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

Solution:

Step 1: Calculate the number of electrons in the ion.

Let the number of electrons in the ion be x.

Since the ion has one unit of negative charge, it must have one more electron than the number of protons. Therefore, the number of protons in the ion is (x - 1).

The atomic number of the ion is the number of protons, which is (x - 1).

Step 2: Calculate the number of neutrons in the ion.

The mass number of the ion is 37, which is the sum of the number of protons and neutrons in the ion.

Mass number = number of protons + number of neutrons

37 = (x - 1) + number of neutrons

number of neutrons = 38 - x

Since the ion contains 11.1% more neutrons than electrons, we can write:

(number of neutrons) = 1.111 * (number of electrons)

Substituting the value of (number of neutrons) from above, we get:

38 - x = 1.111x

Solving for x, we get:

x = 14.5

Therefore, the number of electrons in the ion is 14.5.

Step 3: Determine the symbol of the ion.

The atomic number of the ion is (x - 1) = 13.5

Since the atomic number of an element determines its identity, we can conclude that the ion is of aluminum (Al), which has an atomic number of 13.

Therefore, the symbol of the ion is Al^-.