Chapter 2, Problem 92AP

Interpretation Introduction

Interpretation:

The elements that are neutral, positivelycharged, or negativelycharged are to be identified among the given elements and the conventional symbols of the same are to be calculated.

Concept introduction:

Atoms are neutral if they contain an equal number of protons and electrons.

Atoms are negativelycharged if they contain a higher number of electrons than the number of protons, i.e., number of negative charges are more than the positive charges.

Atoms are positivelycharged if they contain a higher number of protons than the number of electrons i.e. number of positive charges are more than the negative charges.

The conventional symbol for any element ‘X’ with atomic number (Z) and mass number (A) is given by ${}_Z{}^{A}X$.

The atomic number (Z) is the number of protons in the nucleus of each atom, which also indicates the number of electrons in the atom. The mass number (A) is the sum of neutrons and protons present in the nucleus of a particular atom. The mass number (A) is given by:

$\text{Mass number (}A\text{)}=\text{Number of protons (}Z\text{)}+\text{Number of neutrons}$

Answer to Problem 92AP

Solution:

a) A, F, and G.

b) B and E.

c) C andD.

d) $\text{B},\text{N}{}^{3-},\text{K}{}^{+},\text{Z}{\text{n}}^{2+},\text{B}{\text{r}}^{-},\text{B},\text{F}$.

Explanation of Solution

Atoms are neutral if they contain the same number of protons and electrons.

a)The species which are neutral.

For element A,

The number of electrons as well as the number of protons $\text{is}=\text{5}$.

For element F,

The number of electrons as well as the number of protons $\text{is}=\text{5}$.

For element G,

The number of electrons as well as the number of protons $\text{is}=9$.

Hence, the elements A, F, and G are neutral.

b)The species which are negatively charged.

Atoms are negativelycharged if they contain a higher number of electrons than the number of protons.

For element B,

The number of electrons $\text{is}=36$. The number of protons $\text{is}=35$. Since $36>35$, the element B is negativelycharged.

For element F,

The number of electrons $\text{is}=10$. The number of protons $\text{is}=7$. Since $10>7$, the element F is negativelycharged.

c) The species that are positively charged.

Atoms are positivelycharged if they contain a higher number of protons than the number of electrons.

For element C,

The number of electrons $\text{is}=18$. The number of protons $\text{is}=19$. Since $19>18$, the element C is positivelycharged.

For element D,

The number of electrons $\text{is}=28$. The number of protons $\text{is}=30$. Since $30>28$, the element D is positivelycharged.

d) The conventional symbol for all.

The conventional symbol for any element ‘X’ with the atomic number (Z) and the mass number (A) is given by ${}_Z{}^{A}X$.

For element A,

The number of protons (Z) $\text{is }35$. The number of neutrons $\text{is }35$.

Refer to the expression to calculate the mass number (A):

$\text{Mass number (A)}=\text{Number of protons (Z)}+\text{Number of neutrons}$

Substitute the known values in the above expression,

$\begin{array}{c}\text{Mass number (A)}=5+5\\ =10\end{array}$

The conventional symbol for element A is given by ${}_5{}^{10}A$.

For element B,

The number of protons (Z) $\text{is }7$. The number of neutrons $\text{is }7$.

Refer to the expression to calculate the mass number (A):

$\text{Mass number (A)}=\text{Number of protons (Z)}+\text{Number of neutrons}$

Substitute the known values in the above expression,

$\begin{array}{c}\text{Mass number (A)}=7+7\\ =14\end{array}$

The conventional symbol for element B is given by ${}_7{}^{14}B$.

For element C,

The number of protons (Z) $\text{is }19$. The number of neutrons $\text{is }20$.

Refer to the expression to calculate the mass number (A):

$\text{Mass number (A)}=\text{Number of protons (Z)}+\text{Number of neutrons}$

Substitute the known values in the above expression,

$\begin{array}{c}\text{Mass number (A)}=19+20\\ =39\end{array}$

The conventional symbol for element C is given by ${}_{19}{}^{39}C$.

For element D,

The number of protons (Z) $\text{is }30$. The number of neutrons $\text{is }36$.

Refer to the expression to calculate the mass number (A):

$\text{Mass number (A)}=\text{Number of protons (Z)}+\text{Number of neutrons}$

Substitute the known values in the above expression,

$\begin{array}{c}\text{Mass number (A)}=30+36\\ =66\end{array}$

The conventional symbol for element D is given by ${}_{30}{}^{66}D$.

For element E,

The number of protons (Z) $\text{is }35$. The number of neutrons $\text{is }46$.

Refer to the expression to calculate the mass number (A):

$\text{Mass number (A)}=\text{Number of protons (Z)}+\text{Number of neutrons}$

Substitute the known values in the above expression,

$\begin{array}{c}\text{Mass number (A)}=35+46\\ =81\end{array}$

The conventional symbol for element E is given by ${}_{35}{}^{81}E$.

For element F,

The number of protons (Z) $\text{is }5$. The number of neutrons $\text{is }6$.

Refer to the expression to calculate the mass number (A):

$\text{Mass number (A)}=\text{Number of protons (Z)}+\text{Number of neutrons}$

Substitute the known values in the above expression,

$\begin{array}{c}\text{Mass number (A)}=5+6\\ =11\end{array}$

The conventional symbol for element F is given by ${}_5{}^{11}F$.

For element G,

The number of protons (Z) $\text{is }9$. The number of neutrons $\text{is }10$.

Refer to the expression to calculate the mass number (A):

$\text{Mass number (A)}=\text{Number of protons (Z)}+\text{Number of neutrons}$

Substitute the known values in the above expression,

$\begin{array}{c}\text{Mass number (A)}=9+10\\ =19\end{array}$

The conventional symbol for element G is given by ${}_9{}^{19}G$.