In a Dobereiner triad containing elements A, B and C, if the atomic mass of A is 7 and that of C is 39, what is the atomic mass of the middle element B?

Introduction / Context:
This question deals with Dobereiner triads, an early attempt to classify elements before the modern periodic table. The idea was that certain groups of three elements had similar properties and a simple numerical relationship between their atomic masses. Understanding how to use the Dobereiner triad rule helps you appreciate the development of periodic classification and shows how chemists first noticed patterns among the elements.

Given Data / Assumptions:

• Elements A, B and C form a Dobereiner triad.
• Atomic mass of element A = 7.
• Atomic mass of element C = 39.
• We are asked to find the atomic mass of the middle element B.
• We assume standard definition of Dobereiner triads from school level chemistry.

Concept / Approach:
In a Dobereiner triad, three elements with similar chemical properties are arranged in order of increasing atomic mass. The characteristic property is that the atomic mass of the middle element is approximately the arithmetic mean of the atomic masses of the first and third elements. In simple terms, mass of B = (mass of A + mass of C) / 2. This relation is only approximate for real elements, but in exam questions the numbers are chosen to work out exactly.

Step-by-Step Solution:
Step 1: Write down the given masses: A has atomic mass 7, C has atomic mass 39. Step 2: Apply the Dobereiner triad rule, which states that the mass of B is the average of the masses of A and C. Step 3: Use the formula: atomic mass of B = (atomic mass of A + atomic mass of C) / 2. Step 4: Substitute the given values: atomic mass of B = (7 + 39) / 2. Step 5: Compute the numerator: 7 + 39 = 46. Step 6: Divide by 2: 46 / 2 = 23. Step 7: Therefore the atomic mass of the middle element B is 23.

Verification / Alternative check:
You can verify by checking that the three numbers 7, 23 and 39 form a sequence where the middle value is exactly halfway between the first and the last. The difference between 7 and 23 is 16, and the difference between 23 and 39 is also 16. This symmetric spacing is another way to see the arithmetic mean relation. This property is exactly what Dobereiner observed in some sets of elements, such as lithium, sodium and potassium, where the atomic mass of sodium is roughly the average of lithium and potassium masses.

Why Other Options Are Wrong:
Option A, 20, does not satisfy the arithmetic mean relation because (7 + 39) / 2 is not 20. Option B, 40, is even larger than the mass of C, which contradicts the requirement that B is the middle element. Option D, 12, is too close to A and far from the average of 7 and 39. Option E, 28, also does not equal the computed mean of 23. Only 23 fits the rule that the mass of B equals the average of the masses of A and C.

Common Pitfalls:
A frequent mistake is to confuse Dobereiner triads with other periodic trends and simply choose the numerical middle of the options rather than calculate the arithmetic mean. Another pitfall is adding the masses and forgetting to divide by 2, which would give 46 instead of 23. Always remember that triads are based on an average, not on any geometric or harmonic mean in this syllabus. Writing out the simple formula before substituting values can help avoid errors.

Final Answer:
By applying the Dobereiner triad rule, the atomic mass of the middle element B comes out to be 23.