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

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, $$x-1=(2 \times Z) + 1$$.

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.

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 $${ }_{17}^{37} \ce{Cl}^{-1}$$