Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state. Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula. a) both Sc3+ and Zn2+ ions are colourless and form white compounds. state & by the loss of one more electron from the 3d-orbital, it acquires. In case of Fe2+ ion, the third electron is taken out from 3d6 configuration which results in more stable 3d5 configuration. MnO is basic whereas Mn2O7 is acidic. The atomic radii of second and third series are larger than 3d series. Comment on the possible oxidation state of this element. Most of the elements of the first transition series form ions with a charge of 2+ or 3+ that are stable in water, although those of the early members of the series can be readily oxidized by air. Why? Potassium permanganate (KMnO4) is prepared by the fusion of a mixture of pyrolusite (MnO2), potassium hydroxide and oxygen, first green coloured potassium manganate is formed. The stability of Cu2+(aq) rather than Cu+(aq) is due to the much more negative ∆hyd H°of Cu2+(aq) than Cu+, which more than compensates for the second ionisation enthalpy of Cu. When ethyl bromide is boiled with aqueous potassium hydroxide then ethyl alcohol is formed. H 2 O, NH 3 ). (iv) Chemical reactivity : Actinoids are far more reactive than lanthanoids. However, loss of a further electron from the 'd' shell leaves a configuration of [Ar]4s 0 3d 5. Give examples and suggest reasons for the following features of the transition metal chemistry : The element has the configuration [Ar]4s 2 3d 6. Iron, for example has two common oxidation states, +2 and +3. Elemental iron occurs in meteoroids and other low-oxygen environments but is reactive to oxygen and water. The common oxidation state of 3d series elements is + 2 which arises due to participation of only 4s electrons. Maintenance & improvements. Difference between Lanthanides and Actinides. Question 21. Illustrate your answer with examples. The cumulative effect of the contraction of the lanthanoid series, known as lanthanoid contraction, causes the radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. Compare the general characteristics of the first series of transition metals with those of the second and third series metals in the respective vertical columns. Choose the best answer: 1. Question 28. The electronic configuration of Ce3+ is 4f1. Justify this statement by giving some examples from the oxidation state of these elements. These series are also referred to as 3d, 4d, 5d and 6d series, respectively. (iii) The d1 configuration is very unstable in ions. Carbon monoxide is a strong reducing agent because it is easily oxidised to carbon dioxide - where the oxidation state is the more thermodynamically stable +4. Why? For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Question 33. Uncombined iron, cobalt, and nickel can be found in meteors. scandium show only +2 and +3 oxidation states Due to the loss of two electrons from the 4s orbital, Sc acquires +2 oxidation state & by the loss of one more electron from the 3d-orbital, it acquires +3 oxidation state which has extra stable orbital. In other words the 5f electrons themselves provide poor shielding from element to element in the seriest. One of the most striking features of the transition elements is that the elements usually exist in several different oxidation states. Solution: Question 23. Thus in the series Sc(II) does not exist, Ti(II) is less stable than Ti(IV). Except scandium, the most common oxidation state shown by the elements of first transition series is +2. Try to correlate this type of behaviour with the electronic configurations of these elements. Why are Mn2+ compounds more stable than Fe2+ towards oxidation of their +3 state? The frequency of the light absorbed is determined by the nature of the ligand. What are the characteristics of the transition elements and why are they called transition elements? Solution: This table is based on Greenwood's, with all additions noted. Mn2+ is more stable than Mn3+ due to half filled d-orbitals. Question 26. (ii) A transition metal exhibits highest oxidation state in oxides and fluorides. (iii) d1 configuration is very unstable in ions because after losing one more electron it will become more stable due to vacant d-orbital. Solution: (ii) Mn has electronic configuration (Ar) 4s2 3d5 and all the electrons in ‘s’ as well as ‘d’ orbitals can take part in bond formation, therefore, it shows + 7 highest oxidation state… Manganese shows oxidation state of +7 in its oxometal anion MnO4– which is equivalent to its group number 7. (ii) Cobalt(ll) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. The actinoids show in general +3 oxidation state. The most common oxidation state for ions of the inner transition elements is +3. For example : Cr forms CrO42- and Cr2O72-, both contain chromium in +6 oxidation state. Copper exhibits +1 oxidation state in the first series of transition metals because when one electron is lost, the configuration becomes stable due to fully filled d10 configuration. (iii) Oxidation state : The most common oxidation state of lanthanoids is +3 while actinoids show more variable oxidation states than lanthanoids ranging from +3 to +7. They are called transition elements due to their incompletely filled d-orbitals in ground state or any stable oxidation state and they are placed between s and p-block elements. Solution: Solution: Question 37. Question 32. When the ethylenediaminetetraacetate ion (EDTA4-) forms a complex with a transition metal ion, how many electrons does it normally donate to the metal? Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why? What may be the stable oxidation state of the transition element with the following d-electron configurations in the ground state of their atoms: 3d3, 3d5, 3d8 and 3d4 ? 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2 List of oxidation states of the elements 1 List of oxidation states of the elements This is a list of all the known oxidation states of the chemical elements, excluding nonintegral values. Compare the chemistry of the actinoids with that of lanthanoids with reference to (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity. Solution: But some types of atoms such as chlorine form various oxidation numbers like -1, 0, +1, +3, +5, +7 oxidation numbers in compounds. Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures. What are alloys? There is a greater range of oxidation states, which is attributed to the fact that the 5f, 6d and 7s levels are of comparable energies. Both energy levels can be utilized as a part of bond development. Video Explanation. In the formation of metallic bonds, no electrons 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. This oxidation state arises from the loss of two 4s electrons. The valence electrons of the transition elements are in (n-1) d and ns orbitals which have a little distinction in energies. common) oxidation state”. The relative stabilities of various oxidation states of 3d series elements can be correlated with the extra stability of 3d 0, 3d 5 and 3d 10 configurations to some extent. Formation of complex compounds – Due to small size and high charge density of metal ions. What is meant by ‘disproportionation’? Explain why Cu+ ion is not stable in aqueous solutions? Solution: Question 31. Thus, Ti 4+ ion with 3d 0 configuration is more stable than a Ti 3+ ion with 3d 1 configuration. Ce4+, Tb4+, Eu2+, Yb2+, etc. Atomic and ionic size – Ions of same charge in a given series show progressive decrease in radius with increasing atomic number. Sodium dichromate is more soluble than potassium dichromate. Among the elements of 4d-Series Ruthenium belonging to 8th group exhibits maximum oxidation state. Answer. (i)Electronicconfiguration : Lanthanoids have general electronic configuration of [Xe] 4f1-14 5d0-1 6s2 and actinoids have general electronic configuration of [Rn]5f1-14 6d0-1 7s2. The most common oxidation states are in bold. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with. When ligands approach to central metal, metals five degenerate (same energy orbitals) orbitals gets splits into different energy levels as eg & t 2 g. This removes the degeneracy. Write the ionic equations for the reactions. e.g. Thus in the series Sc(II) does not exist, Ti(II) is less stable than Ti(IV). Because the distribution of oxidation states among the actinoids is so uneven and so different for the earlier and latter elements. Solution: Solution: Solution: The reason why Manganese has the highest oxidation state is because the number of unpaired electrons in the outermost shell is … 2Cu+(aq) → Cu2+(aq) + Cu(s) Question 12. Clearly, the +2 oxidation state arises from the loss of the 4s electrons. ii) Ligands are negatively charged ions or neutral molecules, having lone pair of electrons (i.e. (iv) chemical reactivity. What are inner transition elements? Give special emphasis on the following points The For example, the common oxidation numbers of the alkaline metals are all 1. Vanadium(V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in r Catalytic Hydrogenation) are some of the examples.Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst. Cr in Cr2O72- and CrO42- show oxidation state +6 which is equivalent to its group number 6. Question 34. Manganese is the 3d series transition element shows the highest oxidation state. Indicate the steps in the preparation of. The element has the configuration [Ar]4s 2 3d 6. Question 7. 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