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The bond order in \(\mathrm{NO}\) is \(2.5\) while that in \(\mathrm{NO}^{+}\)is 3 . Which of the following statements is true for these two species? (a) bond length in \(\mathrm{NO}^{+}\)is greater than in \(\mathrm{NO}\) (b) bond length in \(\mathrm{NO}\) is greater than in \(\mathrm{NO}^{+}\) (c) bond length in \(\mathrm{NO}^{+}\)is equal than in NO (d) bond length is unpredictable

Short Answer

Expert verified
The true statement is (b): bond length in NO is greater than in NO^+.

Step by step solution

01

Understanding Bond Order

The bond order of a molecule is a measure of the number of chemical bonds between a pair of atoms. It helps to predict the stability and length of a bond. Higher bond orders indicate stronger bonds and typically shorter bond lengths.
02

Analyze Given Data for NO

The bond order for \( ext{NO}\) is given as 2.5. This represents an intermediate bond strength and length.
03

Analyze Given Data for NO^+

The bond order for \( ext{NO}^{+}\) is given as 3, which indicates a stronger and likely shorter bond compared to a bond order of 2.5.
04

Comparing Bond Orders

Since the \( ext{NO}^{+}\) has a higher bond order (3) than \( ext{NO}\) (2.5), it is expected that the bond in \( ext{NO}^{+}\) is shorter owing to the stronger bond.
05

Conclusion on Bond Length

A higher bond order results in a shorter bond length, so the bond length in \( ext{NO}\) is longer than in \( ext{NO}^{+}\). This leads to the conclusion that statement (b) is true: bond length in \( ext{NO}\) is greater than in \( ext{NO}^{+}\).

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Key Concepts

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

Bond Order
Bond order is a fundamental concept that highlights the strength and stability of a bond between two atoms. It effectively represents the number of bonds between a pair of atoms. For instance, a bond order of 1 implies a single bond, 2 represents a double bond, and so forth.

Calculating bond order involves subtracting the number of anti-bonding electrons from the number of bonding electrons and dividing by two: \[ \text{Bond order} = \frac{\text{Number of bonding electrons} - \text{Number of anti-bonding electrons}}{2}\]

Here's why it matters:
  • A higher bond order usually corresponds to a more stable and stronger bond.
  • For example, in \( \text{NO} \), the bond order is 2.5, indicating a bond somewhat between a double and a triple bond.
  • In \( \text{NO}^{+} \), the bond order is 3, denoting a triple bond scenario, and typically implies a stronger bond than one with a lower bond order.
Understanding bond order helps in predicting molecular properties, such as bond length and molecular stability.
Bond Length
Bond length is the distance between the nuclei of two bonded atoms. It's a crucial factor that can influence the physical and chemical properties of molecules.

The bond length is influenced by several factors:
  • Bond Order: Generally, a higher bond order leads to a stronger attraction between the atoms, resulting in shorter bond lengths. This is why \( \text{NO}^{+} \) with a bond order of 3 has a shorter bond length than \( \text{NO} \) with a bond order of 2.5.
  • Atomic Size: Larger atoms usually lead to longer bond lengths. However, in similar sized atoms like \( \text{N} \) and \( \text{O} \) in different electronic states, bond order plays a more significant role.
Compared to \( \text{NO} \), the bond in \( \text{NO}^{+} \) is expected to be shorter due to its higher bond order, indicating a stronger bond relationship between nitrogen and oxygen.
Molecular Stability
Molecular stability is a concept that indicates how likely a molecule is to maintain its structure without undergoing a chemical reaction. Bond order directly impacts stability.

Key points about molecular stability include:
  • High Bond Order, High Stability: A higher bond order correlates with a more stable molecule. \( \text{NO}^{+} \) with its bond order of 3 is typically more stable than \( \text{NO} \) with a bond order of 2.5.
  • Energy Considerations: Molecules with higher bond orders require more energy to break the bonds, thus often exhibiting higher stability.
  • Electron Distribution: Stability is also related to how electrons are distributed in a molecule; fewer unpaired electrons usually mean higher stability.
Therefore, understanding bond order not only helps in comprehending bond length but also offers insights into the stability of molecules, highlighting why \( \text{NO}^{+} \) might be more stable than \( \text{NO} \). This stability is crucial in predicting how a molecule might react with others.

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Most popular questions from this chapter

Nitrogen forms \(\mathrm{N}_{2}\), but phosphorous is converted into \(\mathrm{P}_{4}\) from \(\mathrm{P}\), the reason is (a) triple bond is present between phosphorous atom (b) \(\mathrm{p} \pi-\mathrm{p} \pi\) bonding is strong (c) p \(\pi-\mathrm{p} \pi\) bonding is weak (d) multiple bond is formed easily

Among the following species, identify the isostructural pairs \(\mathrm{NF}_{3}, \mathrm{NO}_{3}^{-}, \mathrm{BF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{HN}_{3}\) (a) \(\left[\mathrm{NF}_{3}, \mathrm{NO}_{3}^{-}\right]\)and \(\left[\mathrm{BF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\) (b) \(\left[\mathrm{NF}_{3}, \mathrm{HN}_{3}\right]\) and \(\left[\mathrm{NO}_{3}-\mathrm{BF}_{3}\right]\) (c) \(\left[\mathrm{NF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\)and \(\left[\mathrm{NO}_{3}-\mathrm{BF}_{3}\right]\) (d) \(\left[\mathrm{NF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\)and \(\left[\mathrm{HN}_{3}, \mathrm{BF}_{3}\right]\)

Which of the pairs have identical values of bond order? (a) \(\mathrm{F}_{2}\) and \(\mathrm{Ne}_{2}\) (b) \(\mathrm{N}_{2}^{+}\)and \(\mathrm{O}_{2}^{+}\) (c) \(\mathrm{O}_{2}^{2-}\) and \(\mathrm{B}_{2}\) (d) \(\mathrm{C}_{2}\) and \(\mathrm{N}_{2}\)

In which one of the following pairs, molecules/ions have similar shape? (a) \(\mathrm{CCl}_{4}\) and \(\mathrm{PtCl}_{4}\) (b) \(\mathrm{NH}_{3}\) and \(\mathrm{BF}_{3}\) (c) \(\mathrm{BF}_{3}\) and \(\mathrm{t}\)-butyl carbonium ion (d) \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\)

The pair of species having identical shapes for molecules of both species is (a) \(\mathrm{CF}_{4}, \mathrm{SF}_{4}\) (b) \(\mathrm{XeF}_{2}, \mathrm{CO}_{2}\) (c) \(\mathrm{BF}_{3}^{4}, \mathrm{PCl}_{3}\) (d) \(\mathrm{PF}_{5}, \mathrm{IF}_{5}\)

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