Question

3
(a) Given below are some groups of three elements with their atomic
masses within parentheses. These elements are arranged downwards
in order of increasing atomic masses. Find out which of these groups
form Dobereiner triads?
Group A elements Group B elements Group C elements
Li (7)
N (14)
CI (35.5)
Na (23)
P (31)
K (39)
As (75)
I (127)
Br (80) =​

Answer 1

Answer:

group A Group B elements Groups c elements

Answer 2

Answer: The elements in the Dobereiner triads are as follows:

Li, Na, K and Cl, Br, I

Explanation:

The chemist Johann Wolfgang Dobereiner discovered Dobereiner's triads, a set of elements with similar properties.

• He noticed that triads (groups of three elements) could be constructed with all of the elements having comparable physical and chemical properties.

The arithmetic mean of the atomic masses of the first and third elements in a triad is approximately equal to the atomic masses of the second element in the triad, according to the law of triads.

In the first triad Li, Na, K

• The atomic mass of Li= 7 and the atomic mass of K =39
• Therefore the arithmetic mean = $\frac{7+39}{2}$ =23 which is equal to the atomic mass of Na

In the second triad Cl, Br, I

• The atomic mass of Cl=35.5 and the atomic mass of I= 127
• Therefore the arithmetic means = $\frac{35.5+127}{2}$ =81.25 which is approximately equal to the atomic mass of Br.

So the Doberenier triads of this question are Li, Na, K and Cl, Br, I