Read the given passage and answer the questions number 1 to 5 that follow : The lowest common oxidation state of transition metals is + 2. To form the M^(2+) ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required in addition to the enthsualpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M^(+) ions have the d^(5) and d^(10) configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of 1s electron which results in the formation of stable d^(10) configuration. 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 d^(5) (Mn^(2+)) and d^(10) (Zn^(2+)) ions. In general, the third ionisation enthalpies are quite high. Also, the high values for third ionisation enthalpies of copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements. Which is the lowest common oxidation state of transition metals?