Download Main Group and Transition Metal Chemistry: Reading: Moore

Main Group and Transition Metal Chemistry:
Reading: Moore chapter 22, sections 22.1, 22.6
Questions for Review and Thought: 14, 16, 24, 26, 30, 34, 36, 42, 48, 50, 58, 60.
Key Concepts and Skills:
•
definition of bidentate, monodentate, coordination number, lanthanide
contraction, chelating ligands, coordinate covalent bond, hexadentate ligand.
•
Be able to write electron configurations for transition metals and their ions;
explain trends in the sizes of transition metal atomic radii; explain coordinate
covalent bonding in coordination compounds and complexes; understand the
different types of isomerism.
Lecture Topics:
I. Review of the properties of the main group elements (Chapter 7)
The Aufbau principle allows one to build up the ground-state electronic configurations
of atoms in the periodic table.
1s2
2 electrons
2s2 2p6
8 electrons (octet rule)
2
6
10
3s 3p 3d
18 electrons
4s2 4p6 4d10 4f14
5s2 5p6 5d10 5f145g18
6s2 6p6 6d10 6f146g18….
Principal quantum number n (1-7)- integer assigned to each of the main electron energy
levels in an atom. Number increases as we move down the periodic table
Angular quantum number , l (l=n-1) – refers to the s (l=0), p (l=1), d (l=2), f (l=3)
subshells within a shell. Each subshell has (l….0…-l) orbitals. Note: the transition series
is filling d orbitals (d block elements); the alkali and alkaline earth metals (groups 1A and
2A, s block elements) are filling s orbitals; groups 3A-8A are filling p orbitals (p –block
elements); the lanthanides and actinides are filling f orbitals (f-clock elements)
There are 1 s orbital, 3 p orbitals, 5 d orbitals and 7 f orbitals. Know the orbital shapes.
Spin quantum number: +1/2 and –1/2: each orbital within a subshell can contain up to 2
electrons; if two electrons fill an orbital they must have opposite spins (up and down)
(Pauli exclusion principle)
Hund’s Rule: electrons pair only after each orbital in a subshell is occupied by a single
electron.
Example: Nitrogen: [He] 2s2 2p3 vs. Oxygen: [He] 2s2 2p4
What is the ground state electronic configuration of Si, Mg2+, F-, S2-, Xe, As, Pb, Ra?
Check your knowledge:
Si: [Ne] 3s2 3p2 What do the unfilled p orbitals look like?
Mg2+: [Ne] 3s0
F-: [He] 2s2 2p5 What do the unfilled p orbitals look like?
S2-: [Ne] 3s2 3p4 What do the unfilled p orbitals look like?
Xe: [Kr] 5s2 (4d10)5p6
As: [Ar] 4s2 (3d10) 4p3 What do the unfilled p orbitals look like?
Pb: [Xe] 6s2 (4f14) (5d10) 6p2 What do the unfilled p orbital look like?
Ra: [Rn] 7s2
II. Atomic/ionic radii
Atomic radius decreases across a row and increases down the periodic table. Why?
Going down the periodic table the principal quantum number increases, and we know
that higher quantum numbers correspond to higher energy levels (shells) which are
further from the nucleus.
Going across the periodic table, as electrons are added to a shell (filling the subshells),
each electron avoids other electrons (due to electron-electron repulsion) and thus fails to
effectively shield the nuclear charge from other added electrons; each added electron thus
feels a greater net nuclear charge (Zeff), causing the atom to contract and radii to decrease.
Note: electrons in p, d, f orbitals are further from the nucleus than electrons in s orbitals
and thus are less effective at shielding the nuclear charge
Compare radii:Al3+, B3+, He, Ne, Li+, Be2+.
Electronegativity, a measure of the ability of an atom in a molecule to attract bonding
electrons to itself, increases across a period until the noble gases which have very small
electronegativities. Why should this be so? Why would you expect electronegativity to
decrease down a group in the periodic table?
III. Chemical reactivity
Group 1A and 2 A metals form ionic compounds since they are relatively electropositive,
giving up electrons to obtain a stable noble gas core (octet)
Na
+
Cl

NaCl
Mg
+
2Br

MgBr2
[Ne]3s1 [Ne]3s23p5
[Ne]•[Ar]
[Ne]3s2
[Ar]4s24p5
[Ne][Kr]2
Hybridization is the mixing of s and p orbitals to maximize the number of unpaired
electrons on an atom available for sharing. Sp, sp2, and sp3 orbitals are slightly higher in
energy than s orbitals and lower in energy than p orbitals.
Consider: BF3,, CH4, H3N, H2O, HF
BF3: 2s22p1  sp2 p0 ; boron has three sp2 orbitals each with one unpaired electron
(available for sigma bonding with an electron from each F atom) and one empty p orbital.
CH4: 2s22p2  sp3; carbon has four equivalent sp3 orbitals each with one unpaired
electron (available for sigma bonding with an electron from each H atom).
H3N: 2s22p3  sp3; nitrogen has four equivalent sp3 orbitals, three with one unpaired
electron (available for sigma bonding with an electron from each H atom) and one with a
lone pair.
H2O: 2s22p4  sp3; oxygen has four equivalent sp3 orbitals, two with one unpaired
electron (available for sigma bonding with an electron from each H atom) and two with
lone pairs.
HF: 2s22p5  sp3; fluorine has four equivalent sp3 orbitals, one with an unpaired electron
(available for sigma bonding with an electron from an H atom) and three with lone pairs.
Geometry: sp3 – tetrahedral; sp2 – trigonal planar; sp – linear. Hybrid (sp) orbitals
participate in sigma (σ) bonding; p-orbitals participate in pi (π) bonding
IV. Transition Metals – Electron configurations
Filling order: 4s2 3d10 4p6; energies of 4s and 3d levels are quite close despite the different
quantum numbers. Ordinarily 4s fills before 3d.
K: [Ar]4s1
Ca: [Ar] 4s2
Sc: [Ar] 4s23d1
Ti: [Ar] 4s23d2
V: [Ar] 4s23d3
Cr: [Ar] 4s13d5 note: spin pairing in the 4s orbital is a higher energy situation than having all spins
Mn: [Ar] 4s23d5
parallel
Fe: [Ar] 4s23d6
Co: [Ar] 4s23d7
Ni: [Ar] 4s23d8
Cu: [Ar] 4s13d10 note: upon filling up with electrons, 3d subshell is lower in energy than 4s
Zn: [Ar] 4s23d10
•Thus, Cu, Ag, Au have filled d10 orbitals, just like Zn, Cd, and Hg.
For Cr, two half-filled subshells (4s13d5) is more stable than 4s23d4
•When a transition metal loses its electrons to form an ion, the s electrons are lost before
the d electrons; example: Ti3+ [Ar]4s03d1; Cr2+ [Ar] 4s03d4; Cu+1 [Ar] 4s03d10
b. How do we know that s electrons are lost first? Magnetic measurements confirm
paramagnetism. Substances that have unpaired spins are paramagnetic and are attracted
into a magnetic field because of alignment of some unpaired spins with the magnetic
field. Diamagnetic substances have all electrons spin-paired so their magnetic fields
effectively cancel one another; they therefore experience negliglible attraction to a
magnetic field.
•The extent of attraction into a magnetic field, measured by the apparent mass of the
sample in the field, is an indication of the number of unpaired electrons.
Consider: Fe2+: [Ar] 4s03d6 (four unpaired spins) vs. Fe3+: [Ar] 4s03d5 (5 unpaired spins)
Alternatively, removing d electrons first: Fe2+: [Ar] 4s23d4 (four unpaired spins) vs.
Fe3+:4s23d3 (3 unpaired spins). It is found that Fe3+ has a greater apparent mass in a
magnetic field than Fe2+
c. Atomic radii of the transition elements
Atomic radii decrease across a period due to the increasing effective nuclear charge felt
by the valence electrons (Zeff, see above). However, down a group an interesting trend is
observed: Ti<Zr =~ Hf; V<Nb =~ Ta; Cr<Mo=~ W. Why? Beginning at Lanthanum,
the 4f subshell starts to fill. In the lanthanide series of elements, effective nuclear charge
builds up, causing a decrease in atomic radii because all additional electrons go into 4f
orbitals, which are so far from the nucleus that they do not screen other valence electrons
from the nuclear charge. The increased nuclear charge pulls the valence electrons closer
to the nucleus, offsetting the size increase on going from quantum number 5 to 6. This
phenomenon is known as the Lanthanide contraction
d. Oxidation states
Multiple oxidation states are possible for most of the transition elements: remove s
electrons first, then paired d electrons, then unpaired d electrons. The maximum
oxidation state for the first five elements of a transition series is the sum of the ns and nd
electrons of the uncharged atom: Ti: [Ar] 4s23d2  4+, 3+, 2+
V: [Ar] 4s23d3  5+, 4+, 3+, 2+; Cr: [Ar] 4s13d5  6+, 5+, 4+, 3+, 2+
Mn: [Ar] 4s23d5  7+, 6+, 5+, 4+, 3+, 2+
Note: compounds in high oxidation states are good oxidants and form covalent complex
ions: example: MnO4-, Cr2O72Compounds in low oxidation states form ionic compounds. Examples: FeCl2, FeCl3,
MnCl2
Electron gain and electron loss: Fe2+  Fe3+ + e-;
Fe3+ + e-  Fe2+
V. Bonding in transition metal complexes.
The most common form of bonding in transition metal complexes is coordinate covalent
bonding, in which one atom contributes both electrons for the shared pair in a bond.
• Typical example: F3B  NH3 complex. The lone pair on nitrogen donates to the
mepty p orbital on boron.
• Transition metals have incompletely filled d shells that can accept lone pairs.
• The molecules bonded to the central metal atom are ligands; ligands may be charged
(CN-, Cl-) or uncharged (H2O, NH3, (CH3)3P)
Consider: [Ni(OH2)6]2+ and [Ni(Cl)4]2-, both are complex ions- formed when several
molecules or ions are connected to a central metal ion or atom by coordinate covalent
bonds.
• Compounds in which complex ions are combined with oppositely charged ions to form
neutral compounds are called coordination compounds: [Ni(H2O)6]2+• 2Cl• The number of coordinate covalent bonds between the ligands and the central metal ion
is the coordination number (CN) of the metal ion
CN=2 examples: [Ag(NH3)2]+, [AuCl2]CN=4 examples: [NiCl4]2-, [Pt(NH3)4]2+
CN=6 examples: [Fe(H2O)6]2+, [Co(NH3)6]3+
•Ligands that can form only one coordinate covalent bond to a metal are monodentate
ligands. Examples: H2O, NH3, H2S, HO-, Cl-, NC-, CO
•Some compounds can form two or more coordinate covalent bonds to the same metal
ion because they possess two or more atoms with lone-pairs separated by several
intervening atoms. These are polydentate ligands.
Examples: bidentate ligands- ethylenediamine: H2NCH2CH2NH2
Carbonate ion: CO32Oxalate ion: -O2CCO2A Hexadentate ligand can form a claw-like complex or chelate with a metal ion,
occupying all possible ligand sites (for an octahedral complex) with lone pairs from a
single ligand molecule. Example: EDTA (-O2CCH2)2N CH2CH2N(CH2CO2-)2
•EDTA complexes metal ions so strongly that it is used to treat lead and mercury
poisoning.
VI. Geometry of coordination compounds (see appendix below)
The geometry of coordination compounds is dictated by the arrangement of electrondonor atoms of the ligands attached to the central metal ion.
Consider the situations with monodentate ligands:
CN=2 linear geometry (bond angle 180°) : H3N•••Ag+•••NH3; also [CuCl2]-, [Au(CN)2]CN=4 either tetrahedral (bond angles 109°) or square planar (bond angles 90°)
Tetrahedral: [Zn(NH3)4]2+
Square planar: [Pt (NH3)4]2+ difference: sp3 vs. sp2d hybr.
CN=6 octahedral (bond-angles are 90° and 180°): [Co(NH3)6]3+, [Fe(H2O)6]2+ (sp3d2)
VII. Isomerism
Constitutional isomers are molecules with the same molecular formula but different
arrangements of atoms
3 different types :
Linkage isomers: a ligand can bond to a metal atom using two different electron-donating
atoms: (H2O)5Co•••:SCN: vs. (H2O)5Co•••:NCS:
Geometric isomerism: positional isomerism that occurs both in square planar and
octahedral complexes: cis vs. trans isomers:
Examples: square planar- Pt(NH3)2Cl2
H 3N
NH3
H 3N
Cl
Pt
Pt
Cl
Cl
cis
Ocatahedral:
Cl
NH3
trans
Co(NH3)4Cl2+
and
Cl
H 3N
Co(NH3)3Cl3
NH3
NH3
NH3
H 3N
Cl
Co
H 3N
NH3
H 3N
Cl
Cl
H 3N
H 3N
cis
cis
N
H3
Co
Cl
Cl
NH3
trans
Cl
Co
Co
H 3N
Cl
H 3N
Cl
Cl
cis
Optical isomerism: enantiomeric compounds are mirror images. Enantiomers are nonsuperimposable molecules which have mostly identical properties but can be
distinguished by enzymes containing chiral receptor sires or binding pockets. Optical
isomerism is not possible in square planar or linear geometries but is possible in
tetrahedral and octahedral geometries:
OH2
H 3N
OH2
Cl
Cl
Co
H 3N
NH3
Co
OH2
H2O
Cl
NH3
Cl
mirror
Additonal problems:
1. Will methylamine be a mono- or bidentate ligand? With which of its atoms will it
bind to a metal ion?
2. Show how glycinate ion (H2N–CH2COO-) can act as a bidentate ligand. Wbhich
atoms in the glycinate ion will bind to a metal ion?
3. Draw the structures of all possible isomers for the following complexes. Indicate
which isomers are enantiomer pairs.
(a) (NH3)2 Br Cl Pt(II) (square planar)
(b) [(H2O)2 Cl (CN)3 Co(III)]- (octahedral)
(c) [(-O2CCO2-)3 V(III)]3- (octahedral)
4. Iron (III) forms octahedral complexes. Sketch the structures of all the distinct
isomers of [Fe(en)2Cl2]+, indicating which pairs of structures are mirror images of
each other. en= ethylenediamine = H2NCH2CH2NH2
5. Consider the ions Mn3+, Fe3+, Co3+, Ni3+. Which ion in its ground state has the
greatest number of unpaired electrons?
6. Explain why the atomic radius of Rh is essentially equal to that of Ir.
7. How many isomers would you expect for the complex [Ni(NH3)3Cl3]- if the
coordination geometry is octahedral? Draw the isomers.
Appendix: hybridization in transition metal complexes and ligand geometry about
a metal.
Consider several cases:
1.) linear geometry for [CuCl2]-, [Au(CN)2]-, Ag(NH3)2+:
All of these metals are +1 charge. The electron configuration on the metal in each
case is 4s03d104p0. Since all d-orbitals are filled, to make empty orbitals available
for coordinate covalent bonding, hybridization must take place between s and p
orbitals only. To accept two ligands, you need only hybridize one s and one p
orbital to make two sp hybrid orbitals which are empty and capable of accepting
lone pairs from ligands. sp hybridization is always indicative of a linear
geometry! Note there are also 5 completely filled (unhybridized) d orbitals on the
metal center.
2). Tetrahedral geometry: [Zn(NH3)4]2+
Zn2+ has an electron configuration 4s03d104p0 . To accept lone pairs from four
ligands, zinc must open up four empty hybrid orbitals. The only way to do this,
since all of the d orbitals are filled, is to hybridize one s orbital and all three porbitals to make four equivalent empty sp3 orbitals. sp3 hybridization is always
indicative of tetrahedral geometry! Note there are also 5 completely filled
(unhybridized) d orbitals on the metal center.
3.) Square planar geometry: [Pt (NH3)4]2+
Pt2+ has an electron configuration 4s03d84p0. To accept lone pairs from four
ligands, platinum must open up four empty hybrid orbitals. hybridization is easier
for orbitals which are closer in energy, and certainly 4s and 3d orbitals are close
in energy. Since there are 8 3d electrons, spin-pairing all of these electrons in four
d orbitals leaves one empty d orbital which can hybridize with one s orbital and 2
p orbitals, giving a total of four empty sp2d hybrid orbitals. sp2d hybridization is
always indicative of a square-planar geometry. Note there are also 4
completely filled (unhybridized) d orbitals on the metal center.
4.) Trigonal bipyramidal: CuCl52-, Fe(CO)5
Cu3+ has an electron configuration 4s03d8; Fe has an electron configuration 4s23d6.
To accept five lone pairs from five ligands, each metal must make available 5
empty hybrid orbitals. For copper, the eight d electrons can be spin-paired in 4 d
orbitals, leaving one d orbital available for hybridization with one s orbital an all
three p orbitals, giving 5 empty sp3d hybrid orbitals. For iron, 2 4s electrons can
easily be promoted to a 3d orbital, and now the 8 d electrons can spin-pair in 4 d
orbitals, leaving one d orbital available for hybridization with the now empty s
orbital and all three p orbitals to make five empty sp3d hybrid orbitals. sp3d
hybridization is always indicative of trigonal bipyramid geometry! Note there
are also 4 completely filled (unhybridized) d orbitals on each metal center.
5.) Octahedral: [Co(NH3)6]3+, [Fe(H2O)6]2+
Fe3+ has an electron configuration of 4s03d64p0; Co3+ has an electron configuration
of 4s03d6 4p0 In both cases, 6 empty orbitals must be available to accept 6 lone
pairs from the ligands. The 6 d electrons can be spin-paired in three d orbitals,
leaving two empty d orbitals available for hybridization. Combining one s, two d,
and three p orbitals results in six equivalent, empty sp3d2 hybrid orbitals. sp3d2
hybridization is always indicative of octahedral geometry! Note there are also
3 completely filled (unhybridized) d orbitals on each metal center.
6.) Dodecahedral: Mo(CN)84Mo4+ has an electron configuration of 4s03d24p0. Mo must open up 8 empty
orbitals to accept 8 lone-pairs from the ligands. The 2 d-electrons can be spinpaired in a single d orbital, leaving 4 empty d orbitals available for hybridization.
Combining one s, 4 d, and 3 p orbitals gives 8 equivalent, empty sp3d4 hybrid
orbitals. sp3d4 hybridization is always indicative of dodecahedral geometry!
Note there is also 1 completely filled (unhybridized) d orbital on the metal center