Hybridization

CH₄

C ground state:

C valence state:

The formation of the valence state costs 406 kJ.

The C valence state suggests that the angles will be 90^o instead of the observed 109.5^o.

CH₄ can achieve the best bond angles if the s & p orbitals mix to form new __________________.

s + p_x + p_y + p_z --->

Rules for forming hybrids:

1. Orbitals must have _____________________ (usually means same n, but cf. 3d & 4s) in order to form hybrids.

2. Different hybridizations allow ________________________.

Hybrid	Geometry	Bond angles
sp	linear	180o
sp2	trigonal planar	120o
sp3	tetrahedral	109.5o
dsp2	square planar	90o
dsp3	trigonal bipyramidal	90o, 120o
d2sp3	octahedral	90o

cos q = s/(s - 1) = (p - 1)/p

CH₄, sp³

3. Hybrid orbitals are more ________________ than pure p or s orbitals & provide more efficient overlap with other orbitals.

efficient of overlap:

sp > sp² > sp³ > > s > p

good vs poor overlap (chalkboard)

Molecule	Hybridization	C-H bond energy, kJ/mol
HCC-H	sp	506
H2CCH-H	sp2	444
CH4	sp3	410
C-H radical	p	335

Hybridization & Structure

1. Bond angles

NH₃ : H-N-H angle =

PH₃ : H-P-H angle =

H₂O : H-O-H angle =

H₂S : H-S-H angle =

Why are the angles smaller for hydrides of second row elements?

Factors:

 2. Bent's rule

PCl₅ sp³d = sp_xp_y + d_z2p_z

PCl₅:

PCl₃F₂:

Bent's rule: More electronegative substituents "prefer" hybrid orbitals having ________ s character, and less electronegative substituents "prefer" hybrid orbitals having ___________ s character.

Explanation:

CH₂F₂:

F-C-F angle = ______

H-C-H angle = _____

C-F bonds have less __ character

C-H bonds have more __ character

Hybridization and Electronegativity

c depends on the environment of the atom:

1. its charge/oxidation state:

2. its hybridization:

acidity:

HCC-H H₂CCH-H CH₄

....sp............sp²...........sp³

basicity:

RNH₂ pyridine R-CN:

25%s.....33%s.........50%s

pyridine:

Bond Energy

Bond energies are stronger when:

1. Bond is __________

2. Bond is ____________

Bond	BE, kJ/mol	Polarity
H-H	431	nonpolar
F-F	155	nonpolar
H-F	565	polar

Ionic component strengthens bond.

Trends in Bond Energies

Main Group Elements

Bond energies tend to _________________ as one descends a group.

p block:

X-Y no lone pairs on X (central atom)

Bond	C-I	C-Br	C-Cl	C-F
BE, kJ/mol	240	276	339	485

X-X no lone pairs on X

Bond	Ge-Ge	Si-Si	C-C
BE, kJ/mol	188	226	347
M.P., oC	937	1412	3827

X-X lone pairs on X

Bond	I-I	Br-Br	Cl-Cl	F-F
BE, kJ/mol	149	193	239	154

Multiple bonds

Bond	CC (triple)	C=C	C-C
Bond length, A	1.20	1.34	1.54
BE, kJ/mol	839	614	347

1^st row (X) vs 2^nd row (Y)

X=X X=Y Y=Y

Bond	BE, kJ/mol
N-N	159
NN (triple)	945
P-P	214
PP (triple)	477

Compare N₄ vs P₄. (chalkboard)

s block

Bond	Li-Li	Na-Na	K-K	Rb-Rb	Cs-Cs
BE, kJ/mol	105	72	49	45	43
MP, oC	181	98	63	40	29

Reactivity:

Transition Metals

Vertical Trend

Bond energies ______________ as one descends a group (4B - 8B).

5d 4d 3d

Bond	Cr-Cr	Mo-Mo	W-W
MP, oC	1857	2617	3407

Bond Energy trends:

Cr-Cr Mo-Mo W-W

Cr-L Mo-L W-L

Horizontal Trend

(see graph)

Na:

W:

Hg:

Cr₂⁴⁺

Hg₂²⁺

Patterns in Oxidation States (OS)

1. Group no. =

e.g., 6+ for Cr, 5+ for N

2. Even-numbered groups tend to have even-numbered OS's (s & p block)

3. Odd-numbered groups tend to have odd-numbered OS's (s & p block)

4. OS's tend to differ by _______ for p block elements.

SO₄^2-

SO₃^2-

5. Negative OS's of p block elements =

N:

6. Stability of high OS's ______________ for p block elements when descending a group:

stability: CCl₄, SiCl₄ SnCl₄ PbCl₄

PbCl₄ ---> PbCl₂ + Cl₂

7. Stability of high OS's _______________ for d block elements when descending a group:

stability: WO₃ MoO₃ CrO₃

Oxidizing abilities?

8. Stability of high OS's decreases for d block elements when moving left to right.

(chalkboard)