Question

Choose the correct order of second ionization enthalpy:

• A. C > N > O > F
• B. O > N > F C
• C. O > F > N > C
• D. F > O > N > C

Answer 1

The correct option is C O > F > N > C

After the removal of an electron from C, N, O and F their electronic configurations are:
$C^{+}$ - $1s^2$ $2s^2$ $2p^1$
$N^{+}$ - $1s^2$ $2s^2$ $2p^2$
$O^{+}$ - $1s^2$ $2s^2$ ${2p_x}^1$ ${2p_y}^1$ ${2p_z}^1$
$F^{+}$ - $1s^2$ $2s^2$ ${2p_x}^2$ ${2p_y}^1$ ${2p_z}^1$

Since $O^{+}$ gives a stable electronic configuration with exactly half-filled 2 p-sub shell, therefore, 2nd ionization enthalpy of O is higher that of F. The 2nd ionization enthalpy of C is lower than that of N because of its larger size.

Thus, the overall order is O > F > N > C .