Question

Use the molecular orbital diagram shown to determine which of the following is most stable. A. F22+ B. Ne22+ C. F22- D. O22+ E. F2

Answer 1

Answer:

$O^{2+} _{2}$

Explanation:

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In the given problem, we will calculate the number of valence electrons and the bond order to determine the most stable:

Bond order (BO) = 0.5*(electrons in the bonding molecules - electrons in the antibonding molecule).

A. For $F^{2+} _{2}$

The valence electrons = 12; BO = 0.5*(8-4) = 2

B. For  $Ne^{2+} _{2}$

The valence electrons = 14; BO = 0.5*(8-6) = 1

C. For  $F^{2-} _{2}$

The valence electrons = 16; BO = 0.5*(8-8) = 0

D. For  $O^{2+} _{2}$

The valence electrons = 10; BO = 0.5*(8-2) = 3

E. For  $F_{2}$

The valence electrons = 14; BO = 0.5*(8-6) = 1

The bond order is commonly used to signify the bond stability. Higher bond order indicates more stability and vice versa. Thus,  $O^{2+} _{2}$ is the most stable.

Answer 2

The one that is most stable is F2 or E