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1,2-DIIMINE LIGANDS (CHELATING AGENTS)

An example of a 1,2-diimine is the dimethylglyoximato ligand (which somewhat resembles a conjugated diene for Diels-Alder reactions).

Name:

bis-(DMG)nickel(II)

As a general rule, "bis" is used to indicate two of the same, complicated ligand.

Chemical formula:

#"Ni"("DMG")_2#

If we call one of the imine carbons "carbon-1", then the other imine carbon is one atom away (meaning, on "carbon-2"). Hence, we can call this a 1,2-diimine.

Each dimethylglyoximato ligand (abbreviated DMG in compound names) has a net charge of #\mathbf(-1)#, and each nitrogen acts as a binding site due to their lone pairs of electrons.

If you recall the chelate effect, it essentially states that the binding of more than one "tooth" is entropically favored, relative to binding with one "tooth".

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This two-"tooth" binding mode requires that the ligand's binding sites be cis (same-side). Therefore, DMG is a bidentate ligand, binding preferentially cis.

1,3-DIIMINE LIGANDS (CHELATING AGENTS)

An example of this that I can find in my book is the N,N'-diphenyl-2,4-pentanediiminato ligand.

Name:

dichloro(nacnac)nickel(II)

As a general rule, complicated ligands tend to be in parentheses.

Chemical formula:

#["Ni"("nacnac")"Cl"_2]^(-)#

Fortunately, this ligand has a simple 'nickname': nacnac.

Like DMG, nacnac is a bidentate ligand, and its two binding sites are the two nitrogens, each using their lone pair of electrons. However, its imine carbons are 1,3 to each other instead.

This ligand also has a net charge of #\mathbf(-1)#.

It substantially favors binding cis to the metal, again due to the chelate effect favoring the binding of both "teeth" rather than only one of them.

Answer:

#"Pb"#, #"Zn"#, and #"Cu"#.

Explanation:

The idea here is that you need to use the standard reduction potentials, #E^@#, given to you to determine the identity of the three metals.

You know that you have

#"Zn"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Zn"_ ((s))" "E^@ = -"0.76 V"#

#"Pb"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Pb"_ ((s))" "E^@ = -"0.13 V"#

#"Cu"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Cu"_ ((s))" "E^@ = +"0.34 V"#

Now, the standard reduction potentials measure the tendency of a chemical species to release electrons and form cations when compared to that of hydrogen.

A negative #E^@# means that the chemical species loses electrons more readily than hydrogen, and that its reduction equilibrium lies to the left.

A positive #E^@# means that the chemical species loses electrons less readily than hydrogen, and that its reduction equilibrium lies to the right.

Now, chemical species that lose electrons more readily are stronger reducing agents than chemical species that tend to lose electrons less readily.

When you compare the #E^@# values for two reduction equilibria, you can say that

  • the equilibrium with the less positive / more negative #E^@# value will lie further to the left
  • the equilibrium with the less negative / more positive #E^@# value will lie further to the right

Let's take the first two reduction equilibria. You have

#E^@("Zn"^(2+), "Zn") = -"0.76 V"#

#E^@("Pb"^(2+), "Pb") = -"0.13 V"#

Here #E^@ = -"0.76 V"# is more negative than #E^@ = -"0.13 V"#, which means that the first equilibrium lies further to the left than the second one.

You can thus say that zinc metal, #"Zn"#, will reduce #"Pb"^(2+)# because zinc loses electrons more readily than lead metal does.

Consequently, zinc will also reduce #"Cu"^(2+)#, since #E^@ = -"0.76 V"# is more negative than #E^@ = +"0.34 V"#.

In other words, when zinc metal is placed in a solution that contains #"Pb"^(2+)# and #"Cu"^(2+)# cations, the cations will be reduced to lead metal and copper metal, respectively, and zinc will be oxidized to #"Zn"^(2+)#.

This implies that zinc is metal #"B"#.

Finally, notice that metal #"C"# cannot reduce #"A"^(2+)#, since it will not be oxidized when placed in a solution that contains #"A"^(2+)#. This means that the #E^@# of metal #"C"# is more positive than #E^@# for metal #"A"#. As a result, metal #"C"# will be copper and metal #"A"# will be lead.

#"Metal A "-> " Lead"#

#"Metal B " -> " Zinc"#

#"Metal C " -> " Copper"#

So, to answer such problems quickly, list the reduction half-reactions in order of increasing #E^@# values

#"Zn"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons color(blue)("Zn"_ ((s)))" "E^@ = -"0.76 V"#

#color(red)("Pb"_ ((aq))^(2+)) + 2"e"^(-) rightleftharpoons "Pb"_ ((s))" "E^@ = -"0.13 V"#

#color(red)("Cu"_ ((aq))^(2+)) + 2"e"^(-) rightleftharpoons "Cu"_ ((s))" "E^@ = +"0.34 V"#

and keep in mind that the chemical species listed in the top right can reduce the chemical species listed in the bottom left.

For example, zinc metal is #color(blue)("top right")#, so it can reduce #"Pb"^(2+)# and #"Cu"^(2+)# because they are located in the #color(red)("bottom left")#.

Similarly, you can say that the chemical species located in the bottom left can oxidize the chemical species located to the top right.

Here #"Cu"^(2+)# and #"Pb"^(2+)# are #color(red)("bottom left")#, so they can oxidize zinc metal, which is located in the #color(blue)("top right")#.

#{(color(red)("Bottom left")color(white)(a)"oxidizes" color(white)(a)color(blue)("top right ")), (color(blue)("Top right")color(white)(a)"reduces" color(white)(a)color(red)("bottom left")) :}#

Answer:

See below:

Explanation:

List the standard electrode potentials in increasing order:

#" "# #sf(E^@" "(V))#

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#

#sf(Zn_((aq))^(2+)+2erightleftharpoonsZn_((s))" "-0.76)#

#sf(Cu_((aq))^(2+)+2erightleftharpoonsCu_((s))" "+0.34)#

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)#

The 1/2 cell with the most +ve #sf(E^@)# value will take in the electrons.

So we can see that the copper 1/2 cell will be driven left to right and the zinc 1/2 cell right to left as indicated by the arrows.

So the two 1/2 equations are:

#sf(Zn_((aq))rarrZn_((aq))^(2+)+2e)#

#sf(Cu_((aq))^(2+)+2erarrCu_((s)))#

Adding these gives the cell reaction when it is working:

#sf(Zn_((s))+Cu_((aq))^(2+)rarrZn_((aq))^(2+)+Cu_((s)))#

When the zinc atoms lose electrons the zinc ions go into solution and the 2 electrons flow away from this electrode into the external circuit.

When they arrive at the copper electrode, they are picked up by the copper(II) ions to become copper atoms.

The two 1/2 cells are connected by a salt bridge or a membrane. The salt bridge consists of some filter paper soaked in a suitable electrolyte such as saturated potassium nitrate solution.

The function of the salt bridge is to maintain electrical neutrality in each half cell.

As zinc ions go into solution #sf(NO_3^-)# ions flood in to the half cell.

As copper(II) ions leave the solution in the other 1/2 cell, #sf(K^+)# ions flood in.

To find the emf of the cell, subtract the least +ve #sf(E^@)# value from the most +ve:

#sf(E_(cell)^(@)=+0.34-(-0.76)=+1.1V)#

This arrangement is known as "The Daniel Cell".

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Answer:

#K_ p = 25.3#

Explanation:

The equilibrium reaction given to you looks like this

#"Xe"_ ((g)) + color(red)(2)"F"_ (2(g)) rightleftharpoons "XeF"_ (4(g))#

You know that the mixture initially contains #"2.24 atm"# of xenon, #"Xe"#, and #"4.27 atm"# of fluorine gas, #"F"_2#.

It's worth mentioning that the problem provides you with pressures instead of moles because when the volume of the container remains unchanged, and the temperature of the reaction is kept constant, the pressure of a gas is directly proportional to the number of moles of gas present.

Now, notice that the pressure of xenon is decreasing to an equilibrium value of #"0.34 atm"#. This is a significant decrease, which can only mean that the equilibrium constant, #K_p#, is greater than #1#.

In other words, the forward reaction is favored at the temperature at which the reaction takes place.

Consequently, you can expect the equilibrium pressure of fluorine gas to be significantly lower than its initial value.

So, use an ICE table to find the equilibrium pressures of fluorine gas and xenon tetrafluoride

#" ""Xe"_ ((g)) " "+" " color(red)(2)"F"_ (2(g)) " "rightleftharpoons" " "XeF"_ (4(g))#

#color(purple)("I")color(white)(aaaaacolor(black)(2.24)aaaaaaaacolor(black)(4.27)aaaaaaaaaaacolor(black)(0)#
#color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaacolor(black)((-color(red)(2)x))aaaaaaaaacolor(black)((+x))#
#color(purple)("E")color(white)(aaacolor(black)(2.24-x)aaaacolor(black)(4.27-color(red)(2)x)aaaaaaaaacolor(black)(x)#

Now, you know that the equilibrium concentration of xenon is #"0.34 atm"#. This means that you have

#2.24 - x = 0.34 implies x = 1.90#

The equilibrium pressures of fluorine gas and xenon tetrafluoride will thus be

#("F"_2) = 4.27 - color(red)(2) * 1.90 = "0.47 atm"#

#("XeF"_4) = "1.90 atm"#

By definition, #K_p# will be equal to

#K_ p = (("XeF"_4))/(("Xe") * ("F"_2)^color(red)(2))#

Plug in your values to find

#K_p = (1.90 color(red)(cancel(color(black)("atm"))))/(0.34color(red)(cancel(color(black)("atm"))) * ("0.47 atm")^color(red)(2)) = "25.3 atm"^(-2)#

Equilibrium constant are usually given without added units, which means that your answer will be

#K_p = color(green)(|bar(ul(color(white)(a/a)color(black)(25.3)color(white)(a/a)|))) -># rounded to three sig figs

As predicted, the equilibrium constant turned out to be grater than #1#.

Answer:

#"42.9 mL"#

Explanation:

!! LONG ANSWER !!

The idea here is that adding sodium hydroxide, #"NaOH"#, to your hydrochloric acid solution will neutralize some, if not all, depending on how much you've added, of the acid.

You know that you start with

#"100.0 mL "-> 6.00 * 10^(-2)"M HCl"#

#"100.0 mL " -> 5.00 * 10^(-2)"M NaOH"#

Now, you know that after you realize your error, you're left with #"81.0 mL"# of hydrochloric acid and #"89.0 mL"# of sodium hydroxide. This means that you've added to the resulting solution

#"100.0 mL " - " 81.0 mL" = "19.0 mL HCl"#

#"100.0 mL " - " 89.0 mL" = "11.0 mL NaOH"#

When you mix hydrochloric acid, a strong acid, and sodium hydroxide, a strong base, a neutralization reaction takes place

#"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#

Because hydrochloric acid and sodium hydroxide produce hydrogen cations, #"H"^(+)#, and hydroxide anions, #"OH"^(-)#, respectively, in a #1:1# mole ratio, you will have

#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#

Notice that this reaction consumes hydrogen cations and hydroxide anions in a #1:1# mole ratio, which means that for every mole of hydrochloric acid present it takes one mole of sodium hydroxide to neutralize it.

Use the molarities and volumes of the solutions you've mixed to calculate how many moles of each were added

#19.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace((6.00 * 10^(-2)"moles HCl")/(1color(red)(cancel(color(black)("L")))))^(color(blue)(=6.00 * 10^(-2)"M")) = "0.00114 moles H"^(+)#

#11.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace((5.00 * 10^(-2)"moles NaOH")/(1color(red)(cancel(color(black)("L")))))^(color(blue)(=6.00 * 10^(-2)"M")) = "0.000550 moles OH"^(-)#

So, you know that you've accidentally added #0.00114# moles of hydrogen cations and #0.000550# moles of hydroxide anions to your solution.

Since you have fewer moles of hydroxide anions present, you can say that the hydroxide anions will be completely consumed by the neutralization reaction, i.e. they will act as a limiting raegent.

Your resulting solution will thus contain

#0.000550 - 0.000550 = "0 moles OH"^(-) -># completely consumed

#0.00114 - 0.000550 = "0.000590 moles H"^(+)#

Now, the volume of this solution will be equal to the volume of hydrochloric acid and the volume of sodium hydroxide solutions you've mixed

#V_"sol" = "19.0 mL" + "11.0 mL" = "30.0 mL"#

This means that the concentration of hydrogen cations in the resulting solution will be

#["H"^(+)] = "0.000590 moles"/(30.0 * 10^(-3)"L") = "0.01967 M"#

The pH of the solution is given by

#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"^(+)])color(white)(a/a)|)))#

In your case, you have

#"pH" = - log(0.01967) = 1.71#

So, you know that your target solution must have a volume of #"1.00 L"# and a pH of #2.50#. Use the above equation to find the concentration of hydrogen cations needed to have this target solution

#"pH" = - log(["H"^(+)]) implies ["H"^(+)] = 10^(-"pH")#

#["H"^(+)] = 10^(-2.50) = "0.003162 M"#

For a volume equal to #"1.00 L"#, this gives you #0.003162# moles of hydrogen cations -- remember that when dealing with a liter of solution, molarity and number of moles of solute are interchangeable.

Your solution contains #0.000590# moles of hydrogen cations and needs #0.003162# moles, which means that you must add

#n_("H"^(+)"needed") = 0.003162 - 0.000590 = "0.002572 moles H"^(+)#

Use the molarity of the stock hydrochloric acid solution to see what volume would contain this many moles of acid

#0.002572 color(red)(cancel(color(black)("moles H"^(+)))) * "1.0 L"/(6.00 * 10^(-2)color(red)(cancel(color(black)("moles H"^(+))))) = "0.0429 L"#

This is equivalent to #"42.9 mL"#, which means that in order to prepare your target solution, you must add another #"42.9 mL"# of stock hydrochloric acid solution to the existing #"30.0 mL"# solution.

This will give you a volume of #"72.9 mL"# of solution. To get your target solution, you must add enough distilled water to get the total volume to #"1000 mL"#.

#color(green)(|bar(ul(color(white)(a/a)color(black)("volume of HCl needed " = " 42.9 mL")color(white)(a/a)|)))#

The answer is rounded to three sig figs.

Answer:

#sf([Co(NH_3)_6]^(3+))# is an inner orbital complex because it adopts a low spin electronic configuration.

I am assuming you mean #sf(NH_3)# as the ligand in question in the 1st example.

Explanation:

One characteristic of the transition elements is their ability to form complex ions. These consist of a central metal ion surrounded by electron - donating species called ligands.

Cobalt(III) ions have the electron configuration:

Fig 1 (a)

MF Docs

The empty 4s, 4p and 4d orbitals are quite close in energy to the 3d orbitals.

This means that ligands with available lone pairs such as ammonia molecules are able to donate electrons into these empty orbitals and form co - ordinate bonds.

Quite a common stable arrangement is with 6 ligands.

An example is hydrated #sf(Fe^(3+)#:

www.chemguide.co.uk

The d orbitals look like this:

Fig 1 (b)

upload.wikimedia.org

Because of the octahedral symmetry of the complex the #sf(d_(z^2))# and #sf(d_(x^2-y^2))# orbitals point directly at negative lone pairs on the ligand and are destabilised.

The other 3d orbitals have lobes which project between the ligands and are relatively more stable. This is shown here:

MF Docs

These orbitals are given the symmetry terms #sf(e_g)# and #sf(t_(2g))# and the energy gap is denoted by the symbol #sf(Delta)#.

The important point to note here is that the value of #sf(Delta)# varies with the nature of the ligand.

The spectrochemical series lists the ligands in order of #sf(Delta)#:

Some are listed here:

I- < Br- < S2- < SCN- < Cl- < NO3- < F- < OH- < C2O42- < H2O < NCS- < CH3CN < NH3 < en < bipy < phen < NO2- < PPh3 < CN- < CO

This tells us that #sf(NH_3)# has a larger #sf(Delta)# value than #sf(F^(-))# which has implications for the way the electrons occupy the 3d orbitals.

For the ammonia complex the value of #sf(Delta)# is relatively large and it is energetically more favourable for the 6 electrons to occupy the #sf(t_(2g))# sub - levels with spins paired:

enter image source here

This is termed a "low - spin" complex.

You can see from Fig 3 that the orbitals shown in red are able to each accept a pair of electrons from each #sf(NH_3)# ligand and form #sf([Co(NH_3)_6]^(3+))#.

The 3d, 4s and 4p orbitals effectively reorganise themselves to form 6 equivalent orbitals which are described, therefore, as #sf(d^2sp^3)# hybrid orbitals

Because inner 3d orbitals are used this can be referred to as an inner orbital complex .

In the case of the #sf([CoF_6]^(3-))# complex the relatively small value of #sf(Delta)# means that coulombic repulsions between the electrons means it is more energetically favourable for them to occupy the orbitals singly in a "high spin" arrangement:

MF Docs

Because the #sf(e_g)# orbitals are now partially occupied the ligands can donate their electrons into the empty 4s, 4p and 4d orbitals which can reorganise themselves into #sf(sp^3d^2)# hybrid orbitals.

Because the outer 4d orbitals are now being used this can be described as an outer orbital complex.

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