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(a) Write balanced equations to represent what happens when (i) Cu^(2+) is treated with KI. (ii) Acidified potassium dichromate solution is reacted with iron (II) solution (ionic equation). (b) (i) The figure given below illustrates the first ionisation enthalpies of first, second and third series of transition elements. Answer the questions that follow : Which series amongst the first, second and third series of transition elements have the highest first ionisation enthalpy and why ? (ii) Separation of lanthanoid elements is difficult. Explain. (iii) Sm^(2+), Eu^(2+) and Yb^(2+) ions in solution are good reducing agents but an aqueous solution of Ce^(4+) is a good oxidising agent. Why? |
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Answer» Solution :(a) (i) `2Cu^(+)+4I^(-)toCu_(2)I_(2)(s)+I_(2)` In this case `Cu^(2+)` oxidises `I^(-)` to `I_(2)`. (ii) `Cr_(2)O_(7)^(2-)+14H^(+)+6Fe^(2+)to2Cr^(3+)+6Fe^(+)+7H_(2)O` (b) (i) The third SERIES of transition elements have the highest first ionisation enthalpy. This is due to weak screening effect of f-orbitals and higher nuclear charge of the third series of transition elements. (ii) Separation of lanthanoid elements is difficult because of the identical IONIC radii of these elements. (iii) Lanthanoids show +3 oxidation state PREDOMINANTLY. That is why `Sm^(2+),Eu^(2+)` and `Yb^(2+)` act as REDUCING agents, they are themselves oxidised to `Sm3^(+),Eu^(3+)` and `Yb^(3+)` RESPECTIVELY which is the preferred oxidation state of lanthanoids. `Ce^(4+)` is a strong oxidising agent and reverts to +3 state. The `E^(@)` value for `Ce^(4+)//Ce^(3+)` is 1.74 V. |
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