A salt (AB) when dissolved in water yields A⁺ and B^- ions. If both the ions have the same number of electrons and the molecular weight of salt is 74.5, then the positions of A and B in the periodic table are:

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A= Group 1, 4^th period B= Group 17, 3^rd period

A= Group 2, 4^th period B= Group 17^th, 4^th period

A= Group 17, 3^rd period B= Group 1, 4^th period

A= Group 17, 4^th period B= Group 2, 3^rd period

answer is A.

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Detailed Solution

Element A is present in group 1 and 4th period whereas, B is present in group 17th and 3rd period.
Valency of element A (cation) = 1
Valency of element B (anion) = 1
So, A loses one electron, and B gains one electron to achieve noble gas configuration. Therefore, A should be present in group 1 and B should be present in group 17.
To achieve the nearest noble gas configuration, A and B form a stable salt. The molecular weight of the salt when divided by 2, gives an idea of the near noble gas.
So,

\frac{74.5}{2}

= 37. 25 which is close to the molecular weight of Argon (40).
So, element A is present in group 1 and 4th period whereas, B is present in group 17th and 3rd period.

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