The carbonate ion CO_3^{2-} has one carbon atom bonded to three oxygen atoms with an overall 2- charge. If we draw a single Lewis structure, we get one C=O double bond and two C-O single bonds, which would predict two different bond lengths. However, experimentally, all three C-O bonds in CO_3^{2-} are identical at approximately 1.30 Å (intermediate between a C-O single bond at 1.43 Å and a C=O double bond at 1.23 Å). This is explained by the existence of three equivalent resonance structures, each differing only in the position of the double bond. The true structure is a resonance hybrid: the pi electron (from the C=O bond) is delocalised over all three C-O bonds equally. Since one pi bond (bond order contribution = 1) is shared equally among three bonds, each C-O bond has bond order = 1 (sigma) + 1/3 (pi) = 4/3 = 1.33. The three oxygen atoms in CO_3^{2-} are all equivalent and have the same hybridisation (sp^2). Carbon in CO_3^{2-} is sp^2 hybridised (not sp^3), with one unhybridised p orbital participating in pi delocalisation. Ionic bonding does not apply here since C-O bonds in carbonate are covalent. This is a key NCERT resonance concept tested in JEE. Plausibility: the equal C-O bond lengths in CO_3^{2-} have been confirmed by X-ray crystallography and confirm complete delocalisation, directly supporting a bond order of 1.33.