Chapter 8, Problem 105AP

Which of the following species are isoelectronic: ${\text{NH}}_4^{+},{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{,CO,CH}}_{\text{4}}{\text{,N}}_{\text{2}}{\text{,B}}_{\text{3}}{\text{N}}_{\text{3}}{\text{H}}_{\text{6}}\text{?}$

Interpretation Introduction

Interpretation:

Whether the given species are isoelectronic or notis to be determined.

Concept introduction:

Isoelectronic species are those molecules or compounds that have an equal number of electrons.

To find the total number of electrons, add the atomic number of elementsand subtract 1 for each positive charge and add 1 for each negative charge.

The total number of electrons in ${\text{H}}_2\text{O}$ can be calculated by:

Total electrons = $\begin{array}{l}\text{(number of element of}\text{H }\times \text{atomic number of H)}\\ \text{+(number of element of}\text{O }\times \text{atomic number of O)}\end{array}$

Answer to Problem 105AP

Solution:

${\text{NH}}_4{}^{+}{\text{ and CH}}_4$, ${\text{CO and N}}_2$, and ${\text{C}}_6{\text{H}}_6{\text{ and B}}_3{\text{N}}_3{\text{H}}_6$ are isoelectronic in nature.

Explanation of Solution

For ${\text{NH}}_4{}^{+}$

The atomic number of nitrogen is seven and that of hydrogen is one. To form ${\text{NH}}_4{}^{+}$, nitrogen loses one electron.

Calculate the total number of electrons as

Total electrons = $\begin{array}{l}\text{(number of element of}\text{N }\times \text{atomic number of N)}\\ \text{+(number of element of}\text{H }\times \text{atomic number of H)}\\ -(1\text{electron for positive charge)}\end{array}$

$\begin{array}{c}\text{Total electrons}=e^{-}\left(\text{N}\right)+4\times e^{-}\left(\text{H}\right)-1\\ =7+4\left(1\right)-1\\ =7+4-1\\ =10\end{array}$

For ${\text{C}}_6{\text{H}}_6$,

The atomic number of carbon is six and that of hydrogen is one.

Total electrons = $\begin{array}{l}\text{(number of element of}\text{C }\times \text{atomic number of C)}\\ \text{+(number of element of}\text{H }\times \text{atomic number of H)}\end{array}$

Calculate the total number of electrons as

$\begin{array}{c}\text{Total electrons}=6\times e^{-}\left(\text{C}\right)+6\times e^{-}\left(\text{H}\right)\\ =6\left(6\right)+6\left(1\right)\\ =36+6\\ =42\end{array}$

For $\text{CO}$,

The atomic number of carbon is six and that of oxygen iseight.

Total electrons = $\begin{array}{l}\text{(number of element of}\text{C }\times \text{atomic number of C)}\\ \text{+(number of element of}\text{O }\times \text{atomic number of O)}\end{array}$

Calculate the total number of electrons as

$\begin{array}{c}\text{Total electrons}=e^{-}\left(\text{C}\right)+e^{-}\left(\text{O}\right)\\ =6+8\\ =14\end{array}$

For ${\text{CH}}_4$,

The atomic number of carbon is six and that of hydrogen is one.

Total electrons = $\begin{array}{l}\text{(number of element of}\text{C }\times \text{atomic number of C)}\\ \text{+(number of element of}\text{H }\times \text{atomic number of H)}\end{array}$

Calculate the total number of electrons as

$\begin{array}{c}\text{Total electrons}=e^{-}\left(\text{C}\right)+4\times e^{-}\left(\text{H}\right)\\ =6+4\left(1\right)\\ =6+4\\ =10\end{array}$

For ${\text{N}}_2$,

The atomic number of nitrogen is seven.

Total electrons = $\text{(number of element of}\text{N }\times \text{atomic number of N)}$

Calculate the total number of electrons as

$\begin{array}{c}\text{Total electrons}=2\times e^{-}\left(\text{N}\right)\\ =2\left(7\right)\\ =14\end{array}$

For ${\text{B}}_3{\text{N}}_3{\text{H}}_6$,

The atomic number of boron is five, that of nitrogen is seven, and that of hydrogen is one.

Calculate the total number of electrons as

Total electrons = $\begin{array}{l}\text{(number of element of}\text{B }\times \text{atomic number of B)}\\ \text{+(number of element of}\text{N }\times \text{atomic number of N)}\\ \text{+(number of element of}\text{H }\times \text{atomic number of H)}\end{array}$ $\begin{array}{c}\text{Total electrons}=3\times e^{-}\left(\text{B}\right)+3\times e^{-}\left(\text{N}\right)+6\times e^{-}\left(\text{H}\right)\\ =3\left(5\right)+3\left(7\right)+6\left(1\right)\\ =15+21+6\\ =42\end{array}$

Since ${\text{NH}}_4{}^{+}{\text{ and CH}}_4$ have 10 electrons, ${\text{CO and N}}_2$ have 14 electrons and ${\text{C}}_6{\text{H}}_6{\text{ and B}}_3{\text{N}}_3{\text{H}}_6$ have 42 electrons;they are isoelectronic.

Conclusion

Hence, ${\text{NH}}_4{}^{+}{\text{and CH}}_4$, ${\text{CO and N}}_2$, and ${\text{C}}_6{\text{H}}_6{\text{ and B}}_3{\text{N}}_3{\text{H}}_6$ are isoelectronic in nature.