Step 1: Drawing the resonance structures of CO\(_3\)\(^{2-}\).
The Lewis structure of the carbonate ion shows a central carbon atom double-bonded to one oxygen atom and single-bonded to two other oxygen atoms, each carrying a negative charge. The double bond can be placed on any of the three oxygen atoms, leading to three equivalent resonance structures.
Step 2: Evaluating each statement.
(a) All C-O bonds in CO\(_3\)\(^{2-}\) are equivalent.
The actual structure of the carbonate ion is a resonance hybrid of the three structures. The \(\pi\)-electrons of the double bond are delocalized over the entire ion. This makes all three C-O bonds identical, with a bond order of 1.33. They have the same length and strength, intermediate between a single and a double bond. Thus, statement (a) is correct.
(b) There are three resonance structures possible for CO\(_3\)\(^{2-}\) ion.
As shown in Step 1, we can draw three distinct, equivalent resonance structures by delocalizing the double bond and negative charges. Thus, statement (b) is correct.
(c) The position of carbon and oxygen should change in every resonance structure.
This violates a fundamental rule of resonance. In resonance structures, only the positions of electrons (lone pairs and \(\pi\) bonds) change; the arrangement of the atomic nuclei must remain the same. Thus, statement (c) is incorrect.
(d) The formal charge on carbon atom is -2.
The formula for formal charge is: FC = [Valence e\(^-\)] - [Non-bonding e\(^-\)] - \(\frac{1}{2}\)[Bonding e\(^-\)].
For the central carbon atom in any resonance structure:
Valence electrons of C = 4.
Non-bonding electrons = 0.
Bonding electrons = 8 (it forms 4 bonds).
FC(C) = 4 - 0 - \(\frac{1}{2}\)(8) = 4 - 4 = 0.
The formal charge on the carbon atom is 0. Thus, statement (d) is incorrect.
Step 3: Final Answer.
Statements (a) and (b) are correct.