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https://pubs.acs.org/doi/10.1021/acsreagents.4235

in naphthol green b, sulphate ion is the ligand, there are 3 sulphate ion, the charge of sulphate ion is -2, totally -2*3=-6

so i put -6 is the charge and multiplicity is 7.

it would not work out in gaussian 03

Likewise i put charge equals -3 and multiplicity equals 4 for naphthol green b. but the link died for both with #2070

can you provide any solution

Naphtol green b is the sodium salt of naohtol green y. The coordination is supposed to be performed by th N in the NO substituent and the deprotonated phenol function. Charge is zero, multiplicity depends on whether Fe(iii) is high-spin or low-spin so either 2 or 6