2.3b: MO theory of bonding in H₂⁺

Last updated
Save as PDF

Page ID: 2563

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

MO for H2.jpg — Figure 3: Schematic represenation of antibonding molecular orbital σ*(1s)

Note that there is a nodal plane in the anti-bonding MO.

Bond order

Bond order = 1/2 (#e- in bonding MO - #e- in antibonding MO)

For H₂, bond order = 1/2 (2-0) = 1, which means H₂has only one bond. The antibonding orbital is empty. Thus, H₂ is a stable molecule.

Again, in the MO, there is no unpaired electron, so H₂ is diamagnetic.

References

Chang, Raymond. Physical Chemistry for the Biosciences. Sausalito, CA: University Science Books, 2005. 458 - 460.
Housecroft, Catherine E. and Alan G. Sharpe. Inorganic Chemistry. 3rd ed. England: Pearson - Prentice Hall, 2008. 33 - 36.

Problems

What does the MO of H₂⁺ look like? What is its bond order? What is its magnetic property? Explain.
What does the MO of H₂^- look like? What is its bond order? What is its magnetic property? Explain.
Which one is the most stable: H₂, H₂⁺, or H₂^-? Why?
When a hydrogen atom accepts an electron, it becomes a hydride H^-. Theoretically would it be possible to form a molecule from two hydrides, that is to form H₂²^-? Why?

Answers:

MO for H2+.jpg

Bond order = 1/2 (1-0) = 1/2

Paramagnetic because it has one unpaired e- in the σ(1s) orbital.

MO for H2-.jpg

Bond order = 1/2 (2-1) = 1/2

Paramagnetic because it has one unpaired e- in the σ*(1s) orbital.

3- H₂ is the most stable because it has the highest bond order (1), in comparison with the bond orders (1/2) of H₂⁺ and H₂^-.

4- Theoretically it would not be possible to form a molecule from two hydrides because the anti-bonding and bonding orbitals would cancel each other out. So, the bond order is zero. Because the antibonding ortibal is filled, it destabilizes the structure, making the "molecule" H₂^2- very non-stable.

MO for H2 2-.jpg

Bond order = 1/2 (2-2) = 0 ---> no bond formation. Thus, this molecule doesn't exist.