Answer:
C2 2+ stable and should exist
Be2 2+ stable and should exit
Li2 stable and should exit
Li2 2- unstable and doesn't exist
Explanation:
The first step in predicting the stability of a specie is knowing its molecular orbital configuration and bond order. If the Specie has a bond order of one or more, it is expected to exist in a stable form. The image attached shows the bond order and molecular orbital configuration of all the species mentioned in the question. This will aid you in understanding the stability of each specie.
The ions or molecule C2 2+ Be2 2+ Li2 exists in stable form but Li2 2 does not exist in stable form when using molecular orbital theory to predict their stability.
The production of molecular orbitals arises by combining atomic orbitals in a linear array. Based on the electronic configuration of molecular orbitals, the stability of molecules or ions can be determined by calculating the bond order for each molecule.
The bond order can be expressed as: [tex]\mathbf{= \dfrac{1}{2} \Big[no \ of \ electrons \ in \ B.O - No \ of \ electrons \ in \ Anti \ B.O \Big]}[/tex]
where:
For CāĀ²āŗ:
The carbon atom has an electronic configuration of 1sĀ²2sĀ²2pĀ². It has 12 electrons. For the formation of Cā Ā from CāĀ²āŗ, there is the removal of 2 electrons.
As such, CāĀ²āŗ has 10 electrons.
Now, the electronic configuration for the molecular orbital for CāĀ²āŗ can be written as:
[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*2} < \pi _{2py}^{1}=\pi _{2pz}^{1} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{0} < \pi _{2px}^{0} }[/tex]
where;
Bond Order [tex]= \mathbf{\dfrac{1}{2} (6-4)}[/tex]
[tex]= \mathbf{\dfrac{1}{2} (2)}[/tex]
= 1
Thus, since Bond order = 1, the CāĀ²āŗ Ā ion exist Ā in a stable form
For BeāĀ²āŗ is a beryllium atom with a configuration is 1sĀ² 2sĀ². It has 8 electrons, for the formation of Beā from BeāĀ²āŗ, there is the removal of 2 electrons. As such, BeāĀ²āŗ has 6 electrons, the electronic configuration for the molecular orbital can be expressed as:
[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*0} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{0} < \pi _{2px}^{0} }[/tex]
Bond Order [tex]= \mathbf{\dfrac{1}{2} (4-2)}[/tex]
[tex]= \mathbf{\dfrac{1}{2} (2)}[/tex]
= 1
Thus, since Bond order = 1, the BeāĀ²āŗ Ā ion exist in a stable form
Lithium Liā has an atomic number 3 with an electronic configuration 1sĀ² 2sĀ¹. Thus, it comprises 6 electrons. The electronic configuration of its molecular orbital is expressed as;
[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*0} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{*0} < \pi _{2px}^{*0} }[/tex]
Bond Order[tex]= \mathbf{\dfrac{1}{2} (4-2)}[/tex]
[tex]= \mathbf{\dfrac{1}{2} (2)}[/tex]
= 1
Thus, since Bond order = 1, the Liā Ā ion exists in a stable form
In LiāĀ²ā», the number of electrons for its formation is 8; Since it needs 2 more electrons to its initial 6 electrons in Liā. so, the electronic configuration of its molecular orbital can be expressed as;
[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*2} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{*0} < \pi _{2px}^{*0} }[/tex]
Bond Order [tex]= \mathbf{\dfrac{1}{2} (4-4)}[/tex]
[tex]= \mathbf{\dfrac{1}{2} (0)}[/tex]
= 0
Thus, since Bond order = 0, the LiāĀ²ā» Ā ion does not exist in a stable form
Learn more about molecular orbital theory here:
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