Download Assignment 30 STRUCTURE OF MOLECULES AND MULTI

Assignment 30
STRUCTURE OF MOLECULES AND MULTI-ATOM IONS - Part II
In Assignment 10, you predicted the three-dimensional shapes of molecules and multi-atom ions based on
electron dot diagrams, which explicitly depicted the number of electron pairs surrounding a central atom. In
this Assignment, only the chemical formula will be given. From this formula, you will develop an electron dot
diagram, which will allow you to predict the electron pair shape and molecular shape as in Assignment 10.
The following step-by-step procedure explains how to deduce molecular shape from a given formula. As an
example, the right hand column shows how each step would be used to determine the shape of permanganate
ion, MnO4─.
Deducing the shape of single center molecules and multi-atom ions from a given formula.
1. From the given formula, you should identify a central atom
which is surrounded by ligands.
(MnO4─ contains the center Mn atom
and four surrounding O atoms.)
2. To construct the electron dot structure, an electron
counting mechanism is employed:
• Determine the total number of electrons in the formula by
adding the valence shell electrons for each neutral atom
(the valence shell electrons are found from the electron
dot diagram for the neutral atom). Adjust this total by
adding or subtracting electrons to account for the charge
on the formula.
(MnO4─ contains one Mn atom, four
surrounding O atoms, and has an
overall –1 charge, giving a total
electron pool of 2e + 4x6e + 1e =
27e.)
• From this pool of electrons, each ligand takes 8 electrons
to complete an octet (except for H, which takes only 2
electrons). Any electrons remaining from the pool
constitute lone pairs on the central atom. Note that the
central atom in a hypervalent compound may have more
than 8 electrons ( see example 5 ).
(To octet each oxygen ligand atom
requires 4 x 8e = 32e. Subtracting the
32 required electrons from the
available pool of 27e leaves a
deficiency of 5 electrons (27e – 32e = 5e). Since there are no ‘leftover’
electrons, there are no lone pairs on the
Mn atom.)
• If the pool does not contain enough electrons to provide
an octet on each ligand, then additional d or f inner shell
electrons (not shown on electron dot diagrams) must be
provided from the central atom.
(For MnO4─ the five additional
electrons required to completely octet
the ligand atoms must come from the
3d orbitals of Mn.)
3. Next draw an electron dot diagram for the formula by
taking the central atom -- including any lone pairs -- and
surrounding it with the correct number of ligands (each
having an octet).
Note: the lone pairs on the central atom are NEVER used
for bonding with the ligands -- they remain completely
with the central atom.
4. Once you have the electron dot diagram, analyze it for
electron pair shape and for molecular (or complex ion)
shape as described in Assignment 10.
tetrahedral
tetrahedral
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5. If your predicted electron dot diagram shows the central atom with less than four electron pairs surrounding it,
recall the treatment of this situation using π-bonding and/or resonance ideas from Assignment 10.
For second row atoms -- especially C, N and O -- formation of π-bonds is a frequent and important phenomenon for
completing octet structures for the central atom. For larger atoms there is only a limited or partial π-bonding
contribution.
EXAMPLES:
1)
NH4+ ammonium ion
Center atom? N
Total electrons?
Ligand atoms? 4 H
Charge? +1 (one electron gone)
1xN = 1x 5e = 5e
4xH = 4 x 1e = 4e
Ion charge (one less e) = -1e
Total pool
= 8e
To "octet" ligands?
4 x (H) = 4 x 2e = 8e (Of the total pool of 8e, 8e are used to "octet" ligands)
Electrons on central atom? N = 8e - 8e = no e
Electron dot diagram?
2)
(N, with no lone pairs, and 4 "octeted" ligands)
Electron pair shape?
Molecular shape?
Tetrahedral
Tetrahedral
BrF3
bromine trifluoride
Center atom? Br
Ligand atoms?
3F
Total electrons?
1 x Br = 1 x 7e = 7e
Charge? zero
3 x F = 3 x 7e = 21e
To octet ligands?
Ion charge
= no e
Total pool
= 28e
3x(F) = 3 x 8e = 24e
(Of the total pool of 28e, 24e are used to
octet ligands)
Electrons on central atom?
Br = 28e  - 24e = 4e, or 2 lone pairs
Electron dot diagram?
Electron Pair Shape?
Molecular shape?
trigonal bipyramidal
T-shaped
(note position of lone pairs)
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HYBRID ORBITALS
In lecture you have learned about atomic orbitals (AOs)—the regions of space surrounding an atom’s nucleus
that ‘house’ that atom’s electrons. A carbon atom has four available atomic orbitals--one 2s AO, and three 2p
AOs (2px, 2py, 2pz)— to house its four valence electrons. Take a
look at the shapes and orientations of carbon’s valence AOs
shown at right. If a bond is formed by the overlap of atomic
orbitals, then to form a molecule of methane, CH4, each of these
C atomic orbitals must overlap with an H 1s atomic orbital to
form four C-H bonds. The problem is that such an overlap would result in two different ‘types’ of C-H bond
(one C(2s) - H(1s) overlap and three C(2p) – H(1s) overlaps) with the three C(2p) – H(1s) overlaps producing
C-H bonds mutually perpendicular to each other. Neither of these attributes is observed experimentally! Recall
that for a CH4 molecule, the procedures of this Assignment would predict a tetrahedral geometry in which four
identical C-H bonds are spaced as far apart from each other as possible (109.5° bond angles). We can’t explain
CH4’s known geometry using our localized electron model of bonding unless we allow for the possibility that
the central carbon atom is capable of rearranging its individual atomic orbitals as needed to attain the most
stable (lowest energy) condition for the molecule.
Although the atomic orbitals in an isolated C atom are
not the best for bonding, the four C atomic orbitals
can be combined and rearranged (hybridized) to form
four new hybrid atomic orbitals (or hybrid orbitals,
HOs) which do possess the appropriate orientation for
bonding (Fig. 1). In this case, the four orbitals
produced are named sp3 hybrid atomic orbitals
because they result from mixing one “s-type” AO and
three “p-type” AOs of the C atom. The number of
AOs used to create hybrid atomic orbitals must equal
the number of HOs produced! Thus, there are four
sp3 hybrid orbitals. This phenomenon is called
orbital conservation.
Figure 1
Note that in the CH4 example, no π-bonds are
expected in the molecule because the central atom has
a complete octet. The expectation of no π-bonding
also makes sense from an atomic orbital standpoint—
if all of carbon’s p-AOs are involved in creating the
bonding HOs, there are no empty p-atomic orbitals
left to participate in π-bonding.
The idea that individual atoms can rearrange their atomic orbitals as needed to achieve the minimum energy for
the molecule is an assumption of this model of bonding. The five common orbital hybridization names and
the corresponding electron pair shapes are listed in the following table:
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Electron pair shape
Orbital
Hybridization
1
(Not appropriate)
(Not appropriate)
2
Linear
sp
3
Trigonal coplanar
sp2
4
Tetrahedral
sp3
5
Trigonal bipyramidal
sp3d
6
Octahedral
sp3d2
Number of electron pairs
(σ-bonds and lone)
NOTE: Each orbital hybridization name listed above carries built-in clues that describe the respective
configuration. Merely from a hybridization name you can determine:
• the atomic orbitals (of the central atom) that are most affected by that hybridization
• the spatial dimensionality the resulting arrangement will have (is it one-, or two-, or three- dimensional?)
• the name given to each of the hybrid orbitals in the final arrangement
• in how many directions (orientations out from the nucleus) the hybrid orbitals are arranged
EXAMPLES:
3)
NO2-
nitrite ion
Center atom? N
Ligand atoms?
Total electrons?
2O
Charge? -1 (one electron extra)
1xN = 1 x 5e = 5e
2xO = 2 x 6e = 12e
Ion charge (one extra e) = + 1e
Total pool = 18e
To octet ligands?
2 x (O) = 2 x 8e = 16e (Of the total pool of 18e, 16e are used to octet ligands)
Electrons on central atom?
N = 18e - 16e = 2e, or 1 lone pair
Electron dot diagram?
(N, with its lone pair, and 2 octeted oxo-ligands)
Electron pair shape and hybridization?
Molecular shape?
trigonal coplanar
angular
sp2
Note: The trigonal-planar electron pair shape indicates that nitrogen’s atomic orbitals must account for
three bonds in two dimensions. This requires three N AOs (2s, 2p, 2p) to combine and
rearrange to form three new sp2 hybrid atomic orbitals. Thus the hybridization is sp2 and one
π-bond is expected though it is not explicitly shown in the diagram. (Review the resonance
treatment of nitrite in Assignment 10 if necessary.)
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4)
SnBr2
Tin(II) bromide (stannous bromide)
Center atom? Sn
Ligand atoms? 2 Br
Total electrons?
1xSn = 1 x 4e = 4e
2xBr = 2 x 7e
= 14e
Ion charge
= no e
Total pool
= 18e
Charge? zero
To octet?
2x(Br) = 2 x 8e = 16e
Electrons on central atom?
Sn = 18e - 16e = 2e, or one lone pair
Electron dot diagram?
Electron pair shape and hybridization?
trigonal coplanar
Molecular shape?
angular
sp2
5)
SnCl62-
hexachlorostannate(IV) ion
Center atom? Sn
Ligand atoms? 6 Cl
Total electrons?
4e + 6 x 7e + 2e = 48e
To octet?
6x(Cl) = 6 x 8e = 48e
Central electrons?
Sn = 48e - 48e = no e
Charge? -2 (two electrons extra)
(no lone pairs)
Electron dot diagram?
Electron pair shape and hybridization?
octahedral
Molecular shape?
octahedral
3 2
sp d
Note: No π-bonds expected because no available empty p atomic orbitals on the central atom. All of
Sn’s valence p AOs, as well as its valence s AO and two of its n-1 d AOs, are involved in a
rearrangement to produce six hybrid sp3d2 hybrid orbitals.
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6)
I3-
Triiodide ion
Center atom? I
Ligand atoms? 2 I
Total electrons?
3x7e + 1e = 22e
To octet ligands?
2x(I) = 2 x 8e = 16e
Electrons on central atom?
I = 22e - 16e = 6e
Charge?
-1 (one electron extra)
(three lone pairs)
Electron dot diagram?
Electron pair shape and hybridization?
Molecular shape?
trigonal bipyramidal
sp3d
linear
(lone pairs MUST be placed equatorially
in order to obtain correct shape)
7)
UO2+2
uranyl(VI) ion
(dioxouranium(VI) ion)
Center atom? U
Ligand atoms? 2 O
Total electrons?
2e + 2 x 6e - 2e = 12e
To octet ligands?
2x(O) = 2 x 8e = 16e
Electrons on central atom?
U = 12e - 16e
Charge? +2 (two electrons gone)
= -4e
(NO lone pairs)
4 electrons MUST come from n-1 d-sub-shells of the uranium atom
Electron dot diagram?
Electron pair shape and hybridization?
linear
Molecular shape?
linear
sp
Note: two π-bonds predicted (but their effectiveness is limited by the large size of uranium).
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8)
SbCl4-
tetrachloroantimonate(III) ion
Center atom? Sb
Ligand atoms? 4 Cl
Total electrons?
5e + 4x7e + 1e = 34e
To octet ligands?
4x(Cl) = 4 x 8e = 32e
Electrons on central atom?
Sb = 34e - 32e = 2e
Charge?
-1
(one lone pair)
Electron dot diagram?
Electron pair shape and hybridization?
Molecular shape?
trigonal bipyramidal
sp3d
teeter
The lone pair on Sb is equatorial, NOT at a
axial (up or down) position.
9)
SO3
sulfur trioxide
Center atom? S
Ligand atoms? 3 O
Total electrons?
6e + 3 x 6e = 24e
To octet ligands?
3x(O) = 3 x 8e = 24e
Electrons on central atom?
Charge? zero
S = 24e - 24e = no e
(no lone pairs)
Electron dot diagram?
Electron pair shape and hybridization?
Molecular shape?
trigonal coplanar
trigonal coplanar
sp
2
Note: One π-bond is likely. Each O ligand will share more of its electron octet with the S to provide
the central atom with an octet without alteration of the shape. The π-bond is delocalized over all four
atoms; one-third of the π-bond comes from each oxygen ligand.
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Problem Set 30
Be able to define the following terms:
hybrid atomic orbital
orbital conservation
hybridization name
For each of the following formulas:
a) construct the electron dot diagram; b) tell what hybridization applies to central
atom; c) give diagram and name of both the electron pair shape and the molecular
shape; d) tell how many π-bonds, if any, the actual molecule should contain.
(Follow the procedures outlined in the examples. Mark electron dots clearly.)
1)
NF3 , nitrogen trifluoride
2)
VO2+ , vanadyl(V) ion
3)
IF4− , tetrafluoroiodate(III) ion
4)
SnCl62− , hexachlorostannate(IV) ion
5)
HgCl3− , trichloromercurate(II) ion
6)
TeCl4 , tellurium tetrachloride
7)
HOCl (O is central) , hypochlorous acid
8)
ClNO (N is central) , nitrosyl chloride
9)
H2CO (C is central) , formaldehyde
Do the same for these multi-centered species:
H
CH3CH2OH (bonding sequence is H3C
C
10)
OH ) , ethanol.
H
Determine the number of electrons in the electron pool of these molecules & complex ions:
CO2
HCN (C is central)
XeF6
NO2
VOCl3 (V is central)
NH4+
PO43−
PCl4+
NO3−
BH4−
SO32−
N3−
01-111-30-8
Partial answers to P.S. 30:
Check your accuracy against the following conclusions. Note: These are not complete answers; yours should also show
your reasoning process ─ e.g., the calculation of the electron pool, and accurate depictions of the dot-picture and the
electron-pair arrangement (in appropriate geometric style) and the geometric rendition of the entire molecule's shape.
The diagrams then serve as basis for deriving the conclusions below. Complete answers may be viewed on the posted
answer key on the Chemistry 111 web page.
1) NF3 has total of 26 electrons in the valence pool, resulting in 1 lone pair on the N;
electron-pair shape is tetrahedral (corresponding to sp3 hybridization);
molecular shape is pyramidal;
no π-bonding in the final molecule (for the central N already has an octet).
2) VO2+ has a total of only 13 electrons in the valence pool, so no lone pairs can exist on the V; indeed,
just to satisfy ligand octets, V must contribute three d electrons (from its inner shell) to the bonding pool;
resulting electron-pair shape is linear (sp hybridization);
molecular shape is linear;
2 π-bonds are plausible (but of limited effectiveness, since vanadium isn't a "second row" element).
3) IF4 has a total of 36 electrons in the valence pool, thus 2 lone pairs remain on the I;
electron-pair shape is octahedral (sp3d2 hybridization);
molecular shape is square co-planar, with the lone pairs of iodine located opposite from each other;
no π-bonding in the final molecule (the central I already has more than an octet).
4) SnCl62 has 48 e; no lone pr on Sn; octahedral e-prs (sp3d2 hybridization); octahedral mol.shape;
no π-bonding.
5) HgCl3 has 24 e; no lone pr on Hg; trigonal co-planar e-prs (sp2); trigonal co-planar mol.shape;
1 π-bond is plausible (but of limited effect, since neither element is on"second-row" of P.Table).
6) TeCl4 has 34 e; 1 lone pr on Te; trigonal bipyramidal e-prs (sp3d); teeter mol.shape; no π-bonding;
the lone pair on Te must be placed equatorially, not at a polar position.
7) HOCl has 14 e; 2 lone prs on O; tetrahedral e-prs (sp3); angular mol.shape; no π-bonding.
8) ClNO has 18 e; 1 lone pr on N; trigonal co-planar e-prs (sp2); angular mol.shape;
1 π-bond is necessary.
9) H2CO has 12 e; no lone pr on C; trigonal co-planar e-prs (sp2); trigonal co-planar mol.shape;
no π-bonding.
10) CH3CH2OH has three central atoms: C1(at left), C2(at center), O(at right).
Around C1 , no lone pr on C1 ; tetrahedral e-prs (sp3); tetrahedral mol.shape; no π-bonding.
Around C2 , no lone pr on C2 ; tetrahedral e-prs (sp3); tetrahedral mol.shape; no π-bonding.
Around O , 2 lone prs on O; tetrahedral e-prs (sp3); angular mol.shape; no π-bonding.
(To design an overall molecular diagram, tilt the tetrahedral shapes so that C-C-O "backbone" lies flat.)
Electron pools of the molecules:
Electron pools of complex ions:
16e; 10e; 50e; 17e(there is an unpaired e on N here); 32e.
8e; 32e; 26e; 32e; 24e; 16e; 8e.
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