Respuesta :
Answer:
Bond order = 1
Explanation:
The 10 valence electrons can be added across bonding and antibonding molecular orbitals (MO) in a manner shown below:
- two 1s orbitals combine to form σ1s and σ*1s with 2 electrons in each MO
- the 3 p-orbitals combine to form σ2pz, σ*2pz with 2 electrons in each MO. In addition a degenerate π2py, π2px and π*2py, π*2px with a total of 4 electrons in each set of degenerate orbitals
Based on the energy level ordering the 10 electrons can be distributed across the bonding MO's as follows
(σ1s)²<(σ*1s)²<(σ2s)²<(σ*2s)²<(π2py)¹ = (π2px)¹
Bond order = 1/2(Bonding electrons - non-bonding electrons) = 6-4/2 = 1
The bond order in a molecule or ion with 10 valence electrons is 3.
Bond order is defined as the number of chemical bonds that exist between two atoms and it determines the bond's stability.
Valence electrons are the electrons located at the outer shell or energy level of an atom that is accountable for the chemical characteristics of the atom.
The formula for determining the bond order of a molecule can be expressed as:
[tex]\mathbf{Bond \ order = \dfrac{electrons \ in \ bonding \ orbitals - electrons \ i n \ antibonding \ orbitals}{2}}[/tex]
In a molecule or ion with 10 valence electrons, the outer electronic configuration is:
[tex]\mathbf{\sigma _{2s} ^2}[/tex] [tex]\mathbf{\sigma _{2s} ^2}[/tex] [tex]\mathbf{\sigma_{2p_z} ^2}[/tex] [tex]\mathbf{\pi _{2p_x} ^2}[/tex] [tex]\mathbf{\pi _{2p_y} ^2}[/tex]
Here;
- the bonding molecular orbital = 8
- the antibonding molecular orbital = 2
∴
[tex]\mathbf{Bond \ order = \dfrac{8 - 2}{2}}[/tex]
[tex]\mathbf{Bond \ order = \dfrac{6}{2}}[/tex]
Bond Order = 3
Therefore, the bond order in a molecule or ion with 10 valence electrons is 3.
Learn more about Bond order here:
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