Question:
In $ SO_2 $, $ NO_2^- $ and $ N_3^- $ the hybridizations at the central atom are respectively :
In $ SO_2 $, $ NO_2^- $ and $ N_3^- $ the hybridizations at the central atom are respectively :
Show 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.
Updated On: Feb 3, 2026
- \( sp^2 \), \( sp^2 \) and \( sp \)
- \( sp^2 \), \( sp \) and \( sp \)
- \( sp^2 \), \( sp^2 \) and \( sp^2 \)
- \( sp \), \( sp^2 \) and \( sp \)
Show Solution
Verified By Collegedunia
The Correct Option is A
Approach Solution - 1
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.
- **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:
- Two double bonds with oxygen
- One lone pair
- This totals three regions of electron density, which corresponds to \(sp^2\) hybridization.
- **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:
- One double bond with oxygen
- One single bond with oxygen
- One lone pair
- This results in three regions of electron density, suggesting \(sp^2\) hybridization.
- **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:
- 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.
Was this answer helpful?
0
0
Show Solution
Verified By Collegedunia
Approach Solution -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).
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).
Was this answer helpful?
0
0
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
View More Questions
Top JEE Main Chemical bonding and molecular structure Questions
- Statement I: Since fluorine is more electronegative than nitrogen, the net dipole moment of NF3 is greater than NH3.
Statement II: In NH3, the orbital dipole due to the lone pair and the dipole moment of NH bonds are in opposite directions, but in NF3, the orbital dipole due to the lone pair and the dipole moments of N-F bonds are in the same direction.
In light of the above statements, choose the most appropriate from the options given below: - What is the structure of Mn2(CO)10?
- Consider the following two statements Statement I: \(π\) \(2p\) bonding molecular orbital has low electron density above & below internuclear axis Statement II: \(π\times2p\) antibonding molecular orbital has only one nodal plane
- Consider the following two statements Statement I: \(SiO_2\) and GeO₂\(GeO_2\) are acidic, \(SnO\) and \(PbO\) are amphoteric Statement II: Allotropes of carbon are formed due to catenation and \(dπ - ρπ\) bond
- Number of compounds with one lone pair of electrons on central atom amongst following is
\(CIF_3, XeO_3, BrF_5, XeF_4, O_3, NH_3\),\( H_2O\)
View More Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions