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Lewis structures for the perchlorate ion (ClO4−) can be drawn with all single bonds or with one, two, or three double bonds. Draw each of these possible resonance forms, including any nonbonding electrons. Include the values of any nonzero formal charges. Use formal charges to determine the most important resonance structure and calculate its average bond order. slader

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Answer:

The most important resonance structure is 4 (attached picture). Its bon order is [tex]\frac{7}{4}[/tex] or [tex]1\frac{3}{4}[/tex].

Explanation:

A picture with 4 forms of the perchlorate structure is attached. The first structure has simple bonds. The second structure contains a double bond, the third structure has two double bonds and the fourth structure has three double bonds.

Formal charge = group number of the periodic table - number of bonds (number of bonding electrons / 2) - number of non-shared electrons (lone pairs)

The formal charges in the first structure is +3 in chlorine and -1 in oxygen.

The formal charges in the second structure is +2 in chlorine, -1 in oxygen and 0 in the double bond oxygen.

The formal charges in the third structure is +1 in chlorine, -1 in the single bond oxygens and 0 in the double bond oxygens.

The formal charges in the fourth structure is 0 in chlorine, -1 in the single bond oxygen and 0 in the double bond oxygens.

The most important resonance structure is given by:

  • Most atoms have 0 formal charge.
  • Lowest magnitude of formal charges.
  • If there is a negative formal charge, it's on the most electronegative atom.

Hence, the fourth structure is the mosr important.

The bond order of the structure is:

Total number of bonds: 7

Total number of bond groups: 4

Bond order= [tex]\frac{7}{4} =1\frac{3}{4}[/tex]

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The most stable resonance structure is the structure in which there are three double bonds.

Sometimes, one Lewis structure does not suffice in explaining the bonding and properties of a given ion or molecule. In such cases, we must invoke the idea of resonance. We must suppose that the actual structure of the molecule lies between two or more bonding extremes called resonance structures.

Looking at the canonical structures shown in the image, we can see that the most stable structure is the structure in which there exists the lowest formal charge and the negative charge is on the more electronegative oxygen atom.

The bond order in the structure that contains all single bonds is;

Bond order = Number of bonds/Number of bonded atoms or groups

Bond order = 4/4 = 0

For the structure that contains one double bond;

Bond order = Number of bonds/Number of bonded atoms or groups

Bond order = 5/4 = 1.25

For the structure that contains two double bonds;

Bond order = Number of bonds/Number of bonded atoms or groups

Bond order = 6/4 = 1.5

For the structure that contains three double bonds;

Bond order = Number of bonds/Number of bonded atoms or groups

Bond order = 7/4 = 1.75

Note that the higher the bond order the greater the stability of the bond. Hence the structure in which there are three double bonds is the most stable structure.

Learn more: https://brainly.com/question/6505878

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