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Use molecular orbital theory to complete the ground state electron configuration for each of the molecules:
a) NF bond orders
b) NF+ bond orders
c) NF- bond orders
Diagmetic or Paragmetic
Molecular orbital theories determines the bond’s order and the electronic structure of homonuclear and heteronuclear molecules. The ability to predict the right order of energies is very difficult for heteronuclear diatomic molecules. This is due to the fact that the electron orbitals of various energies are mixed to create molecular orbitals.♦ Relevant knowledgeHeteronuclear molecules:
This problem was solved using molecular orbital theory.
This theory states that molecular orbitals are formed from the combination of an atom orbital and electrons distributed among molecular orbitals. Calculating the bond order for any molecule can be done by using electrons that are distributed in a molecular orbital diagram.
If the orbital contains no unpaired electrons, then the molecule is diamagnetic. Paramagnetic is when there are no unpaired electrons within the orbital.
This is the formula for bond orders:
[katex]B_O = \frac12\left( n_b – n_a \right)[/katex]
Here, [katex]{n_{\rm{b}}}[/katex] represents the an amount of electrons that are present in bonding orbital, [katex]{n_{\rm{a}}}[/katex] the number of electrons in the antibonding orbital.
Part 1.1
Fluorine has a 9 and nitrogen has 7. The total number of electrons found in NF is 16 The 16 electrons in NF are thus filled as follows:
[katex]\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^2}[/katex]
Part 1.2
The total number of electrons in bonding orbitals consists of 10 and the total number of electrons within antibonding orbitals consists of 6.
Substitute 10 for [katex]{n_{\rm{b}}}[/katex] and 6 for [katex]{n_{\rm{a}}}[/katex]
[katex]B_O = \frac12\left( 10 – 6 \right)\\\\ = \frac42\\\\ = 2\\\\[/katex]
Part 1.3
This is the ground state electronic configuration for NF:
[katex]\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^2}[/katex]
This electronic configuration contains two unpaired electrons in [katex]\pi \left( {2{p^ * }} \right)[/katex]. So, NF is paramagnetic.
Part 2.1
Fluorine has a 9 and nitrogen has 7. [katex]{\rm{N}}{{\rm{F}}^ + }[/katex] has 15 electrons. So, 15 electrons are used to fill in the blanks, according to the following.
[katex]\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^1}[/katex]
Part 2.2
In [katex]{\rm{N}}{{\rm{F}}^ + }[/katex], the total number of electrons within bonds orbitals is 10, and the number of electrons in antibonding orbitals is 5.
Substitute 10 for [katex]{n_{\rm{b}}}[/katex] and 5 for [katex]{n_{\rm{a}}}[/katex]
[katex]\begin{array}{c}\\{B_{\rm{O}}}{\rm{ = }}\frac{1}{2}\left( {10 – 5} \right)\\\\ = \frac{5}{2}\\\\ = 2.5\\\end{array}[/katex]
Part 2.3
The [katex]{\rm{N}}{{\rm{F}}^ + }[/katex] ground state electronic configuration
[katex]\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^1}[/katex]
This electronic configuration has one unpaired electron in [katex]\pi \left( {2{p^ * }} \right)[/katex]. Thus, [katex]{\rm{N}}{{\rm{F}}^ + }[/katex] is paramagnetic.
Part 3.1
Fluorine has a 9 and nitrogen has 7. [katex]{\rm{N}}{{\rm{F}}^ – }[/katex] has 17 electrons. So, 17 electrons are used to fill in the blanks, according to the following.
[katex]\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^3}[/katex]
Part 3.2
In [katex]{\rm{N}}{{\rm{F}}^ – }[/katex], the total number of electrons within bonding orbitals is 10 . The the number of electrons in antibonding orbitals is 7.
Substitute 10 for [katex]{n_{\rm{b}}}[/katex] and 7 for [katex]{n_{\rm{a}}}[/katex]
[katex]\begin{array}{c}\\{B_{\rm{O}}}{\rm{ = }}\frac{1}{2}\left( {10 – 7} \right)\\\\ = \frac{3}{2}\\\\ = 1.5\\\end{array}[/katex]
Part 3.3
The [katex]{\rm{N}}{{\rm{F}}^ – }[/katex] ground state electronic configuration
[katex]\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^3}[/katex]
This electronic configuration contains one unpaired electron in [katex]\pi \left( {2{p^ * }} \right)[/katex]. So, [katex]{\rm{N}}{{\rm{F}}^ – }[/katex] is paramagnetic.
Answer
Part 1.1
This is the ground state electronic configuration for NF:
[katex]\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^2}[/katex]
Part 1.2
The bond order for NF is 2.
Part 1.3
Paramagnetic is the molecule NF.
Part 2.1
The [katex]{\rm{N}}{{\rm{F}}^ + }[/katex] ground state electronic configuration
[katex]\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^1}[/katex]
Part 2.2
The bond order [katex]{\rm{N}}{{\rm{F}}^ + }[/katex] is 2.5
Part 2.3
The molecule [katex]{\rm{N}}{{\rm{F}}^ + }[/katex] is paramagnetic.
Part 3.1
The [katex]{\rm{N}}{{\rm{F}}^ – }[/katex] ground state electronic configuration
[katex]\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^3}[/katex]
Part 3.2
The bond order [katex]{\rm{N}}{{\rm{F}}^ – }[/katex] is 1.5
Part 3.3
The molecule [katex]{\rm{N}}{{\rm{F}}^ – }[/katex] is paramagnetic.