Question

Use molecular orbital theory to complete this table BY FILLING FLANKS, 0,1,2,3, or 4

NF = (?1s) ___(?1s*) ___(?2s) ___(?2s*) ___(? 2p)___ (?2p) ___(? 2p*) Bonding order=

NF+ =(?1s) ___(?1s*) ___(?2s)___ (?2s*) ___(? 2p) ___(?2p) ___(? 2p*) Bonding order=

NF- = (?1s)___ (?1s*)___ (?2s) ___(?2s*)___ (? 2p)___ (?2p) ___(? 2p*) Bonding order=

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.

Fundamentals

The bond order of any molecule is calculated by using electrons which are distributed in molecular orbital diagram.

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

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 a1

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 in the blanks as follow.

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

Part a2

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 b1

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 b2

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 c1

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 c2

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 ${n_{\rm{a}}}$ .

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

Ans: Part a1

The electronic configuration of NF is as follow.

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

Part a2

The bond order of NF is 2.

Part b1

The 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 b2

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

Part c1

The 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 c2

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