Chapter 4.4, Problem 4.6WE

For carbon and nitrogen, use the effective nuclear charges given in Figure 4.7 and the atomic radii given in Figure 4.6 to compare the attractive force between the nucleus in each atom and the valence electron that would be removed by the first ionization.

Interpretation Introduction

Interpretation: For the two given atoms the force of attraction and the first ionization energy between them should be compared.

Concept Introduction:

Coulombs law: It states that the force of attraction between two point charges is directly proportional to the product of the point charges and inversely proportional the square of the distances between the charges.

$\begin{array}{l}F\propto \frac{Q_1\times Q_2}{d^{\text{2}}}\\ Q_1,Q_2\to \text{Given point charges}\\ d\to \text{distance between the charges}\end{array}$

First ionization energy:

The ionization energy is the minimum energy required to remove the electron from an isolated atom which is in the gaseous state results to give gaseous ion with one positive charge.

${\text{atom}}_{\left(\text{g}\right)}\to {\text{ion with positive charge}}_{\left(\text{g}\right)}\text{ + electron}$

Valence electron: The electron is considered as valence electron if it present in outermost shell of atom which gets involved in the formation of chemical bond.

Effective nuclear charge: It is the overall positive charge experienced by the outermost electrons present in the atom from the nucleus of the atom.

Screening Effect: The core electrons present near the nucleus shields the outermost electrons (valence electrons) from the charge of the nucleus.

To compare: The force of attraction and the first ionization energy for the two given atoms.

Answer to Problem 4.6WE

Answer

The force of attraction in $\text{C}$ atom is $-5.3\times {10}^{-4}$ and the attraction force for $\text{N}$ atom is $-6.8\times {10}^{-4}$.

The force for $\text{N}$ is larger than that of $\text{C}$ since the negative sign represents the attractive force hence results to the smaller size of $\text{N}$, where the electron can be easily removed from $\text{C}$ with smaller first ionization energy.

Explanation of Solution

Determine the force of attraction for the two given atoms.

$\begin{array}{l}\text{Given Data,}\\ \text{Effective nuclear charge for C = 3}\text{.14}\\ \text{Effective nuclear charge for N = 3}\text{.83}\\ \text{ Radii for C = 77 pm}\\ \text{ Radii for N = 75 pm}\end{array}$

$\begin{array}{l}\text{The force of attraction for C atom,}\\ F_{\text{C}}\propto \frac{3.14\times \left(-1\right)}{{\left(77\text{pm}\right)}^2}\\ =-5.3\times {10}^{-4}\\ \text{The force of attraction for N atom,}\\ F_{\text{N}}\propto \frac{3.83\times \left(-1\right)}{{\left(75\text{pm}\right)}^2}\\ =-6.8\times {10}^{-4}\end{array}$

The force of attraction for given atoms are determined by substituting the effective nuclear charge and the radii values to the given formula results to give $-5.3\times {10}^{-4}$ and $-6.8\times {10}^{-4}$ values for $\text{C}$ and $\text{N}$ respectively.

Analyze the obtained values for the given atoms.

Examining the obtained values shows us that attraction forces are stronger in $\text{N}$ than $\text{C}$ since the negative sign is just to represent attractive forces.

Stronger attraction forces in $\text{N}$ serves as reason for the smaller size of $\text{N}$ atom compared to $\text{C}$ hence it is easy to remove the first electron from the carbon atom as its size is comparatively large results in smaller first ionization energy for $\text{C}$ atom.

Conclusion

Conclusion

The force of attraction and first ionization energy for the given atoms are compared using the coulombs law and the size.