Three elements $A, B$ and $C$ belong to the same group of the modern periodic table such that the difference in atomic weight of $A$ and $B$ is equal to the difference in atomic weight of $B$ and $C$ . The sum of atomic weights of elements $A$ and $C$ is 46 and that of $A$ and $B$ is 30. Identify elements $A, B$ and $C$ .

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Solution

Take A, B, and C in ascending order of atomic weights.
Difference between weights (atomic) of A and B= Difference between atomic weights of B and C.
$A_{w}, B_{w}, C_{w}$ represent their atomic weights
$B_{w} - A_{w} = C_{w} - B_{w}$
$2 B_{w} = A_{w} + C_{w}$
$2 B_{w} = 46$ [Given, sum of atomic weight of A and C is 46]
$B_{w}$ = 23
B=Sdium
$A_{w} + B_{w} = 30$ $A_{w} + C_{w} = 46$
$A_{w}$ +23=30 7+ $C_{w}$ =46
$A_{w}$ =7 $C_{w}$ =39
A=Lithium C=Potassium
A=Lithium B=Sodium C=Potassium

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Three elements A,B and C belong to the same group of the modern periodic table such that the difference in atomic weight of A and B is equal to the difference in atomic weight of B and C. The sum of atomic weights of elements A and C is 46 and that of A and B is 30. Identify elements A,B and C.