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Question
An element ‘A’ has mass number 23 and atomic number 11. State the –
- no. of neutrons in its shell,
- electronic configuration of the element ‘A’.
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Solution
Mass number A = n + p (atomic number)
A = n + Z
23 = n + 11
∴ n = 23 – 11 = 12
(1) Number of neutrons = 12
(2) e = p = 11
∴ Number of electrons e = 11
∴ Electron configuration of A = 2, 8, 1
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With the help of given Table, find out the mass number of oxygen.
| 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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