E G I R P

voltage source conductance current resistance power

[volts, V] [siemens, S] [amps, A] [ohms, W] [watts]

V X Y Z

voltage drop reactance admittance impedance

[volts, V] [ohms, W] [siemens, S] [ohms, W]

Similarly, when a voltage E is applied across an impedance Z, the resulting current I through the impedance is equal to the voltage E divided by the impedance Z.

Current = Voltage / Impedance

I = E / Z

Similarly, when a current I is passed through an impedance Z, the resulting voltage drop V across the impedance is equal to the current I multiplied by the impedance Z.

Voltage = Current * Impedance

V = IZ

Alternatively, using admittance Y which is the reciprocal of impedance Z:

Voltage = Current / Admittance

V = I / Y

Similarly, at any instant the algebraic sum of all the currents at any circuit node is zero:
SI = 0

Kirchhoff's Voltage Law
At any instant the sum of all the voltage sources in any closed circuit is equal to the sum of all the voltage drops in that circuit:
SE = SIZ

Similarly, at any instant the algebraic sum of all the voltages around any closed circuit is zero:
SE - SIZ = 0

Any linear voltage network which may be viewed from two terminals can be replaced by a voltage-source equivalent circuit comprising a single voltage source E and a single series impedance Z. The voltage E is the open-circuit voltage between the two terminals and the impedance Z is the impedance of the network viewed from the terminals with all voltage sources replaced by their internal impedances.

Any linear current network which may be viewed from two terminals can be replaced by a current-source equivalent circuit comprising a single current source I and a single shunt admittance Y. The current I is the short-circuit current between the two terminals and the admittance Y is the admittance of the network viewed from the terminals with all current sources replaced by their internal admittances.

The open circuit, short circuit and load conditions of the Norton model are:
V_oc = I / Y
I_sc = I
V_load = I / (Y + Y_load)
I_load = I - V_loadY

Th?venin model from Norton model

Voltage = Current / Admittance Impedance = 1 / Admittance

E = I / Y Z = Y -1

Norton model from Th?venin model

Current = Voltage / Impedance Admittance = 1 / Impedance

I = E / Z Y = Z -1

When performing network reduction for a Th?venin or Norton model, note that:
- nodes with zero voltage difference may be short-circuited with no effect on the network current distribution,
- branches carrying zero current may be open-circuited with no effect on the network voltage distribution.

In a linear network with multiple voltage sources, the current in any branch is the sum of the currents which would flow in that branch due to each voltage source acting alone with all other voltage sources replaced by their internal impedances.

If a voltage source E acting in one branch of a network causes a current I to flow in another branch of the network, then the same voltage source E acting in the second branch would cause an identical current I to flow in the first branch.

If the impedance Z of a branch in a network in which a current I flows is changed by a finite amount dZ, then the change in the currents in all other branches of the network may be calculated by inserting a voltage source of -IdZ into that branch with all other voltage sources replaced by their internal impedances.

If any number of admittances Y₁, Y₂, Y₃, ... meet at a common point P, and the voltages from another point N to the free ends of these admittances are E₁, E₂, E₃, ... then the voltage between points P and N is:
V_PN = (E₁Y₁ + E₂Y₂ + E₃Y₃ + ...) / (Y₁ + Y₂ + Y₃ + ...)
V_PN = SEY / SY

The short-circuit currents available between points P and N due to each of the voltages E₁, E₂, E₃, ... acting through the respective admitances Y₁, Y₂, Y₃, ... are E₁Y₁, E₂Y₂, E₃Y₃, ... so the voltage between points P and N may be expressed as:
V_PN = SI_sc / SY

When a current I is passed through a resistance R, the resulting power P dissipated in the resistance is equal to the square of the current I multiplied by the resistance R:
P = I²R

By substitution using Ohm's Law for the corresponding voltage drop V (= IR) across the resistance:
P = V² / R = VI = I²R

Note that power is zero for an open-circuit (zero current) and for a short-circuit (zero voltage).

Voltage Source
When a load resistance R_T is connected to a voltage source E_S with series resistance R_S, maximum power transfer to the load occurs when R_T is equal to R_S.

Under maximum power transfer conditions, the load resistance R_T, load voltage V_T, load current I_T and load power P_T are:
R_T = R_S
V_T = E_S / 2
I_T = V_T / R_T = E_S / 2R_S
P_T = V_T² / R_T = E_S² / 4R_S

Current Source
When a load conductance G_T is connected to a current source I_S with shunt conductance G_S, maximum power transfer to the load occurs when G_T is equal to G_S.

Under maximum power transfer conditions, the load conductance G_T, load current I_T, load voltage V_T and load power P_T are:
G_T = G_S
I_T = I_S / 2
V_T = I_T / G_T = I_S / 2G_S
P_T = I_T² / G_T = I_S² / 4G_S

Complex Impedances
When a load impedance Z_T (comprising variable resistance R_T and variable reactance X_T) is connected to an alternating voltage source E_S with series impedance Z_S (comprising resistance R_S and reactance X_S), maximum power transfer to the load occurs when Z_T is equal to Z_S^* (the complex conjugate of Z_S) such that R_T and R_S are equal and X_T and X_S are equal in magnitude but of opposite sign (one inductive and the other capacitive).

When a load impedance Z_T (comprising variable resistance R_T and constant reactance X_T) is connected to an alternating voltage source E_S with series impedance Z_S (comprising resistance R_S and reactance X_S), maximum power transfer to the load occurs when R_T is equal to the magnitude of the impedance comprising Z_S in series with X_T:
R_T = |Z_S + X_T| = (R_S² + (X_S + X_T)²)^*
Note that if X_T is zero, maximum power transfer occurs when R_T is equal to the magnitude of Z_S:
R_T = |Z_S| = (R_S² + X_S²)^*

When a load impedance Z_T with variable magnitude and constant phase angle (constant power factor) is connected to an alternating voltage source E_S with series impedance Z_S, maximum power transfer to the load occurs when the magnitude of Z_T is equal to the magnitude of Z_S:
(R_T² + X_T²)^* = |Z_T| = |Z_S| = (R_S² + X_S²)^*

Similarly, using admittances:
Y_AB = Y_ANY_BN / (Y_AN + Y_BN + Y_CN)
Y_BC = Y_BNY_CN / (Y_AN + Y_BN + Y_CN)
Y_CA = Y_CNY_AN / (Y_AN + Y_BN + Y_CN)

In general terms:
Z_delta = (sum of Z_star pair products) / (opposite Z_star)
Y_delta = (adjacent Y_star pair product) / (sum of Y_star)

Similarly, using admittances:
Y_AN = Y_CA + Y_AB + (Y_CAY_AB / Y_BC) = (Y_ABY_BC + Y_BCY_CA + Y_CAY_AB) / Y_BC
Y_BN = Y_AB + Y_BC + (Y_ABY_BC / Y_CA) = (Y_ABY_BC + Y_BCY_CA + Y_CAY_AB) / Y_CA
Y_CN = Y_BC + Y_CA + (Y_BCY_CA / Y_AB) = (Y_ABY_BC + Y_BCY_CA + Y_CAY_AB) / Y_AB

In general terms:
Z_star = (adjacent Z_delta pair product) / (sum of Z_delta)
Y_star = (sum of Y_delta pair products) / (opposite Y_delta)

y₁ - y₄ represent solutions of increasing energy. In three dimensions, the equation determines the energy and defines the spatial distribution of each electron. Solutions of the wave equations in three-dimensions allows calculation of the "shape" of each orbital. The first five solutions of the wave equation for an electron associated with a proton can be shown in the figure below:

In the hydrogen atom, the 1s atomic orbital has the lowest energy, while the remainder (2s, 2p_x, 2p_y and 2p_z) are of equal energy (ie.degenerate), but for all other atoms, the 2s atomic orbital is of lower enegry than the 2p_x, 2p_y and 2p_z orbitals, which are degenerate.
In atoms, electrons occupy atomic orbitals, but in molecules they occupy similar molecular orbitals which surround the molecule. The simplest molecule is hydrogen, which can be considered to be made up of two seperate protons and electrons. There are two molecular orbitals for hydrogen, the lower energy orbital has its greater electron density between the two nuclei. This is the bonding molecular orbital - and is of lower energy than the two 1s atomic orbitals of hydrogen atoms making this orbital more stable than two seperated atomic hydrogen orbitals. The upper molecular orbital has a node in the electronic wave function and the electron density is low between the two positively charged nuclei. The energy of the upper orbital is greater than that of the 1s atomic orbital, and such an orbital is called an antibonding molecular orbital.
Normally, the two electrons in hydrogen occupy the bonding molecular orbital, with anti-parallel spins. If molecular hydrogen is irradiated by ultra-violet (UV) light, the molecule may absorb the energy, and promote one electron into its antibonding orbital (s^*), and the atoms will seperate. The energy levels in a hydrogen molecule can be represented in a diagram - showing how the two 1s atomic orbitals combine to form two molecular orbitals, one bonding (s) and one antibonding (s^*). This is shown below - by clicking upon either the s or s^* molecular orbital in the diagram - it will show graphically in a window to the right:

These are molecules in which all valence electrons are involved in the formation of single bonds. There are no non-bonded lone pairs. These molecules are generally less reactive than either electron-rich or electron-deficient species, with all occupied orbitals having relatively low energies.

Methane:

The valence molecular orbitals of methane are delocalized over the entire nuclear skeleton - that is, it is not easy to assign any one orbital to a particular C-H bond. It is possible to see how complex the orbital structure becomes with the increase in energy. Methane has four valence molecular orbitals (bonding), consisting of one orbital with one nodal plane (lowest occupied) and three degenerate (equal energy) orbitals that do have a nodal plane.
For the energy diagram and pictorial view of the orbitals - please see below:

Ethane:

The ethane molecule has fourteen valence electrons occupying seven bonding molecular orbitals. As can be seen from the energy diagram - four of the molecular orbitals occur as degenerate pairs. Like in methane - the molecular orbitals of ethane show increasing nodal structure with increasing orbital energy.
For the energy diagram and pictorial view of the orbitals - please see below:

Ethene:

The simplest alkene is ethene. Its chemistry is dominated by two "frontier orbitals", that is the Highest Occupied Molecular Orbital (HOMO) and the Lowest Unoccupied Molecular Orbital (LUMO). For the ethene orbital energy diagram these are shown as p_CC for the HOMO, and p^*_CC for the LUMO.
An important property of the ethene molecule, and alkenes in general is the existence of a high barrier to rotation about the C=C which tends to hold the molecule flat.
For the energy diagram and pictorial view of the orbitals - please see below:

For the energy diagram and pictorial view of the orbitals - please see below:

A simple diatomic molecule is Hydrogen fluoride. There are eight valence electrons which occupy four molecular orbitals. The two highest energy MO's are degenerate, are p-type and have no electron density associated with the hydrogen atom, ie. they are Non-Bonding Orbitals (NBO) and in Lewis Theory are represented as two "Lone Pairs". Another important difference between Hydrogen Fluoride and previous molecules is that the electron density is not equally distributed about the molecule. There is a much greater electron density around the fluorine atom. This is because fluorine is an exremely electronegative element, and in each bonding molecular orbital, fluorine will take a greater share of the electron density.

For the energy diagram and pictorial view of the orbitals - please see below:

Water:

In the water molecule the highest occupied orbital, (1b₁) is non-bonding and highly localized on the oxygen atom, similar to the non-bonding orbitals of hydrogen fluoride. The next lowest orbital (2a₁) can be thought of as a non-bonding orbital, as it has a lobe pointing away from the two hydrogens. From the lower energy bonding orbitals, it is possible to see that oxygen also takes more than its "fair share" of the total electron density.

Ammonia:

Ammonia has two pairs of degenerate orbitals, one bonding and one antibonding, and like hydrogen fluoride and water has a non-bonding orbital (2a₁). This highest occupied orbital has a lobe pointing away from the three hydrogens, and corresponds to a lone pair orbital localized upon the nitrogen, whereas the three lowest energy MO's lead to the description of the three N-H bonds of the Lewis structure. The lone pair is relatively high in energy, and is responsible for the well known Lewis base properties of ammonia.

The next molecule in the series HF, H₂O and H₃N, is H₄C (methane) - which was discussed earlier - and unlike the other three molecules has no non-bonding orbitals.

Aromatic molecules exhibit a wide range of reactivity patterns toward both electron rich and electron deficient species. These mainly depend on the structures and energies of the frontier p-type molecular orbitals, the HOMO and LUMO. Except for non-bonded lone pairs, the s framework plays little role in the overall reactivity.

trans-1,3-Butadiene:

The energies of the p-molecular orbitals of conjugated molecules like butadiene, (see below) - occur in pairs, with their energies equal to (a±xb), where a and b are constants. For each bonding orbital of and energy a-xb there is a corresponding antibonding orbital of energy a+xb. The p-molecular orbitals are extended over the whole molecule.
For butadiene, the p manifold contains four electrons, leading to an electronic configuration of p₁²p₂².
For the energy diagram and pictorial view of the p-molecular orbitals - please see below:

Allyl radical

The radicals allyl:

and pentadienyl:

have the same arrangement of p-orbitals, (ie. the occur in pairs of energy a±xb), but because there is an odd number of carbon atoms in the conjugate chain, there must be a non-bonding orbital with energy x=0. Also, because of the pairing properties of the p-molecular orbitals of conjugated chains, there will be a node at every alternate carbon atom in the non-bonding orbital. This is important for the unpaired electron of allyl, which will occupy this non-bonding orbital. If an electron is added to the allyl radical to form the anion, the negative charge will appear at the terminal carbon atoms. If the unpaired electron is removed forming the cation, the resulting positive charge is also spread over the termial carbon atoms.
There are three p-molecular orbitals for allyl, the p₁ is bonding, the p₂ orbital is non-bonding and the p₃ is anti-bonding. In the neutral allyl species - there are a total of seventeen valence electrons - of which three fill the p-orbital manifold. A pictorial representation of the energy diagram for the neutral, cationic and anionic allyl species are shown below - (orbitals are shown only for the cationic species):

In the pentadienyl anion, the negative charge is centred on the carbon atoms in the 1,3 and 5 position - similarly with the positive charge for the cation.

These ions are represented in resonance theory as two or three canonical forms:

The delocalisation of p-electrons is associated with a lowering of the orbital energy. Therefore the total energy of the occupied p-orbitals of butadiene is lower in energy then two isolated ethene-type double bonds. Further delocalisation of p-electrons occurs in aromatic hydrocarbons. A figure showing the comparative energy levels of the p-orbitals of the cyclic molecules C_nH_n, for n=3-6, is shown below:

In all cyclic polyenes (C_nH_n), the p-molecular orbitals occur in degenerate pairs, except for the lowest p-orbital, and for the cyclic polyenes with even numbers of carbon atoms, the highest p-orbital (see above).

Cyclobutadiene:

From the cyclic polyene diagram - the square molecule cyclobutadiene (C₄H₄) has four p-orbitals, a bonding orbital (p₁), two degenerate non-bonding orbitals (p₂ and p₃) and an anti-bonding orbital (p₄). Four electrons are placed into these four orbitals; twon into the bonding orbital, and one each with parallel spins into the degenerate non-bonding orbitals (Hund's rule) - see below:

There is no reason to expect cyclobutadiene to be square, theoretically calulations show an oblong with two double bonds structre has lower energy. Experimentally it is shown that cyclobutadiene acts a very strained cyclo-olefin, rather than as a bi-radical species.

Cyclopentadiene:

This molecule is a relatively acidic hydrocarbon, and the anion is formed by the treatment of cyclopentadiene with a strong base. From the cyclic polyene diagram it can be seen that cyclopentadiene has three p bonding orbitals which are delocalised over the five carbon atoms. The uppermost p bonding orbitals are a degenerate pair, and are the highest occupied molecular orbitals (HOMO's).
These orbitals are much higher in energy than those in neutral aromatic species such as benzene, indicating that this anion is far more susceptible t attack by electrophiles. The anion is far more capable of coordinating to transition metals with available empty d-orbitals. The anion has six p-electrons, making the system aromatic. The six electrons are arranged as in the diagram below:

Benzene:

Benzene is the archetypal aromatic compound. It has a symmetrical p system and so is not over reactive on any one site. From the cyclic polyene diagram it can be seen that benzene has six p-molecular orbitals, (which contain the six p-electrons), three bonding and three anti-bonding. The upper bonding degenerate pair of orbitals are the HOMO's of benzene. The p orbital manifold, is shown below - but also of interst is that the pattern of the p orbitals is repeated within the s system. The s functions - like the p orbitals are delocalized throughout the carbon skeleton.
The six p electrons are arranged as in the diagram below:

Bonding in benzene

From the above diagram it can be seen that the lowest lying orbital, p₁, the orbital coefficients are such that the bonding charachter between each pair of adjacent carbon atoms is equal. In p₂ bonding only occurs between atoms C₂ and C₃ and between C₅ and C₆ since the coefficients on C₁ and C₄ are zero. In p₃, C₁, which is bonded to C₂ and C₆ and C₄ is bonded to C₃ and C₅, there are anti-bonding interactions between C₂ and C₃ and between C₅ and C₆. Therefore if we consider the pair of orbitals p₂ and p₃ the contribution to the C-C p bonding is equal for each bond. Since there are three occupied bonding orbitals and six CC linkages - the p bond order is ¹/₂. This description is in accord with the two resonating mesomeric forms (or Kekulé structures in a) below in which single and double bond characters alternate around the ring. Conventionally, the diagram in b) is used to show that the six electrons are delocalized around the ring:

Nitrogen:

This molecule has ten electrons. The atomic orbitals combine to produce the following molecular orbital diagram:

Here the 2p_g orbital is occupied by two electrons to give a total bond order of three. This corresponds well with the Lewis structure (), although the orbital approach tells us that there is one s and two p.

Oxygen:

This molecule has twelve electrons, two more than nitrogen - and these extra two are placed in a pair of degenerate p_g orbitals. The atomic orbitals combine to produce the following molecular orbital diagram:

Comparison of the above energy level diagram wit hthat for nitrogen - you can see that the 2s_g level lies lower than p_u. Here, we are starting to fill the anti-bonding orbitals originating from the p orbital interactions and so the bond order decreases from three to two.
The lowest energy arrangment (Hund's rule) - has a single electron, each with parallel spins, in each of the p_g^x and p_g^y orbitals. This produces a paramagnetic molecule, with a double bond and has two unpaired electrons.

SOURCE:http://www.ch.ic.ac.uk/vchemlib/course/mo_theory/main.html