Question

Of the following, the species with a bond order of 1 is (a) \(\mathrm{H}_{2}^{+} ;\) (b) \(\mathrm{Li}_{2} ;\) (c) \(\mathrm{He}_{2} ;\) (d) \(\mathrm{H}_{2}^{-}\)

Short answer

None of the molecules \(\mathrm{H}_{2}^{+}\), \(\mathrm{Li}_{2}\), \(\mathrm{He}_{2}\), \(\mathrm{H}_{2}^{-}\) has a bond order of 1.

Step-by-step solution

Introduction to bond order

In Molecular Orbital Theory, the bond order is calculated as half the difference between the number of electrons in bonding and antibonding molecular orbitals. The greater the bond order, the more stable the chemical species.

Determine bond orders

(a) For \(\mathrm{H}_{2}^{+}\), it has one electron in its bonding orbital and none in the antibonding, resulting in a bond order of 1/2. (b) For \(\mathrm{Li}_{2}\), it has two electrons each in bonding and antibonding orbitals, rendering a bond order of 0. (c) \(\mathrm{He}_{2}\) has two electrons each in bonding and antibonding orbitals as well, so its bond order is also 0. (d) Finally, \(\mathrm{H}_{2}^{-}\) has two electrons in the bonding orbital and one in the antibonding orbital, giving it a bond order of 1/2.

Identify the species with bond order of 1

From the computed bond orders, none of the species \(\mathrm{H}_{2}^{+}\), \(\mathrm{Li}_{2}\), \(\mathrm{He}_{2}\), \(\mathrm{H}_{2}^{-}\) is found to have a bond order of exactly 1, as calculated. Thus, none of the options is correct in this case.

Key concepts

Bond Order

The concept of bond order is essential in understanding molecular stability within Molecular Orbital Theory (MOT). Bond order is computed as half the difference between the number of electrons in bonding orbitals and antibonding orbitals. Mathematically, it can be expressed as: \[\text{Bond Order} = \frac{1}{2} \times (\text{Number of electrons in bonding orbitals} - \text{Number of electrons in antibonding orbitals})\] A higher bond order indicates more bonded pairs of electrons, which generally leads to a more stable molecule. Conversely, a bond order of zero suggests no net bond formation, implying the molecule does not exist or is very unstable. When considering molecular species such as \(\mathrm{H}_{2}^{+}\) or \(\mathrm{H}_{2}^{-}\), calculating the bond order helps predict molecular behavior, stability, and reactivity.

Bonding Orbitals

In Molecular Orbital Theory, bonding orbitals play a crucial role by holding the electrons that contribute to the bond formation between atoms. These orbitals form when atomic orbitals combine constructively.

• These orbitals have lower energy than the original atomic orbitals, making them attractive for electrons seeking a stable state.

Electrons present in bonding orbitals decrease the overall energy of the molecule, thus making it more stable. Examples include the \(\sigma\) bonding orbital in hydrogen molecules, which holds electrons that contribute to the H-H bond. Calculating the filling of these orbitals gives insights into the bond order and thus the potential strength and length of a bond.

Antibonding Orbitals

Antibonding orbitals are formed alongside bonding orbitals, but unlike bonding orbitals, they increase a molecule's energy when populated by electrons. These orbitals are a result of the destructive interference of atomic orbitals.

• Identified by a star symbol (\(^*\)) as in \(\sigma^*\), they signify the opposite phase combination compared to bonding orbitals.

Presence of electrons in antibonding orbitals diminishes the bond order because they counteract the stabilizing effect of bonding electrons. They can make a molecule less stable and more reactive. Calculating how many electrons occupy these antibonding levels is crucial in predicting the molecule's overall stability and reactivity.

Chemical Stability

Understanding chemical stability through Molecular Orbital Theory involves analyzing bond order, bonding, and antibonding orbitals. A molecule's stability is significantly influenced by the bond order; higher bond orders mean more bonded electron pairs, contributing to greater overall stability.

• The presence of electrons in antibonding orbitals usually reduces stability.
• For a species like \(\mathrm{Li}_{2}\), a bond order of zero implies a lack of stability.

Thus, finding the balance between the population of electrons in bonding versus antibonding orbitals is key in assessing molecular stability and predictability. A molecule with a high bond order, such as \(\mathrm{N}_{2}\) with a bond order of 3, is incredibly stable, which reflects in its kinetic inertness and strong triple bond.