Suppose number of electrons in the ion, `M^(3+) = x :.` No. of neutrons `= x + (30.4)/(100) x = 1.304 x`
No. of electrons in the nuetral atom `= x + 3 " " :.` No. of protons `= x + 3`
Mass no. = No. of protons + No. of neutrons
`56 = x + 3 + 1.304 x or 2.304 x = 53 or x = 23 " " :.` No. of protons = Atominc no. `= x + 3 = 23 + 3 = 26`
Hence, the symbol of the ion will be `._(26)^(56)Fe^(3+)`