Explain Electronegativity of Si and Ge: 65 Characters

[struggles]

Edit: thread moved by mod from non technical forum, hence lack of the template.

Using arguments from MO theory, state why the trends in homonuclear bond energies for rows 1 and 3 of the periodic table in graph D differ so markedly.

S
State, with reasons, whether Si or Ge has the higher electronegativity

So I know that the bond energies in general tend to increase across the period as the atoms get smaller and there is better orbital overlap. I also understand that N/O/F have extra repulsion dues to the fact that π* orbitals are more anti bonding than π orbitals are bonding due to asymmetry in the MO diagrams so they are all weaker than expected. However why is this effect only seen in row 1 and not in row 3?

 

[Asim Riaz]

The difference in the trends of bond energies between rows 1 and 3 of the periodic table can be attributed to the differences in the size, shape, and electronegativity of the atoms involved. In row 1, the atoms are smaller and more symmetrical, meaning that their orbitals have better overlap and form stronger bonds. In contrast, the atoms in row 3 are larger and less symmetrical, leading to weaker bonds due to poorer orbital overlap. Additionally, the atoms in row 3 are also more electronegative, meaning that they will attract electrons more strongly, further weakening the bonds. As for which of Si or Ge has the higher electronegativity, Ge has the higher electronegativity. This is because Ge is larger than Si, meaning its valence electrons are held more tightly by the nucleus, thus making it more electronegative.