A) \[Rb>K\]
B) \[Na>Mg\]
C) \[Cr>Mn\]
D) \[S>P\]
Correct Answer: A
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\[\underset{\begin{smallmatrix} \text{(2,8)} \\ \text{(High}\,\text{IE)} \end{smallmatrix}}{\mathop{\text{N}{{\text{a}}^{\text{+}}}}}\,\text{}\underset{\text{(2,}\,\text{8,}\,\text{1)}}{\mathop{\text{M}{{\text{g}}^{\text{+}}}}}\,\] \[\underset{\begin{smallmatrix} 3{{d}^{5}} \\ (High\,IE) \end{smallmatrix}}{\mathop{C{{r}^{+}}}}\,>\underset{3{{d}^{5}}4{{s}^{1}}}{\mathop{M{{n}^{+}}}}\,\]\[\underset{\begin{smallmatrix} 3{{p}^{3}} \\ (half\,filled. \\ thus\,high\,IE) \end{smallmatrix}}{\mathop{{{S}^{+}}}}\,>\underset{3{{p}^{2}}}{\mathop{{{P}^{+}}}}\,\]
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