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Explain the variation in ionization enthalpies of transition elements in 3d series. |
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Answer» Solution :(i) First ionization enthalpy: The first ionization enthalpy follows a irregular trend. This is because, the first electron when removed alters the relative energies of 3d and 4s orbitals. The electrons are first removed from 4s and then from 3d orbitals. In first transition series going from scandium to zinc, the nuclear charge increases with the increase in atomic number and the electrons are added in 3d orbitals. The increase in nuclear charge is opposed by the sheilding effect of 3d electrons, as a result the atomic radii decreases less rapidly and so there is slight increase in ionization enthalpy in 3d series. (ii) Successive ionization enthalpies: In general the third ionization enthalpy is HIGHER than second ionization enthalpy which in turn is higher than first ionization enthalpy. The high values of successive ionization enthalpies is attributed to high effective nuclear charge and poor sheilding of one d-electron by other. The second ionization enthalpy of CHROMIUM is higher than MANGANESE while the third ionization enthalpy of manganese is higher than chromium. In case of chromium, the second electrons is to be removed from half FILLED `(d^(5))` subshell which is extra stable and hence requires greater energy for ionization. In case of manganese, the third electron is to be removed from half filled `(d^(5))` subshell and so third ionization enthalpy is higher for manganese than chromium `""_(24)Cr^(+) : [Ar] 3d^(5) 4s^(0) rarr ""_(24)Cr^(2+) : [Ar] 3d^(4)` `""_(25)Mn^(2+) : [Ar] 3d^(5) 4s^(0) rarr ""_(25)Mn^(3+) : [Ar] 3d^(4)` (More DIFFICULT to attain) Thus, for manganese, it is difficult to remove the third electrons. Similary, third ionization enthalpy of Mn is higher than Fe because `Mn^(2+)` has `d^(5)` configuration while `Fe^(2+)` has `d^(6)` configuration. So, the removal of electron from `Mn^(2+)` is difficult as the subshell is half filled. `Fe^(2+) : [Ar] 3d^(6) 4s^(0) rarr Fe^(3+) : [Ar] 3d^(5)` `Mn^(2+) : [Ar] 3d^(5) 4s^(0) rarr Mn^(3+) : [Ar] 3d^(4)` (More difficult to attain) The First ionization enthalpy of Cu is less than zinc because the removal of 4s electron in copper will result in `d^(10)` configuration (fully filled) while the second ionization of Zn is lower than copper as removal of second electron from zinc results in `d^(10)` configuration (fully filled). `""_(29)Cu: [Ar] 3d^(10) 4s^(1) rarr ""_(20)Cu^(+): [Ar] 3d^(10) 4s^(0)` [Easy to attain] `""_(30)Zn^(+) : [Ar] 3d^(10) 4s^(1) rarr ""_(30)Zn^(2+): [Ar] 3d^(10) 4s^(0)` [Easy to attain] (iii) Stability of `M^(2+)` ions in gaseous state: The stability of `M^(2+)` ions in gaseous state depends on the summation of first and second ionization enthalpies and enthalpy of atomization. Lower is the sum of ionization enthalpy, greater is the thermodynamic stability. The dominant factor is second ionization enthalpy. This explains why the `Zn^(2+) and Mn^(2+)` ions are formed easily adn removal of third electron in difficult. the high values of third ionization enthalpies of Ni, Cu and Zn indicates why it is difficult to obtain oxidation states higher than two for these elements. |
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