The reaction between N₂ and F₂ gives Nitrogen trifluoride as the product. The balanced equation is;
N₂ + 3F₂ → 2NF₃
The stoichiometric ratio between N₂ and NF₃ is 1 : 2
Hence,
moles of N₂ / moles of F₂ = 1 / 2
moles of N₂ / 25 mol = 0.5
moles of N₂ = 0.5 x 25 mol = 12.5 mol
Hence N₂ moles needed = 12.5 mol
At STP (273 K and 1 atm) 1 mol of gas = 22.4 L
Hence needed N₂ volume = 22.4 L mol⁻¹ x 12.5 mol
= 280 L
Unfortunately you did not specify the electronic configuration in the question, however since one of the answers must be a halogen, i took the liberty to attach an image with the configuration (both the simple numeric and spdf form) for all the halogen and all you have to do is match the electronic configuration you have in your question to the one in the table attached and you can then deduce the answer.
Hope this helps.
Your answer is probably
Vaporization point
1 to 1. Most small atoms have the same number of protons and neutrons