Question

Fill in the gaps in the following table: $$ \begin{array}{|l|l|c|c|c|} \hline \text { Symbol } & { }^{59} \mathrm{Co}^{3+} & & & \\ \hline \text { Protons } & & 34 & 76 & 80 \\ \hline \text { Neutrons } & & 46 & 116 & 120 \\ \hline \text { Electrons } & & 36 & & 78 \\ \hline \text { Net charge } & & & 2+ & \\ \hline \end{array} $$

Short answer

The completed table is: $$ \begin{array}{|l|l|c|c|c|} \hline \text { Symbol } & {}^{59} \mathrm{Co}^{3+} &\mathrm{Se} &\mathrm{Os}^{2+} &\mathrm{Hg} \\ \hline \text { Protons } & 27 & 34 & 76 & 80 \\ \hline \text { Neutrons } & 32 & 46 & 116 & 120 \\ \hline \text { Electrons } & 24 & 36 & 74 & 78 \\ \hline \text { Net charge } & 3+ & 0 & 2+ & 0 \\ \hline \end{array} $$

Step-by-step solution

Identify the given information.

From the symbol \({}^{59}\mathrm{Co}^{3+}\), we can determine that the species has \(59\) nucleons (protons and neutrons) and a net charge of \(3+\).

Determine the number of protons for the remaining species.

The atomic number, or the number of protons, is unique for each element. Here are the elements and their respective atomic numbers: - Cobalt (Co): 27 - Selenium (Se): 34 - Osmium (Os): 76 - Mercury (Hg): 80

Determine the number of neutrons for each species.

To find the number of neutrons, we will subtract the atomic number (number of protons) from the total number of nucleons (protons and neutrons). In the case of \({}^{59}\mathrm{Co}^{3+}\), we subtract 27 (number of protons) from 59 (total nucleons) to find that there are 32 neutrons.

Determine the number of electrons for the remaining species.

For a species to have a net charge, there must be an imbalance between its number of protons and electrons. To find the number of electrons for a species with a net charge, we will add the value of the net charge (in elementary charges) to the number of protons. - For the species with a net charge of \(2+\), we add the number of protons (76) and subtract the net charge (2) to find that there are 74 electrons.

Determine the net charge for the remaining species.

If the numbers of protons and electrons in a species are the same, it has a net charge of zero. With all this information, we can fill in the gaps in the table: $$ \begin{array}{|l|l|c|c|c|} \hline \text { Symbol } & {}^{59} \mathrm{Co}^{3+} &\mathrm{Se} &\mathrm{Os}^{2+} &\mathrm{Hg} \\ \hline \text { Protons } & 27 & 34 & 76 & 80 \\ \hline \text { Neutrons } & 32 & 46 & 116 & 120 \\ \hline \text { Electrons } & 24 & 36 & 74 & 78 \\ \hline \text { Net charge } & 3+ & 0 & 2+ & 0 \\ \hline \end{array} $$