Chapter 2, Problem 99AP

Interpretation Introduction

Interpretation:

The blanks in the table are to be filled.

Concept introduction:

The atomic number $(Z)$ is the number of protons in the nucleus of each atom of an element. It also indicates the number of electrons in an atom. Mathematically, it can be represented as follows:

$\text{Atomic number}(Z)=\text{ No}\text{. of electrons or protons}$

The number of electrons in a neutral atom is exactly equal to the number of protons.

The mass number of an atom $\left(\text{A}\right)$ is equal to the sum of the number of protons and neutrons in the nucleus. The mass number is calculated as follows:

$\left(A\right)=\left(Z\right)+\text{number of neutrons}$

Here, $A$ is the mass number and $Z$ is the atomic number.

Answer to Problem 99AP

Solution:

$\begin{array}{cccccc}\text{Symbols}& \text{B}& \text{F}\text{e}& \text{P}{}^{-3}& \text{A}\text{u}& \text{R}\text{n}\\ \text{Protons}& 5& 26& 15& 79& 86\\ \text{Neutrons}& 6& 28& 16& 117& 136\\ \text{Electrons}& 5& 24& 18& 79& 86\\ \text{Net charge}& 0& +2& -3& 0& 0\end{array}$

Explanation of Solution

For the first element,

Number of protons $=5$.

Number of electrons $=5$.

Number of neutrons $=6$.

For the given element, thenumbers of protons and electrons are equal, so it is a neutral atom. Its atomic number is the number of protons. So, the atomic number of this element is $5$. So, Boron has atomic number $5$. The mass number of Boron can be calculated by using the equation given below:

$\left(A\right)=\left(Z\right)+\text{number of neutrons}$

Substitute $5$ for $Z$ and $6$ for neutrons

$\begin{array}{c}A=5+6\\ =11\end{array}$

The mass number is $11$.

Hence, thesymbol is $\text{B}$ and it has no charge because it is a neutral atom.

For the second element, the symbol given is $\text{F}{\text{e}}^{2+}$.

The atomic number of iron is $26$. The number of protons are equal to the atomic number. So, the number of protons is $26$. It is a cation formed by losing two electrons. The total number of electrons is $24$. The mass number is $54$. Neutrons can be calculated as follows:

$\text{number of neutrons}=\left(A\right)-\left(Z\right)$

Substitute $5$ for $A$ and $6$ for neutrons

$\begin{array}{c}\text{neutrons}=54-26\\ =28\end{array}$

The number of neutrons is $28$.

For the third element,

Number of neutrons $=16$.

Number of electrons $=18$.

Net charge on the element is $-3$.

This element has gained three electrons, so the number of protons becomes $15$. So, the atomic number of this element is $15$. The element is phosphorus. The mass number can be calculated as follows:

$\left(A\right)=\left(Z\right)+\text{number of neutrons}$

Substitute $15$ for $Z$ and $16$ for neutrons

$\begin{array}{c}A=15+16\\ =31\end{array}$

The mass number is $31$.

Hence, the symbol is $\text{P}{}^{-3}$.

For the fourth element,

Number of protons $=79$.

Number of electrons $=79$.

Number of neutrons $=117$.

Thenumbers of protons and electrons are equal, so it is a neutral atom. The atomic number is equal to the number of protons. So, the atomic number of this element is $79$. The atomic number ofgold is $79$. The mass number of gold can be calculated by using the equation given below:

$\left(A\right)=\left(Z\right)+\text{number of neutrons}$

Substitute $79$ for $Z$ and $117$ for neutrons

$\begin{array}{c}A=79+117\\ =196\end{array}$

The mass number is $196$.

Hence, the symbol is $\text{A}\text{u}$ and it has no charge, because it is a neutral atom.

For the fifth element,

Number of protons $=86$.

Number of neutrons $=136$.

Net charge $=0$

The net charge is zero, so it is a neutral atom. The number of electrons is equal to the number of protons, that is, $86$. The atomic number is equal to the number of protons. So, the atomic number of this element is $86$. The elementwith atomic number $86$ is radon. The mass number of radon can be calculated by using the equation given below:

$\left(A\right)=\left(Z\right)+\text{number of neutrons}$

Substitute $136$ for $Z$ and $6$ for neutrons

$\begin{array}{c}A=136+86\\ =222\end{array}$

The mass number is $222$.

Hence, the symbol is $\text{R}\text{n}$ and it has no charge, because it is a neutral atom.

Conclusion

The blanks in the given table are filled as shown below:

$\begin{array}{cccccc}\text{Symbols}& \text{B}& \text{F}\text{e}& \text{P}{}^{-3}& \text{A}\text{u}& \text{R}\text{n}\\ \text{Protons}& 5& 26& 15& 79& 86\\ \text{Neutrons}& 6& 28& 16& 117& 136\\ \text{Electrons}& 5& 24& 18& 79& 86\\ \text{Net charge}& 0& +2& -3& 0& 0\end{array}$