Problem 80
Question
Draw Lewis structures for the \(\mathrm{AsCl}_{4}^{+}\) and \(\mathrm{AsCl}_{6}^{-}\) ions. What type of reaction (acid-base, oxidation- reduction, or the like) is the following?$$2 \mathrm{AsCl}_{5}(g) \longrightarrow \mathrm{AsCl}_{4} \mathrm{AsCl}_{6}(s)$$.
Step-by-Step Solution
Verified Answer
The Lewis structures for \(\mathrm{AsCl}_{4}^{+}\) and \(\mathrm{AsCl}_{6}^{-}\) ions are \(_{\large{\bullet}}^{\large{\bullet}}\text{As}(_{\large{\bullet}}^{\large{\bullet}}\text{Cl}\ _{\large{\bullet}}^{\large{\bullet}}\text{Cl}\ _{\large{\bullet}}^{\large{\bullet}}\text{Cl}\ _{\large{\bullet}}^{\large{\bullet}}\text{Cl}\ _{\large{\bullet}}^{\large{\bullet}}|\) and \(\text{As}(-1)(0\text{Cl}¢0\text{Cl}¢0\text{Cl}¢0\text{Cl}¢0\text{Cl}¢0\text{Cl})\), respectively. The given reaction, $$2 \mathrm{AsCl}_{5}(g) \longrightarrow \mathrm{AsCl}_{4}^{+} \mathrm{AsCl}_{6}^{-}(s)$$, is a **ligand-exchange reaction** as it involves the transfer of a chlorine atom (Cl) between two \(\mathrm{AsCl}_{5}\) molecules without changes in proton transfer or oxidation states.
1Step 1: Draw the Lewis structures for each ion
First, we need to draw the Lewis structures for \(\mathrm{AsCl}_{4}^{+}\) and \(\mathrm{AsCl}_{6}^{-}\) ions. In these ions, arsenic (As) is the central atom surrounded by chlorine (Cl) atoms.
For \(\mathrm{AsCl}_{4}^{+}\):
1. Determine the total number of valence electrons: \(5\) (from As) + \(4 \times 7\) (from 4 Cl atoms) + \(1\) (due to positive charge) = \(33\).
2. Place As in the center and add single bonds to each Cl atom: \(_{As}|\text{Cl}¢\text{Cl}¢\text{Cl}¢\text{Cl}=\).
3. Place remaining electrons as lone pairs, following the octet rule, yielding \(_{\large{\bullet}}^{\large{\bullet}}\text{As}(_{\large{\bullet}}^{\large{\bullet}}\text{Cl}\ _{\large{\bullet}}^{\large{\bullet}}\text{Cl}\ _{\large{\bullet}}^{\large{\bullet}}\text{Cl}\ _{\large{\bullet}}^{\large{\bullet}}\text{Cl}\ _{\large{\bullet}}^{\large{\bullet}}|\).
4. Calculate the formal charge on each atom to confirm the best structure: As (\(0\)), all Cl atoms (\(-1\)), and check that the sum is equal to the overall charge of \(\mathrm{AsCl}_{4}^{+}\).
For \(\mathrm{AsCl}_{6}^{-}\):
1. Determine the total number of valence electrons: \(5\) (from As) + \(6 \times 7\) (from 6 Cl atoms) - \(1\) (due to negative charge) = \(47\).
2. Place As in the center and add single bonds to each Cl atom.
3. As As can accommodate more than 8 electrons (expanded octet), add remaining electrons as lone pairs, following the octet rule for Cl atoms.
4. Calculate the formal charge on each atom to confirm the best structure. In this case, As has a formal charge of \(-1\) and all Cl atoms have a formal charge of \(0\). The sum is equal to the overall charge of \(\mathrm{AsCl}_{6}^{-}\).
2Step 2: Classify the type of reaction
Now, let's examine the given reaction: $$2 \mathrm{AsCl}_{5}(g) \longrightarrow \mathrm{AsCl}_{4}^{+} \mathrm{AsCl}_{6}^{-}(s)$$. To classify this reaction, we observe that:
- There is no transfer of protons (H+), so it's not an acid-base reaction.
- There is no change in oxidation states of As or Cl, so it's not an oxidation-reduction reaction.
This reaction involves the transfer of a chlorine atom (Cl) between two \(\mathrm{AsCl}_{5}\) molecules, forming one \(\mathrm{AsCl}_{4}^{+}\) and one \(\mathrm{AsCl}_{6}^{-}\) species. Therefore, this is a **ligand-exchange reaction**.
Key Concepts
Valence ElectronsFormal ChargeExpanded OctetReaction Classification
Valence Electrons
Valence electrons are the number of electrons in the outermost shell of an atom, crucial in forming bonds. For example, in the ion \(\mathrm{AsCl}_{4}^{+}\), arsenic (As) and chlorine (Cl) have valence electrons of 5 and 7, respectively.
Arsenic typically forms bonds using its 5 valence electrons, while chlorine has 7 valence electrons, needing one more to complete an octet.
To find the total valence electrons for a molecule, sum up the valence electrons from all atoms, adjusting for the ionic charge.
Arsenic typically forms bonds using its 5 valence electrons, while chlorine has 7 valence electrons, needing one more to complete an octet.
To find the total valence electrons for a molecule, sum up the valence electrons from all atoms, adjusting for the ionic charge.
- Positive charge: subtract that number of electrons.
- Negative charge: add that number of electrons.
Formal Charge
Formal charge helps determine the most stable structure of a molecule by estimating the charge distribution across atoms.
Calculate it using this formula:\[ \text{Formal charge} = \text{Valence electrons} - \text{Non-bonding electrons} - \frac{\text{Bonding electrons}}{2} \] This method assumes electrons in a bond are shared equally.
For \(\mathrm{AsCl}_{4}^{+}\), As should ideally hold a formal charge of 0, fitting its central position.
Each Cl aims for a formal charge of -1 due to the positive charge of the ion.
Calculate it using this formula:\[ \text{Formal charge} = \text{Valence electrons} - \text{Non-bonding electrons} - \frac{\text{Bonding electrons}}{2} \] This method assumes electrons in a bond are shared equally.
For \(\mathrm{AsCl}_{4}^{+}\), As should ideally hold a formal charge of 0, fitting its central position.
Each Cl aims for a formal charge of -1 due to the positive charge of the ion.
- Formal charge on As: \(5 - 0 - \frac{8}{2} = 1\)
- Adjust by adding bonds so that formal charges reflect stability.
Expanded Octet
An expanded octet refers to a central atom holding more than eight electrons, possible for elements in or beyond the third period of the periodic table, such as arsenic.
In \(\mathrm{AsCl}_{6}^{-}\), arsenic has six bonds to chlorine that total more than an octet.
This is feasible because arsenic can utilize empty \(3d\) orbitals, exceeding the basic 8-electron octet rule.
In \(\mathrm{AsCl}_{6}^{-}\), arsenic has six bonds to chlorine that total more than an octet.
This is feasible because arsenic can utilize empty \(3d\) orbitals, exceeding the basic 8-electron octet rule.
- Expanded octets often increase stability in transition metal complexes and heavier element compounds.
- Ensure all peripheral atoms like Cl still follow the octet rule.
Reaction Classification
Classifying a reaction involves identifying what changes occur. Consider types like acid-base or oxidation-reduction reactions.
In our exercise, the reaction \(2 \mathrm{AsCl}_{5}(g) \rightarrow \mathrm{AsCl}_{4}^{+} \mathrm{AsCl}_{6}^{-}(s)\) doesn't feature proton or electron transfer.
So, it's neither an acid-base nor redox process. Here, \(\mathrm{AsCl}_{5}\) molecules exchange a chlorine atom (Cl), forming new species but without changing oxidation numbers.
In our exercise, the reaction \(2 \mathrm{AsCl}_{5}(g) \rightarrow \mathrm{AsCl}_{4}^{+} \mathrm{AsCl}_{6}^{-}(s)\) doesn't feature proton or electron transfer.
So, it's neither an acid-base nor redox process. Here, \(\mathrm{AsCl}_{5}\) molecules exchange a chlorine atom (Cl), forming new species but without changing oxidation numbers.
- Ligand-exchange reactions like this involve swapping atomic positions.
- They often occur in coordination chemistry.
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