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Be sure to answer all parts. Use MO diagrams to place C2−, C2, C2+ in order of the following properties: (a) increasing bond energy C2− < C2 < C2+ C2− < C2+ < C2 C2 < C2− < C2+ C2 < C2+ < C2− C2+ < C2 < C2− C2+ < C2− < C2 (b) increasing bond length C2− < C2 < C2+ C2− < C2+ < C2 C2 < C2− < C2+ C2 < C2+ < C2− C2+ < C2 < C2− C2+ < C2− < C2

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Answer:

Bond Order = ½ [# bonded e--# antibonded e-]

C2= ½ [6-2] = 2

C2+= ½ [5-2] = 1.5

C2- = ½ [7-2] = 2.5

Bond Length: C2+> C2> C2-

Bond Energy: C2-> C2> C2+

Explanation:

Bond order is a measurement of the number of electrons involved in bonds between two atoms in a molecule.

How to calculate bond order

1. Draw the Lewis structure.

2. Count the total number of bonds.

3. Count the number of bond groups between individual atoms.

4. Divide the number of bonds between atoms by the total number of bond groups in the molecule.

Bond energy (E) is defined as the amount of energy required to break apart a mole of molecules into its component atoms. It is a measure of the strength of a chemical bond.

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The higher the bond order, the shorter the bond length and the greater the bond energy.

The molecular orbital configuration of a chemical specie shows the arrangement of electrons in such a specie into molecular orbitals in accordance with the Aufbau principle.

For C2;

σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz2

Bond order = 1/2(8 - 4) = 2

For C2^-;

σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz2 σ2px1

Bond order = 1/2(9 - 4) = 2.5

For C2^+;

σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz1

Bond order = 1/2(7 - 4) = 1.5

We must recall that the higher the bond order, the shorter the bond length and the greater the bond energy. Hence;

a) In order of increasing bond energy; C2^+ <  C2 < C2^-

b) In order of increasing bond length; C2^- < C2 < C2^+

Learn more: https://brainly.com/question/20906233

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