Question

Write the symbol for each of the following ions: (a) the ion with a $3+$ charge, $28$ electrons, and a mass number of $71$(b) the ion with $36$ electrons, $35$ protons, and $45$ neutrons (c) the ion with $86$ electrons, $142$ neutrons, and a $4+$ charge (d) the ion with a $2+$ charge, atomic number $38,$ and mass number $87$

Short answer

The symbols are:

a. ${}_{31}^{71}G{a}^{3+}$

b. ${}_{35}^{80}B{r}^{-1}$

c. ${}_{90}^{232}T{h}^{4+}$

d. ${}_{38}^{87}S{r}^{2+}$

Step-by-step solution

Determine the symbol.

It is given that the ion has a $3+$ charge, $28$electrons, and a mass number of $71$.

The atomic numberof an atom $=$ Number electrons $+$atomic charge

$\begin{array}{rl}& =28+3\\ & =31\\ & =Ga\left(\mathrm{g}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{u}\mathrm{m}\right)\end{array}$

Therefore, the symbol of the ion: ${}_{31}^{71}G{a}^{3+}$

Determine the symbol

It is given that the ion has$36$electrons,$35$protons, and$45$neutrons.

The atom with number of protons$=$atomic number

$\begin{array}{rl}& =35\\ & =Br(\mathrm{B}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{e})\end{array}$

Mass number ofbromine$(Br)=$Number of neutrons$+$Number of protons

$\begin{array}{rl}& =35+45\\ & =80\end{array}$

The charge on bromine$(Br)=$Number of protons $-$Number of electrons

$\begin{array}{rl}& =35-36\\ & =-1\end{array}$

Therefore, the symbol of the ion: ${}_{35}^{80}B{r}^{-1}$

Determine the symbol.

It is given that the ion has$86$electrons,$142$neutrons, and a$4+$charge.

The atomic number$=$Electrons$+$Charge on atom

$\begin{array}{rl}& =86+4\\ & =90\\ & =Th(\mathrm{T}\mathrm{h}\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{u}\mathrm{m})\end{array}$

Mass number ofThorium$(Th)=$Number of neutrons$+$Number of protons

$\begin{array}{rl}& =90+142\\ & =232\end{array}$

Therefore, the symbol of the ion: ${}_{90}^{232}T{h}^{4+}$

Determine the symbol.

It is given that the ion has a $2+$ charge, atomic number $38$, and mass number $87$.

The atom with the atomic number $38$ and mass number $87$ is strontium$\left(Sr\right)$.

Therefore, the symbol of the ion: ${}_{38}^{87}S{r}^{2+}$