in +2 oxidation states are not very stable and immediately get oxidised to +3 oxidation state. Hence, they act as good reducing agents in aqueous solutions. Lanthanoid metal ions in +4 oxidation states are not very stable and immediately get reduced to +3 oxidation state. Hence, they act as good oxidising agents in aqueous solutions. Magnetic behaviour. Most of the lanthanoid metal atoms and ions are paramagnetic due to the presence of unpaired electrons. La3+ and Lu3+ do not contain any unpaired electrons and therefore they are diamagnetic in nature. In lanthanoids paramagnetism is due to the spin motion of electron and orbital motion of electrons. Colour of ions. Most of the Tri-positive ions of lanthanoids are coloured due to f-f transition. Tri positive ions containing ‘n’ electrons in 4f subshell have the same colour as the tri-positive ions containing (14-n) electrons in 4f subshell. The colours of the cations from La3+ to Gd3+ repeat themselves from Lu3+ back to Gd3+. Ce3+ and Gd3+ are colourless because they absorb UV radiation and Yb3+ is colourless due to the absorption of IR radiation. Lanthanoid ions having f0, f7 and f14 configurations are colourless. Examples La3+ 4f0 Lu3+ 4f14 Colourless Ce3+ 4f1 Yb3+ 4f13 colourless Pr3+ 4f2 Tm3+ 4f12 green Nd3+ 4f3 Er3+ 4f11 reddish Pm3+ 4f4 Ho3+ 4f10 Pink yellow Sm3+ 4f5 Dy3+ 4f9 Pale yellow Eu3+ 4f6 Tb3+ 4f8 Pale pink/ nearly colourless Gd3+ 4f7 Gd3+ 4f7 Colourless Lanthanoid contraction. The slow and steady decrease in atomic size and ionic size of lanthanoids with increase in atomic number is called lanthanoid contraction Cause of Lanthanoid contraction. Om moving from cerium to lutetium, the nuclear charge increases with increase in atomic number and the differentiating electrons are added to the 4f subshell. 4f electrons have poor shielding effect. Therefore, effective nuclear charge increases with increase in atomic number. This brings the valence electrons closer to the nucleus and therefore size of atoms goes on decreasing steadily on moving from cerium to lutetium. Consequences of lanthanoid contraction. 1. Atomic size of elements of 2nd and 3rd transition series are almost same due to lanthanoid contraction. The pairs of elements Zr-Hf, Nb-Ta, Mo-W etc possess almost same atomic size due to lanthanoid contraction in elements of 3rd transition series. As a result, elements of second and third transition series resemble each other and possess almost similar chemical properties. Question. Zirconium and Hafnium exhibit similar chemical properties. Why? Answer. Zirconium belongs to 2nd transition series and hafnium belongs to third transition series. Hafnium contains filled 4f subshell. As a result of lanthanoid contraction in Hafnium, the atomic sizes of hafnium and zirconium are almost similar and hence they show similar chemical properties. 2. Basic character of hydroxides of lanthanides decreases from cerium to lutetium. Ce(OH)3 is most basic and Lu(OH)3 is least basic. Due to lanthanoid contraction, size of cations decreases from Ce3+ to Lu3+ and polarising power of cations increases. As a result, ionic character of hydroxides (ability to produce OH- ions) decreases and basic character decreases from Ce(OH)3 to Lu(OH)3.  3. Lanthanides are very difficult to be separated from their mixtures. Since the change in the atomic and ionic radii due to lanthanoid contraction is very small, their chemical properties are almost similar. Thus, it is very difficult to separate these elements in the pure state. 4. Yttrium (Y) occurs with heavy lanthanoids like Holmium and Erbium. Y3+ has almost same ionic size as that of Ho3+ and Er3+. This is due to lanthanoid contraction in Ho3+ and Er3+.Due to similar ionic size they have similar chemical properties and hence they occur together in nature. Formation of complexes Lanthanoids form only few complexes compared to transition elements. This is due to the larger size of lanthanoid cations and unavailability of f orbitals for hybridisation. Chemical reactivity. Due to similar outer electronic configuration, all the lanthanides have almost similar chemical reactivity. Lanthanoids are generally more reactive than transition elements. Uses of lanthanides Pure lanthanoids have no specific uses. Therefore, lanthanoids are extracted as mixtures or alloys known as Misch metals. Misch metal is an alloy of lanthanoids/rare earth elements with iron and traces of S, C, Ca and Al. It consists of 94-95% lanthanoids, 5% Fe and traces of S, C, Ca and Al. The main lanthanoids present in misch metals are Cerium and Neodymium. 1. Misch metals are used for making tracer bullets, shells, flints for lighters (spark producing materials/ used in cigarette lighters) 2. Magnesium alloys containing 30% misch metal and 1% Zirconium is used for making parts of jet engine. 3. CeO2 is used for making incandescent gas mantles and used as pigment for glasses 4. Cerium salts are used as catalyst in lead storage batteries. ACTINOIDS. The 14 elements of 7th period from Thorium (90) to Lawrencium (103) which involves the filling of 5f orbitals are called Actinoids. These elements closely resemble actinium and therefore known as actinoids/actinones/actinides. Thorium, protactinium and uranium are the only natural elements found on earth crust. The remaining elements have been prepared artificially and are referred to as synthetic elements or trans-uranic elements. Example of synthetic or trans-uranic elements.  Neptunium, Plutonium, Americium, Curium, Berkelium, californium, Einsteinium, fermium, mendelevium, Nobelium, Lawrencium. Electronic Configuration. General outer electronic configuration of actinoids may be written as 5f1-14 6d⁰-I7s2. Oxidation States. The most common and stable oxidation state of actinoids is +3. In addition to +3, some actinoids show less common oxidation states like +2, +4, +5, +6 and +7. Actinoids show a greater number of oxidation states as compared to lanthanoids. Explanation. In actinoids, 5f, 6d and 7s orbitals have comparable energies. Hence 4f electrons can also take part in bond formation along with 7s and 6d electrons. In lanthanoids, the energy difference between 4f and 5d subshell is quite large. Hence 4f electrons rarely take part in chemical reaction. Comparison of Lanthanoids and Actinoids Similarities. 1. The most common oxidation state of both is +3. 2. Both involves the filling of (n-2)f orbitals. 3. Both exhibits decrease in atomic radii and ionic radii with increase in atomic number. (lanthanoid contraction and actinoid contraction) 4. Both show paramagnetic properties 5. The nitrates, sulphates and perchlorates of are soluble in water and carbonates, hydroxides and fluorides of both are insoluble in water. Differences Lanthanoids Actinoids 1 Show less number of oxidation Show more number of oxidation states. states. (+2, +3, +4) (+2, +3, +4, +5, +6 and +7) 2 Non-radioactive (except Promethium) All are radioactive. 3. Do not form complexes easily Form complexes easily 4. Para magnetism can be easily Para magnetism cannot be easily explained explained 5 Hydroxides are less basic Hydroxides are more basic Uses of Actinoids. 1. U-233, U-235 and Pu-239 are used as fuel in nuclear reactors. 2. ThO2 mixed 1% CeO2 is used for making incandescent gas mantles. 3. Thorium salts are used for the treatment of cancer.