Chemistry, 03.03.2020 05:03 Cassandoo

Use molecular orbital theory to predict whether or not each of the following molecules or ions should exist in a relatively stable form. Drag the appropriate items to their respective bins. C2 2+ Be2 2+ Li2 Li2 2

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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: $\mathbf{= \dfrac{1}{2} \Big[no \ of \ electrons \ in \ B.O - No \ of \ electrons \ in \ Anti \ B.O \Big]}$

where:

B.O = Bonding Orbital

For C₂²⁺:

The carbon atom has an electronic configuration of 1s²2s²2p². It has 12 electrons. For the formation of C₂ from C₂²⁺, there is the removal of 2 electrons.

As such, C₂²⁺ has 10 electrons.

Now, the electronic configuration for the molecular orbital for C₂²⁺ can be written as:

$\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*2} < \pi _{2py}^{1}=\pi _{2pz}^{1} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{0} < \pi _{2px}^{0} }$

where;

the orbitals with (*) = antibonding orbitals the orbitals without (*) = bonding orbitals

Bond Order $= \mathbf{\dfrac{1}{2} (6-4)}$

$= \mathbf{\dfrac{1}{2} (2)}$

= 1

Thus, since Bond order = 1, the C₂²⁺ ion exist in a stable form

For Be₂²⁺ is a beryllium atom with a configuration is 1s² 2s². It has 8 electrons, for the formation of Be₂ from Be₂²⁺, there is the removal of 2 electrons. As such, Be₂²⁺ has 6 electrons, the electronic configuration for the molecular orbital can be expressed as:

$\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*0} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{0} < \pi _{2px}^{0} }$

Bond Order $= \mathbf{\dfrac{1}{2} (4-2)}$

$= \mathbf{\dfrac{1}{2} (2)}$

= 1

Thus, since Bond order = 1, the Be₂²⁺ ion exist in a stable form

Lithium Li₂ has an atomic number 3 with an electronic configuration 1s² 2s¹. Thus, it comprises 6 electrons. The electronic configuration of its molecular orbital is expressed as;

$\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*0} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{*0} < \pi _{2px}^{*0} }$

Bond Order $= \mathbf{\dfrac{1}{2} (4-2)}$

$= \mathbf{\dfrac{1}{2} (2)}$

= 1

Thus, since Bond order = 1, the Li₂ ion exists in a stable form

In Li₂²⁻, the number of electrons for its formation is 8; Since it needs 2 more electrons to its initial 6 electrons in Li₂. so, the electronic configuration of its molecular orbital can be expressed as;

$\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*2} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{*0} < \pi _{2px}^{*0} }$

Bond Order $= \mathbf{\dfrac{1}{2} (4-4)}$

$= \mathbf{\dfrac{1}{2} (0)}$

= 0

Thus, since Bond order = 0, the Li₂²⁻ ion does not exist in a stable form

Learn more about molecular orbital theory here:

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Use molecular orbital theory to predict whether or not each of the following molecules or ions shoul

Answer from: Woodsydal2390

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.

Answer from: Quest

the answer is c.) 1575 j.

explanation:

Answer from: Quest

gas

explanation:

smoke is the matter of a gas

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