# The method of bleaching paper pulp with chlorine is replaced by a green chemical method using non toxic chemicals. Chlorine is highly oxidizing, producing toxic by products such as furans and dioxins in organic chemicals. find a), b) and c)?

## a) Write the electron configuration of the ground state chlorine atom and the most common form of chlorine ion. Does the neutral atom of an element have the same electronic configuration as this chlorine ion? b) When a chlorine atom is heated or shaken by light, the valence electrons of the chlorine can be energized at high energy levels. Predict the electron configuration of the lowest energy state among the excited states of the chlorine atoms. c) Estimate the wavelength (in nm) of the photon that must be absorbed in order for the electron to transfer to the state obtained in (b), using the following equation for the potential energy of an electron on energy level $n$: the effective nuclear charge of chlorine $\text{Z"_"eff}$ is assumed to be $6$. ${E}_{n} = \left(- {Z}_{\text{eff}}^{2}\right) \cdot \frac{2.18 \times {10}^{-} 18 J}{n} ^ 2$

Jun 14, 2018

(a) The two electron configurations differ by a single electron.

• $\text{Cl}$ atoms: $1 {s}^{2} \textcolor{w h i t e}{l} 2 {s}^{2} \textcolor{w h i t e}{l} 2 {p}^{6} \textcolor{w h i t e}{l} 3 {s}^{2} \textcolor{w h i t e}{l} 3 {p}^{5}$
• ${\text{Cl}}^{-}$ ions : $1 {s}^{2} \textcolor{w h i t e}{l} 2 {s}^{2} \textcolor{w h i t e}{l} 2 {p}^{6} \textcolor{w h i t e}{l} 3 {s}^{2} \textcolor{w h i t e}{l} 3 {p}^{6}$

(b) $1 {s}^{2} \textcolor{w h i t e}{l} 2 {s}^{2} \textcolor{w h i t e}{l} 2 {p}^{6} \textcolor{w h i t e}{l} 3 {s}^{2} \textcolor{w h i t e}{l} 3 {p}^{\textcolor{p u r p \le}{4}} \textcolor{w h i t e}{l} 4 {s}^{\textcolor{p u r p \le}{1}}$

(c) $\lambda = 52.1 \textcolor{w h i t e}{l} \text{nm}$, compared to the true value of $\text{138.97 nm}$.

#### Explanation:

Each ground-state chlorine atom (atomic number $17$) contains 17 electrons. Chlorine (a halogen) lies in the third period of the $p$-block on the periodic table, meaning that among all its occupied electron orbitals, $3 p$ orbitals shall have the highest potential energy.

Like other halogens, chlorine tends to gain one electron to form the chloride ion ${\text{Cl}}^{-}$. This process adds one electron to its $3 p$ orbital so that it achieves the noble gas electron configuration of argon $\text{Ar}$.

Electrons in orbitals of higher principal quantum number $n$ possess more potential energies than their counterparts of lower $n$ values. For example, an excited electron in this chlorine atom possesses more potential energy in a $\textcolor{n a v y}{5} s$ orbital than in a $\textcolor{n a v y}{4} s$ orbital. Minimizing total potential energy of electrons in an atom increases the stability of the atom as a whole.

Electrons in the outermost main energy level of this chlorine atom- which includes both the $3 \textcolor{n a v y}{s}$ and $3 \textcolor{n a v y}{p}$ orbitals- are considered valence electrons. Promoting a $3 \textcolor{n a v y}{s}$ electron the the $4 s$ orbital would lead to the electron configuration

$1 {s}^{2} \textcolor{w h i t e}{l} 2 {s}^{2} \textcolor{w h i t e}{l} 2 {p}^{6} \textcolor{w h i t e}{l} 3 {s}^{\textcolor{p u r p \le}{1}} \textcolor{w h i t e}{l} 3 {p}^{\textcolor{b l a c k}{5}} \textcolor{w h i t e}{l} 4 {s}^{\textcolor{p u r p \le}{1}}$

that introduces a vacancy in the $3 \textcolor{p u r p \le}{s}$ orbital. $3 \textcolor{p u r p \le}{s}$ orbitals are of lower potential energies than $3 p$ orbitals. Thus filling the $3 s$ orbital with a $3 p$ electron would increase the stability of the atom, which implies that the $1 {s}^{2} \textcolor{w h i t e}{l} 2 {s}^{2} \textcolor{w h i t e}{l} 2 {p}^{6} \textcolor{w h i t e}{l} 3 {s}^{\textcolor{p u r p \le}{1}} \textcolor{w h i t e}{l} 3 {p}^{\textcolor{b l a c k}{5}} \textcolor{w h i t e}{l} 4 {s}^{\textcolor{p u r p \le}{1}}$ configuration doesn't gurantee maximum stability. Filling the gap with a $3 p$ electron, on the other hand, produces the configuration

$1 {s}^{2} \textcolor{w h i t e}{l} 2 {s}^{2} \textcolor{w h i t e}{l} 2 {p}^{6} \textcolor{w h i t e}{l} 3 {s}^{2} \textcolor{w h i t e}{l} 3 {p}^{\textcolor{p u r p \le}{4}} \textcolor{w h i t e}{l} 4 {s}^{\textcolor{p u r p \le}{1}}$

Promoting a $3 p$ electron to the $4 s$ orbital would directly yield this configuration- likely the most stable setting that satisfies conditions in (b).

There exist an inverse relationship between the energy and wavelength of electromagnetic radiations. Start by calculating the energy change associated with this particular transition:

Delta"PE"="PE"("Final")-"PE"("Initial")

The electron in question was in a $\boldsymbol{3} p$ orbital and possess principle quantum number $n = \boldsymbol{3}$; its final position $\boldsymbol{4} s$ features $n = \boldsymbol{4}$. Thus according to the question:

$\Delta \text{PE} = {E}_{\boldsymbol{4}} - {E}_{\boldsymbol{3}}$
$\textcolor{w h i t e}{\Delta \text{PE}} = \left(- {6}^{2}\right) \cdot \frac{2.18 \times {10}^{-} 18 J}{4} ^ 2 - \left(- {6}^{2}\right) \cdot \frac{2.18 \times {10}^{-} 18 J}{3} ^ 2$
color(white)(Delta "PE")=3.82 xx 10^(-18) color(white)(l) "J"

Note that this equation does not make any physical sense, as it only works for hydrogen-like atoms, i.e. if for chlorine, we would have ${\text{Cl}}^{16 +}$.

But we proceed with this hypothetical calculation...

${E}_{\text{photon}} = \frac{h \cdot c}{\lambda}$ where

• The Planck's Constant $h = 6.62607004 \times {10}^{- 34} \textcolor{w h i t e}{l} m \cdot k g \cdot {s}^{- 2}$
• $c$ the speed of light
• $\lambda$ the wavelength of the radiation

Taking $c = 3.0 \cdot {10}^{8} \textcolor{w h i t e}{l} {\text{m"* "s"^(-1)=3.0*10^17"nm"* "s}}^{- 1}$,

lambda=(h*c)/(E_"photon")=52.1 color(white)(l) "nm"

whereas the true value is $\text{138.97 nm}$ using the excited-state data from NIST (search $\text{Cl I}$).