Question

Photoelectron spectroscopy studies have determined the orbital energies for chlorine atoms to be

$$\begin{gathered} \mathbf{1}\mathbf{s}\mathbf{}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{835}\mathbf{}\mathbf{eV} \\ \mathbf{2}\mathbf{s}\mathbf{}\mathbf{-}\mathbf{273}\mathbf{}\mathbf{eV} \\ \mathbf{2}\mathbf{p}\mathbf{}\mathbf{-}\mathbf{205}\mathbf{}\mathbf{eV} \\ \mathbf{3}\mathbf{s}\mathbf{}\mathbf{-}\mathbf{21}\mathbf{}\mathbf{eV} \\ \mathbf{3}\mathbf{p}\mathbf{}\mathbf{-}\mathbf{10}\mathbf{}\mathbf{eV} \end{gathered}$$

Estimate the value of${\mathit{Z}}_{\mathbf{eff}}$for$\mathbf{Cl}$in each of these orbitals.

Short answer

The $Z_{\mathrm{eff}}$for chlorine atom in $1s,2s,2p,3s$and $3p$is $14.4341$, $8.9582$, $7.7628$, $3.7268$and $2.5717$respectively.

Step-by-step solution

Formula for calculating the effective nuclear charge

A particular electron is shielded by inner electrons that are near the nucleus by the full charge$\mathit{Z}$of the nucleus by effectively canceling some of the positive nuclear charges. So the net reduced nuclear charge experienced by a particular electron, due to inner electrons, is called${\mathit{Z}}_{\mathbf{eff}}$.

The formula for the effective nuclear charge${\mathit{Z}}_{\mathbf{eff}}$is as follows:

${\mathit{Z}}_{\mathbf{eff}}\mathbf{=}\mathit{Z}\mathbf{-}\mathit{S}$

where

- $\mathit{Z}$is a nuclear charge.

- $\mathit{S}$is shielding or screening constant for an orbital with a principal quantum number$\mathit{n}$.

Calculating the effective nuclear charge for orbital 1s

The energy of $1s,2s,2p,3s$and $3s$is $-2835\mathrm{eV},-273\mathrm{eV},-205\mathrm{eV},-21\mathrm{eV}$and $-10\mathrm{eV}$respectively.

The energy of the orbitals and $Z_{\mathrm{eff}}$are related to each other by the expression as follows:

$E=-{\left(\frac{{\mathrm{Z}}_{\mathrm{eff}}}{n}\right)}^2$ …… (1)

where

- $E$is the energy of the orbitals in Rydberg.

Rearrange the above equation (1) to calculate the $Z_{\mathrm{eff}}$.

$Z_{\mathrm{eff}}=-n\sqrt{E}$ …… (2)

For orbital localid="1663477977291" role="math" $1s$:

The energy of $1s$is $-2835\mathrm{eV}$.

The value of $n$is 1.

Substitute these values in equation (2).

$\begin{array}{l}{Z}_{\mathrm{eff}}=-n\sqrt{E}\\ Z_{\mathrm{eff}}=-1\sqrt{-2835\mathrm{eV}\left(\frac{0.07349\mathrm{Rydberg}}{1\mathrm{eV}}\right)}\\ Z_{\mathrm{eff}}=-1\sqrt{-208.344\mathrm{Rydberg}}\\ Z_{\mathrm{eff}}=14.4341\end{array}$

Calculating the effective nuclear charge for orbital 2s

For orbital $2s$:

The energy of $2s$is role="math" localid="1663478120770" $-273\mathrm{eV}$.

The value of $n$is 2.

Substitute these values in equation (2).

$\begin{array}{l}{Z}_{\mathrm{eff}}=-n\sqrt{E}\\ Z_{\mathrm{eff}}=-2\sqrt{-273\mathrm{eV}\left(\frac{0.07349\mathrm{Rydberg}}{1\mathrm{eV}}\right)}\\ Z_{\mathrm{eff}}=-2\sqrt{-20.0627\mathrm{Rydberg}}\\ Z_{\mathrm{eff}}=8.9582\end{array}$

Calculating the effective nuclear charge for orbital 2p

For orbital $2p$:

The energy of $2p$is $-205\mathrm{eV}$.

The value of $n$is 2.

Substitute these values in equation (2).

$\begin{array}{l}{Z}_{\mathrm{eff}}=-n\sqrt{E}\\ Z_{\mathrm{eff}}=-2\sqrt{-205\mathrm{eV}\left(\frac{0.07349\mathrm{Rydberg}}{1\mathrm{eV}}\right)}\\ Z_{\mathrm{eff}}=-2\sqrt{-15.0654\mathrm{Rydberg}}\\ Z_{\mathrm{eff}}=7.7628\end{array}$

Calculating the effective nuclear charge for orbital 3s

For orbital $3s$:

The energy of $3s$is $-21\mathrm{eV}$.

The value of $n$is 3.

Substitute these values in equation (2).

$\begin{array}{l}{Z}_{\mathrm{eff}}=-n\sqrt{E}\\ Z_{\mathrm{eff}}=-3\sqrt{-21\mathrm{eV}\left(\frac{0.07349\mathrm{Rydberg}}{1\mathrm{eV}}\right)}\\ Z_{\mathrm{eff}}=-3\sqrt{-1.54329\mathrm{Rydberg}}\\ Z_{\mathrm{eff}}=3.7268\end{array}$

Calculating the effective nuclear charge for orbital 3p

For orbital $3p$:

The energy of $3p$is $-10\mathrm{eV}$.

The value of $n$is 3.

Substitute these values in equation (2)

$\begin{array}{l}{Z}_{\mathrm{eff}}=-n\sqrt{E}\\ Z_{\mathrm{eff}}=-3\sqrt{-10\mathrm{eV}\left(\frac{0.07349\mathrm{Rydberg}}{1\mathrm{eV}}\right)}\\ Z_{\mathrm{eff}}=-3\sqrt{-0.7349\mathrm{Rydberg}}\\ Z_{\mathrm{eff}}=2.5717\end{array}$$\begin{array}{l}{Z}_{\mathrm{eff}}=-n\sqrt{E}\\ Z_{\mathrm{eff}}=-3\sqrt{-10\mathrm{eV}\left(\frac{0.07349\mathrm{Rydberg}}{1\mathrm{eV}}\right)}\\ Z_{\mathrm{eff}}=-3\sqrt{-0.7349\mathrm{Rydberg}}\\ Z_{\mathrm{eff}}=2.5717\end{array}$

Thus, the $Z_{\mathrm{eff}}$for chlorine atom in role="math" localid="1663478503380" $1s,2s,2p,3s$and $3p$is $14.4341$, $8.9582$, $7.7628$, $3.7268$and $2.5717$respectively.