Q11E - Chemistry (OpenStax)

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

Step-by-Step Solution

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

Step 2: 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

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

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

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

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

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

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

Step 3: 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: \({}_{15}^{31}{P^{ - 3}}\)

Step 4: 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

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

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

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

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