Question

Use the molecular orbital energy diagram below to answer the questions about bond order for the positive ion B2* Number of Bonding Number of Antibonding B2 Valence Electrons Valence Electrons Bond Ord

Answer 1

The molecular orbital energy diagram indicates that the positive ion B⁺₂ has 2 bonding electrons and 1 antibonding electron, resulting in a bond order of 0.5.

Step-by-step explanation:

In the molecular orbital energy diagram for

2

+

B

2

+

​

, we observe that there are a total of 3 electrons (since

�

2

+

B

2

+

​

is a positive ion). The molecular orbital diagram shows the distribution of these electrons among the bonding and antibonding orbitals. In this case, there are 2 electrons in the bonding molecular orbitals and 1 electron in the antibonding molecular orbital. The bond order is calculated as the difference between the number of bonding and antibonding electrons divided by 2.

Now, let's break down the calculation:

Bond Order

=

Number of Bonding Electrons

−

Number of Antibonding Electrons

2

Bond Order=

2

Number of Bonding Electrons−Number of Antibonding Electrons

​

Substituting the values:

Bond Order

=

2

−

1

2

=

1

2

Bond Order=

2

2−1

​

=

2

1

​

So, the bond order for

�

2

+

B

2

+

​

is 0.5, indicating a weak bond. The presence of more bonding electrons than antibonding electrons contributes to a stable molecular structure, while excess antibonding electrons would decrease the stability of the molecule. In this case, the positive ion configuration results in a bond order of 0.5, suggesting a relatively weaker bond in comparison to neutral molecules.

Answer 2

The bond order for the positive ion
`\(B_2^*\)` is 0.5. This is derived from subtracting the number of antibonding electrons from the number of bonding electrons and dividing the result by 2.

Step-by-step explanation:

The molecular orbital energy diagram provides a visual representation of the distribution of electrons in molecular orbitals for the
`\(B_2^*\)` positive ion.

In this context, bonding molecular orbitals (lower energy) are filled first, followed by antibonding molecular orbitals (higher energy). The bond order
`(\(BO\))` is calculated using the formula
`\(BO = (1)/(2)(N_b - N_(ab))\)`, where
`\(N_b\)` is the number of bonding electrons and
`\(N_(ab)\)` is the number of antibonding electrons.

For
`\(B_2^*\)`, the molecular orbital diagram shows two electrons in bonding orbitals and four electrons in antibonding orbitals. Plugging these values into the formula yields
`\(BO = (1)/(2)(2 - 4) = 0.5\).`

The positive bond order of 0.5 suggests a stable configuration, as it indicates a net bonding interaction between the two boron atoms in
`\(B_2^*\).` This positive bond order signifies that there is a partial sharing of electrons, resulting in a stable molecular ion.

Understanding the bond order provides insights into the strength and stability of the bond between the boron atoms, influencing the overall reactivity and behavior of the \(B_2^*\) positive ion in chemical reactions.