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Question

Which of the following forces of attraction acts among the atoms in a molecule of type B-A-B, where A and B belong to group 16 and group 17, respectively?


A

Dipole-dipole interaction

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B

Hydrogen bonding

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C

Van der Waals force

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D

Interionic attractions

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Solution

The correct option is C

Van der Waals force


Explanation for the correct option:

  • Group 16 elements are the Oxygen family in which Oxygen (O), Sulphur (S), Selenium(Se), Tellurium (Te) and Polonium (Po) are present. They are non-metals and are also known as Chalcogens.
  • Group 17 elements are the Halogens in which Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I) and Astatine (At). They are also non-metals and are highly electronegative elements.

(C) Van der Waals force: The type of the molecule is B-A-B, where A and B belongs to group 16 and group 17. Since both, groups are non-metals, the force of attraction between A and B will be Van der Waals forces because between two non-metals Van der Waals forces are present.

Explanation for the incorrect options:

(A) Dipole-dipole interaction: These forces of attraction occurs among the polar molecules. So, dipole-dipole interactions are formed between the atoms having a large difference in the electronegativities. Since the electronegativity of group 16 and group 17 have only small difference therefore, B-A-B cannot have dipole-dipole interaction.

(B) Hydrogen bonding: Hydrogen bonding is formed between the Hydrogen (H) atom and highly electronegative elements like Fluorine (F), Oxygen (O), and Nitrogen (N). Therefore, B-A-B cannot have Hydrogen bonding.

(D) Interionic attractions: Interionic attractions are formed between ions i.e., between cation and anion. The cation is a metal ion and the anion is a non-metal ion. For example, Sodium chloride (NaCl) contains interionic interactions as the cation is Sodium ion (Na+) and the anion is Chloride ion (Cl-).

Therefore, the correct option is (C).


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