Question

ground state electron configuration for NF, NF+, NF-

bond orders

diagmetic or paragmetic

Answer 1

Concepts and reason

The concept used to solve this problem is based on molecular orbital theory.

According to this theory molecular orbitals of a molecule is formed by combination of its atomic orbitals and electrons are distributed among the molecular orbitals. The bond order of any molecule is calculated by using electrons which are distributed in molecular orbital diagram.

If there are no unpaired electrons in the orbital, then molecule is diamagnetic. If there are unpaired electrons in orbital, then molecule is paramagnetic.

Fundamentals

The formula for bond order is written as follows:

${B_{\rm{O}}}{\rm{ = }}\frac{1}{2}\left( {{n_{\rm{b}}} - {n_{\rm{a}}}} \right)$

Here, ${n_{\rm{b}}}$ is total number of electrons present in bonding orbital and ${n_{\rm{a}}}$ is total number of electrons present in antibonding orbital.

Part 1.1

The atomic number of nitrogen is 7 and fluorine is 9. Thus, total number of electrons in NF is 16. Thus, 16 electrons are filled as follows:

$\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^2}$

Part 1.2

Total number of electrons in bonding orbitals is 10 and total number of electron in antibonding orbital is 6.

Thus, substitute 10 for ${n_{\rm{b}}}$ and 6 for ${n_{\rm{a}}}$ .

$\begin{array}{c}\\{B_{\rm{O}}}{\rm{ = }}\frac{1}{2}\left( {10 - 6} \right)\\\\ = \frac{4}{2}\\\\ = 2\\\end{array}$

Part 1.3

The ground state electronic configuration of NF is as follows:

$\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^2}$

From this electronic configuration, there are two unpaired electrons present in $\pi \left( {2{p^ * }} \right)$ orbital. Thus, NF is paramagnetic.

Part 2.1

The atomic number of nitrogen is 7 and fluorine is 9. Thus, total number of electrons in ${\rm{N}}{{\rm{F}}^ + }$ is 15. Thus, 15 electrons are filled in the blanks as follow.

$\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^1}$

Part 2.2

In ${\rm{N}}{{\rm{F}}^ + }$ , total number of electrons in bonding orbitals is 10 and total number of electron in antibonding orbital is 5.

Thus, substitute 10 for ${n_{\rm{b}}}$ and 5 for ${n_{\rm{a}}}$ .

$\begin{array}{c}\\{B_{\rm{O}}}{\rm{ = }}\frac{1}{2}\left( {10 - 5} \right)\\\\ = \frac{5}{2}\\\\ = 2.5\\\end{array}$

Part 2.3

The ground state electronic configuration of ${\rm{N}}{{\rm{F}}^ + }$ is as follows:

$\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^1}$

From this electronic configuration, there are one unpaired electron present in $\pi \left( {2{p^ * }} \right)$ orbital. Thus, ${\rm{N}}{{\rm{F}}^ + }$ is paramagnetic.

Part 3.1

The atomic number of nitrogen is 7 and fluorine is 9. Thus, total number of electrons in ${\rm{N}}{{\rm{F}}^ - }$ is 17. Thus, 17 electrons are filled in the blanks as follow.

$\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^3}$

Part 3.2

In ${\rm{N}}{{\rm{F}}^ - }$ , total number of electrons in bonding orbitals is 10 and total number of electron in antibonding orbital is 7.

Thus, substitute 10 for ${n_{\rm{b}}}$ and 7 for ${n_{\rm{a}}}$ .

$\begin{array}{c}\\{B_{\rm{O}}}{\rm{ = }}\frac{1}{2}\left( {10 - 7} \right)\\\\ = \frac{3}{2}\\\\ = 1.5\\\end{array}$

Part 3.3

The ground state electronic configuration of ${\rm{N}}{{\rm{F}}^ - }$ is as follows:

$\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^3}$

From this electronic configuration, there are one unpaired electron present in $\pi \left( {2{p^ * }} \right)$ orbital. Thus, ${\rm{N}}{{\rm{F}}^ - }$ is paramagnetic.

Ans: Part 1.1

The ground state electronic configuration of NF is as follows:

$\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^2}$

Part 1.2

The bond order of NF is 2.

Part 1.3

The molecule NF is paramagnetic.

Part 2.1

The ground state electronic configuration of ${\rm{N}}{{\rm{F}}^ + }$ is as follow.

$\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^1}$

Part 2.2

The bond order of ${\rm{N}}{{\rm{F}}^ + }$ is 2.5

Part 2.3

The molecule ${\rm{N}}{{\rm{F}}^ + }$ is paramagnetic.

Part 3.1

The ground state electronic configuration of ${\rm{N}}{{\rm{F}}^ - }$ is as follow.

$\sigma {\left( {1s} \right)^2}\sigma {\left( {1{s^ * }} \right)^2}\sigma {\left( {2s} \right)^2}\sigma {\left( {2{s^ * }} \right)^2}\pi {\left( {2p} \right)^4}\sigma {\left( {2p} \right)^2}\pi {\left( {2{p^ * }} \right)^3}$

Part 3.2

The bond order of ${\rm{N}}{{\rm{F}}^ - }$ is 1.5

Part 3.3

The molecule ${\rm{N}}{{\rm{F}}^ - }$ is paramagnetic.