Question

Identify the neutral element represented by this excited-state electron configuration, then write the ground-state electron configuration for that element 1s22s22p23s1

Answer 1

$\boxed{{\mathbf{Nitrogen}}}$ is the neutral element whose excited electronic configuration is $1{s^2}2{s^2}2{p^2}3{s^1}$ . Its ground-state electronic configuration is $\boxed{{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{p}}^{\mathbf{3}}}}$

Further Explanation:

The electronic configuration is the distribution of electrons of an atom in the atomic orbitals. There are two states for an electron: ground as well as the excited state. The configuration of the atom in the lowest possible energy levels is called the ground-state electronic configuration. When an electron jumps from the stable ground state to some higher level, that state is called the excited state and the electronic configuration corresponding to this state is known as the excited-state electronic configuration.

The filling of electrons in different energy levels or orbitals is done in accordance with the following three rules:

1. Aufbau principle: “Aufbau” is a German word that means “build-up”. So this principle states that the electrons are filled in various orbitals in the increasing order of their energies.

2. Hund’s rule: Electron pairing will not start until each orbital is singly occupied.

3. Pauli’s exclusion principle: According to this principle, all the four quantum numbers $\left( {n,\;l,\;{m_l},\;{m_s}} \right)$ for any two electrons can never be the same. In an orbital, the spin of two electrons has to be different. If one electron has the clockwise spin, the other would have the anticlockwise spin and vice-versa.

The number of electrons in both the ground-state and excited-state electronic configuration is the same. So the given element has 7 electrons as well as 7 protons. This indicates its atomic number is 7. For the ground state, the valence shell is 2. So the element belongs to the 2nd period of the periodic table.

According to the periodic table, the element that belongs to period 2 and has the atomic number 7 is nitrogen. Its ground-state electronic configuration is $1{s^2}2{s^2}2{p^3}$ .

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Answer details:

Grade: High School  

Subject: Chemistry  

Chapter: Electronic configuration of the elements

Keywords: neutral element, atomic orbitals, electronic configuration, ground state, excited state., electrons, protons, energy levels, Aufbau principle, Pauli's exclusion principle, Hund's Rule.

Answer 2

The number of electrons in ground state and excited state remains the same in this case. So, from electronic configuration;

                                                1s², 2s², 2p², 3s¹

The number of electrons in this element are 7, so for neutral element for seven electrons there are seven protons. Seven protons show that element has atomic number 7. Also the valence shell is 2, means element is present in period 2, and when electron de-excites it will have following configuration,

                                                1s², 2s², 2p³

As the number of valence electrons in this element are 5, so it belongs to group 5.

Result:
          Atomic Number  =  7

          Period  =  2

          Group  =  5

So, element with Atomic number 7, present in Period 2 and Group 5 is Nitrogen.