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The nitrosonium ion, [tex]NO^{+}[/tex], forms a number of interesting complexes with nickel, cobalt, and iron. According to molecular orbital theory, which statement about [tex]NO^{+}[/tex] is correct?

Select one:
A. [tex]NO^{+}[/tex] has a bond order of 3 and is paramagnetic.
B. [tex]NO^{+}[/tex] has a bond order of 2 and is paramagnetic.
C. [tex]NO^{+}[/tex] has a bond order of 3 and is diamagnetic.
D. [tex]NO^{+}[/tex] has a bond order of 2.5 and is neither paramagnetic nor diamagnetic.
E. [tex]NO^{+}[/tex] has a bond order of 2 and is diamagnetic.

Sagot :

GinnyAnsweridn
To determine the correct statement about the nitrosonium ion [tex]\(NO^+\)[/tex] according to molecular orbital theory, follow these steps:

1. Determine the Electron Configuration of the Nitrogen Monoxide Molecule (NO):
- The total number of valence electrons for NO is the sum of the valence electrons of nitrogen (N) and oxygen (O).
- Nitrogen has 5 valence electrons.
- Oxygen has 6 valence electrons.
- Total [tex]\( = 5 + 6 = 11 \)[/tex] valence electrons.

2. Form the Valence Electron Configuration of [tex]\(NO^+\)[/tex]:
- Since the nitrosonium ion ([tex]\(NO^+\)[/tex]) has lost one electron, it has [tex]\(11 - 1 = 10\)[/tex] valence electrons.

3. Construct the Molecular Orbital (MO) Diagram:
- For diatomic molecules involving elements in the second period (like N and O), we use the standard molecular orbital diagram for homonuclear diatomic molecules, adapting it slightly for heteronuclear situations.

- The appropriate MO filling order for [tex]\(NO^+\)[/tex] is:
[tex]\[ (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x}, \pi_{2p_y})^4 \][/tex]
- This results in the first two MOs ([tex]\(\sigma_{2s}\)[/tex] and [tex]\(\sigma_{2s}^*\)[/tex]) being filled with 4 electrons.
- The [tex]\( \sigma_{2p_z} \)[/tex] bond is filled with 2 electrons.
- The rest 4 electrons are filled in the degenerate [tex]\( \pi_{2p_x} \)[/tex] and [tex]\( \pi_{2p_y} \)[/tex] bonds.

4. Calculate the Bond Order:
- The bond order formula is:
[tex]\[ \text{Bond Order} = \frac{\text{(Number of bonding electrons)} - \text{(Number of antibonding electrons)}}{2} \][/tex]
- From the MO diagram:
- Bonding electrons: [tex]\( 8 \)[/tex] (2 from [tex]\(\sigma_{2s}\)[/tex], 2 from [tex]\(\sigma_{2p_z}\)[/tex], and 4 from [tex]\(\pi_{2p_x,2p_y}\)[/tex])
- Antibonding electrons: [tex]\(2\)[/tex] (from [tex]\(\sigma^_{2s}\)[/tex] and none in antibonding [tex]\( \pi^\)[/tex]).
- Therefore, the bond order is:
[tex]\[ \text{Bond Order} = \frac{8 - 2} {2} = 3 \][/tex]

5. Determine Magnetism:
- Diamagnetic species have all electrons paired.
- Paramagnetic species have unpaired electrons.
- In [tex]\(NO^+\)[/tex], there are no unpaired electrons.

Therefore, [tex]\(NO^+\)[/tex] has a bond order of 3 and is diamagnetic. The correct answer is:

C. [tex]\(NO^+\)[/tex] has a bond order of 3 and is diamagnetic.
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