Question

Among the following which has highest Ionization energy:
A.C
B.N
C.O
D.Fe
E.Ne

Answer 1

Hint: Ionization energy is the amount of energy required to remove an electron from the outermost shell of an isolated gaseous atom which results in the formation of positive ions. It measures the ease of conversion of a neutral atom into cation. Larger ionization energy, lesser the tendency of an atom to become cation. Here we have to check the electronic configuration of given elements.
If an atom has fully or half- filled electronic configuration in the valence shell, the ionization energy will be higher.

Complete step by step answer:
The electronic configuration of given elements are as follows;
 \[\begin{array}{*{20}{l}}
  {_6C - 1{s^2}2{s^2}2{p^2}} \\
  {_7N - 1{s^2}2{s^2}2{p^3}} \\
  {_8O - 1{s^2}2{s^2}2{p^4}} \\
  {_{10}Ne - 1{s^2}2{s^2}2{p^6}_{}} \\
  {_{26}Fe - \left[ {Ar} \right]3{d^6}4{s^2}^{}}
\end{array}\]
Here we can understand that among C, N, O and Ne (2nd period elements) the highest ionization energy will be for Ne. This is because the Ne atom has completely filled p subshell. Next higher ionization energy is for N since it has a half- filled 2p orbital. Both completely filled and half- filled states have higher ionization energy, i.e. it is difficult to remove an electron. For O ionization energy will be lesser than nitrogen, because the removal of one electron makes a stable half- filled configuration. Carbon has least ionization energy among these 2nd period elements. When compared to these elements Fe has lower ionization. This is because its size is larger than 2nd period elements and has more inner shell electrons (more shielding effect).
 Along a period the size gets decreased, since electrons are being added to the same shell. If the size is smaller, the outer electron will be more tightly attracted to the nucleus and will have higher ionization energy. But along a group, the size increases. So the outer electron will be farther from the nuclear attractive force. Hence the ionization energy will be smaller.
Among the elements Neon has highest ionization energy. So the order of ionization energy of given elements are \[Ne > N > O > C > Fe\] .

Hence option E is correct.

Note:

Ionization energy is mainly expressed in \[kj/mol\] . There are different factors that determine the ionization energy. They are;
Atomic size: larger the atomic size, smaller the ionization energy.
Nuclear charge: The force of attraction between the nucleus and outermost electron increases as the nuclear charge increases. Ionization energy increases with increase in nuclear charge.
Number of electrons in the inner shell: Larger the number of electrons in the inner shell lesser the ionization energy. The inner electrons act as the shield between the outer electron and nucleus.
Penetration effect: The penetrating power of orbitals are \[s > p > d > f\] . Energy required to remove electrons are larger than other orbitals.