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Chapter 11: Problem 94

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

Expert verified
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

01

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.
02

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.
03

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.

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Most popular questions from this chapter

Write Lewis structures for the following molecules, and then label each \(\sigma\) and \(\pi\) bond. (a) \(\mathrm{HCN} ;\) (b) \(\mathrm{C}_{2} \mathrm{N}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CHCHCCl}_{3} ;\) (d) HONO.

Delocalized molecular orbitals are found in (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{HS}^{-} ;\) (c) \(\mathrm{CH}_{4} ;\) (d) \(\mathrm{CO}_{3}^{2-}\) .

For each of the species \(\mathrm{C}_{2}^{+}, \mathrm{O}_{2}^{-}, \mathrm{F}_{2}^{+},\) and \(\mathrm{NO}^{+}\) (a) Write the molecular orbital diagram (as in Example \(11-6)\) (b) Determine the bond order, and state whether you expect the species to be stable or unstable. (c) Determine if the species is diamagnetic or paramagnetic; and if paramagnetic, indicate the number of unpaired electrons.

Explain why it is necessary to hybridize atomic orbitals when applying the valence-bond method that is, why are there so few molecules that can be described by the overlap of pure atomic orbitals only?

In which of the following molecules would you expect to find delocalized molecular orbitals: (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{SO}_{2} ;\) (c) \(\mathrm{H}_{2} \mathrm{CO}\) ? Explain.
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