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Resonance structures can be written for (1) \(\mathrm{O}_{3}\) (2) \(\mathrm{NH}_{3}\) (3) \(\mathrm{CH}_{4}\) (4) \(\mathrm{H}_{2} \mathrm{O}\)

Short Answer

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\( \mathrm{O}_3 \) (ozone)

Step by step solution

01

Identify Resonance Structures

Resonance structures are possible for molecules where there is delocalization of electrons. Delocalization occurs in molecules with conjugated systems of p-orbitals.
02

Analyze \( \mathrm{O}_3 \) (Ozone)

\( \mathrm{O}_3 \) has resonance structures because it can have multiple Lewis structures with delocalized electrons.
03

Analyze \( \mathrm{NH}_3 \) (Ammonia)

\( \mathrm{NH}_3 \) does not have resonance structures because it does not possess delocalized p-orbital electrons.
04

Analyze \( \mathrm{CH}_4 \) (Methane)

\( \mathrm{CH}_4 \) does not have resonance structures because it lacks delocalized p-orbital electrons.
05

Analyze \( \mathrm{H}_2 \mathrm{O} \) (Water)

\( \mathrm{H}_2 \mathrm{O} \) does not have resonance structures because it lacks delocalized p-orbital electrons.
06

Conclusion

Based on the analysis, ozone (\( \mathrm{O}_3 \)) is the only molecule among the given options that has resonance structures.

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Key Concepts

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

delocalization of electrons
Delocalization of electrons refers to the spreading of electron density across multiple atoms, rather than being confined between a single pair of atoms. This phenomenon often occurs in molecules with resonance structures. Here are a few points to understand delocalization more clearly:

  • It involves electrons moving within a molecule's p-orbitals.
  • Helps stabilize molecules that have conjugated systems.
  • Indicated by the presence of multiple Lewis structures for the same molecule.
In essence, delocalization helps in distributing electron pairs over several adjacent atoms, which leads to greater stability and often influences the chemical reactivity of the molecule.
Lewis structures
Lewis structures, also known as Lewis dot structures, are diagrams that represent the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule. They are essential tools for understanding chemical bonding. Here are some key points:

  • Shows connectivity between atoms using dots to represent electrons and lines for bonds.
  • Helps in identifying regions of electron density within compounds.
  • Used to visualize possible resonance structures.
For example, in ozone (\text{O\(_3\)}), Lewis structures reveal that electrons can be delocalized over the oxygen atoms, leading to resonance hybrids. This makes understanding molecule stability and reactivity easier.
conjugated systems
Conjugated systems are formations within a molecule where p-orbitals overlap, allowing electrons to be delocalized across adjacent atoms. These systems are significant for several reasons:

  • They provide stability through delocalized electrons.
  • Present in molecules that exhibit resonance structures.
  • Can impact the chemical and physical properties of compounds.
A conjugated system typically consists of alternating single and multiple bonds, which allows for a continuous overlap of p-orbitals and the free movement of electrons within this system. This property explains why molecules like ozone (\text{O\(_3\)}) have resonance structures and are more stable due to electron delocalization.

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Most popular questions from this chapter

Bond order in the MO theory depends on the number of bonding electrons and anti-bonding electrons. The bond order (1) always has an integer value (2) can have only fractional valuc (3) is always equal to onc (4) can be zero, integral or fractional

Which of the following combinations is not allowed in the LC \(\Lambda O\) method for the formation of a MO (consider the \(z\) -axis as the molecular axis?? (1) \(\mathrm{s}+\mathrm{p}_{2}\) (2) \(\mathrm{s}-\mathrm{p}_{x}\) (3) \(\mathrm{p}_{x}-\mathrm{p}_{x}\) (4) \(\mathrm{p}_{2}+\mathrm{p}_{z}\)

\(\mathrm{H}_{2} \mathrm{O}\) has net dipole moment while \(\mathrm{BeF}_{2}\) has zero dipole moment because (1) \(\mathrm{H}_{2} \mathrm{O}\) molecule is linear while \(\mathrm{Be} \mathrm{F}_{2}\) is bent (2) \(\mathrm{BeF}_{2}\) molecule is linear while \(\mathrm{H}_{2} \mathrm{O}\) is bent (3) Fluorine is more electronegative than oxygen (4) Be is more electronegative than oxygen

The II-O-II bond angle in \(\mathrm{II}_{2} \mathrm{O}\) is \(104.5^{\circ}\) and not \(109^{\circ} 28^{\prime}\) because of (1) high electronegativity of oxygen (2) bond pair-bond pair repulsion (3) Lonc pair-lone pair repulsion (4) Lonc pair-bond pair repulsion

The correct order of the arrangement of bond length is (1) \(\mathrm{F}_{2}>\mathrm{N}_{2}>\mathrm{Cl}_{2}>\mathrm{O}_{2}\) (2) \(\mathrm{Cl}_{2}>\mathrm{F}_{2}>\mathrm{O}_{2}>\mathrm{N}_{2}\) (3) \(\mathrm{O}_{2}>\mathrm{Cl}_{2}>\mathrm{N}_{2}>\mathrm{F}_{2}\) (3) \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{N}_{2}>\mathrm{O}_{2}\)

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