Enter the electron configuration for phosphorus. Express your answer in order of increasing orbital energy as a string without blank space between orbitals. For example, the electron configuration of Li could be entered as 1s^22s^1 or [He]2s^1.

₁₅P = 1s² 2s² 2p⁶ 3s² 3p³

Explanation:

Phosphorus is the second element of group (v) with atomic number 15 and the electrons per shell are arrange as thus; 2, 8, 5.

The electronic configuration of phosphorus is

₁₅P = 1s² 2s² 2p⁶ 3s² 3p³

Or

₁₅P [Ne] 3s² 3p³

In its first shell I.e 1s shell, it has 2 electrons filling it to s-orbital requirement. In its second shell, electrons fill it in 2s² 2p⁶ filling the second orbital and attaining an octet configuration. The third shell contains 5 electrons filling 3s² 3p³ orbital.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :