/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 98 The nitrosyl ion, \(\mathrm{NO}^... [FREE SOLUTION] | 91Ó°ÊÓ

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The nitrosyl ion, \(\mathrm{NO}^{+}\), has an interesting chemistry. (a) Is \(\mathrm{NO}^{+}\) diamagnetic or paramagnetic? If paramagnetic, how many unpaired electrons does it have? (b) Assume the molecular orbital diagram for a homonuclear diatomic molecule applies to \(\mathrm{NO}^{+} .\) What is the highest-energy molecular orbital occupied by electrons? (c) What is the nitrogen-oxygen bond order? (d) Is the \(\mathrm{N}-\mathrm{O}\) bond in \(\mathrm{NO}^{+}\) stronger or weaker than the bond in \(\mathrm{NO}\) ?

Short Answer

Expert verified
(a) NO^{+} is diamagnetic. (b) Highest-energy MO: π_{2p_x} or π_{2p_y}. (c) Bond order is 3. (d) N-O bond is stronger in NO^{+}.

Step by step solution

01

Determine magnetic properties

To find if NO^{+} is diamagnetic or paramagnetic, examine its electronic configuration. NO^{+} has 14 electrons. Using the molecular orbital filling sequence for diatomic molecules (similar to N_{2}), we place these electrons in the following order: σ_{1s}^{2}, σ_{1s}^{'2}, σ_{2s}^{2}, σ_{2s}^{'2}, σ_{2p_z}^{2}, π_{2p_x}^{2}, π_{2p_y}^{2}, π_{2p_x}^{2}, π_{2p_y}^{2}. Each orbital is fully paired, leaving no unpaired electrons. Hence, NO^{+} is **diamagnetic**.
02

Identify the highest-energy molecular orbital

Given the molecular orbital configuration: σ_{1s}^{2}, σ_{1s}^{'2}, σ_{2s}^{2}, σ_{2s}^{'2}, σ_{2p_z}^{2}, π_{2p_x}^{2}, π_{2p_y}^{2}, π_{2p_x}^{2}, π_{2p_y}^{2}. The π_{2p_x} and π_{2p_y} orbitals are the highest-energy molecular orbitals occupied by electrons in NO^{+}.
03

Calculate bond order

The bond order can be calculated using the formula: Bond Order = \(\frac{1}{2}(\text{number of bonding electrons} - \text{number of antibonding electrons}) \). For NO^{+}, there are 10 bonding electrons and 4 antibonding electrons, giving a bond order of \(\frac{1}{2}(10 - 4) = 3 \). Thus, the bond order for NO^{+} is **3**.
04

Compare bond strength with NO

To determine the bond strength, compare the bond order of NO^{+} (which is 3) with that of NO, which has a bond order of 2.5. A higher bond order indicates a stronger bond. Hence, the N-O bond in NO^{+} is **stronger** than in NO.

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Key Concepts

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

Diamagnetism
Diamagnetism is the property of a molecule where all electrons are paired, causing it to be slightly repelled by a magnetic field. In contrast, paramagnetic substances contain unpaired electrons and are attracted to magnetic fields. In the case of the nitrosyl ion, \( \mathrm{NO}^{+} \), it contains 14 electrons. By following the molecular orbital sequence similar to that of \( \mathrm{N_2} \), the electrons fill up the orbitals as follows: \( \sigma_{1s}^{2}, \sigma_{1s}^{'2}, \sigma_{2s}^{2}, \sigma_{2s}^{'2}, \sigma_{2p_z}^{2}, \pi_{2p_x}^{2}, \pi_{2p_y}^{2}, \pi_{2p_x}^{2}, \pi_{2p_y}^{2} \). Each orbital has pairs of electrons, leaving no unpaired electrons. Therefore, \( \mathrm{NO}^{+} \) is diamagnetic. - Diamagnetic: All electrons are paired.- Paramagnetic: Unpaired electrons present.- \( \mathrm{NO}^{+} \): No unpaired electrons, so it is diamagnetic.
Bond Order
Bond order is a concept used to determine the strength and stability of a bond between two atoms. The bond order gives us an idea of how many bonds are present between two atoms. For diatomic molecules, it can be calculated using the formula:\[\text{Bond Order} = \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of antibonding electrons})\]In \( \mathrm{NO}^{+} \), there are 10 bonding electrons and 4 antibonding electrons. Plugging these numbers into the formula gives:\[\text{Bond Order} = \frac{1}{2} (10 - 4) = 3\]The bond order of 3 indicates a triple bond between the nitrogen and oxygen atoms. This high bond order suggests a strong bond, which correlates with the stability of the molecule. - Higher bond orders indicate stronger bonds.- \( \mathrm{NO}^{+} \) has a bond order of 3, indicating a strong triple bond.
Bond Strength
The bond strength is influenced by the bond order in a molecule. A higher bond order usually means a stronger bond, as there are more electrons holding the atoms together. Let's consider the bond strength in \( \mathrm{NO}^{+} \) compared to \( \mathrm{NO} \). In \( \mathrm{NO}^{+} \), we calculated a bond order of 3, whereas \( \mathrm{NO} \) has a bond order of 2.5. Because \( \mathrm{NO}^{+} \) has a higher bond order, its \( \mathrm{N-O} \) bond is stronger than that of \( \mathrm{NO} \).- Higher bond orders = Stronger bonds.- \( \mathrm{NO}^{+} \) (Bond order 3) > \( \mathrm{NO} \) (Bond order 2.5) in bond strength.
Homonuclear Diatomic Molecules
Homonuclear diatomic molecules are made of two identical atoms bonded together. Common diatomic molecules include \( \mathrm{N_2} \), \( \mathrm{O_2} \), etc. Although \( \mathrm{NO}^{+} \) is not homonuclear because it consists of nitrogen and oxygen, we use the molecular orbital diagram of homonuclear diatomic molecules to analyze its electron configuration. This approach is a simplification used in chemistry to help predict the behavior of heteronuclear diatomic molecules (like \( \mathrm{NO}^{+} \)). The reasoning is that the electron clouds of \( \mathrm{N} \) and \( \mathrm{O} \) are similar enough in nature to allow such an approximation.- Homonuclear: Two identical atoms bonded.- Example: Using homonuclear models can simplify analysis of heteronuclear diatomics like \( \mathrm{NO}^{+} \).

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