Question

Four \(\mathrm{Fe}^{2+}\) ions are key components of hemoglobin, the protein that transports oxygen in the blood. Assuming that these ions are \(^{53} \mathrm{Fe}^{2+}\), how many protons and neutrons are present in each nucleus, and how many electrons are present in each ion?

Short answer

In each nucleus of the \(^{53}\mathrm{Fe}^{2+}\) ion, there are 26 protons and 27 neutrons. Each ion has 24 electrons.

Step-by-step solution

Determine the atomic number and mass number for \(^{53}\mathrm{Fe}^{2+}\)

First, let us determine the atomic number (Z) and mass number (A) for \(^{53}\mathrm{Fe}^{2+}\). The atomic number of iron (Fe) is Z = 26, which represents the number of protons in the nucleus. The mass number (A) is the sum of protons and neutrons in the nucleus. The given isotope of iron is \(^{53}\mathrm{Fe}\), so A = 53.

Determine the number of neutrons

Now that we have both the atomic number (Z) and the mass number (A), we can determine the number of neutrons present in the nucleus. The number of neutrons (N) can be found using the following formula: \(N = A - Z\) Substitute the values for A and Z: \(N = 53 - 26\) \(N = 27\) There are 27 neutrons in the nucleus of each \(^{53}\mathrm{Fe}\) isotope.

Determine the number of electrons in \(^{53}\mathrm{Fe}^{2+}\) ion

The given ion is \(\mathrm{Fe}^{2+}\), which means that two electrons have been removed to produce a positive charge of +2. To find the number of electrons (e) in each ion, subtract the charge from the atomic number (Z) as following: \(e = Z - Charge\) \(e = 26 - 2\) \(e = 24\) There are 24 electrons in each \(^{53}\mathrm{Fe}^{2+}\) ion.

Final Answer

In each nucleus of the \(^{53}\mathrm{Fe}^{2+}\) ion, there are 26 protons and 27 neutrons. Each ion has 24 electrons.