Chapter 2, Problem 2.56QP

Interpretation Introduction

Interpretation: The missing information corresponding to the given monoatomic ions are to be represented.

Concept introduction: The electrons, protons, and neutrons are sub atomic particles. The number of protons and electrons present in an element are equal and termed as atomic number $\left(Z\right)$. The relation between atomic number, atomic mass $\left(A\right)$ and number of neutrons $\left(N\right)$ is stated by the formula, $A=N+Z$

To determine: The missing information from the given table.

Answer to Problem 2.56QP

Solution

The missing information is stated as follows.

Explanation of Solution

Explanation

The electrons, protons, and neutrons are sub atomic particles. The number of protons and electrons present in an element are equal and termed as atomic number $\left(Z\right)$. The relation between atomic number, atomic mass and number of neutrons is stated by the formula,

$A=N+Z$

Where,

• $A$ is mass number.
• $N$ is number of neutrons.
• $Z$ is atomic number.

The missing information are given below in the form of (I), (II), (III) and (IV).

(I)

For ${}^{137}{\text{Cs}}^{+}$,

The given symbol ${}^{137}{\text{Cs}}^{+}$  represents that cesium has $\underset{\_}{137}$ mass number $\left(A\right)$.

The atomic number $\left(Z\right)$ of cesium is $55$.

Hence, number of protons is $\underset{\_}{55}$. As it consist a positive charge so the number of electrons is $55-1=\underset{\_}{54}$.

The number of neutrons is calculated by the formula,

$\begin{array}{l}A=N+Z\\ N=A-Z\end{array}$

Where,

• $A$ is the mass number.
• $Z$ is the atomic number.

Substitute the values of mass number $\left(A\right)$ and atomic number $\left(Z\right)$ in the above expression.

$\begin{array}{c}N=A-Z\\ N=137-55\\ =\underset{\_}{82}\end{array}$

(II)

For given unknown element the number of protons and electrons $30$ and $28$. The number of neutrons $\left(N=34\right)$.

The number of protons is $30$.

Therefore, $Z=30$. Since, the atomic number is unique to an element; it can be identified from the periodic table.

The element that has a atomic number $30$ is zinc $\text{Zn}$.

The number of electrons is $28$ that is $2$ less than the number of protons.

Therefore, the given atom has lost two electrons to form a positively charged ion, ${\text{Zn}}^{\text{2+}}$.

The mass number $\left(A\right)$ is calculated by the formula,

$A=N+Z$

Where,

• $N$ is the number of neutrons.
• $Z$ is the atomic number.

Substitute the values of number of neutrons $\left(N\right)$ and atomic number $\left(Z\right)$ in above expression.

$\begin{array}{c}A=N+Z\\ A=30+34\\ =\underset{\_}{64}\end{array}$

Hence, the symbol is $\underset{\_}{{}^{64}Z{n}^{2+}}$.

(III)

For given unknown element the number of electrons $18$, mass number $\left(A=32\right)$ and the number of neutrons is $16$.

The number of proton is $32-16=\underset{\_}{16}$.

Hence, the number of proton is $16$ therefore atomic number $\left(Z\right)$ is $16$.

Since the atomic number is unique to an element; it can be identified from the periodic table.

The element that has a atomic number $16$ is sulfur $\text{S}$.

The number of electrons is $18$ that is $2$ more than the number of protons.

Therefore, the given atom has gained two electrons to form a negatively charged ion, $\underset{\_}{S^{2-}}$

Hence, the symbol is $\underset{\_}{{}^{32}{S}^{2-}}$.

(IV)

For given unknown element the number of protons $40$, mass number $\left(A=90\right)$ and the number of electrons is $36$.

The number of proton is $40$ therefore atomic number $\left(Z\right)$ is $40$.

The number of neutrons is calculated by the formula,

$\begin{array}{l}A=N+Z\\ N=A-Z\end{array}$

Where,

• $A$ is the mass number.
• $Z$ is the atomic number.

Substitute the values of mass number $\left(A\right)$ and atomic number $\left(\text{Z}\right)$ in the above expression.

$\begin{array}{c}N=A-Z\\ N=90-40\\ =\underset{\_}{50}\end{array}$

Since the atomic number is unique to an element; it can be identified from the periodic table.

The element that has a atomic number $40$ is zirconium $\text{Zr}$.

The number of electrons is $36$ that is $4$ less from the number of protons.

Therefore atom has lost four electron to form a positively charged ion, ${\text{Zn}}^{\text{4+}}$.

Hence, symbol is $\underset{\_}{{}^{90}Z{n}^{4+}}$.

Conclusion

The given table summarized all the missing information.

Symbol	${}^{137}{\text{Cs}}^{+}$	${}^{64}{\text{Zn}}^{2+}$	${}^{32}{\text{S}}^{2-}$	${}^{90}{\text{Zn}}^{4+}$
Number of Protons	$55$	$30$	$16$	$40$
Number of Neutrons	$82$	$34$	$16$	$50$
Number of electrons	$54$	$28$	$18$	$36$
Mass number	$137$	$64$	$32$	$90$