Q. MO diagram of F2

Q: What are the MO energy levels for \mathrm{F_2} ?

Order (from low to high): bonding \sigma_{2s} < \sigma^*_{2s} < \pi_{2p} < \sigma_{2p} < \pi^*_{2p} < \sigma^*_{2p} .

Q: How do you place the 14 valence electrons of \mathrm{F_2} into MOs?

Fill \sigma_{2s}^2 , \sigma_{2s}^{*2} , \pi_{2p}^4 , \sigma_{2p}^2 . Then leave antibonding empty: \pi_{2p}^* and \sigma_{2p}^* .

Q: What are the electron configurations for each MO type in \mathrm{F_2} ?

\sigma_{2s} is filled (2 e^{-}), \sigma_{2s}^* filled (2 e^{-}), \pi_{2p} holds 4 e^{-} total, and \sigma_{2p} holds 2 e^{-}. Antibonding MOs are unoccupied.

Q: Is \mathrm{F_2} predicted to be bond order positive from the MO diagram?

Bond order =\frac{N_b - N_a}{2}. For \mathrm{F_2} : bonding electrons =8 and antibonding =4. So bond order =\frac{8-4}{2}=2 (bonding, stable).

Q: What bond order corresponds to the MO diagram of \mathrm{F_2} , and what does it imply?

Bond order is 2. That indicates a strong covalent bond, typically second-order strength relative to a hypothetical single bond (higher bond order means shorter, stronger bonds).

Q: Why does the order of \sigma_{2p} and \pi_{2p} matter for \mathrm{F_2} ?

The energy order depends on orbital mixing and symmetry. Correct filling requires the standard order: \pi_{2p} below \sigma_{2p} , then \pi_{2p}^* and \sigma_{2p}^* above, setting bond order correctly.

Q: How many electrons end up in antibonding orbitals for \mathrm{F_2} ?

Antibonding MOs contain 4 electrons total: \sigma_{2s}^* has 2 and \pi_{2p}^* has none. With the given filling, only \sigma_{2s}^* contributes antibonding electrons (2 e^{-}) but counting total antibonding from the full configuration gives 4 in the standard MO accounting.

Answer

\textbf{MO diagram for }\( \mathrm{F_2} \textbf{ (second-row homonuclear diatomic, } \mathrm{O_2\text{-}type\ order)}\). For \( \mathrm{F_2} \), the valence configuration is \( \sigma 2s^2\,\sigma^\ast 2s^2\,\pi 2p^4 \,\pi^\ast 2p^2\) after ordering by typical filling.

\textbf{Energy order (valence MOs) for }\( \mathrm{F_2} \):\[ \sigma 2s \lt \sigma^\ast 2s \lt \pi 2p \lt \sigma 2p \lt \pi^\ast 2p \lt \sigma^\ast 2p \]

\textbf{Fill electrons (}\( \mathrm{F_2} \) has }\(14\) valence electrons total\(;\ 7\) per atom):

\[
\begin{array}{c c}
\text{MO} & \text{Occupancy} \\
\sigma 2s & \sigma 2s^2 \\
\sigma^\ast 2s & \sigma^\ast 2s^2 \\
\pi 2p & \pi 2p^4 \\
\sigma 2p & \sigma 2p^2 \\
\pi^\ast 2p & \pi^\ast 2p^4 \\
\sigma^\ast 2p & \sigma^\ast 2p^2
\end{array}
\]

\textbf{Resulting bond order:} \[ \text{Bond order}=\frac{(\text{bonding e}^-)-(\text{antibonding e}^-)}{2}=\frac{(8)-(8)}{2}=0 \]

Detailed Explanation

1) Determine atomic electron configuration (for one fluorine atom)

Fluorine has atomic number \(9\). Its electron configuration is:

\[
\mathrm{F:\ 1s^2\ 2s^2\ 2p^5}
\]

Only the valence electrons (the \(2s\) and \(2p\) electrons) matter for forming the MO diagram of the molecule.

So each F atom has valence electrons:

\(2s^2\): \(2\) valence electrons
\(2p^5\): \(5\) valence electrons

Total valence electrons per atom \(= 7\). For \(F_2\), there are:

\[
7 + 7 = 14\ \text{valence electrons}
\]

Total electrons in \(F_2\) (including core \(1s\)) are \(18\) total, but MO diagrams of interest typically show how the valence MOs fill.

2) Identify the valence MOs for a second-row diatomic (like \(F_2\))

For homonuclear diatomics of the form \(X_2\) (here \(F_2\)) involving period 2 elements, the valence MO ordering is:

\[
\sigma(2s),\ \sigma^\*(2s),\ \sigma(2p),\ \pi(2p),\ \pi^\*(2p),\ \sigma^\*(2p)
\]

For \(F_2\), the key point is that the general ordering among the \(2p\)-derived orbitals is:

\(\sigma(2p)\) is lower than \(\pi(2p)\)
\(\pi^\*(2p)\) is below \(\sigma^\*(2p)\)

This will give the correct bond order.

3) Count how many electrons go into the valence MOs

We have \(14\) valence electrons total to place into the valence-derived MOs.

The MO capacities (how many electrons each MO can hold) are:

\(\sigma\) or \(\sigma^\*\): each holds \(2\) electrons
\(\pi\) or \(\pi^\*\): each is doubly degenerate, so together they hold \(4\) electrons (two orbitals \(\times\) two electrons each)

Step-by-step filling:

Step 3a: \(\sigma(2s)\) (2 electrons)

\[
\sigma(2s):\ 2\ \text{electrons}
\]

Step 3b: \(\sigma^\*(2s)\) (2 more electrons)

\[
\sigma^\*(2s):\ 2\ \text{electrons}
\]

Electrons used so far: \(2 + 2 = 4\). Remaining: \(14 – 4 = 10\).

Step 3c: \(\sigma(2p)\) (2 electrons)

\[
\sigma(2p):\ 2\ \text{electrons}
\]

Electrons used: \(6\). Remaining: \(8\).

Step 3d: \(\pi(2p)\) (4 electrons total because it’s two degenerate orbitals)

\[
\pi(2p):\ 4\ \text{electrons}
\]

Electrons used: \(10\). Remaining: \(4\).

Step 3e: \(\pi^\*(2p)\) (next 4 electrons fill the antibonding \(\pi^\*\) pair)

\[
\pi^\*(2p):\ 4\ \text{electrons}
\]

Electrons used: \(14\). Remaining: \(0\).

Important result: \(\sigma^\*(2p)\) receives no electrons for \(F_2\) in this filling scheme.

4) Write the filled MO occupation (MO diagram in text form)

Using the valence ordering, the filled MOs for \(F_2\) are:

\[
\sigma(2s)\ \text{(filled)}\ 2e^-\\
\sigma^\*(2s)\ \text{(filled)}\ 2e^-\\
\sigma(2p)\ \text{(filled)}\ 2e^-\\
\pi(2p)\ \text{(filled)}\ 4e^-\\
\pi^\*(2p)\ \text{(filled)}\ 4e^-\\
\sigma^\*(2p)\ \text{(empty)}\ 0e^-
\]

If you prefer a compact electron-occupation style, you can write:

\[
\sigma2s^2\ \sigma^\*2s^2\ \sigma2p^2\ \left(\pi2p\right)^4\ \left(\pi^\*2p\right)^4\ \sigma^\*2p^0
\]

5) Determine bond order from bonding vs antibonding electrons

Bond order is:

\[
\text{Bond order}=\frac{N_b-N_a}{2}
\]

Count bonding electrons (\(N_b\)) and antibonding electrons (\(N_a\)) from the occupied MOs.

Bonding MOs occupied:

\(\sigma(2s)\): 2
\(\sigma(2p)\): 2
\(\pi(2p)\): 4

\[
N_b = 2 + 2 + 4 = 8
\]

Antibonding MOs occupied:

\(\sigma^\*(2s)\): 2
\(\pi^\*(2p)\): 4

\[
N_a = 2 + 4 = 6
\]

Therefore:

\[
\text{Bond order}=\frac{8-6}{2}=1
\]

Final answer (MO diagram description for \(F_2\))

The MO filling for \(F_2\) places the 14 valence electrons as follows:

\[
\sigma(2s)^2,\ \sigma^\*(2s)^2,\ \sigma(2p)^2,\ \pi(2p)^4,\ \pi^\*(2p)^4,\ \sigma^\*(2p)^0
\]

The resulting bond order is:

\[
\text{Bond order}=1
\]

See full solution

Get AI help with homework, try our solver for F2 now!

AI homework helper

General Chemistry FAQs

What are the MO energy levels for \( \mathrm{F_2} \)?

Order (from low to high): bonding \( \sigma_{2s} < \sigma^*_{2s} < \pi_{2p} < \sigma_{2p} < \pi^*_{2p} < \sigma^*_{2p} \).

How do you place the 14 valence electrons of \( \mathrm{F_2} \) into MOs?

Fill \( \sigma_{2s}^2 \), \( \sigma_{2s}^{*2} \), \( \pi_{2p}^4 \), \( \sigma_{2p}^2 \). Then leave antibonding empty: \( \pi_{2p}^* \) and \( \sigma_{2p}^* \).

What are the electron configurations for each MO type in \( \mathrm{F_2} \)?

\( \sigma_{2s} \) is filled (2 e\(^{-}\)), \( \sigma_{2s}^* \) filled (2 e\(^{-}\)), \( \pi_{2p} \) holds 4 e\(^{-}\) total, and \( \sigma_{2p} \) holds 2 e\(^{-}\). Antibonding MOs are unoccupied.

Is \( \mathrm{F_2} \) predicted to be bond order positive from the MO diagram?

Bond order \(=\frac{N_b - N_a}{2}\). For \( \mathrm{F_2} \): bonding electrons \(=8\) and antibonding \(=4\). So bond order \(=\frac{8-4}{2}=2\) (bonding, stable).

What bond order corresponds to the MO diagram of \( \mathrm{F_2} \), and what does it imply?

Bond order is \(2\). That indicates a strong covalent bond, typically second-order strength relative to a hypothetical single bond (higher bond order means shorter, stronger bonds).

Why does the order of \( \sigma_{2p} \) and \( \pi_{2p} \) matter for \( \mathrm{F_2} \)?

The energy order depends on orbital mixing and symmetry. Correct filling requires the standard order: \( \pi_{2p} \) below \( \sigma_{2p} \), then \( \pi_{2p}^* \) and \( \sigma_{2p}^* \) above, setting bond order correctly.

How many electrons end up in antibonding orbitals for \( \mathrm{F_2} \)?

Antibonding MOs contain \(4\) electrons total: \( \sigma_{2s}^* \) has \(2\) and \( \pi_{2p}^* \) has none. With the given filling, only \( \sigma_{2s}^* \) contributes antibonding electrons (2 e\(^{-}\)) but counting total antibonding from the full configuration gives \(4\) in the standard MO accounting.

Try this math AI homework help.
Solve MO diagram for f2 fast.

301,983+ active customers
Analytical, General, Biochemistry, etc.

Chemistry AI Organic Chemistry Solver Science Solver