Question

The first ionization potential of Na, Mg, Al and Si are in the order:
A. $$Na>Mg>Al>Si$$
B.$$Na>Mg>Al<Si$$
C. $$Na<Al<Mg<Si$$
D. $$Na<Mg<Al<Si$$

Answer 1

Hint: The first ionisation potential is an atomic property that represents the outermost electronic state. It is an estimate of the energy needed to remove one valence electron from a neutral atom.

Complete answer:
We know that, as we progress through the period, the atomic size decreases, and thus the ionisation enthalpy rises. That is, first ionisation potential rises from left to right across periods, but owing to full-filled 3s -orbitals, Mg is more stable than Al.
The following are the electronic configurations:
$$N{a}_{11}=1s^2,2s^2,2p^6,3s^1$$
$$M{g}_{12}=1s^2,2s^2,2p^6,3s^2$$
$$A{l}_{13}=1s^2,2s^2,2p^6,3s^2,3p^1$$
$$S{i}_{14}=1s^2,2s^2,2p^6,3s^2,3p^1$$
That is, in the case of Al$$\left(3s^{2}3p^1\right)$$, the electron must be removed from the outer 3 p orbital, while in the case of Mg$$\left(3s^2\right)$$, the electron must be removed from the stable fully filled orbital. As a result, the ionisation energy of Na is lower than that of Mg (Na < Mg).But, the electron must be separated from the 3p-orbital of both Al and Si, but Si has a higher nuclear charge than Al. As a result, Al's ionisation enthalpy is lower than Si's. A p electron requires less energy to remove than a filled s electron from the same principal quantum shell.
Thus, correct order of first ionisation potential is option C ,$$Na<Al<Mg<Si$$

Note:
The energy required to remove an electron from a stable, fully filled orbital is greater than that required to remove an electron from a partially filled orbital. Because of the symmetrical distribution of electrons, orbitals in which the subshell is precisely half-filled or fully filled are more stable. The number of exchanges is greatest when the orbitals are half-filled or fully filled. As a result, it has the highest level of stability.