Chapter 8, Problem 8.60P

(a)

Interpretation Introduction

Interpretation:

The element with the lowest ${\text{IE}}_3$ in the set of $\text{Na}$, $\text{Mg}$, $\text{Al}$ is to be determined.

Concept introduction:

The ionization energy is the amount of energy required to remove the most loosely bound valence electrons from an isolated neutral gaseous atom. It is represented by $\text{IE}$.

Its value varies with the ease of removal of the outermost valence electron. If the outermost electron is removed very easily then the value of ionization energy is very small. If the electron is removed with quite a difficulty then the value of ionization energy will be very high.

When the first electron is removed from a neutral, isolated gaseous atom then the ionization energy is known as the first ionization energy $\left({\text{IE}}_1\right)$. Similarly, when the second electron is removed from the positively charged cation the ionization energy is called the second ionization energy $\left({\text{IE}}_2\right)$ and when the third electron is removed the ionization energy is called the third ionization energy $\left({\text{IE}}_3\right)$ and so on.

Answer to Problem 8.60P

Aluminium $\left(\text{Al};\text{Z}=\text{13}\right)$ has the lowest ${\text{IE}}_3$ among the given elements.

Explanation of Solution

The atomic number of sodium is 11 so its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^1$. The value of its ${\text{IE}}_3$ will be very high because the third electron is to be removed from the inner stable lower energy level which is very difficult.

The atomic number of magnesium is 12 so its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^2$. The value of its ${\text{IE}}_3$ will be high because the third electron is to be removed from the fulfilled $2p$ orbital which is quite difficult.

The atomic number of aluminium is 13 so its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^1$. The value of its ${\text{IE}}_3$ will be the lowest because after the removal of the third electron its configuration becomes $1s^{2}2s^{2}2p^6$ which is the stable noble gas configuration of neon. The ion formation occurs as follows:

$\text{Al}\left(1s^{2}2s^{2}2p^{6}3s^{2}3p^1\right)\to {\text{Al}}^{3+}\left(1s^{2}2s^{2}2p^6\right)+3{\text{e}}^{-}$

Conclusion

Aluminium $\left(\text{Al};\text{Z}=\text{13}\right)$ has the lowest ${\text{IE}}_3$ among the given elements.

(b)

Interpretation Introduction

Interpretation:

The element with the lowest ${\text{IE}}_3$ in the set of $\text{K}$, $\text{Ca}$, $\text{Sc}$ is to be determined.

Concept introduction:

The ionization energy is the amount of energy required to remove the most loosely bound valence electrons from an isolated neutral gaseous atom. It is represented by $\text{IE}$.

Its value varies with the ease of removal of the outermost valence electron. If the outermost electron is removed very easily then the value of ionization energy is very small. If the electron is removed with quite a difficulty then the value of ionization energy will be very high.

When the first electron is removed from a neutral, isolated gaseous atom then the ionization energy is known as the first ionization energy $\left({\text{IE}}_1\right)$. Similarly, when the second electron is removed from the positively charged cation the ionization energy is called the second ionization energy $\left({\text{IE}}_2\right)$ and when the third electron is removed the ionization energy is called the third ionization energy $\left({\text{IE}}_3\right)$ and so on.

Answer to Problem 8.60P

Scandium $\left(\text{Sc};\text{Z}=21\right)$ has the lowest ${\text{IE}}_3$ among the given elements.

Explanation of Solution

The atomic number of potassium is 19 so its electronic configuration is $\left[\text{Ar}\right]4s^1$. The third electron is to be removed from the stable configuration of argon which is very difficult so the value of its ${\text{IE}}_3$ will be very high.

The atomic number of calcium is 20 so its electronic configuration is $\left[\text{Ar}\right]4s^2$. The value of its ${\text{IE}}_3$ will be high because the third electron is to be removed from the stable configuration of argon which is quite difficult.

The atomic number of scandium is 21 so its electronic configuration is $\left[\text{Ar}\right]3d^{1}4s^2$. The value of its ${\text{IE}}_3$ will be the lowest among the given elements because after the removal of three electrons its configuration becomes that of argon which is very stable. The ion formation occurs as follows:

$\text{Sc}\left(\left[\text{Ar}\right]3d^{1}4s^2\right)\to {\text{Sc}}^{3+}\left(\left[\text{Ar}\right]\right)+3{\text{e}}^{-}$

Conclusion

Scandium $\left(\text{Sc};\text{Z}=21\right)$ has the lowest ${\text{IE}}_3$ among the given elements.

(c)

Interpretation Introduction

Interpretation:

The element with the lowest ${\text{IE}}_3$ in the set of $\text{Li}$, $\text{Al}$, $\text{B}$ is to be determined.

Concept introduction:

The ionization energy is the amount of energy required to remove the most loosely bound valence electrons from an isolated neutral gaseous atom. It is represented by $\text{IE}$.

Its value varies with the ease of removal of the outermost valence electron. If the outermost electron is removed very easily then the value of ionization energy is very small. If the electron is removed with quite a difficulty then the value of ionization energy will be very high.

When the first electron is removed from a neutral, isolated gaseous atom then the ionization energy is known as the first ionization energy $\left({\text{IE}}_1\right)$. Similarly, when the second electron is removed from the positively charged cation the ionization energy is called the second ionization energy $\left({\text{IE}}_2\right)$ and when the third electron is removed the ionization energy is called the third ionization energy $\left({\text{IE}}_3\right)$ and so on.

Answer to Problem 8.60P

Aluminium $\left(\text{Al};\text{Z}=\text{13}\right)$ has the lowest ${\text{IE}}_3$ among the given elements.

Explanation of Solution

The atomic number of lithium is 3 so its electronic configuration is $\left[\text{He}\right]2s^1$. The value of its ${\text{IE}}_3$ will be high because the third electron is to be removed from the stable configuration of helium which is quite difficult.

The atomic number of boron is 5 so its electronic configuration is $\left[\text{He}\right]2s^{2}2p^1$. The third electron is to be removed from the inner $2s$ orbital which is tightly bound to the nucleus so its removal will be difficult and therefore the value of ${\text{IE}}_3$ will be high.

The atomic number of aluminium is 13 so its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^1$. The value of its ${\text{IE}}_3$ will be the lowest because after the removal of the third electron its configuration becomes $1s^{2}2s^{2}2p^6$ which is the stable noble gas configuration of neon. The ion formation occurs as follows:

$\text{Al}\left(1s^{2}2s^{2}2p^{6}3s^{2}3p^1\right)\to {\text{Al}}^{3+}\left(1s^{2}2s^{2}2p^6\right)+3{\text{e}}^{-}$

Conclusion

Aluminium $\left(\text{Al};\text{Z}=\text{13}\right)$ has the lowest ${\text{IE}}_3$ among the given elements.