Draw the Lewis structure for \(\mathrm{NO}^{+} .\) Is the nitrogen-oxygen bond in \(\mathrm{NO}^{+}\) longer, shorter, or the same length as the nitrogen- oxygen bond in NO? Explain.

Short Answer

Expert verified
The Lewis structure for \(\mathrm{NO}^+\) has a single bond connecting the nitrogen and oxygen atoms, with 3 lone pairs of electrons around oxygen. The nitrogen-oxygen bond in \(\mathrm{NO}+\) is longer than the bond in the neutral \(\mathrm{NO}\) molecule because the neutral molecule has a double bond, which is shorter and stronger than a single bond.

Step by step solution

01

Calculate the number of valence electrons

First, we need to count the total number of valence electrons in the \(\mathrm{NO}^+\) ion. Nitrogen has 5 valence electrons, oxygen has 6 valence electrons, and since it's a positive ion, we need to subtract one electron from the total count. Thus, the total number of valence electrons in \(\mathrm{NO}^+\) is: \(5 + 6 - 1 = 10 \text{ valence electrons}\).
02

Arrange the atoms and place a single bond

Place the less electronegative atom, nitrogen (N), in the center and the more electronegative atom, oxygen (O), adjacent to it. Between both atoms, place a single bond (which represents two shared electrons) to connect them. At this stage, we have used 2 of the 10 valence electrons.
03

Complete the octets of outer atoms first

Complete the octet of the outer atom, oxygen, by using the remaining valence electrons. We have 8 valence electrons left to use. Oxygen needs 6 more electrons to complete its octet, so place 3 lone pairs of electrons around it. Now all valence electrons have been used, and the octet rule is satisfied for oxygen.
04

Complete the octet of central atom

Count the electrons around the nitrogen atom: - Two electrons in a single bond with oxygen. As nitrogen has only 2 electrons in its valence shell, it does not have an octet. Since all the valence electrons have been used, there's no remaining electron to complete the octet for the nitrogen atom. Thus, the nitrogen atom in \(\mathrm{NO}^+\) will have one less than a complete octet.
05

Determine bond length of nitrogen-oxygen in \(\mathrm{NO}^+\) and compare it with a neutral \(\mathrm{NO}\) molecule

The Lewis structure of \(\mathrm{NO}^+\) has a single bond between the nitrogen and oxygen atoms. In contrast, the neutral \(\mathrm{NO}\) molecule has a double bond between nitrogen and oxygen. As a double bond is shorter and stronger than a single bond, the nitrogen-oxygen bond in \(\mathrm{NO}\) is shorter than the nitrogen-oxygen bond in \(\mathrm{NO}^+\). In conclusion, the nitrogen-oxygen bond in \(\mathrm{NO}^+\) is longer than the nitrogen-oxygen bond in the neutral \(\mathrm{NO}\) molecule.

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Key Concepts

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

Valence Electrons Calculation
Understanding the distribution of valence electrons is crucial when constructing Lewis structures for molecules and ions. Valence electrons are the outermost electrons of an atom and are important in the formation of chemical bonds.

To calculate the number of valence electrons, we must examine the periodic table and count the electrons in the outer shell or look at the group number for main-group elements. For example, nitrogen is in group 15 and has 5 valence electrons, while oxygen is in group 16 and has 6 valence electrons.

When considering ions like (NO+), we must adjust the valence electron count to reflect the ionic charge. A positive charge indicates the loss of electrons. Therefore, in the case of (NO+), we calculate the total valence electrons by adding the valence electrons of nitrogen and oxygen, and then subtracting one electron due to the positive charge: (5 from N) + (6 from O) - 1 = 10 valence electrons.

This step is vital because it dictates how electrons are distributed in the molecule, which in turn determines the molecular shape and bond characteristics.
Electron Octet Rule
The electron octet rule is a fundamental principle in chemistry that helps predict the arrangement of electrons in atoms during compound formation. It states that atoms tend to combine in such a way that they each have eight electrons in their valence shell, giving them the same electron configuration as a noble gas.

However, there are exceptions to this rule. In the Lewis structure of (NO+), we see that while oxygen can satisfy the octet rule by adding three lone pairs of electrons, the nitrogen atom cannot. Nitrogen in (NO+) is left with just six electrons: two shared with oxygen and four in the form of lone pairs. This is an exception to the octet rule since the nitrogen atom has less than eight electrons.

The incompleteness of nitrogen's octet in certain compounds like (NO+) can affect the overall stability and reactivity of the molecule. Notably, these exceptions to the octet rule often occur in molecules with an odd number of valence electrons, with electrons in d or f orbitals, or when bonding elements have a low electronegativity.
Chemical Bond Length Comparison
Chemical bond length is a key aspect of molecular structure. It refers to the distance between the nuclei of two bonded atoms. Bond lengths can vary depending on the type and strength of the bond. Generally, single bonds are longer than double bonds, and double bonds are longer than triple bonds due to the increased electron sharing and subsequently greater electrostatic attraction between atoms in multiple bonds.

In comparing the bond lengths of (NO+) and neutral NO, we observe that (NO+) has a single bond, while NO has a double bond between the nitrogen and oxygen atoms. This is reflective of the differing number of bonding electrons: a double bond in NO has two pairs of shared electrons, while the single bond in (NO+) only has one pair. Consequently, the nitrogen-oxygen bond in NO is indeed shorter than that in (NO+), which aligns with the Lewis structures and predicted bond strengths of these species.

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Most popular questions from this chapter

For each of the following molecules or ions of sulfur and oxygen, write a single Lewis structure that obeys the octet rule, and calculate the oxidation numbers and formal charges on all the atoms: (a) SO \(_{2},(\mathbf{b}) \mathrm{SO}_{3},(\mathrm{c}) \mathrm{SO}_{3}^{2-}\) (d) Arrange these molecules/ions in order of increasing \(S-O\) bond length.

(a) Write the electron configuration for the element titanium, Ti. How many valence electrons does this atom possess? (b) Hafnium, Hf, is also found in group 4 \(\mathrm{B}\) . Write the electron configuration for Hf. (c) Ti and Hf behave as though they possess the same number of valence electrons. Which of the subshells in the electron configuration of \(\mathrm{Hf}\) behave as valence orbitals? Which behave as core orbitals?

Energy is required to remove two electrons from Ca to form \(\mathrm{Ca}^{2+},\) and energy is required to add two electrons to \(\mathrm{O}\) to form \(\mathrm{O}^{2-} .\) Yet \(\mathrm{CaO}\) is stable relative to the free elements. Which statement is the best explanation? (a) The lattice energy of CaO is large enough to overcome these processes. (b) CaO is a covalent compound, and these processes are irrelevant. (c) CaO has a higher molar mass than either Ca or O. (d) The enthalpy of formation of CaO is small. (e) CaO is stable to atmospheric conditions.

In the following pairs of binary compounds, determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{TiCl}_{4}\) and \(\mathrm{CaF}_{2},(\mathbf{b}) \mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3},(\mathbf{c}) \mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3} .\)

Write Lewis structures that obey the octet rule for each of the following, and assign oxidation numbers and formal charges to each atom: (a) OCS, (b) SOCl_ \(_{2}(S\) is the central atom), \((\mathbf{c}) \mathrm{BrO}_{3}^{-},(\mathbf{d}) \mathrm{HClO}_{2}(\mathrm{H}\) is bonded to O)

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