Question

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.

Answer 1

Assume the number of electrons in the ion carrying a negative charge be x.

Then, the  

The number of neutrons present is

x + 11.1% of x

= x + 0.111 x

= 1.111 x

The number of atoms in the neutral atom = (x – 1)

Note: When an ion carries a negative charge, it carries an extra electron

So, Number of protons in the neutral atom = x – 1

Therefore 37 = 1.111x + x -1

Or 2.111 x = 38

Or x = 18

Therefore the no of protons = atomic no = x -1 = 18 – 1 = 17

Therefore, the symbol of the ion is 37(UPPERCASE)Cl-17(LOWERCASE)

Answer 2

As the ion carries one unit of negative charge, it will have one electron more than the number of protons. Let the number of electrons

= x + x × 11.1 / 100 ; 37 = x - 1 + 1.111 x - 1 ; x = 38 / 2.111 = 18

Number of electrons = 18 ; Number of protons = 18 - 1 = 17 ; Number of neutrons = 37 - 17 = 20.

The ion under discussion is chloride and it's symbol is Cl−