Download Chem 323 I) [NiCl4]2

Answers for First Mid Term Exam, Second semester, 1431/1432H (2010/2011G)
Chem 323
I)
[NiCl4]2- is a weak field d8 coordination number-4 complex. Therefore:
a) The complex should be of a tetrahedral symmetry Td.
b) The hyberdization is sp3 namely (s px py pz).
c) The d-orbital configurationis: ( e 4 t24 ) and it is therefore, this complex is
paramagnetic with two unpaired electrons.
d) The value of the crystal field stabilization energy = –3.56 Dq.
[Ni(CN)4]2- is a strong field d8 coordination number-4 complex. Therefore
a) The complex should be of a square planar symmetry D4h.
b) The hyberdization is dsp2 namely (dx2–y2 s px py).
c) The d-orbital configurationis: ( e g4 a12g b22g b10g ) and it is therefore, this
complex is diamagnetic having zero unpaired electrons.
d) The value of the crystal field stabilization energy = –24.56 Dq.
II)
The three complexes :
[W(CN)6]3-
,
[Ir(CN)6]3-
and
[Pt(CN)6]3-
They are all strong field octahedral complexes.
The metal ions in each of them is of the 5-d transition metal series and the
oxidation number is +3, i.e. (W3+, d3), (Ir3+, d6) and (Pt3+, d7).
The ligands are the same, i.e. CN is all three complexes.
Therefore, with all the above mentioned factors are similar for all three
complexes, the factors which will differ and will cause the variation in M-L
bond length are:
a) The ionic radius.
b) The value of the crystal field stabilization energy.
c) The repulsion between the d electrons and the ligands.
For [W(CN)6]3-, the atomic number is 74, its W3+ should have the largest
ionic radius compared to Ir3+ (atomic number 77) and Pt3+ (atomic number
78). For this tungsten complex, the d-orbital configuration is t23g e g0 and the
C.F.S.E. is –12 Dq which is the least compared to the other two.
For [Ir(CN)6]3-: t26g e 0g and –24 Dq.
and for [Pt(CN)6]3-: t26g e1g and –18 Dq.
In the tungesten complex, all of the d3 three electrons will occupy the
three lower t2g orbitals away from M-L direction and therefore, will not
have any significant repulsion with the ligands and thereby will not cause
bond lengthening.
For the irridium complex, the d6 six electrons will also occupy the three
lower t2g orbitals away from M-L direction and therefore, will not have any
significant repulsion with the ligands and thereby will not cause bond
lengthening. Add to that, the irridium complex will enjoy the maximum
C.F.S.E. (–24 Dq) and for smaller Ir3+ ionic radium as compared to W3+.
For the [Ir(CN)6]3-, we should have the shortest M-L bond length, while
for [W(CN)6]3-, we should have the longest M-L bond length.
For the [Pt(CN)6]3-, one of d7 electrons will occupy the higher eg level
( t26g e1g ) where it will be in an orbital directed to the ligands and thereby
causing the M-L bond length to be longer than that in the platinum complex.
Add to that, the value of C.F.S.E. for the platinum complex is –18 Dq.
These two factors, i.e. the eg electron-ligand repulsion and the marked
difference in the value of C.F.S.E. will make the bond length is longer for
the [Pt(CN)6]3- as compared to the [Ir(CN)6]3-, but still shorter than the M-L
bond length for the [W(CN)6]3- which will be the longest M-L bond length
between the three complexes.
The ionic radius which is bit smaller for Pt3+ as compared to that of Ir3+
could not help to overcome the other two factors for the comparison
between the platinum and irridium complexes.
[W(CN)6]3- > [Pt(CN)6]3- > [Ir(CN)6]3-
●
●
Weak field
●
Strong field
[W(CN)6]3- d3
●
d7 [Pt(CN)6]3●
[Ir(CN)6]3- d6
Answers for Second Mid Term Exam, First semester, 1431/1432H (2010/2011G)
Chem 323
I: According to IUPAC rules, write the Werner formula for each of the following
complexes and state the number of ionizable ions when the complex is dissolved.
1) Tetraamminecobalt(III)-µ-amido-µ-nitrotetraamminecobalt(III) Phosphate
NH2
(NH3)4Co Co(NH3)4 (PO4)4
NO2
3
7
ions
5
ions
2
ions
2) Hexaamminecobalt(III) Hexachloroplatinate(IV)
[Co(NH3)6]2[PtCl6]3
3) Potassium Ethylenediaminetetraacetatocobaltate(III)
K[Co(OOCCH2)2NCH2CH2N(CH2COO)2]
II: According to IUPAC, Name each of the following three complexes:
N
1)
C
S
(en)2Co Cr(acac)2 (NO3)2
N
C
S
Bis(ethylenediamine)cobalt(III)-µ-thiocyanato-µ-thiocyanatobis(acetylacetonato)chromium(III) nitrate
2)
CH3
C
O
O
(H2O)3Cu Cu(H2O)3
O
O
C
CH3
Triaquocopper(II)-µ-diacetatotriaquocopper(II)
3) [(SCN)3(H2O)2Cr
H
O
Co(NH3)5]SO4
Diaquotrithiocyanatochromium(III)-µ-hydroxopentaammine cobalt(III) sulfate
III: Write down the d-orbitals configuration and the related number of electrons for
each of the following strong field complexes. Calculate in Dq the ligand field
stabelization energy of each complex.
1) [Pt(CN)6]2-
t 62g e0g
–24
2) [Ni(CN)5]3-
C4v
Dq
S.F.
e4 b 22 a12 b10
–18.28 Dq
3) [Fe(CN)5]2-
D3h
W.F.
e 22 e12 a11
0
Dq
Answers for Second Semister Second Mid Term Exam 1431/1432H (2010/2011G)
Chem 323
I) According to IUPAC, name the following complexes:
H
O
1) [(SCN)3(H2O)2Cr
Co(NH3)5]SO4
Aquotrithiocyanatochromium(III)-µ-hydroxopentaamminecobalt(III) sulfate
2) Al2[Cr(NH2CH2COO)(C2O4)2]3
Aluminum glycinatodioxalatochromate(III)
II) Give the Werner's formula for each of the following complexes and state the
number of ionizable ions in each.
1) Diaquobis(ethylenediamine)cobalt(III) Amminepentanitritocobalt(III)
[Co(NH2CH2CH2NH2)(H2O)2]2[Co(NH3)(ONO)5]3
Number of ionizable ions = 5
2) Diammine(ethylenediamine)chromium(III)µ-bis(dioxygen)
tetramminecobalt(III) bromide
O2
(NH3)2(NH2CH2CH2NH2)Cr Co(NH3)4 Br6
O2
Number of ionizable ions = 7
3) Calcium Dithiosulfatoargentate(I)
Ca3[Ag(S2O3)2]2
Number of ionizable ions = 5
III) According to Slater's rule, calculate the value of Z* of each ion from the
following compounds.
1) [Ni(CO)4]
Ni0
d10
S = 18 + (9 × 0.35) = 21.15
Z* = 28 – 21.15 = 6.85
2) [Pt(NH3)2Cl2]
Pt2+
d8
S = 68 + (7 × 0.35) = 70.35
Z* = 7.55
3) [Rh(NH3)6]PO4
Rh3+
S = 36 + (5 × 0.35) = 37.75
Z* = 45 – 37.75 = 7.25
d6
Model Answers of Final Exam – Second Semester 1431/1432H (2010/2011G)
Chem 323
13/7/1432H
Sheet I
According to Slater’s Rules for electron-shielding, calculate Z* for the following:
a) The 3d electron of Fe2+ ion
Z* = 6.25
b) The 5s electron of Ag atom
Z* = 3.70
c) The 4f electron of Eu atom
Z* = 14.90
d) The 4d electron of Sn4+ ion
Z* = 10.85
e) The 5d electron of Os5+ ion
Z* = 7.30
(10 marks)
Sheet II
According to IUPAC, Name the following five complexes:
[Ag(NH3)2][Mn(H2O)2(C2O4)2]
Diamminesilver(I)diaquodioxalatomanganate(III)
Ba[Cr(NH3)2(NCS)4]2
Barium diamminetetraisocyanatochromate(III)
[Pt(NH2CH2CH2NH2)2][PtCl4]
Bis(ethylenediamine)platinum(II) tetrachloroplatinate(II)
[Pt(CH3NH2)Cl2]
Dichlorobis(methylamine)platinum(II)
H
[(NH3)4Cr
N
O
H
H
Fe(NH2CH2CH2NH2)2]3(PO4)4
Tetraamminechromium(III)-µ-amido-µ-hydroxobis(ethylenediammineiron(III)
phosphate
(10 marks)
Sheet III
Write Werner’s Formula for each of the following complexes:
Tetraamminecobalt(III)-µ-trihydroxotetraamminecobalt(III) Sulfate
OH
(NH3)4Co OH Co(NH3)4 (SO4)3
OH
2
Calcium Dithiosulfatoargentate(I)
Ca3[Ag(S2O3)2]2
Triamminechloro(ethylene)nitroplatinum(IV) Phosphate
[Pt(NO2)(NH3)3(CH2=CH2)Cl]3(PO4)2
Aluminum Hexacyanoplatinate(IV)
Al2[Pt(CN)6]3
Barium Dibromodioxalatocobaltate(III)
Ba3[Co(C2O5)Br2]2
(10 marks)
Sheet IV
For the following seven complexes, write down the electron configuration of the
splitted d-orbitals and also calculate the emperical ∆ value for each complex (in cm-1
unit).
1) [Ni(CN)4]2-
D4h
2
0
Configuration: e 4g a1g
b22g b1g
cm-1
∆ value = 26356
2) [Co(CN)5]2-
C4v
S.F.
Configuration: e4 b12 a11 b10
cm-1
∆ value = 44283
3) [CrCl5]3-
C4v
W.F.
Configuration: e2 b12 a11 b10
cm-1
∆ value = 15465
4) [NiBr5]3-
D3h
W.F.
Configuration: e 42 e13 a11
cm-1
∆ value = 6629
5) [Re(CN)5]1-
D3h
S.F.
Configuration: e 22 e11 a10
cm-1
∆ value = 58310
6) [MoF7]2-
D5h
Configuration: e12 e10 a10
cm-1
∆ value = 49572
7) [Rh(CN)7]4-
D5h
S.F.
Configuration: e 42 e12 a10
∆ value = 46818
cm-1
(10 marks)
Sheet V
For the following Free Ions, calculate the value of µS+L in B.M. unit.
1- Pd3+
: µS+L :
3 5
4 × × + (3 × 4)
2 2
B.M = 5.196
2- Pt4+
: µS+L :
4 × 2 × 3 + ( 2 × 3)
B.M = 5.477
3- Sm3+
: µS+L :
5 7
4 × × + ( 5 × 6)
2 2
B.M = 8.06
4- Ru4+
: µS+L :
2
4 × × 2 + (3 × 4)
3
B.M = 5.471
5- Gd3+
: µS+L :
7 9
4× × +0
2 2
B.M = 7.973
6- W4+
: µS+L :
4 × 1 × 2 + ( 3 × 4)
B.M = 4.47
7- Ho3+
: µS+L :
4 × 2 × 3 + ( 6 × 7)
B.M = 8.124
8- Ir4+
: µS+L :
5 7
4× × +0
2 2
B.M = 5.916
9- Yb3+
: µS+L :
1 3
4 × × + (3 × 4)
2 2
B.M = 3.873
10- Mo3+
: µS+L :
3 5
4 × × + (3 × 4)
2 2
B.M = 5.196
(20 marks)