Question & Answer » Chemistry Questions » What is the Electron Configuration of Nitrogen

What is the Electron Configuration of Nitrogen

What is the electron configuration of nitrogen? Find the answer to this question and access a vast question bank customised for students.

What is the electron configuration of nitrogen?

Nitrogen has an atomic number of 7. This is the number of protons in a nitrogen atom’s nucleus. The number of electrons in a neutral atom is the same as the number of protons. In the ground state, the electron configuration will consist of 7 electrons positioned in the suitable s and p orbitals (state of lowest energy) in the ground state. 

1s 2 2s 2 2p 3 is the complete electron configuration of nitrogen.

[ He ] 2s 2 2p 3 is the abbreviated electron configuration for noble gases

According to Neil Bohr

Niels Bohr, a scientist, was the first to describe the atom’s orbit. In 1913, he presented a model of the atom. There is a comprehensive description of the orbit there. The electrons in an atom travel in a circular path around the nucleus. Orbit refers to these circular pathways (shell). The number n represents these orbits. [n = 1,2,3,4,…, the orbit’s serial number] 

Nitrogen(N) has an atomic number of 7. That is, nitrogen has a total of seven electrons. As a result, the first shell of a nitrogen atom will have two electrons while the second shell will have five. As a result, the number of electrons in each shell of the nitrogen(N) atom is in the order 2, 5.

According to Orbital Method:

Sub-energy levels are subdivided from atomic energy levels. Orbital levels are sub-energy levels. ‘l’ represents the sub energy levels. The range of values for ‘l’ is 0 to (n – 1). S, p, d, and f are the sub-energy levels. The value of ‘l’ for various energy levels is determined by-

  • If n = 1,
    (n – 1) = (1–1) = 0
    Therefore, the orbital number of ‘l’ is 1; And the orbital is 1s.

  • If n = 2,
    (n – 1) = (2–1) = 1.
    Therefore, the orbital number of ‘l’ is 2; the orbital is 2s, 2p.

  • If n = 3,
    (n – 1) = (3–1) = 2.
    Therefore, the orbital number of ‘l’ is 3; the orbital is 3s, 3p, 3d.

  • If n = 4,
    (n – 1) = (4–1) = 3
    Therefore, the orbital number of ‘l’ is 4; the orbital is 4s, 4p, 4d, 4f.

  • If n = 5,
    (n – 1) = (n – 5) = 4.