An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion

Explanation

Suppose, the number of electrons present in the ion is x
No. of neutrons present in the ion `=x+((x xx 11.1)/(100))`
=1.111 x
Since, the ion contains one unit of negative charge, the number of electrons present in the parent atom = x-1
Sinc the ion contains 3 units of positive charge, the number of electrons present in the parent atom = x - 1
and, number of protons present in the parent atom = x -1
No. of protons + No. of neutrons = Mass number
`(x-1)+(1.11x=37`
`2.111x=38`
`x=(38)/(2.111)=18`
No. of protons = Atomic number
= x -1 =18 -1=17
Hence, the symbol of the ion is `""_(17)^(37)Cl^(-)`.