Question

In $ SO_2 $, $ NO_2^- $ and $ N_3^- $ the hybridizations at the central atom are respectively :

• A. \( sp^2 \), \( sp^2 \) and \( sp \)
• B. \( sp^2 \), \( sp \) and \( sp \)
• C. \( sp^2 \), \( sp^2 \) and \( sp^2 \)
• D. \( sp \), \( sp^2 \) and \( sp \)

Hint:

To determine the hybridization of the central atom, use the steric number concept: Steric Number = Number of sigma bonds + Number of lone pairs. For resonance structures, the hybridization of the central atom remains the same across all significant contributing structures.

Answer 1

The Correct Option is A

To determine the hybridization of the central atom in the molecules \(SO_2\), \(NO_2^-\), and \(N_3^-\), we need to understand the bonding and electron arrangements in each molecule.

1. **Hybridization in \(SO_2\):**
   • Sulfur dioxide (\(SO_2\)) has a central sulfur atom bonded to two oxygen atoms.
   • The sulfur atom forms double bonds with each oxygen atom, and it also has a lone pair of electrons.
   • To determine the hybridization, count the regions of electron density around the sulfur atom:
     1. Two double bonds with oxygen
     2. One lone pair
   • This totals three regions of electron density, which corresponds to \(sp^2\) hybridization.
2. **Hybridization in \(NO_2^-\):**
   • The nitrogen dioxide ion (\(NO_2^-\)) has a central nitrogen atom bonded to two oxygen atoms.
   • The nitrogen atom forms one double bond and one single bond with the oxygens, and it has one lone pair of electrons.
   • The regions of electron density around nitrogen are:
     1. One double bond with oxygen
     2. One single bond with oxygen
     3. One lone pair
   • This results in three regions of electron density, suggesting \(sp^2\) hybridization.
3. **Hybridization in \(N_3^-\):**
   • The azide ion (\(N_3^-\)) consists of a linear arrangement of three nitrogen atoms.
   • In this ion, the central nitrogen atom is bonded linearly to the two terminal nitrogen atoms.
   • The linear geometry indicates two regions of electron density:
     1. One bond with each terminal nitrogen atom
   • Two regions of electron density imply that the hybridization is \(sp\).

Thus, the hybridizations at the central atoms in \(SO_2\), \(NO_2^-\), and \(N_3^-\) are \(sp^2\), \(sp^2\), and \(sp\) respectively. Therefore, the correct answer is: sp^2, sp^2, and sp.

Answer 2

Step 1: Determine the hybridization of the central atom in \( SO_2 \).
Central atom: Sulfur (S)
Number of valence electrons of S: 6
Number of bond pairs: 2 (with two oxygen atoms)
Number of lone pairs: To complete the octet of S, we form double bonds with both O atoms. This uses 4 electrons. The remaining 2 electrons form one lone pair.
Steric number = Number of bond pairs + Number of lone pairs = 2 + 1 = 3 Hybridization: \( sp^2 \)
Step 2: Determine the hybridization of the central atom in \( NO_2^- \).
Central atom: Nitrogen (N) Number of valence electrons of N: 5
Number of surrounding atoms: 2 (two oxygen atoms)
Total electrons to accommodate (considering the negative charge): \( 5 + 2 \times 6 + 1 = 18 \) Lewis structure: \( [O=N-O^-] \) or \( [O^-N=O] \) (resonance structures). The nitrogen atom forms one double bond and one single bond with oxygen atoms. Number of bond pairs: 2
Number of lone pairs on N: To satisfy the formal charges and octet rule, nitrogen has one lone pair.
Steric number = Number of bond pairs + Number of lone pairs = 2 + 1 = 3 Hybridization: \( sp^2 \)
Step 3: Determine the hybridization of the central atom in \( N_3^- \).
Central atom: Central Nitrogen (N)
Number of valence electrons of N: 5
Total electrons to accommodate (considering the negative charge): \( 3 \times 5 + 1 = 16 \) Lewis structure: \( [:N=N=N:]^- \) or \( [:\overset{-}{N}-N \equiv N:] \) or \( [N \equiv N-\overset{-}{N}:] \) (resonance structures). The central nitrogen atom forms two double bonds or one single and one triple bond. Number of bond pairs around the central N: 2 (regardless of the type of bonds)
Number of lone pairs on the central N: To satisfy the formal charges and octet rule, the central nitrogen has zero lone pairs in the \( [:N=N=N:]^- \) resonance structure. Steric number = Number of bond pairs + Number of lone pairs = 2 + 0 = 2 Hybridization: \( sp \)
Step 4: Combine the hybridizations.
The hybridizations of the central atoms in \( SO_2 \), \( NO_2^- \), and \( N_3^- \) are \( sp^2 \), \( sp^2 \), and \( sp \) respectively.
Step 5: Match with the given options.
This matches option (1).