Question:
Among \( \text{N}_2\text{O} \), \( \text{ClF}_2^- \), \( \text{SO}_2 \) and \( \text{I}_3^+ \), the species having the linear structures are
Among \( \text{N}_2\text{O} \), \( \text{ClF}_2^- \), \( \text{SO}_2 \) and \( \text{I}_3^+ \), the species having the linear structures are
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For triatomic species, look at the steric number and lone pairs on the central atom. A steric number of 2 (with 0 lone pairs) or 5 (with 3 equatorial lone pairs) always corresponds to a linear geometry.
Updated On: May 28, 2026
- \( \text{N}_2\text{O} \), \( \text{ClF}_2^- \)
- \( \text{ClF}_2^- \), \( \text{I}_3^+ \)
- \( \text{I}_3^+ \), \( \text{SO}_2 \)
- \( \text{N}_2\text{O} \), \( \text{SO}_2 \)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question asks us to identify the chemical species that possess a linear molecular geometry from the given list of molecules and ions.
Step 2: Key Formula or Approach:
We will use the Valence Shell Electron Pair Repulsion (VSEPR) theory to determine the hybridization and molecular geometry of each species:
- Find the steric number of the central atom:
\[ \text{Steric Number} = \text{Number of } \sigma\text{-bonds} + \text{Number of lone pairs} \]
Step 3: Detailed Explanation:
1. \( \text{N}_2\text{O} \) (Nitrous Oxide):
The central nitrogen atom forms one triple bond with another nitrogen atom and one single dative bond with an oxygen atom (represented as \( \text{N}\equiv\text{N}\rightarrow\text{O} \)). It has 2 \( \sigma \)-bonds and 0 lone pairs. Thus, the central atom is \( sp \)-hybridized, resulting in a linear structure.
2. \( \text{ClF}_2^- \):
The central chlorine atom has 7 valence electrons plus 1 additional electron from the negative charge, giving 8 valence electrons. It forms 2 \( \sigma \)-bonds with fluorine atoms, leaving 6 non-bonding electrons (3 lone pairs).
- Steric Number = \( 2 + 3 = 5 \) (\( sp^3d \) hybridization).
- To minimize repulsion, the 3 lone pairs occupy the equatorial positions, and the 2 fluorine atoms occupy the axial positions, yielding a linear geometry.
3. \( \text{SO}_2 \):
The central sulfur atom (6 valence electrons) forms 2 \( \sigma \)-bonds (via double bonds with oxygen) and has 1 lone pair.
- Steric Number = \( 2 + 1 = 3 \) (\( sp^2 \) hybridization).
- This results in a bent V-shaped geometry.
4. \( \text{I}_3^+ \):
The central iodine atom (7 valence electrons minus 1 from the positive charge, so 6 valence electrons) forms 2 \( \sigma \)-bonds with the other iodine atoms, leaving 4 non-bonding electrons (2 lone pairs).
- Steric Number = \( 2 + 2 = 4 \) (\( sp^3 \) hybridization).
- This results in a bent V-shaped geometry.
Step 4: Final Answer:
The linear species are \( \text{N}_2\text{O} \) and \( \text{ClF}_2^- \), which corresponds to option (A).
The question asks us to identify the chemical species that possess a linear molecular geometry from the given list of molecules and ions.
Step 2: Key Formula or Approach:
We will use the Valence Shell Electron Pair Repulsion (VSEPR) theory to determine the hybridization and molecular geometry of each species:
- Find the steric number of the central atom:
\[ \text{Steric Number} = \text{Number of } \sigma\text{-bonds} + \text{Number of lone pairs} \]
Step 3: Detailed Explanation:
1. \( \text{N}_2\text{O} \) (Nitrous Oxide):
The central nitrogen atom forms one triple bond with another nitrogen atom and one single dative bond with an oxygen atom (represented as \( \text{N}\equiv\text{N}\rightarrow\text{O} \)). It has 2 \( \sigma \)-bonds and 0 lone pairs. Thus, the central atom is \( sp \)-hybridized, resulting in a linear structure.
2. \( \text{ClF}_2^- \):
The central chlorine atom has 7 valence electrons plus 1 additional electron from the negative charge, giving 8 valence electrons. It forms 2 \( \sigma \)-bonds with fluorine atoms, leaving 6 non-bonding electrons (3 lone pairs).
- Steric Number = \( 2 + 3 = 5 \) (\( sp^3d \) hybridization).
- To minimize repulsion, the 3 lone pairs occupy the equatorial positions, and the 2 fluorine atoms occupy the axial positions, yielding a linear geometry.
3. \( \text{SO}_2 \):
The central sulfur atom (6 valence electrons) forms 2 \( \sigma \)-bonds (via double bonds with oxygen) and has 1 lone pair.
- Steric Number = \( 2 + 1 = 3 \) (\( sp^2 \) hybridization).
- This results in a bent V-shaped geometry.
4. \( \text{I}_3^+ \):
The central iodine atom (7 valence electrons minus 1 from the positive charge, so 6 valence electrons) forms 2 \( \sigma \)-bonds with the other iodine atoms, leaving 4 non-bonding electrons (2 lone pairs).
- Steric Number = \( 2 + 2 = 4 \) (\( sp^3 \) hybridization).
- This results in a bent V-shaped geometry.
Step 4: Final Answer:
The linear species are \( \text{N}_2\text{O} \) and \( \text{ClF}_2^- \), which corresponds to option (A).
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