To determine which sequence of elements is arranged in order of decreasing atomic radii, we need to compare the atomic radii of the elements within each sequence and decide which one correctly follows the pattern of having a larger atomic radius followed by smaller ones. Let's analyze each option:
### Option (A): [tex]\( Al, Si, P \)[/tex]
- Aluminum (Al): Atomic radius is larger than Silicon (Si).
- Silicon (Si): Atomic radius is larger than Phosphorus (P).
- Phosphorus (P): Has the smallest atomic radius among the three.
So, in this sequence:
[tex]\[ \text{Al} > \text{Si} > \text{P} \][/tex]
This is in decreasing order, so option (A) is correct.
### Option (B): [tex]\( Li, Na, K \)[/tex]
- Lithium (Li): Atomic radius is smaller than Sodium (Na).
- Sodium (Na): Atomic radius is smaller than Potassium (K).
- Potassium (K): Has the largest atomic radius among the three.
In this sequence:
[tex]\[ \text{Li} < \text{Na} < \text{K} \][/tex]
This is in increasing order, so option (B) is incorrect.
### Option (C): [tex]\( Cl, Br, I \)[/tex]
- Chlorine (Cl): Atomic radius is smaller than Bromine (Br).
- Bromine (Br): Atomic radius is smaller than Iodine (I).
- Iodine (I): Has the largest atomic radius among the three.
In this sequence:
[tex]\[ \text{Cl} < \text{Br} < \text{I} \][/tex]
This is in increasing order, so option (C) is incorrect.
### Option (D): [tex]\( N, C, B \)[/tex]
- Nitrogen (N): Atomic radius is smaller than Carbon (C).
- Carbon (C): Atomic radius is smaller than Boron (B).
- Boron (B): Has the largest atomic radius among the three.
In this sequence:
[tex]\[ \text{N} < \text{C} < \text{B} \][/tex]
This is in increasing order, so option (D) is incorrect.
Given the analysis above, the correct sequence of elements arranged in order of decreasing atomic radii is:
[tex]\[
\boxed{1}
\][/tex]
Which corresponds to option (A): [tex]\( Al, Si, P \)[/tex].
