Question

The cyclopentadienide ion has the formula \(\mathrm{C}_{5} \mathrm{H}_{5}^{-}\). The ion consists of a regular pentagon of \(C\) atoms, each bonded to two 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 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?

Short answer

The cyclopentadienide ion (C5H5-) has 5 equivalent resonance structures with delocalized electrons around the carbon ring, which can be represented by a pentagon enclosing a circle. Though our Lewis structure in part (a) indicated different hybridizations for the carbon atoms, chemists commonly consider all carbon atoms to have sp² hybridization. The nonbonding pair of electrons resides within an unhybridized p orbital of the carbon atom. Both benzene and the cyclopentadienide ion contain 6 delocalized π electrons, forming a stable molecule or ion.

Step-by-step solution

(a) Draw a Lewis structure

To draw the Lewis structure for the cyclopentadienide ion (C5H5-), we first place the C atoms in a pentagonal shape, each having a hydrogen atom bonded to it. Each carbon atom has 4 valence electrons, and each hydrogen atom has 1 valence electron. The negatively charged electron gained by the ion is also considered a valence electron. 1. Place the C atoms in a regular pentagonal shape. 2. Connect each C atom to its neighboring C atoms with a single bond. 3. Attach a hydrogen atom to each C atom with a single bond (C-H bond). 4. Now consider the gained electron in the ion, as a non-bonded electron pair. 5. Assign this non-bonded electron pair to one of the carbon atoms. 6. Complete the structure by fulfilling the octet rule for the remaining 4 carbon atoms with a double bond between each adjacent pair. 7. Add brackets and a negative charge to show the cyclopentadienide ion: \[\begin{array}{c|c|c} & \overset{-1}{\begin{array}{c|c|c|c} & \overset{\ominus}{\mathrm{C}} & \phantom{\rule{6pt}{0ex}} \overset{\phantom{0}}{\mathrm{H}} & \overset{\phantom{0}}{\mathrm{C}} \\ \phantom{\rule{6pt}{0ex}} & |\hspace{-1.5mm}=. & \phantom{\rule{6pt}{0ex}} \\ & \overset{\phantom{0}}{\mathrm{C}} & \d\underset{\phantom{0}}{\mathrm{H}} \\ & \phantom{\rule{6pt}{0ex}} & \\ \end{array}} & \\ \end{array}\] According to this Lewis structure, all five C atoms have different hybridization. The carbon atom with a non-bonded electron pair will have sp hybridization (two σ bonds and one unpaired electron), and the other four carbon atoms will have sp² hybridization (three σ bonds).

(b) Compatibility with sp² hybridization

Chemists generally view this ion as having sp² hybridization at each C atom. However, our Lewis structure in part (a) indicates that only four of the carbon atoms have sp² hybridization, while one has sp hybridization. Hence, our answer in part (a) is not entirely consistent with the usual view of having sp² hybridization for all carbon atoms.

(c) Orbital for the nonbonding pair

If we assume sp² hybridization for all carbon atoms as suggested in part (b), there would be an unhybridized p orbital remaining for each carbon atom. The nonbonding pair of electrons must reside within this unhybridized p orbital of the carbon atom.

(d) Resonance structures

Our Lewis structure in part (a) can have resonance structures. To identify the equivalent resonance structures, we shift the position of the double bonds and the nonbonding pair around the pentagonal carbon ring. There are a total of 5 equivalent resonance structures.

(e) Representation consistency

The cyclopentadienide ion is often drawn as a pentagon enclosing a circle. This representation implies the delocalization of the electrons around the carbon ring, which is consistent with our answer of having 5 equivalent resonance structures in part (d).

(f) Six π electrons

Both benzene and the cyclopentadienide ion are often described as systems containing six π electrons. In the context of these molecules, this description means that there are 6 electrons shared between the carbon atoms that are delocalized over the entire carbon ring. These delocalized electrons create a stable molecule or ion in both cases, making them particularly stable compounds. The π electrons are found in the overlapping p orbitals of the carbon atoms.

Key concepts

Resonance Structures

Imagine electrons flowing around a structure like dancers on a stage. In the context of Lewis structures, resonance structures depict different dancing arrangements of these electrons without altering the positions of the atoms. For the cyclopentadienide ion, each resonance structure is like a snapshot of where electrons might be at any given moment. Although each picture looks different, they all represent the same molecule.

• Resonance structures help depict the delocalized distribution of electrons across the molecule.
• There are 5 equivalent resonance structures possible for the cyclopentadienide ion, reflecting the movement of electrons along the pentagon's bonds.
• These structures are not real or separate versions of a molecule but a representation of the electron delocalization.

The concept of resonance assures that the actual state of the molecule is more stable than any single resonance form suggests. Picture a hazy cloud of shared electrons, rather than fixed positions, which allows the structure more energy stability.

Hybridization

In organic chemistry, hybridization explains the shape and bonding capabilities of atoms within a molecule. Each carbon atom in the cyclopentadienide ion is often described as having \( ext{{sp}}^2\) hybridization. This is due to the three sigma bonds it forms: one with a hydrogen atom and two with its neighboring carbons.

• Hybridization involves mixing atomic orbitals to create new hybrid orbitals.
• In \( ext{{sp}}^2\) hybridization, one \(s\) and two \(p\) orbitals combine, forming three equal hybrid orbitals oriented at 120° to each other.
• These \( ext{{sp}}^2\) orbitals are important for planar structures and explain the stability offered by resonance.

However, the initial solution presented a carbon atom with \( ext{{sp}}\) hybridization due to the nonbonding electron pair. Generally, in chemical descriptions, adjusting each carbon to assume \( ext{{sp}}^2\) hybridization aligns better with expected outcomes such as aromaticity and stability from delocalized electrons.

Delocalization

Delocalization is like spreading butter evenly on bread; it distributes electrons across multiple atoms or bonds, rather than confining them to a single bond position. In the cyclopentadienide ion, electron delocalization occurs among the carbon atoms forming a conjugated system. These delocalized electrons lead to enhanced stability.

• Delocalization contributes to the aromatic nature of the ion, resembling characteristics seen in benzene.
• This concept is symbolized by an inscribed circle in the pentagon, suggesting that certain electrons do not belong to one carbon but flow freely around the ring.
• Delocalized electrons result in a lower energy state for the entire ion, explaining its depiction as an 'aromatic ring'.

Understanding delocalization helps in predicting reactivity patterns, stability, and other chemical properties tied to aromatic systems.

Nonbonding Electron Pair

Nonbonding electron pairs are often called "lone pairs" since they don't engage in bonding. These pairs significantly affect molecular behavior. In the case of the cyclopentadienide ion, the lone pair plays a critical role in the structure's resonance and hybridization.

• These electrons reside in an atomic orbital that does not take part in forming sigma or pi bonds.
• In \( ext{{sp}}^2\) hybridized carbons in the cyclopentadienide ion, a lone pair can occupy a p orbital, participating in the aromatic electron system.
• The presence of lone pairs needs careful attention as they influence the geometry and can also determine whether the molecule can serve as a Lewis base or nucleophile.

By including lone pairs correctly in Lewis structures, chemists can better predict molecular geometry and electron density distribution, which further affects physical properties and reactivity.