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The ration of bond orders in \(\mathrm{CO}_{3}^{2-}\) ion and \(\mathrm{SO}_{3}^{2-}\) ion is ___ .

Short Answer

Expert verified
The ratio of bond orders is 1:1.

Step by step solution

01

Understanding the bond order

Bond order is defined as the number of chemical bonds between a pair of atoms. For example, in diatomic nitrogen N₂, which has a triple bond, the bond order is 3. In general, bond order can be calculated using the formula: \( \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of anti-bonding electrons}}{2} \). However, for polyatomic ions, this concept is typically simplified to the average number of bonds between atoms in a species.
02

Calculating bond order for CO₃²⁻

In the carbonate ion \(\text{CO}_3^{2-}\), there are three resonance structures. Each structure consists of one double bond and two single bonds with oxygen atoms. To find the average bond order: bond orders from the resonance structures can be considered and divided by the three bond locations: \[ \text{Average Bond Order for CO}_3^{2-} = \frac{4 \text{ bonds}}{3} = \frac{4}{3} \]
03

Calculating bond order for SO₃²⁻

The sulfate ion \(\text{SO}_3^{2-}\) also has resonance structures similar to carbonate. It consists of one double bond and two single bonds. Analyzing its resonance: \[ \text{Average Bond Order for SO}_3^{2-} = \frac{4 \text{ bonds}}{3} = \frac{4}{3} \]
04

Determining the ratio of bond orders

After calculating the bond orders for both \(\text{CO}_3^{2-}\) and \(\text{SO}_3^{2-}\), we find they are both \( \frac{4}{3} \). Thus, the ratio of bond orders is simply: \[ \text{Ratio} = 1:1 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carbonate Ion
A carbonate ion, symbolized as \( \text{CO}_3^{2-} \), is a polyatomic ion with a charge of -2. It consists of one carbon atom centrally bonded to three oxygen atoms. The structure of this ion is a trigonal planar shape, which means all atoms lie on the same plane, forming a flat triangle with the carbon atom at the center.

Understanding the bonding in the carbonate ion requires looking at its resonance structures. In truly simplistic terms, resonance structures are different ways to represent molecule structures using electron dot diagrams. For carbonate, no single structure adequately describes the electron arrangement, so chemists use resonance to show that the actual electron distribution is a blend of the possible configurations.
  • The carbonate ion has three equivalent resonance structures.
  • In each resonance structure, one oxygen-carbon bond shows a double bond while the other two are single bonds.
  • Collectively, these structures suggest that the bonds are of equal length, shorter than a single bond but longer than a typical double bond.
Through this resonance phenomenon, we understand that the carbonate ion is stabilized by this equal distribution of bonding over the three oxygen atoms.
Sulfate Ion
The sulfate ion, \( \text{SO}_3^{2-} \), shares a similar conception with carbonate ions in terms of resonance and structure. However, its geometry is a bit different due to the central atom being sulfur instead of carbon.

Like the carbonate ion, the sulfate ion exhibits resonance. Resonance helps explain the structure of compounds that cannot be represented by a single Lewis structure. The sulfate ion can swap places of double and single bonds in its numerous resonance forms, helping to stabilize the ion by reducing potential energy.
  • Sulfate has multiple resonance structures.
  • Each structure depicts one sulfur-oxygen double bond while the rest are single bonds.
  • This indicates that, as in the carbonate, none of these resonance forms can fully represent the actual distribution, which is an average.
The sulfur atom can expand its valance shell beyond the octet, as it utilizes the d orbitals. This allows it to form more bonds and contribute to the equivalent bond lengths suggested by the resonance hybrid structure.
Resonance Structures
Resonance structures are alternative ways of arranging electrons in molecules that can't be accurately depicted with a single electron dot Lewis structure. These structures are crucial in understanding complex ions like carbonate and sulfate.

When resonance structures are possible, the true structure of the molecule is not any single one of them but is a hybrid of all possible structures. This results in
  • A stabilization factor for the molecular structure due to the delocalization of electrons across the different bonds.
  • Each resonance structure holding equal importance in determining the chemical properties of the ion.
  • An average bond order that gives a good approximation for the bonds' strengths compared to a theoretical localized version.
In the case of ions like carbonate and sulfate, resonance allows the electron density to be distributed evenly across the different bonds, explaining the uniform bond lengths that are shorter than single but longer than double bonds.
Oxidation State Calculation
Understanding oxidation state calculation is important for figuring out how electrons are distributed in a compound or ion. Oxidation states provide insight into what each atom in a molecule contributes in terms of electron sharing or electron ownership.

Here's how you calculate oxidation states, particularly useful for complex ions like \( \text{CO}_3^{2-} \) and \( \text{SO}_3^{2-} \):
  • Assign oxidation numbers based on commonly known values, e.g., oxygen is typically \( -2 \).
  • Calculate the total oxidation based on the number of atoms, e.g., for \( \text{CO}_3^{2-} \), the three oxygens at \( -2 \) each result in \( -6 \) total.
  • The overall charge of the ion must match the sum of the calculated oxidation states.
  • The remaining charge accounts for the central atom's oxidation state, balancing out that calculation.
For carbonate and sulfate, the central atom calculations will result in +4 for carbon in \( \text{CO}_3^{2-} \) and sulfur in \( \text{SO}_3^{2-} \) having an oxidation state of +6. This guidance allows one to tally the diverse valences that atoms can assume in these ions.

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Most popular questions from this chapter

Match the following $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (a) } \mathrm{Hg}_{2}^{2+} & \text { (p) Nucleophile } \\ \text { (b) } \mathrm{AlCl}_{3} & \text { (q) Catalyst } \\ \text { (c) } \mathrm{Br} & \text { (r) Lewis acid } \\ \text { (d) } \mathrm{H}^{+} & \text {(s) Soft acid } \\ & \text { (t) Electrophile } \\ \hline \end{array} $$

Which of the following undergoes nucleophillic substitution exclusively by \(\mathrm{SN}^{1}\) mechanism? (a) ethyl chloride (b) isopropyl chloride (c) chlorobenzene (d) benzyl chloride

The reaction of HBr with C=C(C)C in the presence of peroxide will give (a) CC(C)(C)C (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) (c) CC(C)CBr (d) CCC(C)C

The order of effectiveness of o- and p- directing groups is (a) \(\mathrm{OH}>\mathrm{NR}_{2}>\mathrm{NH} \cdot \mathrm{COCH}_{3}>\mathrm{Cl}>\mathrm{CH}_{3}\) (b) \(\mathrm{OH}>\mathrm{Cl}>\mathrm{CH}_{3}>\mathrm{NR}_{2}>\mathrm{NH} . \mathrm{COCH}_{3}\) (c) \(\mathrm{OH}>\mathrm{NH} . \mathrm{COCH}_{3}>\mathrm{Cl}>\mathrm{NR}_{2}>\mathrm{CH}_{3}\) (d) \(\mathrm{OH}>\mathrm{CH}_{3}>\mathrm{NH} . \mathrm{COCH}_{3}>\mathrm{Cl}>\mathrm{NR}_{2}\)

The correct order of increasing basic nature for the bases \(\mathrm{NH}_{3}, \mathrm{CH}_{3} \mathrm{NH}_{2}\), and \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\) is (a) \(\mathrm{CH}_{3} \mathrm{NH}_{2}<\mathrm{NH}_{3}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}<\mathrm{NH}_{3}<\mathrm{CH}_{3} \mathrm{NH}_{2}\) (c) \(\mathrm{NH}_{3}<\mathrm{CH}_{3} \mathrm{NH}_{2}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\) (d) \(\mathrm{CH}_{3} \mathrm{NH}_{2}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}<\mathrm{NH}_{3}\)

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