Question

The electron arrangement of atoms of three elements A, B and C is : A (2, 8, 1); B (2, 8, 6); C (2, 8, 18,7).(a) Write down the formula of the molecule of B and its electron dot diagram. Mention the type of bonding.(b) Write down the formula of the compound formed between A and C and type of bonding.​

Answer 1

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Answer 2

(a) THE FORMULA OF MOLECULE OF B IS  "$S_{8}$". IT HAS COVALENT BONDING.

(b) FORMULA OF COMPOUND FORMED BY A AND C IS "AC". IT FORMS IONIC BOND.

Explanation:

• The electronic configurations of A (2, 8, 1); B (2, 8, 6); C (2, 8, 18,7) suggests that A is sodium, B is Sulphur, and C is Bromine.
• Sulphur forms a molecule which is in the shape of a ring, it contains 8 atoms attached to each other through covalent bonds. It forms  $S_{8}$. It has 2 electrons in K shell, 8 in L shell, and 6 in M.
• A can lose 1 electron as it has one valence electron and it will lose it to complete its octet. C has a shortage of electron so it will take it from A, and forms an ionic compound having formula AC.

(a) $S_{8}$ . Covalent bonding.

(b) AC. Ionic bond.