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Question
If an atom ‘A’ has atomic number 19 and mass number 39, state –
- Its electronic configuration.
- The number of valence electrons it possesses.
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Solution
Atom ‘A’ has mass number A = 39 and atomic number Z = 19 = p
∴ A = Z + n
A = p + n
39 = 19 + n
n = 39 – 19 = 20
But e = p = 19
- A (K, L, M, N)
19 = 2, 8, 8, 1
There will be 2 electrons in K-shell or 1st shell
8 electrons in 2nd shell or L-shell
8 electrons in 3rd shell or M-shell
1 electron in 4th shell or nth-shell - The number of valence electrons i.e. in outer most shell = 1 electron.
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| Composition of Atoms of the First Eighteen Elements with Electron Distribution in Various Shells | ||||||||||
| Name of Element | Symbol | Atomic Number | Number of protons | Number of Neutrons | Number of Electrons | Distribution of Electrons | Valency | |||
| K | L | M | N | |||||||
| Hydrogen | H | 1 | 1 | - | 1 | 1 | - | - | - | 1 |
| Helium | He | 2 | 2 | 2 | 2 | 2 | - | - | - | 0 |
| Lithium | Li | 3 | 3 | 4 | 3 | 2 | 1 | - | - | 1 |
| Beryllium | Be | 4 | 4 | 5 | 4 | 2 | 2 | - | - | 2 |
| Boron | B | 5 | 5 | 6 | 5 | 2 | 3 | - | - | 3 |
| Carbon | C | 6 | 6 | 6 | 6 | 2 | 4 | - | - | 4 |
| Nitrogen | N | 7 | 7 | 7 | 7 | 2 | 5 | - | - | 3 |
| Oxygen | O | 8 | 8 | 8 | 8 | 2 | 6 | - | - | 2 |
| Fluorine | F | 9 | 9 | 10 | 9 | 2 | 7 | - | - | 1 |
| Neon | Ne | 10 | 10 | 10 | 10 | 2 | 8 | - | - | 0 |
| Sodium | Na | 11 | 11 | 12 | 11 | 2 | 8 | 1 | - | 1 |
| Magnesium | Mg | 12 | 12 | 12 | 12 | 2 | 8 | 2 | - | 2 |
| Aluminium | Al | 13 | 13 | 14 | 13 | 2 | 8 | 3 | - | 3 |
| Silicon | Si | 14 | 14 | 14 | 14 | 2 | 8 | 4 | - | 4 |
| Phosphorus | P | 15 | 15 | 16 | 15 | 2 | 8 | 5 | - | 3.5 |
| Sulphur | S | 16 | 16 | 16 | 16 | 2 | 8 | 6 | - | 2 |
| Chlorine | Cl | 17 | 17 | 18 | 17 | 2 | 8 | 7 | - | 1 |
| Argon | Ar | 18 | 18 | 22 | 18 | 2 | 8 | 8 | 0 | |
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Phosphorus – \[\ce{^31_15P}\]
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