Bonding and Hybridization in Molecules

This document discusses the process of hybridization in molecules and its impact on bonding properties and molecular geometry. It provides examples of hybridized orbitals in methane, sulfur hexafluoride, and ammonia. The document also covers topics such as the octet rule, resonance Lewis structures, the photoelectric effect, quantum world concepts, atomic structure and orbitals, and molecular orbital theory for diatomic molecules. References are provided for further reading.

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Name of the Student
Course Name
Professor Name
Institution of Affiliation
Date

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1.) Bonding
1) Hybridization is the process by which the atomic orbitals are fu so as to form new hybridized
orbitals. The hybridization process therefore has an influence on the bonding properties and the
molecular geometry of molecules. Hybridization helps to explain more on the valence bond
theory. The newly formed orbitals have equal number of electrons as the original orbitals (Kasap,
Koughia & Ruda, 2017). The characteristics of the new orbitals in terms of energy and properties
are seen as a summation or average of the old unhybridized orbitals. Methane CH4 has a carbon
with a configuration 1s2 2s2 2p2 meaning it has two unpaired electrons in the p orbital. Therefore,
the 1s and the 3p orbitals will be combined forming a hybrid orbital that will be sp3. For sulfur
hexafluoride the orbitals thar are used are the 3s, 3py, 3py, 3pz and 3dx2–y2 and 3dz2. On SF6,
formation S atom in excited state electron pairs in the 3s and 3px orbitals are unpaired and one
pair gets promoted to vacant 3dz2 and 3dx2-y2 orbitals resulting to formation of six sp3d2 hybridized
orbitals. Ammonia has an unbonded pairs of electrons with also presence of three sigma bonds.
Consequently, this makes ammonia sp3 hybridized as there are S and also three P’s which are
hybridized around nitrogen.
2). Octet rule states that atoms have a tendency of having a total of eight electrons in the valence
orbital. In the event that there are fewer than eight, atoms react forming stable. Formal charge is
given by [number electrons in valence shell] – [the non-bonded electrons]- summation of
bonding electrons]/2.
a) NO2-

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Nitrogen 5-3-3= 0, oxygen double bonded 6-4-2= 0 and oxygen singly bonded 6-6-1= -1
0+0-1= -1
b) BSF
c)
Resonance Lewis structures
(a) nitrate ion, NO−
(b) Ozone, O3

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Quantum world
1) . A surface of a metal which is bombarded with monochromatic light of electromagnetic wave
having a frequency above the threshold frequency), the light incident to metal surface causes
ejection of electrons from metal surface. This process is thus referred to as photoelectric effect.
The ejected electrons are referred as photoelectrons.
Einstein resolved that energy possessed by light particles depends on frequency
E = hν
With h being Planck’s constant = 6.6261 × 10-34 Js. Part of the energy is required to remove
electrons from metal surface and the other as kinetic energy of electron. Therefore: Photon
energy= work function + electrons maximum kinetic energy
E = W + K.E

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hv = W + K.E
KE = hv – w
2) The much higher uncertainty in position of a particle, the lesser the uncertainty of its
momentum. Therefore, it not possible to predict both the position and momentum of a body
simultaneously with greater
λ= h
mv
6.6261 ×10−34 Js .
1.0× 1.0 =6.626 10−34 m
6.6261 ×10−34 Js .
6.6464731 ×10−24 ×1000 =9.96910−14 m
The wavelength for the 6.626 10−34 m is so small compared to that of He atom to be easily
detected and this can be related to mass of the bodies. The smaller the mass the greater the
detectability.
4. ΔxΔ px ≥ h/4 π
a) 6.6261 × 10-34 Js/4 π ×1.0×10-6 = 5.2681×10-29
p=mv = 5.2681×10-29/500 = 1.0536×10-31
b) [6.6261 × 10-34 Js/4 π]/1750 = 9.2263×10-34
Schrodinger Equation
1.) Energy levels are increasing by n2 and n is the number of energy states. Hence the particle
will have four energy states

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2.) From the above formulae and substituting n=0 and L = 1nm, n=0, x=0. Sin 0 = 0 hence
probability is equals 0
When n=100, x=0.25 and L=1nm then the probability of finding the particle is 0.98
3.) The zero-point energy (ZPE) is the least possible energy for a system in quantum retains even
when it is at temperature of absolute. Temperature is the measure of the intensity of the
molecular motion of particles in random motion (Zong et al. 2016). It is thus expected that at this
absolute zero temperature the motion of these particles ceases and they come to rest but the
motion of particles does not however come to. This ZPE can thus be defined
E(ZPE) = Minimum energy of quantum state – Energy of classical state. Classically, energy can
be calculated by the natural oscillating frequency of particles required to move or shift position.
4 Atomic Structure and Orbitals
1.) Radial part of a wavefunction explains if there is a low or a high probability at different
distances from the atom’s nucleus an electron is likely to be found. When this probability is
obtained it is then multiplied by area that is available at that particular distance for a given
electron. The representation is significant in atomic orbitals as location of electron can.
2.) Principal quantum number characterizes principal energy level or orbital where electrons
revolve in an atom as it moves round the nucleus. Angular quantum number defines shape of the
orbitals where each value denotes specific s, p or d subshell and hence determines angular

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distribution and depends on principal quantum number. Magnetic quantum number describes the
sum of orbitals and their alignment in a subshell. The electron spin ms describes the spin of
electrons
When n=3, then the angular quantum number l= 0, 1, 2, 3. The magnetic quantum number ml = -
3, -2, -1, 0, 1, 2, 3. The spin quantum number are either ± ½.
Rnl ( r )=2( z
a0
)
3/ 2
e−Zr/ a 0
P ( r ) =r 2 Rnl ( r ) 2
4 ¿
dP ( r )
dr =8( z
a0
¿¿¿ 3( r−Z r2 /a0)e
−Zr
a 0 )
(r − Z r2
a0 )=0
r =a0 / z
for lithium= 52.9
3 =17.6 pm
Radius of the 1s electron is higher than that of lithium suggesting that for a fixed principal
quantum number n the probability decreases from s > p > d > f

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4. Selection rules are used to describe the possibility of a transition occurring from one quantum
state to another state. For a hydrogen atom and considering transition from a 2p orbital, allowed
transitions are the 2p orbital to 1s orbital where Δ n=1 , Δl=−1∧∆ J =0∨1
5.) Molecular Orbital Theory- Diatomic Molecule
1.) N2
Bond order = (8 -2)/2 = 3 (all electrons are paired diamagnetic)
b.) N2 +1

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B.O = (7-2)/2 = 2.5 (unpaired electrons -paramagnetic)
c.) N2+2
B.O = (6-2)/2 = 2 (Paired electrons – diamagnetic)
d.) N2-1

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B.O = (8-3)/2 = 2.5 (unpaired electrons -paramagnetic)
2. Lewis structure for NO
It is hard for NO to achieve Lewis stability

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B.O = (10-5)/2 = 2.5
The octet rule is not achieved in the N-O molecule. The total number of valence electrons is 11
therefore, one atom will not obey the octet rule. From the molecular orbital diagram N-O has a
higher bond order of 2.5 which signifies more stability as electrons are held together tightly.
6. Symmetry
(a) Nitrate anion D3h
(b) HCl C∞ V

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(c) Benzene D6h
(d). CO2 D∞ h
(e). PF3 C3v
(f) Alanine C1
(g) Silane, SiH4 C2V
(h) Ni(CO)4 D4h
Maximum orbital degeneracy of orbitals in the following point groups
(a) C2v 3 degenerate orbitals
(b) Td 5 degenerate orbitals
(c) D6h 2 maximum degenerate orbitals
(d) D2h non degenerate
7. Molecular Orbital Theory
Illustrate the possible group hydrogen orbitals in the following. Indicate to which
symmetry species each group orbital belongs.
(a) Water, H2O has the C2 , σ y ( xz ) ∧σ y ( yz )
(b) NH3 has five symmetry species that is the C3 , E ,∧three σ v

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(c) BH3 (in this case we can pretend that this molecule exists - in reality only B2H6 exists)
C3 , 3x C2 , 2X S3 , 2X σ v
2. MO energy diagram
(a) silane, SiH4

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(b) PH3
8 Bonus Question
1.) En =n2h2/ (8mL2)
13 × (6.6261×10-34)2 / (8× 1.40×10-10 × 9.11 × 10-31)
= 3.0761× 10-23 J
V = ∆ E
h
3.0761×10−23 J
6.6261× 10−34 =4.638× 1010 Hz

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As the number of conjugations of polyene increases so does the frequency. The frequency shifts
to a higher, that is the frequency increases
2. Draw a molecular orbital energy diagram for CS (carbon sulfide). Compare the outcome with
CO, and predict the coordination chemistry of CS when binding to a metal.
CO and CS form almost the same molecular orbitals. S is less electronegative than O so the CO
will be highly stable than CS and with lower in energy. In coordination chemistry, CS will thus
form complexes easily than CO utilizing the free 3p orbitals that are not yet filled.

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Reference
Kasap, S., Koughia, C., & Ruda, H. E. (2017). Electrical conduction in metals and
semiconductors. In Springer Handbook of Electronic and Photonic Materials (pp. 1-1).
Springer, Cham.
Zong, Y., Shao, H., Pang, Y., Wang, D., Liu, K., & Wang, L. (2016). Multicomponent hydrogen-
bonding organic solids constructed from 6-hydroxy-2-naphthoic acid and N-heterocycles:
Synthesis, structural characterization and synthon discussion. Journal of Molecular
Structure, 1115, 187-198.