Question

In periodic table the highest ionization energy are found in:
$(a)$ Upper left corner
$(b)$ Lower left corner
$(c)$ Upper right corner
$(d)$ Lower right corner
$(e)$ Middle of transition elements.

Answer 1

The ionization energy of an atom is the amount of energy required to remove an electron from a mole of atom in the gas phase: $M(g) \to {M^ + }(g) + {e^ - }$. It is possible to remove more electrons from most elements, so this quantity is more precisely known as the first ionization energy.

Complete answer:
The first ionization energy varies in a predictable way across the periodic table. The ionization energy decreases from top to bottom in groups, and increases from left to right across a period. Thus, helium has the largest first ionization energy.
From top to bottom in a group, orbitals corresponding to higher value of the principal quantum number $(n)$ are being added, which are on average further away, they are less strongly affected by the nucleus, and are easier to remove, corresponding to a lower value for the first ionization energy.
From left to right across a period, more protons are being added to the nucleus, but the number of electrons in the inner, lower-energy shells remains the same. The valence electrons feel a higher effective nuclear energy charge- the sum of the charges on the protons in the nucleus and the charges on the inner, core electrons. The valence electrons are therefore held more tightly, the atom decreases in size and it becomes increasingly difficult to remove them, corresponding to a higher value for the first ionization energy.
Thus option C is the correct answer.

Note:
There are some fluctuations in the ionization energy, for example- Oxygen is to the right of the nitrogen in period $2$, its first ionization energy is slightly lower than that of nitrogen. Nitrogen has an electronic configuration of $1{s^2}2{s^2}2{p^3}$, which puts one electron in each $p$ orbital, making it a half-filled set of orbital. Half-filled sets of $p$ orbitals are slightly more stable than those with ${2_{}}o{r_{}}4$ electrons, which makes it slightly harder to ionize a nitrogen atom. Oxygen has an electronic configuration of $1{s^2}2{s^2}2{p^4}$, which puts another electron in one $p$ orbitals; since this is one electron away from being half-filled, it is slightly easier to remove this additional electron.