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What is the Ionization energy of Magnesium?

Answer
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Hint: We can define ionization energy (enthalpy) as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so as to convert it into a gaseous cation.

Complete answer:
In order to find the Ionization energy of Magnesium, we should first know the exact meaning of ionization energy (enthalpy). It is the energy needed to remove the most loosely held electron of the atom.
Also, the ionization energy is represented by ΔiH.
 The ionization energy required to remove first, second, third etc electrons from an isolated gaseous atom are called successive ionization enthalpies.
Now, coming to magnesium. The electronic configuration of Magnesium is;
Mg = 1s22s22p63s2
The first ionization energy of magnesium (removing one 3s electron)
 Mg (g) + ΔiH1Mg++e(g) ΔiH1=737.76KJmol1
Where, Mg (g) and Mg+(g) represents the gaseous magnesium and gaseous magnesium cation.
The first ionization energy of Mg was found to be 737.76 KJmoll1
The second ionization energy of magnesium (removing the other electron from 3s orbital)
Mg+(g)+ΔiH2Mg2+(g)+e(g) ΔiH2=1450.73KJmol1
The second ionization energy of magnesium was found to be 1450.73KJmol1
We can see clearly that the second ionization energy is greater than the first ionization energy. It is because the positively charged ion formed after the removal of the first electron has one electron less than the number of protons in the nucleus. As a result, the effective nuclear charge will increase and the remaining electrons will be held more firmly by the nucleus. Thus, the energy required to remove the second electron from the positively charged ion is much higher than the first.

Note:
The ionization energy depends upon various factors such as Nuclear charge (the ionization energy increases with increase in nuclear charge), Atomic size (ionization enthalpy decreases as the atomic size increases) etc.