Problem 112 - Chemistry

Question

The dichromate ion, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), has neither \(\mathrm{Cr}-\mathrm{Cr}\) nor \(\mathrm{O}-\mathrm{O}\) bonds. (a) Taking both \(4 s\) and \(3 d\) electrons into account, draw an electron-dot structure that minimizes the formal charges on the atoms. (b) How many outer-shell electrons does each \(\mathrm{Cr}\) atom have in your electron-dot structure?

Step-by-Step Solution

Answer

Each Cr atom has 6 outer-shell electrons in the electron-dot structure.

Step 1: Understand the dichromate ion

The dichromate ion (\[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\]) consists of two chromium (Cr) atoms and seven oxygen (O) atoms, with an overall 2- charge (\[2-\]). It is important to ensure that the structure avoids any Cr-Cr or O-O direct bonds.

Step 2: Arrange the atoms

In the common structure of (\[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\]), the two chromium atoms are typically bonded to each other through oxygen bridges, and each chromium atom forms bonds with oxygen atoms.

Step 3: Distribute electrons

Each chromium atom has six electrons total from its "4s" and "3d" orbitals available for bonding. Oxygen typically requires two electrons to complete its valence shell when bonded. The overall charge of 2- must be considered in the distribution of electrons.

Step 4: Draw Lewis structure

1. Connect two Cr atoms with a bridging pair of oxygen atoms. Connect each Cr atom to three additional oxygen atoms. 2. Ensure no direct Cr-Cr or O-O bonds exist. 3. Distribute outer electrons to fulfill the octet rule, giving 12 shared electron pairs in the structure, including double bonds for each terminal oxygen, and single bonds for bridging ones.

Step 5: Calculate formal charges

Minimize the formal charge by adjusting bonds: 1. Assign double bonds where necessary to neutralize oxygen’s formal charges, preferably on the terminal oxygens.2. The electron dots should be arranged such that formal charges are close to zero, but the overall structure respects the ion’s charge (\[2-\]). This involves having two double-bonded oxygens connected to each Cr.

Step 6: Find outer-shell electrons for Cr

In this accurate depiction, each (\[\mathrm{Cr}\]) shares three double-bonded oxygens, effectively contributing 6 electrons from bonding and its hybridization ensures 6 valence outer-shell electrons.

Key Concepts

Chromium Oxidation States

Understanding the oxidation state of chromium is vital when determining the structure and behavior of the dichromate ion (\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\)). In a dichromate ion, each chromium atom typically has an oxidation state of +6. This is higher than some of the other oxidation states that chromium can exhibit, such as +2 or +3 found in different compounds. Here, the +6 oxidation state is necessary to account for the overall charge of the ion.

Consider how these high oxidation states influence bonding: Chromium must utilize electrons from both its "4s" and "3d" orbitals. Each Cr atom gives up six electrons to bond with oxygen. It's fascinating as this demonstrates chromium's ability to wield multiple oxidation states based on its complex electron arrangement. Higher oxidation states, like +6, indicate chromium's loss of electrons, and thus its involvement in forming more stable compounds through greater covalent bond formation.

Recognizing chromium's oxidation state helps predict its bonding behavior and determine how electrons are distributed between chromium and oxygen, minimizing the structure’s overall energy.

Formal Charge Calculation

Minimizing formal charge in a molecule helps achieve a stable and realistic electron-dot structure. When calculating formal charges, it's essential to compare the number of valence electrons in an isolated atom to its electron count in the molecule.

Here's a step-by-step breakdown:

• For each chromium atom, count the electrons it "owns"—these include half of the bonding electrons in its shared bonds with oxygen.
• Assign formal charges to oxygen by considering lone pairs and shared electrons. Terminal oxygens, which directly bond to the chromium atoms, generally have double bonds, minimizing their formal charge, ideally to zero.
• Calculate: the formal charge \(\mathrm{FC} = V - (N + \frac{B}{2})\), where \(V\) is the number of valence electrons, \(N\) is the number of non-bonded electrons, and \(B\) is the number of bonding electrons.

Formal charge should be calculated to ensure that the overall charge of the dichromate ion is \(-2\). Balancing the charges enables the electrons to be most efficiently shared between chromium and oxygen, and guides further adjustments on the electron-dot structure to refine it.

Electron-dot Structures

Devising an electron-dot structure requires comprehension of each element's valence electrons and bonding needs. For the dichromate ion \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), the electron-dot structure is key to visualizing the molecule's geometric arrangement.

Here's how this is typically constructed:

• The chromium atoms are linked by "bridging" oxygen atoms, meaning each Cr is bonded through an intermediary.
• Each chromium bonds with three "terminal" oxygen atoms, forming \(\mathrm{Cr-O} \) double bonds to satisfy oxygen’s octet requirement.
• The remaining oxygens are part of "single" bonds with chromium, but they help bridge the two Cr atoms together.

This structure helps in depicting a full picture of shared and lone electron pairs. Maintaining balance and symmetry in the electron-dot structure aids in keeping formal charges low and reflects the actual chemistry factually. As a result, accurately constructed electron-dot structures are integral in predicting the physical and chemical properties of molecules like the dichromate ion.