Chapter 8, Problem 8.59P

(a)

Interpretation Introduction

Interpretation:

The element with the highest ${\text{IE}}_2$ in the set of $\text{Na, Mg, Al}$ 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 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.

The ionization energy of an element increases along the period because the electrons are held by high effective nuclear charge. The value of ionization energy decreases down the group because the valence electrons are much farther from the nucleus and therefore experience weaker forces of attraction.

Answer to Problem 8.59P

Na will have the highest value of ${\text{IE}}_2$ among the given elements.

Explanation of Solution

The first ionization energy is determined by the removal of an electron from the neutral parent atom. The value of the second ionization for all the elements will be higher than that of ${\text{IE}}_1$ because the second electron is to be removed from the cation which is very difficult.

The atomic number of Na is 11 so it's ground state electronic configuration is $1s^{2}2s^{2}2p^{6}3s^1$.

After the removal of an electron, $\text{Na}$ will be converted into ${\text{Na}}^{+}$ and its electronic configuration is $1s^{2}2s^{2}2p^6$ which represents the stable configuration of $\text{Ne}$.

$\text{Na }\text{}\left(\left[\text{Ne}\right]\text{}3s^1\right)\to {\text{Na}}^{+}\left(\left[\text{Ne}\right]\text{}\right)+\text{}{\text{e}}^{-}\left(\text{low}{\text{IE}}_1\right)$

For the second ionization energy, ${\text{Na}}^{+}$ will have to lose an electron from its stable noble gas configuration of $\text{Ne}$, which is quite difficult so the value of ${\text{IE}}_2$ will be the highest.

${\text{Na}}^{+}\left(\left[\text{Ne}\right]\text{}\right)\to {\text{Na}}^{2+}\left(\left[\text{He}\right]2s^{2}2p^5\right)+{\text{e}}^{-}\left(\text{high}{\text{IE}}_2\right)$

The atomic number of Mg is 12 so it’s ground state electronic configuration is $1s^{2}2s^{2}2p^{6}3s^2$. After the removal of an electron, $\text{Mg}$ will be converted into ${\text{Mg}}^{+}$ and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^1$.

$\text{Mg }\text{}\left(\left[\text{Ne}\right]\text{}3s^2\right)\to {\text{Mg}}^{+}\left(\left[\text{Ne}\right]3s^1\text{}\right)+\text{}{\text{e}}^{-}\left({\text{IE}}_1\right)$

For the second ionization energy, ${\text{Mg}}^{+}$ will have to lose an electron and get converted into the stable noble gas configuration of $\text{Ne}$ so the value of ${\text{IE}}_2$ will be low.

${\text{Mg}}^{+}\left(\left[\text{Ne}\right]3s^1\text{}\right)\to {\text{Mg}}^{2+}\left(\left[\text{Ne}\right]\right)+\text{}{\text{e}}^{-}\left({\text{low IE}}_2\right)$

The atomic number of $\text{Al}$ is 13 so it’s ground state electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^1$.

For the first ionization energy, the change in electronic configuration is as follows:

$\text{Al}\text{}\left(\left[\text{Ne}\right]\text{}3s^{2}3p^1\right)\to {\text{Al}}^{+}\left(\left[\text{Ne}\right]3s^2\text{}\right){\text{+e}}^{-}\left({\text{IE}}_1\right)$

For the second ionization energy, the change in electronic configuration is as follows:

${\text{Al}}^{+}\left(\left[\text{Ne}\right]3s^2\text{}\right)\to {\text{Al}}^{2+}\left(\left[\text{Ne}\right]3s^1\text{}\right)+{\text{e}}^{-}\left({\text{IE}}_2\right)$

The first ionization energy is determined by the removal of an electron from the neutral parent atom. The value of the second ionization for all the elements will be higher than that of ${\text{IE}}_1$ because the second electron is to be removed from the cation which is very difficult.

Conclusion

$\text{Na}$ will have the highest value of ${\text{IE}}_2$ among the given elements.

(b)

Interpretation Introduction

Interpretation:

The element with the highest ${\text{IE}}_2$ in the set of $\text{Na, K, Fe}$ 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 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.

The ionization energy of an element increases along the period because the electrons are held by high effective nuclear charge. The value of ionization energy decreases down the group because the valence electrons are much farther from the nucleus and therefore experience weaker forces of attraction.

Answer to Problem 8.59P

$\text{Na}$ will have the highest value of ${\text{IE}}_2$ among the given elements.

Explanation of Solution

The first ionization energy is determined by the removal of an electron from the neutral parent atom. The value of the second ionization for all the elements will be higher than that of ${\text{IE}}_1$ because the second electron is to be removed from the cation which is very difficult.

The atomic number of $\text{Na}$ is 11 so it's ground state electronic configuration is $1s^{2}2s^{2}2p^{6}3s^1$.

After the removal of an electron, $\text{Na}$ will be converted into ${\text{Na}}^{+}$ and its electronic configuration is $1s^{2}2s^{2}2p^6$ which represents the stable configuration of $\text{Ne}$.

$\text{Na }\text{}\left(\left[\text{Ne}\right]\text{}3s^1\right)\to {\text{Na}}^{+}\left(\left[\text{Ne}\right]\text{}\right)+\text{}{\text{e}}^{-}\left(\text{low}{\text{IE}}_1\right)$

For the second ionization energy, ${\text{Na}}^{+}$ will have to lose an electron from its stable noble gas configuration of $\text{Ne}$, which is quite difficult so the value of ${\text{IE}}_2$ will be the highest.

${\text{Na}}^{+}\left(\left[\text{Ne}\right]\text{}\right)\to {\text{Na}}^{2+}\left(\left[\text{He}\right]2s^{2}2p^5\right)+{\text{e}}^{-}\left(\text{high}{\text{IE}}_2\right)$

The atomic number of $\text{K}$ is 19 so it’s ground state electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^1$.

For the first ionization energy, the change in electronic configuration is as follows:

$\text{K}\left(\left[\text{Ar}\right]4s^1\right)\to {\text{K}}^{+}\left(\left[\text{Ar}\right]\right)+{\text{e}}^{-}\left({\text{IE}}_{\text{1}}\right)$

For the second ionization energy, the change in electronic configuration is as follows:

${\text{K}}^{+}\left(\left[\text{Ar}\right]\right)\to {\text{K}}^{2+}\left(\left[\text{Ne}\right]3s^{2}3p^5\right)+\text{}{\text{e}}^{-}\left({\text{IE}}_2\right)$

For both sodium and potassium, the second electron is removed from a stable noble gas configuration but because the size of sodium is smaller than potassium the second ionization energy of sodium will be greater than potassium.

The atomic number of $\text{Fe}$ is 26 so it's ground state electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^6$.

For the first ionization energy, the change in electronic configuration is as follows:

$\text{Fe}\left(\left[\text{Ar}\right]3d^{6}4s^2\right)\to {\text{Fe}}^{+}\left(\left[\text{Ar}\right]3d^{6}4s^1\right)+{\text{e}}^{-}\left({\text{high IE}}_1\right)$

For the second ionization energy, the change in electronic configuration is as follows:

${\text{Fe}}^{+}\left(\left[\text{Ar}\right]3d^{6}4s^1\right)\to {\text{Fe}}^{2+}\left(\left[\text{Ar}\right]3d^6\right)+{\text{e}}^{-}\left({\text{low IE}}_2\right)$

In case of iron, the first ionization energy is high because the first electron is removed from the fully filled $s$ subshell and second ionization energy is low because the second electron is removed from half-filled $s$ subshell.

Conclusion

$\text{Na}$ will have the highest value of ${\text{IE}}_2$ among the given elements.

(c)

Interpretation Introduction

Interpretation:

The element with the highest ${\text{IE}}_2$ in the set of $\text{Sc, Be, Mg}$ 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 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.

The ionization energy of an element increases along the period because the electrons are held by high effective nuclear charge. The value of ionization energy decreases down the group because the valence electrons are much farther from the nucleus and therefore experience weaker forces of attraction.

Answer to Problem 8.59P

$\text{Be}$ will have the highest value of ${\text{IE}}_2$ among the given elements.

Explanation of Solution

Down the group the ionization energy decreases. Scandium is a placed in the  4^th period, magnesium is placed in the 3^rd  period and beryllium is placed 2^nd period. Therefore, among $\text{Sc, Be, Mg}$, scandium will have the lowest ionization energy.

The atomic number of $\text{Be}$ is 4 so it's ground state electronic configuration is $1s^{2}2s^2$.

For the first ionization energy, the change in electronic configuration is as follows:

$\text{Be}\left(\left[\text{He}\right]2s^2\right)\to \text{}{\text{Be}}^{+}\left(\left[\text{He}\right]2s^1\right)+{\text{e}}^{-}\left({\text{IE}}_1\right)$

For the second ionization energy, the change in electronic configuration is as follows:

${\text{Be}}^{+}\left(\left[\text{He}\right]2s^1\right)\to {\text{Be}}^{2+}\left(\left[\text{He}\right]\right)+{\text{e}}^{-}\left({\text{IE}}_2\right)$

The size of $\text{Be}$ is very small so the removal of an electron from its valence shell is quite difficult. So the value of ${\text{IE}}_2$ will be highest for $\text{Be}$.

The atomic number of $\text{Mg}$ is 12 so it's ground state electronic configuration is $1s^{2}2s^{2}2p^{6}3s^2$.

After the removal of an electron, $\text{Mg}$ will be converted into ${\text{Mg}}^{+}$ and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^1$.

$\text{Mg}\text{}\left(\left[\text{Ne}\right]\text{}3s^2\right)\to {\text{Mg}}^{+}\left(\left[\text{Ne}\right]3s^1\text{}\right)+\text{}{\text{e}}^{-}\left({\text{IE}}_1\right)$

For the second ionization energy, the change in electronic configuration is as follows:

${\text{Mg}}^{+}\left(\left[\text{Ne}\right]3s^1\text{}\right)\to {\text{Mg}}^{2+}\left(\left[\text{Ne}\right]\right)+{\text{e}}^{-}\text{}\left({\text{IE}}_2\right)$

It is expected that $\text{Sc}$ will have the highest value of ionization energy but the size of $\text{Be}$ is very small so the removal of an electron from its valence shell is more difficult as compared to that in case of $\text{Sc}$. So the value of ${\text{IE}}_2$ will be highest for $\text{Be}$.

Conclusion

$\text{Be}$ will have the highest value of ${\text{IE}}_2$ among the given elements.