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

The one that is most stable is F2 or E

Answer 2

Answer:

The most stable element based on the molecular orbital diagram is $\text F_{2} ^{2^{+}}$.

Explanation:

Molecular Orbital Theory (MOT) was proposed by Hund and Mulliken. The theory describes the bonding in molecules, elements, or atoms. The theory uses Molecular Orbital (MO) diagram to explain the bonding between atoms. For example, to determine the stability of an atom, bond order is calculated. Bond order is calculated to determine the strength and length of the bond. The high value of the bond order means the bond is shorter, stronger, and have high bond strength.  

The Bond Order can be calculated by:

$\begin{aligned}\text{Bond order}&=\frac{1}{2}[\text{no. of}\; e^{-} \text{in bonding MO} ]-[\text{no. of}\; e^{-} \text{in antibonding MO}^{*} ] \end{aligned}$

In $\text F_{2} ^{2^{+}}$molecule, the number of valence electrons is 7.

$\text {2 \;atoms\; of\; F} &= 7*2=14 \;\text {electrons}$

Due to positive charge of 2, the number of valence electron is reduced to 12.  

Now, the MO of $\text F_{2} ^{2^{+}}$ is given below in the attachment. The bond order of

$\begin{aligned}\text{Bond order}&=\frac{1}{2}[\text{no. of}\; e^{-} \text{in bonding MO} ]-[\text{no. of}\; e^{-} \text{in antibonding MO}^{*} ] \end{aligned}\\\begin{aligned}\text{Bond order}&=\frac{1}{2}[\text {8}-\text{4}] \end{aligned}\\\begin{aligned}\text{Bond order}&= 2\end{aligned}$  

Thus,  $\text F_{2} ^{2^{+}}$has highest bond order and is the most stable element.  

Reasons for incorrect answers:

B. The element is  molecule, the number of valence electrons is 8.

$\text {2\;atoms\; of \;Ne &= 8*2\;&=16 electrons}$

Due to positive charge of 2, the number of valence electron is reduced to 14.  

Based on the MO diagram, the number of bonding and anti-bonding electrons are 8 and 6, respectively.  

So,

$\begin{aligned}\text{Bond order}&=\frac{1}{2}[\text{no. of}\; e^{-} \text{in bonding MO} ]-[\text{no. of}\; e^{-} \text{in antibonding MO}^{*} ] \end{aligned}\\\begin{aligned}\text{Bond order}&=\frac{1}{2}[\text {8}-\text{6}] \end{aligned}\\\begin{aligned}\text{Bond order}&= 1\end{aligned}$

Thus, the bond order of will be 1.  

C. Similarly, the element $\text {F}_{2}^{2{-}}$ have the number of valence electrons is 8.

$\text {2\;atoms\; of \;F&= 8*2\;&=16 electrons}$  

Due to negative charge of 2, the number of valence electron is increased to 18.  

Based on the MO diagram, the number of bonding and anti-bonding electrons are 8 and 8, respectively.  

So,

$\begin{aligned}\text{Bond order}&=\frac{1}{2}[\text{no. of}\; e^{-} \text{in bonding MO} ]-[\text{no. of}\; e^{-} \text{in antibonding MO}^{*} ] \end{aligned}\\\begin{aligned}\text{Bond order}&=\frac{1}{2}[\text {8}-\text{8}] \end{aligned}\\\begin{aligned}\text{Bond order}&= 0\end{aligned}$

Thus, the element will be highly unstable and will not exist.

D. Also the element $\text {O}_{2}^{2{-}}$ have 6 valence electrons.

$\text {2\;atoms\; of \;O &= 6*2\;&=12 electrons}$

Due to negative charge of 2, the number of valence electron is increased to 14.

Based on the MO diagram, the number of bonding and anti-bonding electrons are 8 and 6, respectively.  

So,

$\begin{aligned}\text{Bond order}&=\frac{1}{2}[\text{no. of}\; e^{-} \text{in bonding MO} ]-[\text{no. of}\; e^{-} \text{in antibonding MO}^{*} ] \end{aligned}\\\begin{aligned}\text{Bond order}&=\frac{1}{2}[\text {8}-\text{6}] \end{aligned}\\\begin{aligned}\text{Bond order}&= 1\end{aligned}$

Thus, the bond order of $\text {O}_{2}^{2{-}}$ is 1.

E. Also the element $\text {F}_{2}$ have 7 valence electrons.

$\text {2\;atoms\; of \;O &= 7*2\;&=14 electrons}$

Based on the MO diagram, the number of bonding and anti-bonding electrons are 8 and 6, respectively.  

So,

$\begin{aligned}\text{Bond order}&=\frac{1}{2}[\text{no. of}\; e^{-} \text{in bonding MO} ]-[\text{no. of}\; e^{-} \text{in antibonding MO}^{*} ] \end{aligned}\\\begin{aligned}\text{Bond order}&=\frac{1}{2}[\text {8}-\text{6}] \end{aligned}\\\begin{aligned}\text{Bond order}&= 1\end{aligned}$

Thus, the bond order of $\text {F}_{2}$ is 1.

For Further Reference:

https://brainly.com/question/11571911