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Answer:

You can do it like this:

Explanation:

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:

ths.talawanda.org

Sodium has 11 protons in its nucleus i.e #sf(Z=11)#. The outer electron in green is attracted to the nucleus by electrostatic forces which depend on the magnitude of the charges and the distance between them.

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 #sf(Z_(eff)#.

So we can write:

#sf(Z_(eff)=Z-s)#

Where #sf(s)# is the screening constant and is the total screening effect of all the other electrons.

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 #sf(n)# and #sf(l)# values. #sf(n)# tells you the number of the energy level and #sf(l)# denotes the sub - shell such as #sf(s, p, d)#. Because #sf(s)# and #sf(p)# electrons are close in energy they are grouped together:

#sf([1s][2s,2p][3s,3p][3d][4s,4p][4d][4f][5s,5p])# and so on.

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:

montessorimuddle.org

However if you look at the probability of finding an #sf(s)# electron at a given distance from the nucleus you get:

intro.chem.okstate.edu

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 #sf(s)# electrons in the same orbital can effectively "screen" each other. The same applies to the #sf(s)# and #sf(p)# electrons in the same groups.

2.

The number and type of electrons in the preceding groups.

The total shielding constant is, therefore, the overall shielding effect from #sf(1)# and #sf(2)#.

For #sf(1)# the shielding constant is the sum of :

0.35 for each other electron within the same group, except for #sf[1s]# when this is 0.3

For #sf(2)# if the group is of the #sf(["ns,np"])# type you add an amount of 0.85 for each electron with a principal quantum number of #sf((n-1))# and an amount of 1.00 for each electron with a quantum number of #sf((n-2))# or less.

For #sf([d])# and #sf([f])# electrons just add 1.00 for each electron which is closer to the nucleus.

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

#sf(""_9F^(-))# is #sf([1s^(2)][2s^(2)2p^(6)])# in which the electrons have been grouped according to the rules.

So #sf(Z=9)#

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.

#:.##sf(s=(7xx0.35)+(2xx0.85)=4.15)#

#:.##sf(Z_(eff)=9-4.15=4.85)#

#sf(""_11Na^(+))# is also #sf([1s^(2)][2s^(2)2p^(6)])#

But this time #sf(Z=11)#

So

#sf(Z_(eff)=11-4.15=6.85)#

This shows that a 2p electron in #sf(Na^+)# experiences a greater effective nuclear charge than a 2p electron in #sf(F^-)#.

Two separate trends are occurring.


EFFECTIVE NUCLEAR CHARGE (LEFT/RIGHT)

Across a period, the number of protons and electrons both increase by #1# for each element as we move from left to right. It 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.

http://classconnection.s3.amazonaws.com/

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.

https://upload.wikimedia.org/

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 #"Mg"# and #"F"#:

#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 #Z_(eff,F) > Z_(eff,Mg)#, so the atomic radius of #"F"# (#"42 pm"#) is smaller than that of #"Mg"# (#"145 pm"#). The large difference is also due to having one more quantum level.

As a general rule then (besides for the transition metals), as #Z_(eff)# increases, atomic radii decrease. Thus, since #Z_(eff)# increases from left to right, atomic radii decrease from left to right.

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 #n#.

https://kaiserscience.files.wordpress.com/

Each quantum level is farther away from the nucleus than the previous, and is higher in energy. For instance, that's why the #3s# orbital is larger than the #2s# orbital---because #n = 3# is higher in energy and farther away from the nucleus than #n = 2#.

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

Answer:

#"pH" = 5.7#

Explanation:

!! VERY LONG ANSWER !!

Your ultimate goal here is to figure out how much weak acid, #"HA"#, and how much conjugate base, #"A"^(-)#, you have in the target solution.

You already know that both buffers have

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

so your starting point here will be to figure out the concentration of #"A"^(-)# in buffer #"X"# and in buffer #"Y"#.

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 #"p"K_a# of the acid by using

#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 #"A"^(-)# in the two buffers

#color(white)(a)#

  • #ul("For buffer X")#

This buffer has #"pH" = 4.0#, which means that you get

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

Rearrange to solve for #["A"^(-)]_"X"/(["HA"])#

#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))#

#color(white)(a)#

  • #ul("For buffer Y")#

This buffer has #"pH" = 6.0#, which means that you get

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

Rearrange to solve for #["A"^(-)]_"Y"/(["HA"])#

#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 #"X"# and of buffer #"Y"#. Right from the start, you could say that because the volume doubles, the concentration of the weak acid remain unchanged. #color(purple)(("*"))#

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 #x# #"L"# to be the volume of the two buffers, and using #color(orange)((1))# and #color(orange)((2))#, you can say that buffer #"X"# will contain

#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 #"Y"# will contain

#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 remained unchanged, 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"# of the resulting solution

#"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.

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 #d# atomic orbitals, not less. From the point of view of crystal field theory, it means that the crystal field splitting energy increases.

This is only because the ligands are assumed to only donate electron density. That is, the #t_(2g)# orbitals can apparently never get any lower in energy than the free-ion #d# orbital energies (which isn't true in real life).

Inorganic Chemistry, Miessler et al., pg. 364


IMPORTANT NOTE:

Inorganic Chemistry, Miessler et al., pg. 371

In terms of #sigma# donors, they then donate more electron density into the metal #d_(x^2-y^2)# and #d_(z^2)# orbitals (#sigma#-bonding atomic orbitals), which increases the energy of the #e_g^"*"# antibonding orbitals, which increases the splitting energy.

This is accounted for in crystal field theory.

In terms of #pi# donors, they then donate electron density into the metal #d_(xy)#, #d_(xz)#, and #d_(yz)# orbitals, which increases their energies relative to no interaction.

With a higher metal oxidation state, more #pi# donation occurs into the metal from the ligand, so this technically means the #t_(2g)# orbital energies 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 #pi# acceptors, they normally accept electron density from the metal #d_(xy)#, #d_(xz)#, and #d_(yz)# orbitals. With a higher metal oxidation state, there is less metal-to-ligand backbonding, which technically means the stabilizing effect of backbonding is less, and thus the #t_(2g)# orbitals are 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.

Answer:

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

Explanation:

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.

www.rkm.com.au

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 #sf(color(red)("in"))# from the outside.

This means that this is the terminal where electrons #sf(color(red)("leave"))# the device.

The CATHODE of the device is the terminal where conventional current flows #sf(color(red)("out")# to the outside.

This means that this is the terminal where electrons #sf(color(red)("enter")# the device.

www.av8n.com

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:

mrtremblaycambridge.weebly.com

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

#sf(2Br^(-)rarrBr_2+2e)#

You can see from the diagram that these electrons flow into the external circuit and #sf(color(red)("leave")# the cell.

By our earlier definition this makes the +ve electrode the #sf(color(red)("anode"))#.

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

#sf(Pb^(2+)+2erarrPb)#

You can see from the diagram that electrons are #sf(color(red)("entering"))# the cell from the external circuit so from our earlier definition we can say that the -ve electrode is the #sf(color(red)("cathode"))#

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

A good example is The Daniel Cell:

orapreferata.wikispaces.com

Zinc atoms go into solution as zinc ions:

#sf(ZnrarrZn^(2+)+2e)#

So zinc forms the -ve side of the cell. You can see that these electrons flow into the external circuit and #sf(color(red)("leave"))# the cell.

By our earlier definition this means that the -ve zinc electrode is now the #sf(color(red)(anode))#.

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:

#sf(Cu^(2+)+2erarrCu)#

You can see from the diagram that electrons are #sf(color(red)("entering"))# the cell at the copper electrode so, from our earlier definition, this makes the +ve copper electrode the #sf(color(red)("cathode"))#.

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.

Answer:

Here's my explanation

Explanation:

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

upload.wikimedia.org

You can see the mathematical equation that describes this proportionality

#I = V/R#

#V = IR#

Where #I# is the current in the units of "amperes"
Where #V# is voltage
and #R# is the resistance of the material in units of "ohms"

The other equation

#E = Jsigma#

Where E is electric field at the given location
J is current density
σ (sigma) is a material-dependent parameter called the conductivity

#R =\rho\frac l A #

R is the electrical resistance of a uniform specimen of the material
#l# is the length of the piece of material
A is the cross-sectional area of the specimen

Simplifying

#\rho = \R\frac A l #

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

If the #A = 1m^2# and #l=1# the resistance of the conductor is equal to the the ρ

ρ depends on the conductor and is a constant

Here's a table of ρ of the conductor at [email protected]#

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

#sigma = 1/ρ#

Where #sigma# is conductivity
ρ is resistivity

There are many version of the Ohm's law

#P = (ΔV^2) / R#

#DeltaV# is the potential difference between the two points in the conductor

#P = I^2 • R#

Now lets come to the main problem.

#ΔV =( ΔPE )/ Q#

Where Q is charge
PE is energy

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

As

#C = V * Q#
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

www.docbrown.info

Suppose that #2C#/s is applied to a solution of #CuSO4# solution for 2 seconds.

The reaction

= #2H_2O + 2CuSO_4 rarr O_2 + 2H_2SO4 + 2Cu#

Net ionic reaction

#2H_2O + Cu^(2+) rarr 2H_2 + 2Cu#

Each mole of #Cu^(2+)# produces one mole of #Cu#

#2H_2O# is reduced to #H_2#

#H_2O + rarr H_2 + 2e^-#
#Cu^(2+) + 2e^-) rarr Cu #

Thus
#2H_2O rarr 2H_2 + 4e^-#
#2Cu^(2+) + 4e^-) rarr Cu #

The whole reaction

#2H_2O + 2Cu^(2+) + 4e^-) rarr 2H_2 + Cu + 4e^-)#
#2H_2O + 2Cu^(2+) rarr 2H_2 + Cu #

As 2C/s is applied for 2s

#(2C)/cancel(s) xx 2cancel(s)#

= #4C#

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

#( 4 cancel"Coulombs")xx ("1 mole of electrons") / (96500 cancel"Coulombs") = "0.00004145077 mole of electrons"#

1 mole of #Cu^(2+)# is reduced per 2 mole electrons

Thus moles of #Cu# formed

#"1 mol of Cu "/(2 molcancel( e^-)) xx "0.00004145077 mol of" cancel(e^-)#

#1/2 xx 0.00004145077 = "0.00002072538 mol of Cu"#

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

#(3C)/cancel(s) xx 2cancel(s)#

= #6C#

#( 6 cancel"Coulombs")xx ("1 mole of electrons") / (96500 cancel"Coulombs") = "0.00006217616 mole of electrons"#

#"1 mol of Cu "/(2 molcancel( e^-)) xx "0.00004145077 mol of" cancel(e^-)=" 0.00003108808 mol of Cu"#

#"0.00003108808 mol of Cu" > "0.00002072538 mol of Cu"#

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

Plotting a graph

enter image source here

If my handwriting is too bad

y = mol of Cu formed from reduction of #Cu^(2+)#
x = mol of electrons used in the process

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