Three different particles have the same electron configuration one is a cation of an alkali metal one is an anion of the halide in the third period and the third particle is an atom of a noble gas what are the identity of the three particles which particle should have the smallest atomic radius which should have the largest and why

The hint here is: anion of the halide in the third period. Halides are elements found in group 7. Thus, the halide referred to here is Chloride, Cl⁻. Thus, the noble gas (elements in group 8) and the alkali metal (elements in group 1) that has a similar electronic configuration are Argon and Potassium ion. 

1. The three different particles having the same electronic configuration are: Potassium ion, Chloride ion and Argon.

2. Anions will always be larger than cations of the same electronic configuration. So, the largest atomic radius would be the Chloride ion, while the smallest would be the Potassium ion. The Argon is right in between.

Adozen is 1221/12 = 1.751.75 x 12 = 21therefore, 1.75 is the answer.