Question

Write the symbol for each of the following ions: (a) the ion with a \(1 + \) charge, atomic number \(55,\) and mass number \(133\)(b) the ion with \(54\) electrons, \(53\) protons, and \(74\) neutrons (c) the ion with atomic number\(15,\) mass number \(31,\) and a \(3 - \) charge (d) the ion with \(24\) electrons, \(30\) neutrons, and a \(3 + \)charge

Short answer

The symbols are:

a. \({}_{55}^{133}C{s^{ + 1}}\)

b. \({}_{53}^{127}{I^{ - 1}}\)

c. \({}_{15}^{31}{P^{ - 3}}\)

d. \({}_{27}^{57}C{o^{ + 3}}\)

Step-by-step solution

Determine the symbol of part (a).

It is given that the ion with a\(1 + \) charge, atomic number \(55\), and mass number \(133\).

The atom with the atomic number \(55\)and mass number \(133\)is Caesium\(\left( {Cs} \right)\).

Therefore, the symbol of the ion: \({}_{55}^{133}C{s^{ + 1}}\)

Determine the symbol of part (b).

It is given that the ion has\(54\)electrons,\(53\)protons, and\(74\)neutrons.

The atom with the number of protons \( = \)atomic number

\(\begin{aligned}{} &= 53\\& = I{\rm{(Iodine)}}\end{aligned}\)

Mass number of Iodine\((I) = \)Number of neutrons\( + \)Number of protons

\(\begin{aligned}{} &= 74 + 53\\ &= 127\end{aligned}\)

The charge onIodine\((I) = \)Number of protons \( - \)Number of electrons

\(\begin{aligned}{} &= 53 - 54\\ &= - 1\end{aligned}\)

Therefore, the symbol of the ion: \({}_{53}^{127}{I^{ - 1}}\)

Determine the symbol of part (c).

It is given that the ion has atomic number \(15,\) mass number \(31,\) and a \(3 - \) charge.

The atom with the atomic number \(15\) and mass number\(31,\) is Phosphorus\(\left( P \right)\).

Therefore, the symbol of the ion: \({}_{15}^{31}{P^{ - 3}}\)

Determine the symbol of part (d).

It is given that the ion has \(24\) electrons, \(30\) neutrons, and a \(3 + \)charge.

The atomic number\( = \)Electrons\( + \)Charge on atom

\(\begin{aligned}{} = 24 + 3\\ = 27\\ = Co{\rm{(Cobalt)}}\end{aligned}\)

Mass number ofCobalt\((Co) = \)Number of neutrons\( + \)Number of protons

\(\begin{aligned}{} = 27 + 30\\ = 57\end{aligned}\)

Therefore, the symbol of the ion: \({}_{27}^{57}C{o^{ + 3}}\)