Answer:
2,1,-1
Explanation:
The electronic configuration of oxygen is 
The valence electrons are in the second principle quantum number, which means that the value of n = 2.
For, n =2,
l can be 0 or 1 which corresponds to s and p orbital respectively.
If l = 1 , which is p orbitals and it exits in 3 degenerate orbitals which has values of m = -1 , 0 , +1
<u>So, The possible set pf quantum number for the outermost valence electron of the oxygen atom is 2, 1, -1.</u>
We can use the ideal gas law equation to find the pressure
PV = nRTwhere
P - pressure
V - volume - 2.6 x 10⁻³ m³
n - number of moles - 0.44 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values into the equation,
P x 2.6 x 10⁻³ m³ = 0.44 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 419 281.41 Pa
101 325 Pa is equivalent to 1 atm
Therefore 419 281.41 Pa - 1/ 101 325 x 419 281.41 = 4.13 atm
Pressure is 4.13 atm