To solve the question regarding the applicability of the law of triads to the given groups of elements, we will analyze each option based on the definition of the law of triads.
### Step-by-Step Solution:
1. **Understanding the Law of Triads**:
The law of triads, proposed by Johann Wolfgang Döbereiner, states that in a group of three elements with similar properties, the atomic weight of the middle element is approximately the average of the atomic weights of the other two elements.
2. **Analyzing Option (a) `Cl, Br, I`**:
- Atomic weights:
- Chlorine (Cl) = 35.5
- Bromine (Br) = 80
- Iodine (I) = 126.9
- Calculate the average of Cl and I:
\[
\text{Average} = \frac{35.5 + 126.9}{2} = \frac{162.4}{2} = 81.2
\]
- The average (81.2) is approximately equal to the atomic weight of Bromine (80). Thus, the law of triads is applicable to this group.
3. **Analyzing Option (b) `C, N, O`**:
- Atomic weights:
- Carbon (C) = 12
- Nitrogen (N) = 14
- Oxygen (O) = 16
- Calculate the average of C and O:
\[
\text{Average} = \frac{12 + 16}{2} = \frac{28}{2} = 14
\]
- The average (14) is equal to the atomic weight of Nitrogen (N). Thus, the law of triads is applicable to this group.
4. **Analyzing Option (c) `Na, K, Rb`**:
- Atomic weights:
- Sodium (Na) = 23
- Potassium (K) = 39
- Rubidium (Rb) = 85.5
- Calculate the average of Na and Rb:
\[
\text{Average} = \frac{23 + 85.5}{2} = \frac{108.5}{2} = 54.25
\]
- The average (54.25) is not equal to the atomic weight of Potassium (39). Thus, the law of triads is not applicable to this group.
5. **Analyzing Option (d) `H, O, N`**:
- Atomic weights:
- Hydrogen (H) = 1
- Oxygen (O) = 16
- Nitrogen (N) = 14
- Calculate the average of H and N:
\[
\text{Average} = \frac{1 + 14}{2} = \frac{15}{2} = 7.5
\]
- The average (7.5) is not equal to the atomic weight of Oxygen (16). Thus, the law of triads is not applicable to this group.
### Conclusion:
Based on the analysis, the law of triads is applicable to the group of elements in option (a) `Cl, Br, I`. Therefore, the correct answer is:
**Answer: a) `Cl, Br, I`**
