Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The incorrect order of second ionization energies in the following is

EAMCETEAMCET 2010

Solution:

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{smallmatrix} \text{(2,8)} \\ \text{(High}\,\text{IE)} \end{smallmatrix}}{\mathop{\text{N}{{\text{a}}^{\text{+}}}}}\,\text{}\underset{\text{(2,}\,\text{8,}\,\text{1)}}{\mathop{\text{M}{{\text{g}}^{\text{+}}}}}\, $ $ \underset{\begin{smallmatrix} 3{{d}^{5}} \\ (High\,IE) \end{smallmatrix}}{\mathop{C{{r}^{+}}}}\,>\underset{3{{d}^{5}}4{{s}^{1}}}{\mathop{M{{n}^{+}}}}\, $ $ \underset{\begin{smallmatrix} 3{{p}^{3}} \\ (half\,filled. \\ thus\,high\,IE) \end{smallmatrix}}{\mathop{{{S}^{+}}}}\,>\underset{3{{p}^{2}}}{\mathop{{{P}^{+}}}}\, $