Which of the following triads does not follow Dobereiner's law of triads? (a) \(\mathrm{Li}, \mathrm{Na}, \mathrm{K}\) (b) \(\mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}\) (c) Be, \(\mathrm{Mg}\), \(\mathrm{Ca}\) (d) \(\mathrm{Cu}, \mathrm{Ag}, \mathrm{Au}\)

We will start by calculating the average atomic mass for each triad and compare it to the atomic mass of the middle element. If the differences are large, it's likely that the triad doesn't follow Dobereiner's law. Triad (a) Li, Na, and K: \\ \(\text{Average Atomic Mass} = \frac{\text{Atomic Mass of Li} + \text{Atomic Mass of K}}{2} = \frac{6.94 + 39.1}{2} = 23.02\) Triad (b) Ca, Sr, and Ba: \\ \(\text{Average Atomic Mass} = \frac{\text{Atomic Mass of Ca} + \text{Atomic Mass of Ba}}{2} = \frac{40.08 + 137.3}{2} = 88.69\) Triad (c) Be, Mg, and Ca: \\ \(\text{Average Atomic Mass} = \frac{\text{Atomic Mass of Be} + \text{Atomic Mass of Ca}}{2} = \frac{9.01 + 40.08}{2} = 24.545\) Triad (d) Cu, Ag, and Au: \\ \(\text{Average Atomic Mass} = \frac{\text{Atomic Mass of Cu} + \text{Atomic Mass of Au}}{2} = \frac{63.55 + 197.0}{2} = 130.275\)

Now, we will compare the calculated average atomic mass with the atomic mass of the middle element in each triad. Triad (a) Li, Na, and K: \\ Difference between average atomic mass and atomic mass of Na = \(|23.02 - 22.99|\) = 0.03 Triad (b) Ca, Sr, and Ba: \\ Difference between average atomic mass and atomic mass of Sr = \(|88.69 - 87.62|\) = 1.07 Triad (c) Be, Mg, and Ca: \\ Difference between average atomic mass and atomic mass of Mg = \(|24.545 - 24.31|\) = 0.235 Triad (d) Cu, Ag, and Au: \\ Difference between average atomic mass and atomic mass of Ag = \(|130.275 - 107.87|\) = 22.405

Based on the above comparisons, we can see that Triad (d) Cu, Ag, and Au has the largest difference between the average atomic mass and the atomic mass of the middle element (Ag) with a difference of 22.405. All other triads have much smaller differences. Therefore, Triad (d) Cu, Ag, and Au does not follow Dobereiner's law of triads.