Question

in the space below, write the noble gas notation/shorthand electron configurations of the following elements: 4) nitrogen

Explanation:

Step1: Identify the noble gas before nitrogen

The noble gas with an atomic number less than nitrogen (atomic number 7) is helium (He), which has an atomic number of 2.

Step2: Determine the remaining electrons

Nitrogen has 7 electrons. Helium has 2 electrons, so the remaining electrons are \(7 - 2=5\). These electrons fill the \(2s\) and \(2p\) orbitals. The electron configuration for the remaining electrons is \(2s^{2}2p^{3}\).

Step3: Write the noble gas notation

Combine the noble gas symbol (He) with the remaining electron configuration. So the noble gas notation for nitrogen is \([\text{He}]2s^{2}2p^{3}\).

Answer:

\([\text{He}]2s^{2}2p^{3}\)