Question

A salt when dissolved in water dissociates into cation and anion. If both the ions consists of the same number of electrons and the molecular weight of salt is 74.5 . Then , identify the position of A and B in the periodic table.

Answer 1

Answer:

A is K (group 1A, 4th period) and element B is Cl(group VII A,3rd period).

Explanation:

• Since, both the ions consist of the same number of electrons and have +1 and -1 charges, hence the ions should belong to 1A(cation i.e.A⁺) and group VII A(anion i.e.B⁻).
• The same number of electrons indicates that their electronic configuration is the same as that of the noble gas whose atomic number lies between that of two elements A and B.
• Dividing the molecular weight (which is the sum of atomic masses of A and B), we get a rough idea about the atomic mass of noble gas which is,

                                              $\frac{74.5}{2}=37.25$

      i.e.near to argon(Ar-40).

• Hence, A is K (group 1A, 4th period) and element B is Cl(group VII A,3rd period).

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Answer 2

We can see that the given ions in the question have the same number of electrons along with +1 and -1 charges.

Therefore, A must belong to group 1A and B must belong to group 7A.

Here, the number of electrons is the same which implies that the electronic configuration is similar to that of the noble gas having an atomic number between these two given elements.

When we divide the molecular weight (total atomic mass of A and B), we will get an approximation of the atomic mass of the noble gas.

So,

$\frac {74.5}{2} = 37.25$

This means that it is nearest to the noble gas Argon.

Therefore, A is K having group 1A and period 4.

B is Cl having roup 7A and period 3.