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Chapter 10: Problem 55

Explain why the bond order of \(\mathrm{N}_{2}\) is greater than that of \(\mathrm{N}_{2}^{+},\) but the bond order of \(\mathrm{O}_{2}\) is less than that of \(\mathrm{O}_{2}^{+}\)

Short Answer

Expert verified
The bond order of N2 is greater than that of N2+ because the removal of an electron in N2+ usually happens from the antibonding orbital and results in a lower bond order. In the case of O2, the bond order is less than that of O2+ because the removal of an electron in O2+ leads to an increase in bond order, meaning the bond in O2+ is more stable than O2.

Step by step solution

01

Understand bond order

Bond order gives information about the stability of a bond. It is calculated as half the difference of the number of electrons in bonding and antibonding orbitals. A larger bond order implies a more stable bond.
02

Analyze the molecular orbital diagram for N2 and N2+

The molecular orbital diagram of nitrogen (N2) considers atomic nitrogen's electron configuration 1s2 2s2 2p3. In N2, there are 10 electrons in bonding orbitals and 4 in antibonding orbitals, leading to a bond order of \(\frac{(10-4)}{2}=3\). For N2+, one electron is lost usually from an antibonding orbital, leaving 10 bonding electrons and 3 antibonding electrons, thus the bond order is \(\frac{(10-3)}{2}=3.5\).
03

Analyze the molecular orbital diagram for O2 and O2+

For oxygen (O2), the electron configuration is 1s2 2s2 2p4 and thus has 12 electrons in bonding orbitals and 8 in antibonding orbitals, yielding a bond order of \(\frac{(12-8)}{2}=2\). For O2+ ion, it will lose an electron, typically from an antibonding orbital, so we have 12 bonding electrons and 7 antibonding electrons. Therefore, the bond order for O2+ ion becomes \(\frac{(12-7)}{2}=2.5\) which means the bond in O2+ is more stable than in O2.

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