Chapter 2, Problem 2.55QP

Interpretation Introduction

Interpretation: The missing information for the given monoatomic ions is to be stated.

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 2.55QP

Solution

The missing information has been rightfully stated.

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.

(I)

For ${}^{35}{\text{Cl}}^{-}$,

The given symbol ${}^{35}{\text{Cl}}^{-}$ represents that chlorine has $35$ mass number $\left(A\right)$.

The atomic number $\left(Z\right)$ of chlorine is $17$.

Hence, number of protons is $17$.

It has a negative charge so the number of electrons $\begin{array}{l}=17+1\\ =18e^{-}\end{array}$

The number of neutrons is calculated by the formula,

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

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

$\begin{array}{c}N=A-Z\\ N=35-17\\ =18\end{array}$

(II)

For given unknown element the number of protons and electrons is $11$ and $10$.

The number of neutrons $\left(N=12\right)$.

The number of protons is $11$; therefore, $Z=11$.

Since the atomic number is unique to an element; it can be identified from the periodic table. The element corresponding to the atomic number $11$ is sodium, $\text{Na}$.

The number of electrons is $10$ which is one less than the number of protons.

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

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

$A=N+Z$

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

$\begin{array}{c}A=N+Z\\ A=12+11\\ =23\end{array}$

Hence, the symbol is ${}^{23}{\text{Na}}^{+}$.

(III)

For given unknown element, the number of electrons is $36$, mass number $\left(A=81\right)$ and the number of neutrons is $46$.

The number of protons $\begin{array}{l}=81-46\\ =35\end{array}$

Hence, the number of proton is $35$; therefore atomic number $\left(Z\right)$ is $35$.

Since the atomic number is unique to an element; it can be identified from the periodic table. The element corresponding to the atomic number $35$ is bromine, $\text{Br}$.

The number of electrons is $36$ which is one more than the number of protons.

Therefore, the atom has gained an electron to form a negatively charged ion, ${\text{Br}}^{-}$.

Hence, the symbol is ${}^{35}{\text{Br}}^{-}$.

(IV)

For given unknown element, the number of protons is $82$, mass number $\left(A=210\right)$ and the number of electrons is $80$.

The number of proton is $82$; therefore, atomic number $\left(Z\right)$ is $82$.

The number of neutrons is calculated by the formula,

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

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

$\begin{array}{c}N=A-Z\\ N=210-82\\ =128\end{array}$

Since the atomic number is unique to an element; it can be identified from the periodic table. The element corresponding to the atomic number $82$ is lead $\text{Pb}$.

The number of electrons is $80$ which is two less than the number of protons.

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

Hence, symbol is ${}^{210}{\text{Pb}}^{2+}$.

Conclusion

The given table summarized all the missing information.

Symbol	${}^{35}{\text{Cl}}^{-}$	${}^{23}{\text{Na}}^{+}$	${}^{81}{\text{Br}}^{-}$	${}^{210}{\text{Pb}}^{2+}$
Number of Protons	$17$	$11$	$35$	$82$
Number of Neutrons	$18$	$12$	$46$	$128$
Number of electrons	$18$	$10$	$36$	$80$
Mass number	$35$	$23$	$81$	$210$