Give the electron configuration of vanadium (V), atomic number 23

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d³ or [Ar] 4s²3d³

Given

Vanadium, atomic number 23

Required

The electron configuration

Solution

In an atom, there are levels of energy in the shell and subshell  

This energy level is expressed in the form of electron configurations.  

Charging electrons in the subshell uses the following sequence:  

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.  

The electron configuration is based on the atomic number which indicates the number of electrons of the atom

Vanadium has the atomic number 23, so the number of electrons = 23

Electron configuration:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d³

or we can write it using the noble gas notation

[Ar] 4s²3d³

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.