Question

The incorrect order of second ionization energies in the following is

• A. $Rb>K$
• B. $Na>Mg$
• C. $Cr>Mn$
• D. $S>P$

Answer 1

Answer: (A)

Rb and K both belong to 1st group and on moving downwards in a group all ionization energies decrease because size increases. Thus, the IE (second) of K > Rb. All other orders of second ionisation energies are correct and can be explained as $\underset{\begin{array}{c}\text{(2,8)}\\ \text{(High}\text{IE)}\end{array}}{\text{N}{\text{a}}^{\text{+}}}\underset{\text{(2,}\text{8,}\text{1)}}{\text{M}{\text{g}}^{\text{+}}}$ $\underset{\begin{array}{c}3d^5\\ (HighIE)\end{array}}{C{r}^{+}}>\underset{3d^{5}4s^1}{M{n}^{+}}$ $\underset{\begin{array}{c}3p^3\\ (halffilled.\\ thushighIE)\end{array}}{S^{+}}>\underset{3p^2}{P^{+}}$