# All About Electronic configuration of atoms

Electronic configuration, also known as electron structure, is the arrangement of energy levels of electrons around the nucleus. As far as finer quantum mechanical models are concerned, the K-Q shell is subdivided into a set of orbits (see orbits), each of which can be occupied by no more than a pair of electrons. The electron configuration of an atom in the shell atom model can be expressed by indicating the number of electrons in each shell starting from the first.

### Example of Electronic Configuration

Formula = 2n2

 Shell and ‘n’ value Max Electrons in the Electron Configuration K shell, n = 1 2 x 12 = 2 L shell, n = 2 2 x 22 = 8 M shell, n = 3 2 x 32 = 18 N shell, n = 4 2 x 42 = 32

1. Types of bonds

Example of ionic or electrovalent bonding

1. Na +          cl                     Na+ + cl

152 252 2P6 351                   152 252 2P6 352 3P5

Nax                  +          cl                     Na+ + xcl

1. Chemical combination
2. Weak bonds e.g hydrogen bond, van-der waals forces, etc.
3. System of Naming compound – conventional IUPAC

1. Chemical combination
2. States of matter – solid, liquid gaseous state
3. The kinetic theory and its application

Atom to determine Empirical formula

1. Begin with the number of grams of each element which you usually find in any given problem
2. Us the molar mass you get by adding up the atomic weight of the elements from the periodic table to convert the mass of each element into moles
3. Divide each mole value by the small number of moles you obtained from your calculation
4. Round each number you get to the nearest whole number are the mole ratio of elements in the compound, which are the subscript numbers that follow the element symbols in the chemical formulas

Note: To determine whole number ration is tricky and you’ll need to use this formula for all numbers ending with 0.5 or .5 multiply by 2 if you get 0.25 or .25 multiply by 4

Example: A compound is analyzed and calculated to consist of 13.5 ca, 10.8g O, and 0.67sg H, find the empirical formula.

Note: Masses for these elements as:. Ca = 40.1

.16.0 for O, .1.01 for 4

Ca                        H                     O

13.5                    0.675             10.8

40                           1                    16

-> .0.337           0.668             0.675

Divide by smallest mole

0.337                 0.675             0.675

0.337                 0.337             0.337

1                          2                      2

:. CaO2H2

1. Symbols, formulae and equation
2. Law of constant composition
3. Law of multiple proportions
• Chemical equations

Example of constant composition: Calculate % composition of each element in each of the following compound

• H2SO4 => where H = 1, S = 32, O = 16

Solution

(1 x 2) + 32 + (16 x 4) = 98

% of H = 1 x 2 x 100% = 2.04%

98

% of O = 16 x 4 x 100% = 65.32%

98

% of S =    32      x 100% = 32.65%

98

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