Certainly! Let's break down the solution step-by-step for the ion \( \mathrm{{}^{103}_{45}Rh^{6+}} \).
1. Identify the number of protons:
- The atomic number (Z) of Rhodium (\(Rh\)) is 45. This number represents the number of protons in the nucleus of an atom of Rhodium.
- Therefore, the number of protons is 45.
2. Determine the mass number and calculate the number of neutrons:
- The mass number (A) of the given isotope is 103. The mass number is the sum of protons and neutrons in the nucleus.
- To find the number of neutrons, we subtract the atomic number (protons) from the mass number:
[tex]\[
\text{Number of neutrons} = \text{Mass number} - \text{Atomic number} = 103 - 45 = 58
\][/tex]
- Hence, the number of neutrons is 58.
3. Calculate the number of electrons:
- The ion in question is a \( \mathrm{Rh^{6+}} \) ion. This means the ion has a +6 charge, indicating it has lost 6 electrons compared to the neutral atom.
- A neutral atom of Rhodium would have the same number of electrons as protons, which is 45.
- Since the ion has a +6 charge, it has 6 fewer electrons than the neutral atom:
[tex]\[
\text{Number of electrons} = \text{Number of protons} - \text{Ion charge} = 45 - 6 = 39
\][/tex]
- Consequently, the number of electrons in the \( \mathrm{Rh^{6+}} \) ion is 39.
So, summarizing the information:
- Number of protons: 45
- Number of neutrons: 58
- Number of electrons: 39
