List the bonding pairs H and I; S and O; K and Br; Si and Cl, H and F; Se and S; C and H in order of increasing covalent character. 1. (S, O) < (Se, S) < (C, H) = (H, I) = (H, F) < (Si, Cl) < (K, Br) 2. (C, H) = (H, I) = (Se, S) < (K, Br) < (S, O) = (Si, Cl) < (H, F) 3. (K, Br) < (Si, Cl) < (H, F) = (H, I) = (C, H) < (Se, S) < (S, O) 4. (K, Br) < (H, F) < (Si, Cl) < (S, O) < (H, I) = (C, H) < (Se, S) 5. (Se, S) < (C, H) < (H, I) < (S, O) < (Si, Cl) = (H, F) < (K, Br) 6. (H, F) < (Si, Cl) = (S, O) < (K, Br) < (Se, S) = (H, I) = (C, H) 7. None of these

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Chemistry

Beetle

5 June, 08:04

List the bonding pairs H and I; S and O; K and Br; Si and Cl, H and F; Se and S; C and H in order of increasing covalent character. 1. (S, O) < (Se, S) < (C, H) = (H, I) = (H, F) < (Si, Cl) < (K, Br) 2. (C, H) = (H, I) = (Se, S) < (K, Br) < (S, O) = (Si, Cl) < (H, F) 3. (K, Br) < (Si, Cl) < (H, F) = (H, I) = (C, H) < (Se, S) < (S, O) 4. (K, Br) < (H, F) < (Si, Cl) < (S, O) < (H, I) = (C, H) < (Se, S) 5. (Se, S) < (C, H) < (H, I) < (S, O) < (Si, Cl) = (H, F) < (K, Br) 6. (H, F) < (Si, Cl) = (S, O) < (K, Br) < (Se, S) = (H, I) = (C, H) 7. None of these

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June · Answer 1

1. (S, O) < (Se, S) < (C, H) = (H, I) = (H, F) < (Si, Cl) < (K, Br)

Explanation:

The covalent character always increases down the group, this is because ionic character decreases down the group and also electronegativity.

In the same way, Covalent character always decreases across a period because electronegativity increases across a period.

The higher the electronegativity values between the two atoms, the more ionic it will be.