# What are the first excited states of "Mg", "Al", "Si", "P", "S", and "Cl"?

Jun 4, 2017

Still magnesium

#### Explanation:

In chemistry, excited state means that an electron from an orbital absorbs incoming radiation or a photon that corresponds to its energy level and hence is elevated to a higher energy level. Hence, when an element is excited, only electrons are concerned, thus not affecting the identity of the element. In order for an element to transmute or decay into another element, such as magnesium into aluminium, it would have to gain protons which is irrelevant to it being in an excited state (in terms of chemistry).

Therefore, excited state of magnesium is still magnesium, but with electrons elevated to higher energy levels.

(In physics, excited state of an element means an unstable nuclide of the element, and in THAT case it is possible for the element to undergo decay into another element. However, as this is in the chemistry section, I assume you are not referring to this definition of excitation)

Jun 4, 2017

I will only do magnesium in full. It would be too long to explain for six atoms, unless you want to do most of the work yourself...

Magnesium experiences a spin-forbidden transition, giving:

$\left[N e\right] 3 {s}^{1} 3 {p}^{1}$

Aluminum can be found here.

I will let you ponder this, and determine that the following transitions are correct. Hopefully you notice the pattern here...

$\text{ground states"" "" ""excited states}$
$\underline{\text{ "" "" "" "" "" "" "" "" "" "" "" "" }}$
$\text{Si}$: $\left[N e\right] 3 {s}^{2} 3 {p}^{2} \to \textcolor{b l u e}{\left[N e\right] 3 {s}^{2} 3 {p}^{1} 4 {s}^{1}}$

$\text{P}$: $\left[N e\right] 3 {s}^{2} 3 {p}^{3} \to \textcolor{b l u e}{\left[N e\right] 3 {s}^{2} 3 {p}^{2} 4 {s}^{1}}$

$\text{S}$: $\left[N e\right] 3 {s}^{2} 3 {p}^{4} \to \textcolor{b l u e}{\left[N e\right] 3 {s}^{2} 3 {p}^{3} 4 {s}^{1}}$

$\text{Cl}$: $\left[N e\right] 3 {s}^{2} 3 {p}^{5} \to \textcolor{b l u e}{\left[N e\right] 3 {s}^{2} 3 {p}^{4} 4 {s}^{1}}$

Each time, it is the unpaired $n p$ electron that transitions into the $\left(n + 1\right) s$ orbital for $\text{Si}$, $\text{P}$, $\text{S}$, and $\text{Cl}$, since there is one unpaired $n p$ electron available in each atom, and it results in a spin-allowed transition, which tends to be favored.

GROUND STATE

Magnesium starts off with a ground state configuration of

$\left[N e\right] 3 {s}^{2}$

$\underline{\uparrow \downarrow}$
$3 s$

Thus, only a $3 s$ electron, which is highest in energy in the atom (and thus easiest to excite), could transition.

ATOMIC SELECTION RULES

Energy transitions follow the selection rules:

• The change in total angular momentum, $\Delta L$, must be $\boldsymbol{\pm 1}$, where $L = | {\sum}_{k} {m}_{l , k} |$, $k$ indicates the $k$th electron, and ${m}_{l}$ is the magnetic quantum number of the orbital.
• There must be no change in total spin, $S = | {\sum}_{i} {m}_{s , i} |$, where $k$ indicates the $k$th electron and ${m}_{s}$ is the spin quantum number of the electron. If there is, then the transition is spin-forbidden. It means the absorption will be weak, but it can happen.

EXPECTED TRANSITIONS

From these rules, we expect a transition into the next energy level, one of the $3 {p}_{\pm 1}$ orbitals (so that $\Delta L \ne 0$):

Option 1

$\underline{\textcolor{w h i t e}{\downarrow} \uparrow} \text{ "ul(color(white)(uarr darr))" } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\underbrace{\text{ "" "" "" "" "" "" "" }}$
$\text{ "" "" } 3 p$

$\underline{\uparrow \cancel{\downarrow}}$
$3 s$

Option 2

$\underline{\textcolor{w h i t e}{\uparrow \downarrow}} \text{ " ul(color(white)(uarr darr))" } \underline{\textcolor{w h i t e}{\downarrow} \uparrow}$
$\underbrace{\text{ "" "" "" "" "" "" "" }}$
$\text{ "" "" } 3 p$

$\underline{\uparrow \cancel{\downarrow}}$
$3 s$

CHECK TO SEE HOW L AND S CHANGE

The initial and final states of $L$ and $S$ for the two electrons are therefore:

• ${L}_{i} = | {m}_{l , 1} + {m}_{l , 2} | = 0 + 0 = 0$
• ${L}_{f} = | {m}_{l , 1} + {m}_{l , 2} | = | 0 \pm 1 | = 1$
• ${S}_{i} = | {m}_{s , 1} + {m}_{s , 2} | = {\overbrace{+ \frac{1}{2} - \frac{1}{2}}}^{\text{same orbital}} = 0$
• ${S}_{f} = | {m}_{s , 1} + {m}_{s , 2} | = {\overbrace{+ \frac{1}{2} + \frac{1}{2}}}^{\text{separate orbitals}} = 1$

Therefore, $\Delta S = 1$ and $\Delta L = + 1$, which is spin-forbidden but the change in orbital angular momentum is allowed.

This transition is spin-forbidden and weak, but is more energetically favorable than the spin-allowed option.

CHECK THE DATA

This is the transition to the ""^3 P_1 state with $\lambda = \text{457.3 nm}$, with a $\text{2.3 ms}$ decay time, and gives $\textcolor{b l u e}{\left[N e\right] 3 {s}^{1} 3 {p}^{1}}$.

The spin-allowed transition to a ""^1 P_1 state would not follow Hund's rule of maximum multiplicity, and thus it is not energetically favorable. In fact, its decay time is $\text{2.0 ns}$, which demonstrates that it is more unstable than the spin-forbidden result.