Question

Identify the set of three elements which can obey the Law of triads.

• A. $\mathrm{Na},\mathrm{K},\mathrm{Rb}$
• B. $\mathrm{Be},\mathrm{Mg},\mathrm{Ca}$
• C. $\mathrm{F},\mathrm{Cl},\mathrm{Br}$
• D. $\mathrm{C},\mathrm{Si},\mathrm{Ge}$

Answer 1

The correct option is B

$\mathrm{Be},\mathrm{Mg},\mathrm{Ca}$

Explanation for correct option: (b)

1. According to Dobereiner's law of triads, when the elements are arranged in the increasing order of atomic masses the atomic mass of the central element will be approximately equal to the arithmetic mean of the atomic masses of the other two elements.
2. Here the atomic number of $\mathrm{Mg}$ is $12$ and it is the arithmetic mean of $\mathrm{Be}$ of atomic number $4$ and $\mathrm{Ca}$ of atomic number $20$.
3. The arithmetic mean of $\mathrm{Mg}$ is: $\frac{20+4}{2}=12$

Explanation for incorrect options:

a)

1. Here the atomic number of $\mathrm{K}$ is $19$ and it is not the arithmetic mean of $\mathrm{Rb}$ of atomic number $37$and $\mathrm{Na}$ of atomic number $11$.
2. The arithmetic mean of $\mathrm{K}$ is: $\frac{37+11}{2}=24$

c)

1. Here $\mathrm{F},\mathrm{Cl},\mathrm{Br}$ does not form a triad. Because the atomic number of $\mathrm{Cl}$ is $17$ and it is not the arithmetic mean of $\mathrm{F}$ of atomic number $9$ and $\mathrm{Br}$ of atomic number $35$.
2. The arithmetic mean of $\mathrm{Cl}$ is: $\frac{35+9}{2}=22$

d)

1. Here the atomic number of $\mathrm{Si}$ is $14$ and it is not the arithmetic mean of $\mathrm{C}$ and $\mathrm{Ge}$.
2. The arithmetic mean of $\mathrm{Si}$ is: $\frac{6+32}{2}=19$

Hence option (b) is correct and the elements $\mathrm{Be},\mathrm{Mg},\mathrm{Ca}$ form the Law of triads.