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Featured 7 months ago

You can do it like this:

Your question does not use Slater's Rules. Since you want to understand these in your sub - note I will use these rules in my answer.

Consider an atom like sodium in the gaseous state:

Sodium has 11 protons in its nucleus i.e

The presence of the other 10 electrons in red has the effect of "shielding" or "screening" the outer electron from the attractive force of the nucleus.

This means that it experiences an attraction for the nucleus which is less than you would expect from the 11 protons.

This is referred to as the "effective nuclear charge" or

So we can write:

Where

In 1930 John Slater worked out a method of calculating this screening effect which are known as "Slater's Rules".

Electrons are grouped together in increasing order of

Each group has its own shielding constant. This depends on the sum of these contributions:

**1.**

The number of electrons in the **same** group.

You might think that the 2 electrons in helium are not able to shield each other since they are the same distance from the nucleus using the Bohr model:

However if you look at the probability of finding an

Point C is referred to as "The Bohr Radius" but you can see that there is some probability of the electron being at points closer to the nucleus such as A and B.

This means that the

**2.**

The number and type of electrons in the preceding groups.

The total shielding constant is, therefore, the overall shielding effect from

For

0.35 for each other electron within the same group, except for

For

For

Lets see how this applies to the examples in your comment:

So

Within the [2s and 2p] group a single 2p electron will be shielded by 7 other electrons each contributing 0.35 and below the group there are 2 x 1s electrons each contributing 0.85.

But this time

So

This shows that a 2p electron in

Featured 5 months ago

Two separate trends are occurring.

**EFFECTIVE NUCLEAR CHARGE (LEFT/RIGHT)**

Across a period, the number of protons and electrons both increase by *appears* as if no net attractive force is attained.

However, the added electron presents a bit of shielding; **each electron can effectively block the others from the positive nucleus**. That means an *effectively-less* negative charge interacts with the positively-charged nucleus, i.e. the net electron charge increases slower than the net nuclear charge does.

So, the nuclear attraction becomes larger from left to right, and the electrons get pulled in more easily.

*We say that the effective nuclear charge increases from left to right on the periodic table for the same row.*

We calculate effective nuclear charge as follows:

#bb(Z_(eff) = Z - S)# where

#Z_(eff)# is the effective nuclear charge,#Z# is the atomic number, and#S# is the shielding constant.

By definition:

#Z_(eff)# is the**average attraction of the electron(s) to the proton(s)**after accounting for the rapid motion of electrons and the*wall of core electrons*in between the nucleus and the valence electron in question.#Z# corresponds to the**number of protons**.#S# is a value based on**degree of "shielding"**of the valence electrons from the positively-charged nucleus. For general chemistry, we approximate#S# as the*number of core electrons*.

For example, comparing

#Z_(eff,Mg) ~~ 12 - (stackrel("1s electrons")overbrace(2) + stackrel("2s electrons")overbrace(2) + stackrel("2p electrons")overbrace(6)) = 2#

#Z_(eff,F) ~~ 9 - (stackrel("1s electrons")overbrace(2)) = 7#

and we have that

As a general rule then (besides for the transition metals), as

**QUANTUM LEVELS (UP/DOWN)**

This is a simpler explanation. Each row of the periodic table corresponds to a new quantum level, denoted by the principal quantum number

Each quantum level is farther away from the nucleus than the previous, and is higher in energy. For instance, that's why the

*Therefore, an element on the same column but a new row is automatically larger, because it has another quantum level.*

Featured 5 months ago

**!! VERY LONG ANSWER !!**

Your ultimate goal here is to figure out how much weak acid,

You already know that both buffers have

#["HA"] = "0.5 M"#

so your starting point here will be to figure out the concentration of

To do that, use the **Henderson - Hasselbalch equation**, which for a weak acid - conjugate base buffer takes the form

#color(blue)(ul(color(black)("pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))))#

Calculate the

#color(blue)(ul(color(black)("p"K_a = - log(K_a))))#

In your case, you will have

#"p"K_a = - log(1.0 * 10^(-5)) = 5.0#

Now, calculate the concentration of

#ul("For buffer X")#

This buffer has

#4.0 = 5.0 + log((["A"^(-)]_"X")/(["HA"]))#

Rearrange to solve for

#log(["A"^(-)]_"X"/(["HA"])) = -1.0#

#10^log(["A"^(-)]_"X"/(["HA"])) = 10^(-1.0)#

This will get you

#["A"^(-)]_"X"/(["HA"]) = 0.10#

which results in

#["A"^(-)]_"X" = 0.10 * ["HA"]#

#["A"^(-)]_"X" = 0.10 * "0.5 M" = "0.050 M" " "color(orange)((1))#

#ul("For buffer Y")#

This buffer has

#6.0 = 5.0 + log((["A"^(-)]_"Y")/(["HA"]))#

Rearrange to solve for

#log(["A"^(-)]_"Y"/(["HA"])) = 1.0#

This will get you

#["A"^(-)]_"Y"/(["HA"]) = 10#

which results in

#["A"^(-)]_"Y" = 10 * ["HA"]#

#["A"^(-)]_"Y" = 10 * "0.5 M" = "5.0 M" " "color(orange)((2))#

Now, you are told that you must mix **equal volumes** of buffer **doubles**, the concentration of the weak acid remain **unchanged**.

That is the case because you're essentially doubling the number of moles of weak acid **and** the volume of the solution.

If you take **volume** of the two buffers, and using

#x color(red)(cancel(color(black)("L buffer"))) * "0.5 moles HA"/(1color(red)(cancel(color(black)("L buffer")))) = (0.5 * x)" moles HA"#

#x color(red)(cancel(color(black)("L buffer"))) * "0.050 moles A"^(-)/(1color(red)(cancel(color(black)("L buffer")))) = (0.050 * x)" moles A"^(-)#

Similarly, buffer

#x color(red)(cancel(color(black)("L buffer"))) * "0.5 moles HA"/(1color(red)(cancel(color(black)("L buffer")))) = (0.5 * x)" moles HA"#

#x color(red)(cancel(color(black)("L buffer"))) * "5.0 moles A"^(-)/(1color(red)(cancel(color(black)("L buffer")))) = (5.0 * x)" moles A"^(-1)#

The **resulting solution**, which has a volume of

#xcolor(white)(.)"L" + xcolor(white)(.)"L" = 2xcolor(white)(.)"L"#

will contain

#{((0.5x)" moles" + (0.5x)" moles" = x" moles HA"), ((0.050x)" moles" + (5.0x)" moles" = 5.05x" moles A"^(-)) :}#

The **concentrations** of the two species in the resulting solution will be

#["HA"] = (color(red)(cancel(color(black)(x)))"moles")/(2color(red)(cancel(color(black)(x)))"L") = "0.5 M" -># the concentration remainedunchanged, as predicted in#color(purple)(("*"))#

#["A"^(-)] = (5.05color(red)(cancel(color(black)(x)))"moles")/(2color(red)(cancel(color(black)(x)))"L") = "2.525 M"#

Finally, use the Henderson - Hasselbalch equation to find the

#"pH" = 5.0 + log( (2.525 color(red)(cancel(color(black)("M"))))/(0.5color(red)(cancel(color(black)("M"))))) = color(darkgreen)(ul(color(black)(5.7)))#

The answer is rounded to *one decimal place*, since that is how many sig figs you have for the concentration of the weak acid.

Featured 4 weeks ago

A **higher** oxidation state means the transition metal is **more** electropositive, so it draws **more** electron density towards itself (relative to a metal with a lower oxidation state).

That means there are **more** repulsions within the **increases**.

*This is only because the ligands are assumed to only donate electron density. That is, the* *orbitals can apparently never get any lower in energy than the free-ion* *orbital energies* (which isn't true in real life).

**IMPORTANT NOTE:**

In terms of **more** electron density into the metal **increases** the energy of the **increases the splitting energy.**

*This is accounted for in crystal field theory.*

In terms of

With a higher metal oxidation state, more *increase*, and **thus the splitting energy decreases.**

*Crystal field theory does not account for this. So, we ignore this in crystal field theory.*

In terms of **less** metal-to-ligand backbonding, which technically means the stabilizing effect of backbonding is **less**, and thus the **less lowered** in energy and **the splitting energy decreases.**

*This is not accounted for in crystal field theory either, and we ignore this until we introduce ligand field theory.*

Featured 4 weeks ago

This is because the terms **cathode** and **anode** describe the **function** of the electrode and not its **charge.**

Before we go into the internal workings of the two types of cell it helps if we adopt a "black box" approach to these devices.

Conventional electric current is said to flow from the positive terminal to the negative one.

This convention was adopted in the 19th Century at the time of Michael Faraday. When the electron was later discovered and shown to have a negative charge you can see from the diagram that the electron flow is from negative to positive.

This, however, is the convention we are left with.

The ANODE of the device is the terminal where conventional current flows

This means that this is the terminal where electrons

The CATHODE of the device is the terminal where conventional current flows

This means that this is the terminal where electrons

Now lets see how this applies to an electrolytic cell and a galvanic cell.

In an electrolytic cell electrical energy is used to cause chemical change.

A good example is the electrolysis of molten lead(II) bromride:

Bromide ions are attracted to the +ve electrode and discharged:

You can see from the diagram that these electrons flow into the external circuit and

By our earlier definition this makes the +ve electrode the

Lead(II) ions are attracted to the +ve electrode and discharged:

You can see from the diagram that electrons are

In a galvanic cell chemical energy is converted into electrical energy.

A good example is The Daniel Cell:

Zinc atoms go into solution as zinc ions:

So zinc forms the -ve side of the cell. You can see that these electrons flow into the external circuit and

By our earlier definition this means that the -ve zinc electrode is now the

These electrons flow round the external circuit and arrive at the copper electrode.

Here copper(II) ions pick up the electrons and a deposit of copper forms:

You can see from the diagram that electrons are

The salt bridge is there to provide electrical neutrality between the two 1/2 cells.

You should be aware that oxidation is the loss of electrons and reduction is the gain of electrons.

A good way of remembering cathode and anode in a cell is to realise that:

"Oxidation occurs at the anode"

and

"Reduction occurs at the cathode"

Care must be taken when physically labelling the electrodes on any electrical device.

In a rechargeable cell such as a lead / acid accumulator the cathode and anode will change identity depending on whether it is discharging or being charged.

In summary, the terms cathode and anode depend on the function of the electrode, not its structure.

Featured 3 weeks ago

Here's my explanation

According to Ohm's law the current passed through a material is directly proportional to the voltage.

You can see the mathematical equation that describes this proportionality

Where

Where

and

The other equation

Where E is electric field at the given location

J is current density

σ (sigma) is a material-dependent parameter called the conductivity

R is the electrical resistance of a uniform specimen of the material

A is the cross-sectional area of the specimen

Simplifying

The resistance of a given material increases with length, but decreases with increasing cross-sectional area

If the

ρ depends on the conductor and is a constant

Here's a table of ρ of the conductor at

http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/rstiv.html

Note that ρ is different for conductors at different temperature

ρ increases as the temperature increases

ρ decreases as the temperature decreases

Where

ρ is resistivity

There are many version of the Ohm's law

Now lets come to the main problem.

Where Q is charge

PE is energy

By this you can understand that Voltage is proportional to charge which is proportional to coulombs

As

There is a relationship between V and C

As the V increases and the Q is thought to be constant the C increases. Means more voltage more coulombs

Now let me show you an example

Suppose that

The reaction

=

Net ionic reaction

Each mole of

Thus

The whole reaction

As 2C/s is applied for 2s

=

Use the Faraday's constant to calculate the moles of electrons lost and gained in total.

1 mole of

Thus moles of

Now see what happens if we apply more charge

Consider the charge as 3C/s for 2s

Do the same calculations

As 3C/s is applied for 2s

=

Thus more the current,voltage and coulomb the more the mol of copper formed thus more the mass of copper formmed

Plotting a graph

If my handwriting is too bad

y = mol of Cu formed from reduction of

x = mol of electrons used in the process

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