Question

goal: to demonstrate knowledge of ionic bonding

1. examine the bohr - rutherford diagrams below. circle the atoms that are stable.
2. the following bohr - rutherford diagram represents an oxygen atom. examine the diagram, then answer the following questions.

a. why is this not a stable electron arrangement?
b. what would make this atom stable?
use a coloured pencil to show a stable electron arrangement on the diagram.
label the charge on this newly formed ion.
bohr - rutherford diagram to show an ionic bond between sodium (na) and chlorine

Explanation:

Step1: Understand stable electron configuration

Atoms are stable when their out - most energy level is full. Oxygen has 6 valence electrons in its outermost shell and a full outer shell for the second energy level requires 8 electrons.

Step2: Answer part a

Since oxygen has 6 valence electrons in its outermost shell and a full outer shell requires 8 electrons, it is not stable as its outer shell is not full.

Step3: Answer part b

Oxygen can gain 2 electrons to fill its outer shell and become stable.

Step4: Show stable electron arrangement

Using a coloured pencil, add 2 more electrons to the outermost shell of the oxygen atom in the diagram.

Step5: Label the charge

When oxygen gains 2 electrons, it forms an ion with a charge of - 2.

Step6: Draw ionic bond between Na and Cl

For sodium (Na), it has 1 valence electron. It will lose this 1 electron to achieve a stable electron configuration (becoming a $Na^+$ ion). Chlorine (Cl) has 7 valence electrons and will gain 1 electron to achieve a stable electron configuration (becoming a $Cl^-$ ion). Draw the Bohr - Rutherford diagrams for Na and Cl, show the transfer of the electron from Na to Cl, and label the ions formed.

Answer:

a. Its outer shell is not full (has 6 instead of 8 valence electrons).
b. Gaining 2 electrons.
Stable electron arrangement: Add 2 electrons to oxygen's outer shell with coloured pencil.
Charge on newly formed ion: - 2.
Ionic bond between Na and Cl: Draw Na losing 1 electron to Cl, forming $Na^+$ and $Cl^-$ ions.