Question

When a Dobernier triad is considered, the sum of the atomic weights of the terminal elements 'X' and 'Z' is 162.5. The atomic weight of 'Z' is 50 units less than 5 times that of X. Calculate the atomic weights of X, Y, and Z. Identify their positions in the present periodic table.

Answer 1

Given X, Y, Z forms Dobernier triad.

So let X${}_W$, Y${}_W$, Z${}_W$ represents atomic weights of X,Y an Z.

Y${}_W$ = $\frac{X_W+Z_W}{2}$

Given X${}_W$ + Z${}_W$ =162.5

So Y${}_W$ =$\frac{162.5}{2}$ = 81.25 =Bromine Atomic weight

And Z${}_W$ is 50 units less than 5 times that of 'x'.

Z${}_W$ = 5X${}_W$ -50

X${}_W$ +Z${}_W$ =162.5

X${}_W$ +5X${}_W$ -50 =162.5

6X${}_W$ =212.5

X${}_W$ = 35.4 = Chlorine Atomic weight

Z${}_W$=127 = Iodine Atomic weight

Position of X is VII A group, 3rd period.

Position of Y is VII A group , 4th period.

Position Z is VII A group, 5th period.