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How many protons, neutrons, and electrons are in this ion?

[tex]\[ ^{103}_{45}\text{Rh}^{6+} \][/tex]

A. [tex]\( 45 \, p^{+}, \, 58 \, n^0, \, 45 \, e^{-} \)[/tex]

B. [tex]\( 45 \, p^{+}, \, 103 \, n^0, \, 45 \, e^{-} \)[/tex]

C. [tex]\( 39 \, p^{+}, \, 58 \, n^0, \, 45 \, e^{-} \)[/tex]

D. [tex]\( 45 \, p^{+}, \, 58 \, n^0, \, 39 \, e^{-} \)[/tex]



Answer :

GinnyAnswer
To determine the number of protons, neutrons, and electrons in the ion [tex]\(103 \text{Rh}^{6+}\)[/tex], we need to consider the following information:

1. Number of Protons (p⁺):
The atomic number of Rhodium (Rh) is 45. This means that a neutral atom of Rhodium contains 45 protons. Since changing the charge of the ion does not affect the number of protons, the ion [tex]\(103 \text{Rh}^{6+}\)[/tex] also contains 45 protons.

2. Number of Neutrons (n⁰ or n):
We are given that this isotope of Rhodium has a mass number of 103. The mass number (A) is the sum of protons (Z) and neutrons (N):

[tex]\[ A = Z + N \][/tex]

For [tex]\(\text{Rh}\)[/tex], [tex]\(A = 103\)[/tex] and [tex]\(Z = 45\)[/tex]:

[tex]\[ 103 = 45 + N \][/tex]

Solving for neutrons (N):

[tex]\[ N = 103 - 45 = 58 \][/tex]

Thus, the number of neutrons is 58.

3. Number of Electrons (e⁻):
In a neutral atom, the number of electrons is equal to the number of protons, which is 45 for Rhodium. However, this ion has a [tex]\(6+\)[/tex] charge, indicating that it has lost 6 electrons. Therefore:

[tex]\[ \text{Number of electrons} = 45 - 6 = 39 \][/tex]

Therefore, the number of protons, neutrons, and electrons in the ion [tex]\(103 \text{Rh}^{6+}\)[/tex] is:

- Protons: 45
- Neutrons: 58
- Electrons: 39

The last option,
- [tex]\(45 \, p^+, 58 \, n^{\circ}, 39 \, e\)[/tex], is the correct one.

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