Question

The first ionization potential of Na, Mg, Al & Si are in the order

• A. Na < Mg > Al < Si
• B. Na > Mg > Al > Si
• C. Na < Mg < Al > Si
• D. Na > Mg > Al < Si

Answer 1

The correct option is A

Na < Mg > Al < Si

Recall how to write electronic configuration.

Na $\to$ 1$s^2$ 2$s^2$ 2$p^6$ 3$s^1$

Mg $\to$ 1$s^2$ 2$s^2$ 2$p^6$ 3$s^2$

Al $\to$ 1$s^2$ 2$s^2$ 2$p^6$ 3$s^2$ 3$p^1$

Si $\to$ 1$s^2$ 2$s^2$ 2$p^6$ 3$s^2$3$p^2$

We have to note that,

In Al and Si, we have to remove an electron from 3p, while in Na and Mg, we have to remove from 3s.

Now, because of penetration effect, the ionization potential will be more for Al than Mg.

Penetration effect of orbitals: The order of energy required to remove electron from s, p, d- and f-orbtial's of a shell is s> p> d >f because the distance of the electron from the nucleus increases. For example- the value of ionization potential of Be(Z = 4, 1$s^2$2$s^2$) and Mg(Z = 12), 1$s^2$2$s^2$2$p^6$3$s^2$) are more than the I.P of B(Z = 5, 1$s^2$2$s^2$2${p_x}^1$) and AI(Z = 12, 1$s^2$2$s^2$2$p^6$3${p_x}^1$) because the penetration power of 2s and 3s electrons is more than 2p and 3p orbitals respectively. More energy will be required to separate the electrons from 2s and 3s orbitals.

Now, in Na and Mg, because the atomic radius of Na > Mg, therefore ionization potential of Mg will be more than Na.

$\to$ Same logic in Al and Si also.

$I{P}_1$(Al) < $I{P}_1$(Si)

Therefore, the correct order of IP, will be

Na > Mg > Al < Si