Question

1. for each of the following elements, circle the element with the highest ionization energy and put a square around the element with the lowest ionization energy.

a) o c be ne b) na rb cs he
c) pb c sn si d) au w s sr

Explanation:

• a)
• Ionization energy generally increases across a period (left - to - right) and decreases down a group.
• In the period \(2\) elements \(Be\), \(C\), \(O\), \(Ne\), \(Ne\) (a noble gas with a full valence shell) has the highest ionization energy as it is very stable. \(Be\) has a lower ionization energy than \(C\) and \(O\) because it has a \(2s^{2}\) configuration.
• b)
• \(He\) is a noble gas with a very stable electron configuration (\(1s^{2}\)), so it has a high ionization energy. Among the alkali metals \(Na\), \(Rb\), \(Cs\), ionization energy decreases down the group. \(Cs\) has the lowest ionization energy among them.
• c)
• \(C\) is in period \(2\), \(Si\) in period \(3\), \(Sn\) in period \(5\) and \(Pb\) in period \(6\) of group \(14\). Ionization energy decreases down the group. \(C\) has the highest ionization energy (smallest atom in the group with the electrons closest to the nucleus). \(Pb\) has a relatively lower ionization energy due to its larger atomic size.
• d)
• \(S\) is a non - metal in period \(3\) with a relatively high ionization energy compared to metals. Among the metals \(Sr\), \(W\), \(Au\), \(Sr\) (an alkaline earth metal) has a lower ionization energy. \(S\) (smaller atom with a higher effective nuclear charge for its valence electrons) has a higher ionization energy.

Answer:

a) Circle \(\text{Ne}\), Square \(\text{Be}\)
b) Circle \(\text{He}\), Square \(\text{Cs}\)
c) Circle \(\text{C}\), Square \(\text{Pb}\)
d) Circle \(\text{S}\), Square \(\text{Sr}\)