Question

An example of Dobereiner’s triad is ----
A. $Li,Al,Ca$
B. $Li,Na,K$
C. $Li,K,Na$
D..$K,Al,Ca$

Answer 1

Dobereiner’s triad requires three elements that have the relation where in the average atomic mass of the first and third elements can find the atomic mass of the central atoms. The three elements also have similar properties.

Complete step by step answer:
Dobereiner attempted to classify the discovered elements at the time. He tried doing this by grouping certain elements in groups of three. This grouping was very specific. This concept of classification of elements in a particular order was a relatively new concept and Dobereiner’s attempt was one of the first attempts.
The elements in a triad were grouped according to their atomic masses. This categorization was done in such a way that if the average of the first and third element from the triad was considered it would give the atomic mass of the second element in the triad. The elements in the triad also had some shared chemical properties.
Now if we look at the question given; we need to find a triad where the atomic masses follow the condition mentioned above.
In the first option the Lithium atomic mass is $7$ . aluminium is $27u$ and Calcium is $40u$ .
Therefore, we have to take the average of the atomic mass of Lithium and calcium. We get,

& m = \dfrac{{7 + 40}}{2} \cr & \Rightarrow \dfrac{{47}}{2} \cr & \Rightarrow 23.5u \cr} $$ This is the atomic mass obtained. But the atomic mass of Aluminium is $27u$ . therefore, this cannot be the answer. In the second option, we have Lithium again whose atomic mass number is already mentioned. Sodium has an atomic mass number of $23u$ . And finally, Potassium which has an atomic number of $39u$ . Therefore, taking the average we get, $m = \dfrac{{7 + 39}}{2}$ $ \Rightarrow \dfrac{{48}}{2}$ $ \Rightarrow 24u$ The atomic mass is close to the atomic mass of sodium. In the third option we get, Lithium and Sodium’s atomic mass must now be added together and the arithmetic mean must be obtained to get a number close to the atomic mass of Potassium. This is demonstrated below: $m = \dfrac{{7 + 23}}{2}$ $ \Rightarrow \dfrac{{30}}{2}$ $ \Rightarrow 15u$ Thus, the atomic mass obtained is nowhere near the atomic mass of potassium. The last option contains Potassium, Aluminium and Calcium. The atomic mass of aluminium is $27u$ . therefore, the average atomic mass of $K$ and $Ca$ is $m = \dfrac{{39 + 40}}{2}$ $ \Rightarrow \dfrac{{79}}{2}$ $ \Rightarrow 39.5u$ Therefore, the average is not equal to the actual atomic mass. Therefore, it will not be the answer. Since the second option is the only option that fulfils the required conditions, we can say that $Li,Na,K$ form a triad. **So, the correct answer is Option B.** **Note:** The arithmetic mean or average of the first and third elements may or may not be equal to the atomic mass of the second element. But the value should be very close. This concept can be used to classify elements into triads. One of the drawbacks in this method was that only a few elements could be classified in this way.