Question

Between Se^2-, I^-, Br^-, O^2- and F^-, which will have the maximum number of valence electrons? Explain.

Answer 1

The electrons present in the outermost or valence shell are valence electrons.

The periodic trend of the number of valence electrons:

1. In a period, the number of valence electrons increases from left to right in a period.
2. In a group, the number of valence electrons remains constant.

Selenium (Se^2-) ion

• Selenium with atomic number $54$ gains $2$ electrons.
• So, its number of electrons becomes $36$.
• Electronic configuration: [Ar] $3d^{10}4s^{2}4p^6$
• Number of valence electrons = $2\left(4\mathrm{s}\right)+6\left(4\mathrm{p}\right)=8$

Iodine (I^-) ion

• Iodine with atomic number $53$ gains $1$ electron.
• So, its number of electrons becomes $54$.
• Electronic configuration: [Kr] $4d^{10}5s^{2}5p^6$
• Number of valence electrons = $2\left(5\mathrm{s}\right)+6\left(5\mathrm{p}\right)=8$

Bromine (Br^-) ion

• Bromine with atomic number $35$ gains $1$ electron.
• So, its number of electrons becomes $36$.
• Electronic configuration: [Ar] $4s^{2}3d^{10}4p^6$
• Number of valence electrons = $2\left(4\mathrm{s}\right)+6\left(4\mathrm{p}\right)=8$

Oxygen (O^2-) ion

• Oxygen with atomic number $8$ gains $2$ electrons.
• So, its number of electrons becomes $10$.
• Electronic configuration: [He] $2s^{2}2p^6$
• Number of valence electrons = $2\left(2\mathrm{s}\right)+6\left(2\mathrm{p}\right)=8$

Fluorine (F^-)

• Flurine with atomic number $9$ gains $1$ electron.
• So, its number of electrons becomes $10$.
• Electronic configuration: [He] $2s^{2}2p^6$
• Number of valence electrons = $2\left(2\mathrm{s}\right)+6\left(2\mathrm{p}\right)=8$

Considering both the periodic trends and electronic configuration for the number of valence electrons, all the ions have the same number of valence electrons, i.e. $8$.