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## What is dipole moment and how to calculate it?

jmleyva
Featured 4 months ago

The dipole moment characterizes a system of two equal but oposite charges by the equation $\vec{p} = q \cdot \vec{d}$

#### Explanation:

An electric dipole is a system formed by two charges with the same (magnitude) but opposite charge, as you can see in the figure:

This system can be characterized by a vectorial magnitude called dipole moment ($\vec{p}$) which depends on the charge of this particles and the distance between them. It's calculated by multiplying the charge of one of the two particles, q (same for them both) by the vector distance($\vec{d}$), which has a magnitude equal to the distance between the two particles, a direction particle-particle and sense that goes from the negative to the positive one. This is represented in the next figure:

Therefore, as you can see, the dipole moment has the mathematical expression:

$\vec{p} = q \cdot \vec{d}$

Sometimes, its measured in Debyes (1D=3.33564×10−30 C·m) and, in chemistry, is usually referred to the dipole moment between atoms. It's a very useful measure , as it tells us how molecules and particles behave in the presence of an electromagnetic field.

Here you have a problem example:

• ¿What is the dipole moment of a pair of one proton (placed in A=(-2,1,4)m) and one electron (placed in B=(3,0,2)m)? *

• We know that electrons and protons have the same charge, e=1.6×10−19 C, positive for the proton and negative for the electron.

• The vector distance is the vector generated by the two positions of these particles (electron-proton sense)

$\vec{d} = \left(3 , 0 , 2\right) m - \left(- 2 , 1 , 4\right) m = \left(5 , - 1 , - 2\right) m$

Applying the equation above, we have:

$\vec{p} = q \cdot \vec{d} = 1.6 \cdot {10}^{-} 19 C \cdot \left(5 , - 1 , - 2\right) m$

$\vec{p} = \left(8.01 , - 1.60 , - 3.20\right) \cdot {10}^{-} 19 C \cdot m$

Therefore, it's magnitude is $8.774 \cdot 10 - 19$ C·m, or $2.63 \cdot {10}^{11}$debyes.

## What determines the probability pattern of an orbital?

Truong-Son N.
Featured 4 months ago

It just turned out that way. We can't define equations to influence nature, only describe it.

You can imagine anything you want for what "determines" the probability distribution, but ultimately, nature turned out that way, and we described these orbitals after-the-fact.

DISCLAIMER: This is going to be a long answer.

WHAT IS THE WAVE FUNCTION FOR AN ORBITAL?

Erwin Schrodinger published the wave function $\psi$, which describes the state of a quantum mechanical system.

For each orbital, its radial density distribution describes the regions with particular probabilities for finding an electron in that particular orbital.

In general, the wave function for spherical harmonics coordinates can be written as:

$\setminus m a t h b f \left({\psi}_{n l {m}_{l}} \left(r , \theta , \phi\right) = {R}_{n l} \left(r\right) {Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)\right)$

where:

• $r$, $\theta$, and $\phi$ are spherical coordinates:

• ${\psi}_{n l {m}_{l}}$ is a wave function that can be constructed to describe what the orbital's electron distribution looks like.

It depends on the quantum numbers $\setminus m a t h b f \left(n\right)$, $\setminus m a t h b f \left(l\right)$, and $\setminus m a t h b f \left({m}_{l}\right)$.

• ${R}_{n l}$ is the radial component of the wave function, describing the variation in the distance from the center of the orbital (the radius!).

It depends on the quantum numbers $\setminus m a t h b f \left(n\right)$ and $\setminus m a t h b f \left(l\right)$.

• ${Y}_{l}^{{m}_{l}}$ is the angular component of the wave function, describing the aspects of the orbital electron distribution that can possibly give it a non-spherical shape.

It depends on the quantum numbers $\setminus m a t h b f \left(l\right)$ and $\setminus m a t h b f \left({m}_{l}\right)$.

OKAY, WHAT DOES IT HAVE TO DO WITH PROBABILITY PATTERNS??

Focus on the ${R}_{n l} \left(r\right)$ portion of $\psi$.

When we plot the radial density distribution (or "probability pattern") of an orbital, we plot $4 \pi {r}^{2} {\left({R}_{n l} \left(r\right)\right)}^{2}$ vs. $r$ in units of ${a}_{0} = 5.29177 \times {10}^{- 11} \text{m}$ (Bohr radii) like so:

The wave functions themselves, as we defined them, generate these "probability patterns".

HOW DO I KNOW YOU'RE NOT MAKING THIS UP?

Well, with an example, I guess. This will give a result that should prove familiar.

The simplest example of an orbital wave function (used by $\text{Real Chemists"^"TM}$, folks!) is the one for the $\setminus m a t h b f \left(1 s\right)$ orbital, which happens to be:

#color(blue)(psi_(1s) = R_(10)(r)Y_(0)^(0)(theta,phi)#

$= \stackrel{{R}_{10} \left(r\right)}{\overbrace{\left(2 {Z}^{\text{3/2")/(a_0^"3/2")e^(-"Zr/}} {a}_{0}\right)}} \cdot \stackrel{{Y}_{0}^{0} \left(\theta , \phi\right)}{\overbrace{\frac{1}{\sqrt{4 \pi}}}}$

#= color(blue)(1/sqrt(pi) (Z/(a_0))^"3/2" e^(-"Zr/"a_0))#

where:

• $Z$ is the atomic number of the atom.
• ${a}_{0}$ is the same Bohr radius we just defined as $5.29177 \times {10}^{- 11} \text{m}$.

In order to plot the "probability pattern", we have to grab what we need and turn it into $4 \pi {r}^{2} {\left({R}_{n l} \left(r\right)\right)}^{2}$.

$4 \pi {r}^{2} {\left({R}_{n l} \left(r\right)\right)}^{2}$

$= 4 \pi {r}^{2} {\left(\stackrel{{R}_{n l} \left(r\right)}{\overbrace{\left(2 {Z}^{\text{3/2")/(a_0^"3/2")e^(-"Zr/}} {a}_{0}\right)}}\right)}^{2}$

$= 4 \pi {r}^{2} \frac{4 {Z}^{3}}{{a}_{0}^{3}} {e}^{- \text{2Zr/} {a}_{0}}$

$\implies \textcolor{red}{4 \pi {r}^{2} {\left({R}_{n l} \left(r\right)\right)}^{2} = 16 \pi {\left(\frac{Z}{{a}_{0}}\right)}^{3} {r}^{2} {e}^{- \text{2Zr/} {a}_{0}}}$

This is the function that corresponds to the red curve in the diagram above.

Now, since we have the plot for $4 \pi {r}^{2} {R}_{n l}^{2} \left(r\right)$ vs. $r$, one thing we can do is find the region of highest electron density, which is just the highest point in the $1 s$ orbital graph above.

So...

1. Take the derivative with respect to $r$.
2. Set it equal to $0$, since the peak has a slope of $0$.

$\implies 16 \pi {\left(\frac{Z}{{a}_{0}}\right)}^{3} \frac{d}{\mathrm{dr}} \left[{r}^{2} {e}^{- \text{2Zr/} {a}_{0}}\right]$

Utilize the product rule:

$\implies 16 \pi {\left(\frac{Z}{{a}_{0}}\right)}^{3} \left[\frac{- 2 Z}{{a}_{0}} {r}^{2} {e}^{- \text{2Zr/"a_0) + 2re^(-"2Zr/} {a}_{0}}\right]$

Now setting it equal to $0$ and crossing out nonzero terms:

$0 = \cancel{16 \pi {\left(\frac{Z}{{a}_{0}}\right)}^{3}} \left[\frac{- 2 Z}{{a}_{0}} {r}^{2} {e}^{- \text{2Zr/"a_0) + 2re^(-"2Zr/} {a}_{0}}\right]$

$0 = \frac{- 2 Z}{{a}_{0}} {r}^{2} {e}^{- \text{2Zr/"a_0) + 2re^(-"2Zr/} {a}_{0}}$

Although ${e}^{- x}$ reaches $0$ at very large $r$, it is nonzero for small $r$, and we are at very small $r$. So...

$0 = \left(\frac{- 2 Z}{{a}_{0}} {r}^{2} + 2 r\right) \cancel{{e}^{- \text{2Zr/} {a}_{0}}}$

$0 = \frac{- 2 Z}{{a}_{0}} {r}^{2} + 2 r$

$\cancel{2} r = \frac{\cancel{2} Z}{{a}_{0}} {r}^{2}$

$r = \frac{Z}{{a}_{0}} {r}^{2} \implies \textcolor{g r e e n}{r = \frac{{a}_{0}}{Z}}$

...for a Hydrogen-like atom (e.g. ${\text{He}}^{+}$, ${\text{Li}}^{2 +}$, etc).

But for the Hydrogen atom, $Z = 1$, so

$\setminus m a t h b f \left(\textcolor{b l u e}{r = {a}_{0} = 5.29177 \times {10}^{- 11} \text{m}}\right)$.

That tells us that the electron in the $1 s$ orbital of Hydrogen atom is most often one bohr radius distance away from the center of the nucleus, just as we would expect. Great!

## Can someone check my work please and please tell me if I made a mistake where I error?

Stefan V.
Featured 4 months ago

Here's my take on this.

#### Explanation:

The problem wants you to predict the precipitate produced by mixing two solutions, one that contains aluminium nitrate, #"Al"("NO"_3)_3#, a soluble ionic compound, and one that contains sodium hydroxide, $\text{NaOH}$, another soluble ionic compound.

The reaction takes place in aqueous solution, so right from the start you should be aware that you're dealing with ions.

Soluble ionic compounds dissociate completely in aqueous solution to form cations, which are positively charged ions, and anions, which are negatively charged ions.

The two solutions can thus be written as

${\text{Al"("NO"_ 3)_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"NO}}_{3 \left(a q\right)}^{-}$

It's absolutely crucial to make sure that you add the charges of the ions to the chemical equation.

${\text{NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

Now, you should be familiar with the solubility rules for aqueous solutions.

The aluminium cations, ${\text{Al}}^{3 +}$, will combine with the hydroxide anions, ${\text{OH}}^{-}$, to form the insoluble aluminium hydroxide, which precipitates out of solution.

Notice that the aluminium cations have a $3 +$ charge and the hydroxide anions have a $1 -$ charge. This means that you're going to need $3$ hydroxide anions in order to balance the positive charge of the cation.

Therefore, the chemical formula for aluminium hydroxide is #"Al"("OH")_3#, .

The other product of the reaction will be aqueous sodium nitrate. The sodium cations have a $1 +$ charge and the nitrate anions have a $1 -$ charge, so the chemical formula for sodium nitrate will be ${\text{NaNO}}_{3}$.

The balanced chemical equation will thus looks like this

${\text{Al"("NO"_ 3)_ (3(aq)) + 3"NaOH"_ ((aq)) -> "Al"("OH")_ (3(s)) darr + 3"NaNO}}_{3 \left(a q\right)}$

The complete ionic equation for this reaction looks like this

${\text{Al"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-) + 3"Na"_ ((aq))^(+) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr + 3"Na"_ ((aq))^(+) + 3"NO}}_{3 \left(a q\right)}^{-}$

To get the net ionic equation, eliminate spectator ions, which are those ions that are found on both sides of the equation

#"Al"_ ((aq))^(3+) + color(red)(cancel(color(black)(3"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(3"Na"_ ((aq))^(+)))) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr + color(red)(cancel(color(black)(3"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)(3"NO"_ (3(aq))^(-))))#

This will get you

#color(green)(|bar(ul(color(white)(a/a)color(black)("Al"_ ((aq))^(3+) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr)color(white)(a/a)|)))#

Now, I'm not really sure if the math formatting is to blame for some of the equations you wrote, but you're missing subscripts and charges in all of them.

Take the first equation, for example. #"Al"("NO"_3)# is actually #"Al"("NO"_3)_3#.

I'm assuming that $\text{Al"^3"OH}$ is actually ${\text{Al"^(3+)"OH}}^{-}$. If that is the case, then the correct version would be

${\text{Al"^(3+)("OH"^(-))_3 implies "Al"^(3+) + 3"OH}}^{-}$

There's no such thing as ${\text{Na}}_{3}$. Sodium cannot form molecules, it can exist either as a solid, $\text{Na}$, or a cation, ${\text{Na}}^{+}$.

In your case, the sodium cation was present in solution, so ${\text{Na}}_{3}$ is actually $3 {\text{Na}}^{+}$.

Once again, remember to always add charges when dealing with ions. An ion without an added charge is not actually an ion.

For example, there's no such thing as ${\text{NO}}_{3}$. Instead, add the $1 -$ charge that belongs to the ions to get ${\text{NO}}_{3}^{-}$ anion.

Also, keep in mind that a coefficient add to a soluble ionic compound gets distributed to all the ions that are produced by said compound in solution.

For example,

$\textcolor{red}{3} {\text{NaOH" = color(red)(3)("Na"^(+) + "OH"^(-)) = color(red)(3)"Na"^(+) + color(red)(3)"OH}}^{-}$

$\textcolor{b l u e}{2} {\text{Al"("NO"_3)_3 = color(blue)(2)("Al"^(3+) + 3"NO"_3^(-)) = color(blue)(2)"Al"^(3+) + 6"NO}}_{3}^{-}$

Notice that the charges must remain balanced at all times.

All in all, you should review ionic compounds before diving into complete and net ionic equations.

## Part C. 8. a). i). Outline briefly how you would identify the following dil. aq solutions by mixing them with one another. KI, Fe2(SO4)3, BaCl2, K4Fe(CN)6. ?

Michael
Featured 3 months ago

See below:

#### Explanation:

$\textsf{8 \left(a\right) . \left(i\right)}$

Potassium iodide solution could be identified by its reaction with iron(III) sulfate solution.

This is a redox reaction where $\textsf{{I}^{-}}$ is oxidised to $\textsf{{I}_{2}}$ by the $\textsf{F {e}^{3 +}}$ ions.

Omitting the spectator ions:

$\textsf{F {e}_{\left(a q\right)}^{3 +} + {I}_{\left(a q\right)}^{-} \rightarrow F {e}_{\left(a q\right)}^{2 +} + \frac{1}{2} {I}_{2 \left(a q\right)}}$

Addition of starch solution would give a blue/black precipitate to confirm the presence of iodine since iron(III) and iodine are both brown in colour.

$\textsf{- - - - - - - - - - - - - - - - - - - - -}$

Barium chloride solution would give a white precipitate with iron(III) sulfate solution of barium sulfate. Again, omitting the spectators:

$\textsf{B {a}_{\left(a q\right)}^{2 +} + S {O}_{4 \left(a q\right)}^{2 -} \rightarrow B a S {O}_{4 \left(s\right)}}$

Strictly speaking you should add an XS of dilute HCl to dissolve other possible precipitates.

$\textsf{- - - - - - - - - - - - - - - - - - - - -}$

If you mix iron(III) sulfate solution with the $\textsf{{K}_{4} F e {\left(C N\right)}_{6}}$ solution you get a dark blue solid commonly known as "Prussian Blue":

$\textsf{F {e}^{3 +} + {\left[F {e}^{I I} {\left(C N\right)}_{6}\right]}^{4 -} \rightarrow {\left[F {e}^{I I I} \left[F {e}^{I I} {\left(C N\right)}_{6}\right]\right]}^{-}}$

$\textsf{- - - - - - - - - - - - - - - - - - - - -}$

$\textsf{\left(a\right) \left(i i\right)}$

Aluminium will dissolve warm sodium hydroxide solution to give sodium aluminate and hydrogen:

$\textsf{2 A {l}_{\left(s\right)} + 2 N a O {H}_{\left(a q\right)} + 2 {H}_{2} {O}_{\left(l\right)} \rightarrow 2 N a A l {O}_{2 \left(a q\right)} + 3 {H}_{2 \left(g\right)}}$

This reflects the amphoteric nature of aluminium.

The same type of reaction occurs with zinc which is also amphoteric:

$\textsf{Z n \left(s\right) + {H}_{2} O \left(l\right) + 2 N a O H \left(a q\right) \rightarrow N {a}_{2} Z n {\left(O H\right)}_{2} \left(a q\right) + {H}_{2} \left(g\right)}$

If ammonium ions are warmed with hydroxide ions, ammonia gas is evolved which will turn red litmus blue:

$\textsf{N {H}_{4 \left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-} \rightarrow N {H}_{3 \left(g\right)} + {H}_{2} {O}_{\left(l\right)}}$

An aqueous solution of chlorine will react with sodium hydroxide solution, removing any green coloration due to elemental chlorine:

$\textsf{C {l}_{2 \left(a q\right)} + 2 N a O {H}_{\left(a q\right)} \rightarrow N a C {l}_{\left(a q\right)} + N a O C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}}$

This is an example of "disproportionation". Chlorine (0) is simultaneously oxidised to $\textsf{O C {l}^{-}}$ (+1) and to $\textsf{C {l}^{-}}$ (-1).

#sf((b)(i)#

$\textsf{A}$ is ammonium dichromate and decomposes on heating to give green chromium(III) oxide:

It is quite spectacular and is known as "The Volcano Experiment".

#sf((NH_4)_2Cr_2O_(7(s))rarrCr_2O_(3(s))+N_(2(g))+4H_2O_((g))#

So $\textsf{M}$ is chromium.

$\textsf{B}$ is chromium(III) oxide $\textsf{C {r}_{2} {O}_{3}}$.

$\textsf{C}$ is nitrogen gas $\textsf{{N}_{2}}$.

$\textsf{{N}_{2}}$ will burn in magnesium to give magnesium nitride:

#sf(3Mg_((s))+N_(2(g))rarrMg_3N_(2(s))#

So $\textsf{D}$ is magnesium nitride.

This reacts with water:

$\textsf{M {g}_{3} {N}_{2 \left(s\right)} + 6 {H}_{2} {O}_{\left(l\right)} \rightarrow M g {\left(O H\right)}_{2 \left(s\right)} + 2 N {H}_{3} \left(g\right)}$

So $\textsf{E}$ is ammonia, which turns red litmus paper blue.

When aqueous $\textsf{A}$ is warmed with carbonate ions an acid base reaction occurs giving $\textsf{E}$ which we know is ammonia:

$\textsf{N {H}_{4 \left(a q\right)}^{+} + C {O}_{3 \left(a q\right)}^{2 -} \rightarrow N {H}_{3 \left(g\right)} + H C {O}_{3 \left(a q\right)}^{-}}$

Green chromium(III) oxide is also amphoteric and will dissolve in aqueous alkali to give $\textsf{{\left[C r {\left(O H\right)}_{6}\right]}^{3 -}}$ which is soluble.

Under these conditions hydrogen peroxide is a powerful oxidising agent and is able to oxidise green $\textsf{C r \left(I I I\right)}$ to give yellow $\textsf{C r \left(V I\right)}$:

$\textsf{2 {\left[C r {\left(O H\right)}_{6}\right]}^{3 -} + 3 {H}_{2} {O}_{2} \rightarrow 2 C r {O}_{4}^{2 -} + 2 O {H}^{-} + 8 {H}_{2} O}$

## In the VESPR theory what is the difference between electron pair geometry and molecular geometry?

Ernest Z.
Featured 3 months ago

Here's my understanding of the difference.

#### Explanation:

To obtain any of these geometries, you must first draw the Lewis structure for the molecule.

Let's say you are dealing with water, $\text{H"_2"O}$.

You draw the Lewis structure:

Then you use VSEPR theory to predict the electron pair geometry.

The electron pair geometry is the arrangement that the electron pairs will assume.

You see two lone pairs and two bonding pairs (4 electron pairs).

VSEPR tells you that the electron pairs will arrange themselves at the corners of a tetrahedron, with angles of about 109.5°.

The actual $\text{H-O-H}$ bond angle is about 104.4°, because the lone pairs repel the bonding pairs.

Nevertheless, we say that the electron pair geometry is tetrahedral.

Our measuring instruments can't "see" the lone pairs. They can "see" only the bonds.

Hence, the molecular geometry is the arrangement of just the bonds (ignoring the lone pairs).

Since the H-O-H bonds are at an angle of 104.4°, we say that the molecular geometry of water is bent.

Here's a table of some common electron pair and molecular geometries.

The names of the electron pair geometries are in the left hand column, and all the names represent the molecular geometries.

Here's an interesting video on VSEPR theory.

## Why is the electron configuration of chromium #1s^2 2s^2 2p^6 3s^2 3p^6 3d^color(red)(5) 4s^color(red)(1)# instead of #1s^2 2s^2 2p^6 3s^2 3p^6 3d^color(red)(4) 4s^color(red)(2)#?

Truong-Son N.
Featured 2 months ago

It's a combination of factors:

• Less electrons paired in the same orbital
• More electrons with parallel spins in separate orbitals
• Pertinent valence orbitals not quite close enough in energy for electron pairing to be unfavorable

DISCLAIMER: Long answer, but it's a complicated issue, so... :)

A lot of people want to say that it's because a "half-filled subshell" increases stability, which is a reason, but not necessarily the only reason. However, for chromium, it's the significant reason.

It's also worth mentioning that these reasons are after-the-fact; chromium doesn't know the reasons we come up with; the reasons just have to be, well, reasonable.

The reasons I can think of are:

• Minimization of coulombic repulsion energy
• Maximization of exchange energy
• Lack of significant reduction of pairing energy overall in comparison to an atom with larger occupied orbitals

COULOMBIC REPULSION ENERGY

Coulombic repulsion energy is the increased energy due to opposite-spin electron pairing, in a context where there are only two electrons of nearly-degenerate energies.

So, for example...

$\underline{\uparrow \downarrow} \text{ " ul(color(white)(uarr darr)) " } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ is higher in energy than $\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(darr color(white)(uarr)) " } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$

To make it easier on us, we can crudely "measure" the repulsion energy with the symbol ${\Pi}_{c}$. We'd just say that for every electron pair in the same orbital, it adds one ${\Pi}_{c}$ unit of destabilization.

When you have something like this with parallel electron spins...

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

It becomes important to incorporate the exchange energy.

EXCHANGE ENERGY

Exchange energy is the reduction in energy due to the number of parallel-spin electron pairs in different orbitals.

It's a quantum mechanical argument where the parallel-spin electrons can exchange with each other due to their indistinguishability (you can't tell for sure if it's electron 1 that's in orbital 1, or electron 2 that's in orbital 1, etc), reducing the energy of the configuration.

For example...

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(uarr color(white)(darr)) " } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ is lower in energy than $\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(darr color(white)(uarr)) " } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$

To make it easier for us, a crude way to "measure" exchange energy is to say that it's equal to ${\Pi}_{e}$ for each pair that can exchange.

So for the first configuration above, it would be stabilized by ${\Pi}_{e}$ ($1 \leftrightarrow 2$), but the second configuration would have a $0 {\Pi}_{e}$ stabilization (opposite spins; can't exchange).

PAIRING ENERGY

Pairing energy is just the combination of both the repulsion and exchange energy. We call it $\Pi$, so:

$\Pi = {\Pi}_{c} + {\Pi}_{e}$

Basically, the pairing energy is:

• higher when repulsion energy is high (i.e. many electrons paired), meaning pairing is unfavorable
• lower when exchange energy is high (i.e. many electrons parallel and unpaired), meaning pairing is favorable

So, when it comes to putting it together for chromium... ($4 s$ and $3 d$ orbitals)

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$

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

compared to

$\underline{\uparrow \downarrow}$

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

is more stable.

For simplicity, if we assume the $4 s$ and $3 d$ electrons aren't close enough in energy to be considered "nearly-degenerate":

• The first configuration has $\setminus m a t h b f \left(\Pi = 10 {\Pi}_{e}\right)$.

(Exchanges: $1 \leftrightarrow 2 , 1 \leftrightarrow 3 , 1 \leftrightarrow 4 , 1 \leftrightarrow 5 , 2 \leftrightarrow 3 ,$
$2 \leftrightarrow 4 , 2 \leftrightarrow 5 , 3 \leftrightarrow 4 , 3 \leftrightarrow 5 , 4 \leftrightarrow 5$)

• The second configuration has $\setminus m a t h b f \left(\Pi = {\Pi}_{c} + 6 {\Pi}_{e}\right)$.

(Exchanges: $1 \leftrightarrow 2 , 1 \leftrightarrow 3 , 1 \leftrightarrow 4 , 2 \leftrightarrow 3 , 2 \leftrightarrow 4 , 3 \leftrightarrow 4$)

Technically, they are about $\text{3.29 eV}$ apart (Appendix B.9), which means it takes about $\text{3.29 V}$ to transfer a single electron from the $3 d$ up to the $4 s$.

We could also say that since the $3 d$ orbitals are lower in energy, transferring one electron to a lower-energy orbital is helpful anyways from a less quantitative perspective.

COMPLICATIONS DUE TO ORBITAL SIZE

Note that for example, $\text{W}$ has a configuration of $\left[X e\right] 5 {d}^{4} 6 {s}^{2}$, which seems to contradict the reasoning we had for $\text{Cr}$.

But, we should also recognize that $5 d$ orbitals are larger than $3 d$ orbitals, which means the electron density can be more spread out for $\text{W}$ than for $\text{Cr}$, thus reducing the pairing energy $\Pi$.

That is, ${\Pi}_{\text{W" < Pi_"Cr}}$.

Since a smaller pairing energy implies easier electron pairing, that is probably how it could be that $\text{W}$ has a $\left[X e\right] 5 {d}^{4} 6 {s}^{2}$ configuration instead of $\left[X e\right] 5 {s}^{5} 6 {s}^{1}$; its $5 d$ and $6 s$ orbitals are large enough to accommodate the extra electron density.

Indeed, the energy difference in $\text{W}$ for the $5 d$ and $6 s$ orbitals is only about $\text{0.24 eV}$ (Appendix B.9), which is quite easy to overcome simply by having larger orbitals.

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