Chapter 2, Problem 52QP

Interpretation Introduction

Interpretation:

The table for the designated ions is to be completed.

Concept Introduction:

Isotopes can be defined as the element that has two or more different forms of atoms containing same number of protons but different number of neutrons.

Explanation of Solution

Given information: The table is provided in the question.

In the first row of the table, the mass number of $\text{Br}$ is $81$ and the atomic number of $\text{Br}$ is $35$ . The number of protons and the number of electrons correspond to the atomic number in a neutral atom. The overall charge is negative, when the number of electrons is greater than the number of protons while the overall charge is positive, when the number of electrons are less than the number of protons. The number of neutrons in ${}_{35}^{81}{\text{Br}}^{-}$ is calculated as follows.

$\begin{array}{c}\text{number of neutrons}=\text{mass number}-\text{atomic number}\\ =\text{81}-35\\ =46\end{array}$

Therefore, the number of protons in ${\text{Br}}^{-}$ is $35$ . As the charge on the anion is $-1$ , the number of electrons is $36$ . The number of neutrons is $46$ .

In the second row of the table, the number of protons is $38$ which corresponds to the atomic number. The number of electrons is $36$ . This means that the number of electrons is less than the number of protons by $2$ . Now, the number of neutrons is $50$ . So, the mass number of the atom is calculated as follows.

$\begin{array}{c}\text{number of neutrons}=\text{mass number}-\text{atomic number}\\ \text{mass number}=50+38\\ =88\end{array}$

Therefore, the atomic number of the ion is $38$ and the mass number is $88$ . Then, the isotope symbol is ${}_{38}^{88}{\text{Sr}}^{2+}$ .

In the third row of the table, the number of protons is $1$ . As the number of protons corresponds to the atomic number in an ion, the atomic number is $1$ . The number of electrons is $2$ . This means that the number of electrons in the element exceeds by one which represents that the isotope symbol is of an anion. The number of neutrons is $1$ . The mass number of the isotope symbol is calculated as follows.

$\begin{array}{c}\text{number of neutrons}=\text{mass number}-\text{atomic number}\\ \text{mass number}=1+1\\ =2\end{array}$

Therefore, the atomic number of the anion is $1$ and the mass number is $2$ . Then, the isotope symbol is ${}_1^2{\text{H}}^{-}$ .

In the fourth row of the table, the isotopic symbol is ${}^1{\text{H}}^{+}$ . The mass number is $1$ . As the atomic number corresponds to the number of protons, the number of protons is $1$ . Since hydrogen contains the positive charge, the number of electrons is less than the number of protons by one. Hence, the number of electrons in ${}^1{\text{H}}^{+}$ is $0$ . The mass number is $1$ . The number of neutrons of ${}^1{\text{H}}^{+}$ is calculated as follows.

$\begin{array}{c}\text{number of neutrons}=\text{mass number}-\text{atomic number}\\ =1-1\\ =0\end{array}$

Therefore, the number of neutrons in ${}^1{\text{H}}^{+}$ is $0$ .

The completed table of the designated atoms is given below.

$\begin{array}{|llll|}\hline \begin{array}{l}\text{Isotope}\\ \text{symbol}\end{array}\hfill & \begin{array}{l}\text{Number of }\\ \text{ Protons}\end{array}\hfill & \begin{array}{l}\text{Number of }\\ \text{ Neutrons}\end{array}\hfill & \begin{array}{l}\text{Number of }\\ \text{ Electrons}\end{array}\hfill \\ {}_{35}^{81}{\text{Br}}^{-}\hfill & \underset{¯}{35}\hfill & \underset{¯}{46}\hfill & \underset{¯}{36}\hfill \\ \underset{¯}{{}_{38}^{88}{\text{Sr}}^{2+}}\hfill & 38\hfill & 50\hfill & 36\hfill \\ \underset{¯}{{}_1^2{\text{H}}^{-}}\hfill & 1\hfill & 1\hfill & 2\hfill \\ {}^1{\text{H}}^{+}\hfill & \underset{¯}{1}\hfill & \underset{¯}{0}\hfill & \underset{¯}{0}\hfill \\ \hline\end{array}$

Conclusion

The completed table of the designated atoms is shown above.