Question

Which of the following is incorrect:
$\left(\text{IE = Ionisation Enthalpy}\right)$

• A. ${\text{IE}}_1\left(\text{Mg}\right)>{\text{IE}}_1\left(\text{Al}\right)$
• B. ${\text{IE}}_2\left(\text{Mg}\right)>{\text{IE}}_2\left(\text{Al}\right)$
• C. ${\text{Order of IE}}_1:\text{Cu < Zn < Ga}$
• D. ${\text{Order of IE}}_1:\text{Li > Na > K}$

Answer 1

Answer: (B), (C)

${\text{Order of IE}}_1:\text{Cu < Zn < Ga}$

When we remove an electron from a fully filled or a half filled subshell, comparatively more energy is required to remove an electron as the system is getting destabilized by the removal of the electron.
$\text{Mg}-1s^{2}2s^{2}2p^{6}3s^2\text{Al}-1s^{2}2s^{2}2p^{6}3s^{2}3p^1$
Here $\text{Mg}$ has fully filled $s$ orbital.
So, ${\text{IE}}_1\left(\text{Mg}\right)>{\text{IE}}_1\left(\text{Al}\right)$

${\text{Mg}}^{+}-1s^{2}2s^{2}2p^{6}3s^1{\text{Al}}^{+}-1s^{2}2s^{2}2p^{6}3s^2$
Here ${\text{Al}}^{+}$ has fully filled $s$ orbital.
So, ${\text{IE}}_2\left(\text{Mg}\right)<{\text{IE}}_2\left(\text{Al}\right)$

$\text{Cu}-\left[Ar\right]4s^{1}3d^{10}\text{Zn}-\left[Ar\right]4s^{2}3d^{10}\text{Ga}-\left[Ar\right]4s^{2}3d^{10}4p^1$
Here the ${\text{IE}}_1$ of $\text{Ga}$ will be lowest as the electron is removed from $4p$ orbital which has only one electron.
In $\text{Cu}$ the electron is removed from $4s$ orbital which is more closure to nucleus than $4p$. So ${\text{IE}}_1$ of $\text{Cu}$ will be more than $\text{Ga}$.
The ${\text{IE}}_1$ of $\text{Zn}$ will be highest among these as the electron is removed from fully filled $4s$ orbital.
So, ${\text{Order of IE}}_1:\text{Zn > Cu > Ga}$

As we move down the group, the atomic size goes on increasing which results in the decrease of effective nuclear charge on the elctron in the outermost orbital. So, ionisation energy decreases down the group.
Hence, ${\text{Order of IE}}_1:\text{Li > Na > K}$