Question

Consider the following orderings:
$I$. $Al>Si>P>Cl$
$II$. $Be<Mg<Ca<Sr$
$III$. $I<Br<Cl<F$
$IV$. ${Na}^{+}<{Mg}^{2+}<{Al}^{3+}<{Si}^{4+}$
Which of these give(s) a correct trend in ionization energy?

• A. $I$ only
• B. $III$ only
• C. $I$ and $II$ only
• D. $I$ and $IV$ only
• E. $I,III$ and $IV$ only

Answer 1

The correct option is E $I,III$ and $IV$ only

First ionization energy decreases as you go down Group 17 because the electron being removed is further from the nucleus

Ionization energy generally increases across Period 3 from left to right as the increasing nuclear charge on successive atoms more tightly binds the electron being removed to the nucleus.With increase of positive charge, it increases.