The cyclopentadienide ion has the formula C5 H5^-. The ion consists of a regular pentagon of C a

step 1 In this problem, we're asked to consider the cyclopenta dionide ion. This ion has the formula of

The cyclopentadienide ion has the formula C5 H5^-. The ion...

The cyclopentadienide ion has the formula $\mathrm{C}_{5} \mathrm{H}_{5}^{-}$. The ion consists of a regular pentagon of $\mathrm{C}$ atoms, each bonded to two $\mathrm{C}$ neighbors, with a hydrogen atom bonded to each $\mathrm{C}$ atom. All the atoms lie in the same plane. (a) Draw a Lewis structure for the ion. According to your structure, do all five $\mathrm{C}$ atoms have the same hybridization? Explain. (b) Chemists generally view this ion as having $s p^{2}$ hybridization at each $C$ atom. Is that view consistent with your answer to part (a)? (c) Your Lewis structure should show one nonbonding pair of electrons. Under the assumption of part (b), in what type of orbital must this nonbonding pair reside? (d) Are there resonance structures equivalent to the Lewis structure you drew in part (a)? If so, how many? (e) The ion is often drawn as a pentagon enclosing a circle. Is this representation consistent with your answer to part (d)? Explain. (f) Both benzene and the cyclopentadienide ion are often described as systems containing six $\pi$ electrons. What do you think is meant by this description?

The cyclopentadienide ion has the formula $\mathrm{C}_{5} \mathrm{H}_{5}^{-}$. The ion consists of a regular pentagon of $\mathrm{C}$ atoms, each bonded to two $\mathrm{C}$ neighbors, with a hydrogen atom bonded to each $\mathrm{C}$ atom. All the atoms lie in the same plane.
(a) Draw a Lewis structure for the ion. According to your structure, do all five $\mathrm{C}$ atoms have the same hybridization? Explain. (b) Chemists generally view this ion as having $s p^{2}$ hybridization at each $C$ atom. Is that view consistent with your answer to part (a)? (c) Your Lewis structure should show one nonbonding pair of electrons. Under the assumption of part (b), in what type of orbital must this nonbonding pair reside?
(d) Are there resonance structures equivalent to the Lewis structure you drew in part (a)? If so, how many? (e) The ion is often drawn as a pentagon enclosing a circle. Is this representation consistent with your answer to part (d)? Explain.
(f) Both benzene and the cyclopentadienide ion are often described as systems containing six $\pi$ electrons. What do you think is meant by this description?

Key Concepts

Constructing Lewis Structures

Lewis structures are diagrams that show the bonding between atoms and the placement of nonbonding electrons. They are used to represent the electron arrangement in molecules and ions, indicating covalent bonds, lone pairs, and formal charges. This method is fundamental for predicting molecular geometry and reactivity.

Hybridization

Hybridization is the concept where atomic orbitals mix to form new hybrid orbitals that are suited for the pairing of electrons to form chemical bonds. In a planar molecule, sp2 hybridization is common, where three sp2 orbitals lie in a plane and one unhybridized p orbital is available for pi bonding. Understanding hybridization is crucial for predicting the shape and bonding properties of molecules.

Molecular Planarity and Delocalized Pi Systems

Planarity in a molecule allows the unhybridized p orbitals to align parallel to each other, facilitating the overlap necessary for pi bonding. This overlap leads to delocalized pi systems where electrons are spread out over several atoms rather than localized between just two, which is key for the stability and reactivity of conjugated and aromatic systems.

Resonance Structures

Resonance structures are different ways of drawing the same molecule or ion that highlight the delocalization of electrons within the structure. These multiple contributing forms represent a hybrid where the actual electron distribution is an average of the forms depicted. This concept helps explain uniform bond lengths and enhanced stability in conjugated systems.

Aromaticity

Aromaticity is a property of cyclic, planar molecules with a conjugated pi electron system that follows Huckel's rule (typically having 4n + 2 pi electrons, where n is an integer). Aromatic compounds exhibit extra stability due to the delocalized nature of their pi electrons, which creates a significant stabilization energy and affects the physical and chemical properties of the molecule.

Transcript

00:01 In this problem, we're asked to consider the cyclopenta dionide ion.

00:06 This ion has the formula of c5h5 with a negative 1 charge.

00:11 We're told that the ion consists of a regular pentagon of c ions, or excuse me, c atoms.

00:19 So i'm going to start with this.

00:35 Regular pentagon of c atoms.

00:38 All atoms are in the same plane.

00:44 A hydrogen atom is bonded to each c.

00:58 There we go.

00:59 So i'm going to draw one of our resonance structures, and i'll draw this one.

01:06 And notice we have an unshared pair of electrons.

01:09 And i'm going to make sure to enclose this and give a negative one charge.

01:15 So a asks us to draw the lewis structure, is the lewis structure, which we just did.

01:29 And i see four atoms on here that appear four are easily identified as sp2 hybridization.

01:46 The atom with a single unshared pair of electrons, the fifth with the unshared pair we'll talk about later.

02:03 It looks like now that it has an sp3, but that's not going to work because we were told it's planer.

02:13 The problem specified it's plainer.

02:18 And you can't have an sp3 if it's plainer.

02:21 Okay, for b.

02:25 I already answered it, i think.

02:27 B said chemistry generally views.

02:30 This ion is having sp2 on each carbon.

02:36 Is that consistent? and i've explained that right there.

02:42 For c, you better switch colors.

02:46 Our lewis structure should show one unbonded pair, and it does.

02:58 What type of orbital must this bonding pair reside in? so based on our assumptions in b, it cannot be an sp3.

03:10 So we have to have an unhybridized.

03:13 It's going to be in the unhybridized 2p orbital.

03:25 Remember, sp2 means one unhybridized p exists.

03:41 So this will be in the unhybridized 2p orbital.

03:48 For d -assus, are there hybridized, let me change colors.

03:56 Are there resonance structures? and the answer, of course, is yes...

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