Chapter 2, Problem 44E

Interpretation Introduction

Interpretation: The given table is to be completed.

Concept introduction: In a neutral atom, the number of electrons present are equal to protons. These two are also equivalent to the atomic number in case of neutral atoms. The mass number is calculated by taking sum of number of protons and number of neutrons.

The variation in number of neutrons causes variation in mass number. This results in isotopes of the elements.

Answer to Problem 44E

The complete table is shown below.

Symbol	Number of Protons in Nucleus	Number of Neutrons in Nucleus	Number of Electrons	Net Charge
${}_{92}^{238}\text{U}$	$92$	$146$	$92$	$0$
${}_{20}^{40}{\text{Ca}}^{2+}$	$20$	$20$	$18$	$2+$
${}_{23}^{51}{\text{V}}^{3+}$	$23$	$28$	$20$	$3+$
${}_{39}^{89}\text{Y}$	$39$	$50$	$39$	$0$
${}_{35}^{79}{\text{Br}}^{1-}$	$35$	$44$	$36$	$1-$
${}_{15}^{31}{\text{P}}^{3-}$	$15$	$16$	$18$	$3-$

Explanation of Solution

In the symbol of an isotopic element or ion, subscript represents the atomic number and superscript represents the mass number.

The given sumbol of the isotope in the first row is ${}_{92}^{238}\text{U}$.

The atomic number is equivalent to number of protons, thus, the atomic number is $92$. Since it’s a neutral species, therefore, number of electrons is also equal to $92$. Net charge on it is zero as it is neutral.

The formula used for the calculation of number of netutrons is shown below.

  $\text{Number}\text{of}\text{neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}$

Substitute the value of number of netrons and atomic number of ion in the above formula.

  $\begin{array}{c}\text{Number}\text{of}\text{neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}\\ =\text{238}-\text{92}\\ \text{=146}\end{array}$

In the second row, number of protons and neutrons is $20$ each. The net charge is $+2$. This indicates that number of electrons is two less than the number of protons that is $\left(20-2\right)=18$.

The atomic number is equivalent to number of protons, thus, the atomic number is $20$. The atomic number of $20$ belongs to calcium $\left(\text{Ca}\right)$.

The formula used for the calculation of mass number is shown below.

  $\text{Mass}\text{number}=\text{Number}\text{of}\text{neutrons}+\text{Atomic}\text{number}$

Substitute the value of number of netrons and atomic number of ion in the above formula.

  $\begin{array}{c}\text{Mass}\text{number}=\text{Number}\text{of}\text{neutrons}+\text{Atomic}\text{number}\\ =20+20\\ =40\end{array}$

Therefore, the atomic symbol for isotope is ${}_{20}^{40}{\text{Ca}}^{2+}$.

In the third row, number of protons, electrons and neutrons is $23,20$ and $28$ respectively. Since number of electrons is three less than number of protons, therefore, charge on the ion is $\left(23-20\right)=+3$.

The atomic number is equivalent to number of protons, thus, the atomic number is $23$. The atomic number of $23$ belongs to vanadium $\left(\text{V}\right)$.

The formula used for the calculation of mass number is shown below.

  $\text{Mass}\text{number}=\text{Number}\text{of}\text{neutrons}+\text{Atomic}\text{number}$

Substitute the value of number of netrons and atomic number of ion in the above formula.

  $\begin{array}{c}\text{Mass}\text{number}=\text{Number}\text{of}\text{neutrons}+\text{Atomic}\text{number}\\ =28+23\\ =51\end{array}$

Therefore, the atomic symbol for isotope is ${}_{23}^{51}{\text{V}}^{3+}$.

The given sumbol of the isotope in the fourth row is ${}_{39}^{89}\text{Y}$.

The atomic number is equivalent to number of protons, thus, the atomic number is $39$. Since it’s a neutral species, therefore, number of electrons is also equal to $39$. Net charge on it is zero as it is neutral.

The formula used for the calculation of number of netutrons is shown below.

  $\text{Number}\text{of}\text{neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}$

Substitute the value of number of netrons and atomic number of ion in the above formula.

  $\begin{array}{c}\text{Number}\text{of}\text{neutrons}=\text{Mass}\text{number}-\text{Atomic}\text{number}\\ =\text{89}-39\\ =\text{50}\end{array}$

In the fifth row, number of protons, electrons and neutrons is $35,36$ and $44$ respectively. Since number of electrons is one more than number of protons, therefore, charge on the ion is $\left(35-36\right)=-1$.

The atomic number is equivalent to number of protons, thus, the atomic number is $35$. The atomic number of $35$ belongs to bromine $\left(\text{Br}\right)$.

The formula used for the calculation of mass number is shown below.

  $\text{Mass}\text{number}=\text{Number}\text{of}\text{neutrons}+\text{Atomic}\text{number}$

Substitute the value of number of netrons and atomic number of ion in the above formula.

  $\begin{array}{c}\text{Mass}\text{number}=\text{Number}\text{of}\text{neutrons}+\text{Atomic}\text{number}\\ =44+35\\ =79\end{array}$

Therefore, the atomic symbol for isotope is ${}_{35}^{79}{\text{Br}}^{1-}$.

In the sixth row, number of protons and neutrons is $15$ and $16$ respectively. The net charge is $-3$. This indicates that number of electrons is three more than the number of protons that is $\left(15+3\right)=18$.

The atomic number is equivalent to number of protons, thus, the atomic number is $15$. The atomic number of $15$ belongs to phosphorus $\left(\text{P}\right)$.

The formula used for the calculation of mass number is shown below.

  $\text{Mass}\text{number}=\text{Number}\text{of}\text{neutrons}+\text{Atomic}\text{number}$

Substitute the value of number of netrons and atomic number of ion in the above formula.

  $\begin{array}{c}\text{Mass}\text{number}=\text{Number}\text{of}\text{neutrons}+\text{Atomic}\text{number}\\ =16+15\\ =31\end{array}$

Therefore, the atomic symbol for isotope is ${}_{15}^{31}{\text{P}}^{3-}$.

Thus, the complete table is shown below.

Symbol	Number of Protons in Nucleus	Number of Neutrons in Nucleus	Number of Electrons	Net Charge
${}_{92}^{238}\text{U}$	$92$	$146$	$92$	$0$
${}_{20}^{40}{\text{Ca}}^{2+}$	$20$	$20$	$18$	$2+$
${}_{23}^{51}{\text{V}}^{3+}$	$23$	$28$	$20$	$3+$
${}_{39}^{89}\text{Y}$	$39$	$50$	$39$	$0$
${}_{35}^{79}{\text{Br}}^{1-}$	$35$	$44$	$36$	$1-$
${}_{15}^{31}{\text{P}}^{3-}$	$15$	$16$	$18$	$3-$