Question

What is the Lewis structure for $Ge{{F}_{4}}$?

Answer 1

Hint: To solve this question we first need to know what Lewis dot structures are. The valence electrons of the atoms of a molecule can be represented using the Lewis dot structure. They are a simplified version of the molecular geometry of a compound and can be used to depict the chemical bonding in the molecules.

Complete answer:
The Lewis dot structure of a compound can be drawn by following steps.
1. We first need to know the valence electrons in the atoms of the molecule.
The atomic number of germanium (Ge) is 32 and has 4 valence electrons.
Whereas the atomic number of fluorine (F) is 9 and has 7 valence electrons.
2. Identify the least electronegative atom. This atom is the central atom of the molecule.
Germanium is the least electronegative atom in $Ge{{F}_{4}}$.
3. Connect the atoms via single bonds to form a skeleton structure.

4. Assign lone pairs to each molecule.
Since a Ge atom is connected to four F atoms, it has no lone pairs.
Whereas each F atom is bonded to one Ge atom, so one electron of the valence electron is used during bonding, and hence the number of lone pairs on F is 3.

5. Check if every atom has an octet configuration. Draw double or triple bonds to satisfy the octet.
Since the octet is satisfied for both germanium and fluorine, the structure drawn above is the Lewis dot structure of $Ge{{F}_{4}}$.

Note:
It should be noted that when the molecule is an ion, the electrons added to or subtracted from the Lewis dot structure are equal to the magnitude of the charge. In the case of cation (positively charged ion) electrons are subtracted, and in the case of an anion (negatively charged ions), electrons are added.