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.

Solution:

Let the number of electrons in the ion carrying a negative charge be$x$

Then Number of neutrons present

$=x+11.1$

$=x+0.111x$

$=1.111x$

Number of electrons in the neutral atom $=(x-1)$

Number of protons in the neutral atom $=(x-1)$

Therefore ,

$37=1.111x+(x-1)$

$2.11x=38$

$x=18$

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

Therefore symbol of the ion is ${}_{17}^{37}{Cl}^{-1}$