Chapter 5, Problem 5.65P

Interpretation Introduction

Interpretation:

Given species below has to be arranged into group of isoelectronic species.

    $\begin{array}{l}{\text{F}}^{-}{\text{Sc}}^{3+}{\text{Be}}^{2+}{\text{Rb}}^{+}{\text{O}}^{2-}{\text{Na}}^{+}{\text{Ti}}^{4+}\\ \text{Ar}{\text{B}}^{3+}\text{He}{\text{Se}}^{2-}{\text{Y}}^{3+}\end{array}$

Concept Introduction:

Electrons of an atom are arranged in orbitals by the order of increasing energy.  This arrangement is known as electronic configuration of atom.  This can be represented using noble-gas shorthand notation also.

Ions are formed from neutral atom either by removal or addition of electrons from the valence shell.

Each orbital has two electrons and they both are in opposite spin.  Electrons are filled up in the orbitals of a sub-shell by following Hund’s rule.  This rule tells that all the orbitals are singly filled in a sub-shell and then the pairing occurs.

Isoelectronic species are the ones that have different number of protons but same number of electrons.  No two neutral atom can be isoelectronic.  A neutral atom can be isoelectronic with an ion.

Explanation of Solution

Electronic configuration of ${\text{F}}^{-}$:

Fluorine has an atomic number of $9$.  Therefore, the electronic configuration of ${\text{F}}^{-}$ is given as $1s^2 2s^2 2p^6$.  The total number of electrons present in ${\text{F}}^{-}$ is calculated by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}\text{in}{\text{F}}^{-}=2+2+6\\ =10\end{array}$

Electronic configuration of ${\text{Sc}}^{3+}$:

Scandium has an atomic number of $21$.  Therefore, the electronic configuration of ${\text{Sc}}^{\text{3+}}$ is given as $1s^2 2s^2 2p^6 3s^2 3p^6$.  The total number of electrons present in ${\text{Sc}}^{\text{3+}}$ is calculated by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}\text{in}{\text{Sc}}^{3+}=2+2+6+2+6\\ =18\end{array}$

Electronic configuration of ${\text{Be}}^{2+}$:

Beryllium has an atomic number of $4$.  Therefore, the electronic configuration of ${\text{Be}}^{2+}$ is given as $1s^2$.  The total number of electrons present in ${\text{Be}}^{2+}$ is calculated by summing up the superscript.

    $\text{Total}\text{number}\text{of}\text{electrons}\text{in}{\text{Be}}^{2+}=2$

Electronic configuration of ${\text{Rb}}^{+}$:

Rubidium has an atomic number of $37$.  Therefore, the electronic configuration of ${\text{Rb}}^{+}$ is given as $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6$.  The total number of electrons present in ${\text{Be}}^{2+}$ is calculated by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}\text{in}{\text{Rb}}^{+}=2+2+6+2+6+2+10+6\\ =36\end{array}$

Electronic configuration of ${\text{O}}^{2-}$:

Oxygen has an atomic number of $8$.  Therefore, the electronic configuration of ${\text{O}}^{2-}$ is given as $1s^2 2s^2 2p^6$.  The total number of electrons present in ${\text{O}}^{2-}$ is calculated by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}\text{in}{\text{O}}^{2-}=2+2+6\\ =10\end{array}$

Electronic configuration of ${\text{Na}}^{\text{+}}$:

Sodium has an atomic number of $11$.  Therefore, the electronic configuration of ${\text{Na}}^{\text{+}}$ is given as $1s^2 2s^2 2p^6$.  The total number of electrons present in ${\text{Na}}^{\text{+}}$ is calculated by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}\text{in}{\text{Na}}^{\text{+}}=2+2+6\\ =10\end{array}$

Electronic configuration of ${\text{Ti}}^{\text{4+}}$:

Titanium has an atomic number of $22$.  Therefore, the electronic configuration of ${\text{Ti}}^{\text{4+}}$ is given as $1s^2 2s^2 2p^{6}3s^{2}3p^6$.  The total number of electrons present in ${\text{Ti}}^{\text{4+}}$ is calculated by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}\text{in}{\text{Ti}}^{\text{4+}}=2+2+6+2+6\\ =18\end{array}$

Electronic configuration of $\text{Ar}$:

Argon has an atomic number of $18$.  Therefore, the electronic configuration of $\text{Ar}$ is given as $1s^2 2s^2 2p^{6}3s^{2}3p^6$.  The total number of electrons present in $\text{Ar}$ is calculated by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}\text{in}\text{Ar}=2+2+6+2+6\\ =18\end{array}$

Electronic configuration of ${\text{B}}^{3+}$:

Boron has an atomic number of $5$.  Therefore, the electronic configuration of ${\text{B}}^{3+}$ is given as $1s^2$.  The total number of electrons present in ${\text{B}}^{3+}$ is calculated by summing up the superscript.

    $\text{Total}\text{number}\text{of}\text{electrons}\text{in}{\text{B}}^{3+}=2$

Electronic configuration of $\text{He}$:

Helium has an atomic number of $2$.  Therefore, the electronic configuration of $\text{He}$ is given as $1s^2$.  The total number of electrons present in $\text{He}$ is calculated by summing up the superscript.

    $\text{Total}\text{number}\text{of}\text{electrons}\text{in}\text{He}=2$

Electronic configuration of ${\text{Se}}^{2-}$:

Selenium has an atomic number of $34$.  Therefore, the electronic configuration of ${\text{Se}}^{2-}$ is given as $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6$.  The total number of electrons present in ${\text{Se}}^{2-}$ is calculated by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}\text{in}{\text{Se}}^{2-}=2+2+6+2+6+2+10+6\\ =36\end{array}$

Electronic configuration of ${\text{Y}}^{3+}$:

Yttrium has an atomic number of $39$.  Therefore, the electronic configuration of ${\text{Y}}^{3+}$ is given as $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6$.  The total number of electrons present in ${\text{Y}}^{3+}$ is calculated by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}\text{in}{\text{Y}}^{3+}=2+2+6+2+6+2+10+6\\ =36\end{array}$

Isoelectronic species are the one that have same number of electrons.  Therefore, the species can be grouped according to isoelectronic species as given below.

Isoelectronic species	${\text{F}}^{-},{\text{O}}^{2-}$ and ${\text{Na}}^{+}$
	${\text{Sc}}^{3+},{\text{Ti}}^{4+}$ and $\text{Ar}$
	${\text{Be}}^{\text{2+}},{\text{B}}^{3+}$ and $\text{He}$
	${\text{Rb}}^{+},{\text{Se}}^{2-}$ and ${\text{Y}}^{\text{3+}}$