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(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Why is this consistent with the electronic structure of NO? (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in \(\mathrm{NO}, \mathrm{NO}^{+}\), and \(\mathrm{NO}^{-}\), and describe the magnetic properties of each (c) With what neutral homonuclear diatomic molecules are the \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

Short Answer

Expert verified
(a) NO readily loses one electron to form NO+ because it has a higher bond order (3) compared to NO (2.5), resulting in a stronger N-O bond. (b) The bond strength order is: NO+ > NO > NO-. NO and NO- are paramagnetic while NO+ is diamagnetic. (c) NO+ is isoelectronic with N2, and NO- is isoelectronic with O2.

Step by step solution

01

Identify the electronic structures of NO, NO+, and NO-

First, we need to find the electronic structure of each species. The atomic numbers for Nitrogen (N) and Oxygen (O) are 7 and 8 respectively. Therefore, the number of valence electrons in the molecule or ion can be calculated as follows: NO: 7 (N) + 8 (O) = 15 electrons NO+: 7 (N) + 8 (O) - 1 (due to the positive charge) = 14 electrons NO-: 7 (N) + 8 (O) + 1 (due to the negative charge) = 16 electrons
02

Determine the molecular orbital diagram

Since both N and O are part of the second row on the periodic table, we will follow the 2s, 2px, 2py, and 2pz atomic orbitals to fill molecular orbitals in the following order using the aufbau principle - σ(2s), σ*(2s), σ(2px), π(2py), π(2pz), π*(2py), π*(2pz) and σ*(2px).
03

Fill the molecular orbitals for NO, NO+, and NO-

Fill electrons in the molecular orbitals for each species, following the aufbau principle and Hund's rule: NO: σ(2s)^2, σ*(2s)^2, σ(2px)^2, π(2py)^2, π(2pz)^2, π*(2py)^1, π*(2pz)^0, σ*(2px)^0 (15 electrons) NO+: σ(2s)^2, σ*(2s)^2, σ(2px)^2, π(2py)^2, π(2pz)^2, π*(2py)^0, π*(2pz)^0, σ*(2px)^0 (14 electrons) NO-: σ(2s)^2, σ*(2s)^2, σ(2px)^2, π(2py)^2, π(2pz)^2, π*(2py)^2, π*(2pz)^2, σ*(2px)^0 (16 electrons)
04

Calculate the bond order for each species

The bond order can be calculated using the formula: Bond Order = (1/2) [(number of electrons in bonding MOs) - (number of electrons in anti-bonding MOs)] NO: Bond Order = (1/2) [(8) - (3)] = 2.5 NO+: Bond Order = (1/2) [(8) - (2)] = 3 NO-: Bond Order = (1/2) [(8) - (4)] = 2
05

Predict the N-O bond strength and magnetic properties

(a) NO readily loses one electron to form the NO+ ion because NO+ has a higher bond order (3) compared to NO (2.5) and therefore has a stronger N-O bond. (b) The bond strength order is: NO+ > NO > NO-. Magnetic properties: NO: Since it has one unpaired electron in the π*(2py) orbital, it is paramagnetic. NO+: No unpaired electrons, so it is diamagnetic. NO-: Two unpaired electrons in the π*(2py) and π*(2pz) orbitals, so it is paramagnetic.
06

Identify the isoelectronic neutral homonuclear diatomic molecules

(c) For the diatomic ions to be isoelectronic with a homonuclear diatomic molecule, they should have the same number of electrons. Therefore: NO+: 14 electrons, isoelectronic with N2 (7 + 7 = 14) NO-: 16 electrons, isoelectronic with O2 (8 + 8 = 16)

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Key Concepts

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

Molecular Orbital Diagram
Molecular orbital (MO) theory provides a means of understanding the electronic structure of molecules. A molecular orbital diagram visualizes the energy levels of molecular orbitals as they are formed from the combination of atomic orbitals. According to the MO theory, atomic orbitals of similar energy from different atoms can overlap to form molecular orbitals, which can be bonding or anti-bonding.

To depict this, we start with separate energy levels for the atomic orbitals of the two atoms and then draw the resulting molecular orbitals in between them. Electrons from atomic orbitals fill the molecular orbitals in order of increasing energy, following the aufbau principle and Hund's rule. In the case of NO, NO+, and NO-, we consider the atomic orbitals from nitrogen and oxygen, specifically the 2s and 2p orbitals, to understand their chemical bonding and properties.
Bond Order Calculation
The bond order is an index that provides insight into the strength and stability of a chemical bond within a molecule. It can be determined by subtracting the number of electrons in anti-bonding molecular orbitals from the number of electrons in bonding molecular orbitals and then dividing by two. Mathematically, it is denoted as:
\[ \text{Bond Order} = \frac{1}{2} \left( \text{{number of electrons in bonding MOs}} - \text{{number of electrons in anti-bonding MOs}} \right) \]
Using the molecular orbital diagrams for NO, NO+, and NO-, we can calculate the bond orders. Higher bond orders correspond to stronger and shorter bonds. For instance, NO+ with a bond order of 3 is more stable than NO with a bond order of 2.5. The bond order calculation helps predict molecular stability, bond length, and bond energy.
Paramagnetism and Diamagnetism
Paramagnetism and diamagnetism are properties that describe how a substance interacts with a magnetic field. Paramagnetic materials have one or more unpaired electrons and are attracted to a magnetic field. In contrast, diamagnetic materials, which have only paired electrons, weakly repel a magnetic field.

For molecules like NO, NO+, and NO-, their magnetic properties can be explained by the presence or absence of unpaired electrons within their molecular orbitals. NO, with one unpaired electron, exhibits paramagnetism, while NO+, which has all electrons paired, is diamagnetic. NO-, with two unpaired electrons, also shows paramagnetic behavior. These properties can be confirmed through experiments such as magnetic susceptibility measurements and provide practical confirmation of the electronic structures predicted by molecular orbital theory.
Isoelectronic Species
Isoelectronic species are atoms, ions, or molecules that have identical electron configurations - they contain the same number of electrons. This concept is crucial for understanding the relative stability and reactivity of various chemical species.

For the charged molecules NO+ and NO-, identifying isoelectronic neutral homonuclear diatomic molecules provides insight into their electronic structures. NO+ has 14 electrons, the same as N2, making them isoelectronic. Similarly, NO- has 16 electrons, which is the same number that O2 has, making them isoelectronic as well. Isoelectronic species often have similar chemical properties, which assists in predicting the behavior of ions based on well-understood neutral molecules.

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Most popular questions from this chapter

In ozone, \(\mathrm{O}_{3}\), the two oxygen atoms on the ends of the molecule are equivalent to one another. (a) What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the \(\pi\) electrons? (d) How many electrons are delocalized in the \(\pi\) system of ozone?

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{HCN}\), (b) \(\mathrm{SO}_{3}{ }^{2-}\), (c) \(S F_{4}\) (d) \(\mathrm{PF}_{6}\), (e) \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\), (f) \(\mathrm{N}_{3}^{-}\).

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) \(\mathrm{CO}_{2} ;\) (b) thiocyanate ion, \(\mathrm{NCS}^{-}\) : (c) formaldehyde, \(\mathrm{H}_{2} \mathrm{CO} ;\) (d) formic acid, \(\mathrm{HCOOH}\), which has one \(\mathrm{H}\) and two \(\mathrm{O}\) atoms attached to \(\mathrm{C}\).

There are two compounds of the formula \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) : The compound on the right, cisplatin, is used in cancer therapy. The compound on the left, transplatin, is ineffective for cancer therapy. Both compounds have a square-planar geometry. (a) Which compound has a nonzero dipole moment? (b) The reason cisplatin is a good anticancer drug is that it binds tightly to DNA. Cancer cells are rapidly dividing, producing a lot of DNA. Consequently cisplatin kills cancer cells at a faster rate than normal cells. However, since normal cells also are making DNA, cisplatin also attacks healthy cells, which leads to unpleasant side effects. The way both molecules bind to DNA involves the \(\mathrm{Cl}^{-}\) ions leaving the Pt ion, to be replaced by two nitrogens in DNA. Draw a picture in which a long vertical line represents a piece of DNA. Draw the \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}\) fragments of cisplatin and transplatin with the proper shape. Also draw them attaching to your DNA line. Can you explain from your drawing why the shape of the cisplatin causes it to bind to DNA more effectively than transplatin?

Use average bond enthalpies (Table 8.4) to estimate \(\Delta H\) for the atomization of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) : $$ \mathrm{C}_{6} \mathrm{H}_{6}(g) \longrightarrow 6 \mathrm{C}(g)+6 \mathrm{H}(g) $$ Compare the value to that obtained by using \(\Delta H_{f}^{\circ}\) data given in Appendix \(C\) and Hess's law. To what do you attribute the large discrepancy in the two values?

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