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.

You are watching: How much negative charge is on each oxygen of the carbonate ion?


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,

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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,

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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