Find the ion with mass number 37 if the ion contains 11.1% more neutrons than the electrons

Question

An ion with mass number 37 possesses one unit of negative charge. The ion contains 11.1% more neutrons than the electrons in its nucleus. An ion with a mass of 37 has % negative charge compared to % positive charge.

With the aid of Periodic table, I doubt if the given ion is Chloride $Cl^{-}$, but I am not sure. My attempt is as follows:

Let the ion be $X^{-}$, and let the number of electrons be $x$, then the number of protons will be $x-1$ as the ion carries one unit of negative charge.

Now I know that $A=(2 \times Z)$ if $Z$ is even and $A=(2 \times Z) + 1$ if $Z$ is odd, where $Z$ and $A$ are atomic number and mass number respectively. Therefore, $x-1=(2 \times Z) + 1$.

Answer 1

You are overly complicating the solution with even or odd atomic numbers.

It is by definition that $Z+N=A$ where $A=$ atomic mass number, $Z=$ atomic number, and $N=$ number of neutrons. Thus,

$$Z+N=37 \tag1$$

If the number of electrons in the ion is $E$, then $$E=Z+1 \tag2$$ because the ion has a negative charge (otherwise $E=Z$ for a neutral element).

And, $$N=(1+0.111)E \tag3$$ This is from the condition.

You have three equations and three variables. If you solve the three linear equations, you will know the value $Z$. This defines the element. Look, it up in the periodic table. $A$ is already given and you have a negative charge. Remember mass number does not define the element, only $Z$ does.

Your intuition about chlorine is right. All you have to do now is to prove it.

Answer 2

Let the number of electrons in the ion carrying a negative charge bex Then Number of neutrons present $$ \begin{array}{l} =\mathrm{x}+11.1 \% \mathrm{of} \mathrm{x} \\ =\mathrm{x}+0.111 \mathrm{x} \\ =1.111 \mathrm{x} \end{array} $$ Number of electrons in the neutral atom $=(x-1)$ Number of protons in the neutral atom $=(x-1)$ Therefore, $$ \begin{array}{l} 37=1.111 x+(x-1) \\ 2.11 x=38 \\ x=18 \end{array} $$ Therefore no of protons $=$ atomic no $=x-1=18-1=17$ Therefore symbol of the ion is ${ }_{17}^{37} \ce{Cl}^{-1}$