Chem Revision

GENERAL CHEMISTRY

I. IDEAL GAS LAW:

Ideal gas: is a theoretical gas composed of randomly moving, non-interacting point particles.

nRT=PV

R = 0.082 atm.l/K (the same constant as R 8.314J/K)

Change of unit goes like this: P= F/S (N/m2 = 1kgm/(sm)2 = 1Pa= 101 325 atm)

=> R= 0.082atm.l/K * 101 325 pa/atm * 0.001m3/l = 8.314 J/K.

Real gas – Van der Waals model:

nRT=(P+a

V 2 )× (V −b )

In which, a and b are parameters that are determined empirically for each gas.

II. DETERMINE THE ATOMIC AND MOL AR MASS:

Boiling point elevation or freezing point depression:

M=1000×mct

k ×∆ t×mdd

Dulong and Petit method:

M×c=6.3 calmol×C

=26 Jmol×C

c: specific heat in J/g.C

Mass spectrometry: used to discover the mass/ charge of atoms. This can also be used to find different isotopes of an element and their percentages.

How it works:

Stage 1: IonizationThe atom is ionized by knocking one or more electrons off to give a positive ion (even for atoms which you normally expect to be negative: chlorine, oxygen). Mass spectrometers always work with positive ions.

Stage 2: AccelerationThe ions are accelerated so that they all have the same kinetic energy

Stage 3: Deflection

The ions are then deflected by a magnetic field according to their masses. The lighter they are the more they are deflected. The amount of deflection also depends on the number of positive charges on the ion – in other words, on how many electrons were knocked off in the first stage. The more ion is charged, the more it gets deflected.

Stage 4: DetectionThe beam of ions passing through the machine is detected electrically.

M atom

n=Ke r2×h×

H 2

V

n: Number of electron separate from the atomK: a constantr: radius of the curveH: Magnetic field intensityV: voltage

III. STRUCTURE OF ATOMS:

Heinsenberg uncertainty principle:

σ x×σ p≥h4 π

Slater’s rules [semi-empirical]: In a many-electron atom, each electron is said to experience less than the actual nuclear charge owing to shielding or screening by the other electrons. For each electron, Slater’s rules provide a value for screening constant, which relates the effective/actual nuclear charges:

Zeff=Z−s

The shielding constant is determined as followed.

Step 1: electrons are arranged into a sequence of group: [1s] [2s,2p] [3s,3p] [3d] [4s,4p] [4d] [4f] [5s, 5p] [5d] etc.Fe: [1s]2 [2s2p]8 [3s3p]8 [3d]6 [4s]2

Step 2:

GroupOther electrons

in the same group

Electrons in group(s) with

principal quantum

number n and azimuthal quantum

number < l

Electrons in group(s) with

principal quantum

number n-1

Electron in all group(s) with

principal quantum

number < n-1

[1s] 0.30 - - -

[ns np] 0.35 - 0.85 1

[nd] or [nf] 0.35 1 1 1

Ex: Fe4s : 0.35×1+0.85×14+1.00×10=22.25=¿Zeff (4 s )=26−22.25=3.753d : 0.35×5+1.00×18=19.75=¿Zeff (3d )=26−19.75=6.253s, 3p : 0.35×7+0.85×8+1.00×2=11.25=¿Zeff (3 s ,3 p )=26−11.25=14.752s, 2p : 0.35×7+0.85×2=4.15=¿Zeff (2 s ,2 p )=26−4.15=21.851s : 0.3×1=0.3=¿Zeff (1 s )=26−0.3=25.7

Application of STO’s (Slater Type Orbitals):

Calculation of the energy of any electron in any atom or ion by the formula:

E=−2.18×10−18×(Z¿

n¿ )2

J=−13.6×( Z¿

n¿ )2

eV

While the value of n* in accord to n:

n 1 2 3 4 5 6n* 1 2 3 3.7 4.0 4.2

Example: Predict the energy of a 4s electron in Fe:

E=−13.6×(3.753.7 )2

eV =−13.97eV

Example: Calculation of the first ionization energy (IE) of Fe

IE=EFe+¿−E Fe

0 ¿

EFe+¿ ¿

4s : 0.35×0+0.85×14+1.00×10=21.9=¿Zeff (4 s )=26−21.9=4.1

EFe0 =−13.6×[(3.753.7 )

2

×2+( 6.253 )2

×6+( 14.753 )2

×7+( 21.852 )2

×7+( 25.71 )2

×1]EFe

+¿=−13.6×[( 4.13.7 )2

× 1+( 6.253 )2

×6+( 14.753 )2

×7+( 21.852 )2

×7+( 25.71 )2

× 1]¿

IE=−13.6×[( 4.13.7 )2

×1−( 3.753.7 )2

×2]=11.24 eVIn the same sense, the electron affinity can be measured.

Relation between principal quantum number and energy level:

Solution for the Schrodinger shows that electron can only have some specific energy state. For atom with only one single electron, its energy equation is as follow:

En=−me4

8 ε02n2h2

×Z2=−2.18×10−18 Z2

n2J=−13.6 Z2

n2eV

IV. STRUCTURE OF MOLECULE:

VB and VSEPR theories:

Lewis theory and Lewis acid/base.

VSEPR theory:

Geometry notes on the theory:

1. Although lone pairs (unbounded electron pairs) are clearly smaller than atoms, they need to be closer to the nucleus of an atom than a bonding pair. Being closer to the central atom causes lone-pairs take up more of the available 'bonding space'.

2. The lone electron takes less space than a bond (Ex: NO2 has the O-N-O angle of 134o

3. Repulsion between Lone pair - Lone pair > Lone Pair - Bond pair > Bond pair - Bond pair => the change of shape when number of bonding electrons decrease (Ex: for d2sp3 octahedral square pyramidal Square planar …)

4. Triple bonds > double bonds > single bonds in taking up space (still not taking up as much space as lone pairs)

5. Electronegativity

The repulsion between electron pairs increases with increase in electronegativity of central atom and hence the bond angle increases. The bond pairs are closer and thus by shortening the distance between them, which in turn increases the repulsion. Hence the bonds tend to move away from each other.

However the bond angle decreases when the electronegativities of ligand atoms are more than that of central atom. There is increase in the distance between bond pairs since they are now closer to ligand atoms. Due to this, they tend to move closer resulting in the decrease in bond angle.

6. Size of atoms:The bond angle decreases with increase in the size of central atom.

The bond angle increases with increase in the size of ligand atoms, which surround the central atom.

7. Equatorial bonds are shorter than axial bonds.

What I’ve come across about MOLECULAR ORBITAL THEORY

Fig. 1: MO model for homogenous molecule in the end of the 2nd period (O2, F2)

Fig. 2: MO model for monatomic molecules in the beginning of the 2nd period (B2, C2, and N2) and diatomic molecules (NO, CO). The energy level of orbital shift because there is hybridization between 2s orbital on one atom to interact with the 2pz orbital on the other, which can affect the energy levels more on those in the beginning of the period because the difference in energy between 2s and 2pz is not too great.

After filling out electrons into the model, we can get the

Bond order:

Bondorder= Bondingelectrons−anitbonding electrons2

Oxygen paramagnetic property : a molecule is paramagnetic when it has unpaired electron(s) and is diamagnetic if it has no unpaired electron. Lewis structure can’t explain the paramagnetic property of oxygen (at least in its liquid form; for gas, the kinetic energy was too large for the magnetic field to have an effect on it) while MO model can.

Geometry of the MO orbitals:

Conjugated system MOs:

First let’s take a look at the formation of π2p and π2p*

Then the π bond MO of ethene from two 2p orbitals: There are two electron on the π2p

In the same sense, conjugated system can be constructed. The energy levels increase as the number of nodes increases.

Examples:

Butadiene MOs

A π bond and a lone pair conjugated system

Benzene MOs

Hexatriene

Hückel method/ How to draw these conjugated MOs correctly: