Consider the resonance frameworks for the carbonate ion. Just how much an unfavorable charge is on every oxygen the the carbonate ion? What is the shortcut order of each carbon-oxygen bond in the rbonate ion?

Carbonate ion acts together a base and it can form three resonance structures. If a compound has more number that resonance frameworks then the is more stable.

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Negative fee on every oxygen atom have the right to be calculated by dividing the charge existing oxygen atom with number of oxygen atoms.

Calculate the shortcut order of each C-O bond by using the following expression.

Bondorder=TotalnumberofbondsNumberofC−Obonds\rmBond\,\rmorder\,\rm = \,\frac\rmTotal\,\rmnumber\,\rmof\,\rmbonds\rmNumber\,\rmof\,\rmC - O\,\rmbondsBondorder=NumberofC−ObondsTotalnumberofbonds​

Part a

The resonance structures of carbonate ion are,


Here, two oxygen atoms have actually a fee of −1 - 1−1 and other one is neutral.

So,thenegativechargeoneachoxygenatom=−2charge3=−0.67\beginarrayc\\\rmSo,\,\rmthe\,\rmnegative\,\rmcharge\,\rmon\,\rmeach\,\rmoxygen\,\rmatom\,\rm = \,\frac - \rm2\,\rmcharge\rm3\\\\ & & & & & \,\,\,\,\,\,\,\,\,\,\,\,\,\rm = \, - \rm0\rm.67\\\endarraySo,thenegativechargeoneachoxygenatom=3−2charge​​​​​​=−0.67​

Part b

The resonance structures of lead carbonate ion are,


Total number of bonds existing in the lead carbonate ion is four.

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Number the C-O (i.e. Solitary bonds) binding in lead carbonate ion is three.

Bondorder=TotalnumberofbondsNumberofC−Obonds=43=1.33\beginarrayc\\\rmBond\,\rmorder\,\rm = \,\frac\rmTotal\,\rmnumber\,\rmof\,\rmbonds\rmNumber\,\rmof\,\rmC - O\,\rmbonds\\\\\,\rm = \,\frac43\\\\ = \,1.33\\\endarrayBondorder=NumberofC−ObondsTotalnumberofbonds​=34​=1.33​

Ans: component a

Thus, the an adverse charge on every oxygen atom of the carbonate ion is −0.67 - \rm0\rm.67−0.67

Part b

Thus, the link order at each carbon-oxygen shortcut in the carbonate ion is 1.331.331.33