Question

Identify the atom with the following ground-state electron configuration for its valence shell 4s23d104p6

Answer 1

The atom with the ground-state valence shell electronic configuration $4{s^2}3{d^{10}}4{p^6}$ is$\boxed{{\text{Krypton}}\left( {{\text{Kr}}} \right)}$.

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.

Filling of electrons in energy levels or orbitals is according to the following three rules.

1. Aufbau principle: The principle states that the electrons are filled in various orbitals in the increasing order of their energies as follow:

$1s$

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 general expression for full electron configuration contains all the symbols of occupied. It is given as follows:

$n{\left( {{\text{symbol for }}l} \right)^{{\text{ number of electrons in orbital}}}}$

For example, Boron has atomic number 5. Therefore, filling of electrons for boron takes place as follows:  

Full Electronic configuration: $1{s^2}2{s^2}3{s^1}$

The given valence shell electronic configuration is $4{s^2}3{d^{10}}4{p^6}$. This indicates the given element is present in the 4th row. The noble gas that comes before this row is argon. So the electronic configuration of the given element is $\left[{{\text{Ar}}}\right]4{s^2}3{d^{10}}4{p^6}$. Argon is a noble gas with atomic number 18, therefore 18 electrons are present in it. So the total number of electrons present in the element is 36 and its atomic number is also 36. So the given element is krypton (Kr).

Learn more:

1. Determine the ion with configuration $\left[ {{\text{Ar}}} \right]\;3{d^2}$: brainly.com/question/7599542

2. Identification of element which has electron configuration $1{s^2}2{s^2}2{p^2}3{s^1}$:brainly.com/question/9616334

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Electronic configuration of the elements.

Keywords: electronic configuration, Aufbau, krypton, Kr, Pauli’s exclusion principle, Hund’s rule, argon, 18, 36, distribution of electrons, row, 4th row.

Answer 2

Since the greatest number in the given electron configuration of the atom is 4, it means that the element is found in the fourth energy level. The number of electrons found in the valence shell is 2+10+6 = 18. From this answer and using the periodic table, we can see that the atom is that of Krypton (Kr).