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The first ionization potential of four consecutive elements in the second period of the periodic table are 8.3, 11.3, 14.5 and 13.6 eV respectively. Which one of the following is the first ionization potential of carbon?
(A) 13.6
(B) 11.3
(C) 8.3
(D) 14.5

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Hint: In chemistry, the term ionization energy refers to the amount of energy which is required to remove an electron from an isolated atom which is in gaseous state in its ground state. When expressed in electron volts, it is called ionization potential.
Complete step by step answer: The ionization energy which is required to remove the first electron from the isolated gaseous atom in the ground state or unexcited state is termed as the first ionization energy or the first ionization potential of that atom.
The ionization energy which is required to remove the ${{\text{2}}^{{\text{nd}}}}$ electron from the isolated gaseous atom is termed as the second ionization energy or the second ionization potential of that atom and that which is required to remove the ${{\text{3}}^{{\text{rd}}}}$ electron from the isolated gaseous atom is called the third ionization energy or the third ionization potential of that atom and so on.
In general, the ionization energy tends to decrease down the group from top to bottom and tends to increase along a period from left to right. But there are some exceptions. When the atom has a stable half filled electron configuration, it will have higher ionization energy than the next elements along a period.
According to the given question, the first ionization energies of four consecutive elements in the second period are 8.3, 11.3, 14.5 and 13.6 eV respectively. Now, four consecutive elements in the second period can be any of the following: B, C, N, O or C, N, O, F or N, O, F, Ne.
Generally, the ionization energies should have increased along the period, but here we can observe that the ionization element of the fourth element is less than that of the third element.
If we consider the case of the two consecutive N and O elements, it is seen that the 2p orbital of N has a stable half filled electronic configuration $\left( {{\text{2}}{{\text{p}}^{\text{3}}}} \right)$ while the 2p orbital of O have 4 electrons, two of which are paired and will feel more repulsion than unpaired electrons. Hence, removal of an electron is harder in case of N as compared to O. So, the ionization energy of N is greater than that of O.
Thus, the third element having 14.5eV ionization energy is N and the fourth element having 13.6eV ionization energy is O. So, the four consecutive elements are B, C, N, O. So the second value of ionization enthalpy (11.3eV) corresponds to Carbon.

So, option B is correct.

Note: Whenever a stable electronic configuration is present, the ionization energy for further ionization increases exceptionally. This means that the ionization energy for an element with a half-filled or a fully-filled configuration will be very high.
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