This particular configuration denotes a particle with ten total electrons. The sodium atom, with eleven electrons, is the only one listed that could not have this configuration. Ionized sodium, however, symbolized as Na+, does apply. (Be careful to distinguish neutral atoms and ions).

This particular configuration denotes a particle with ten total electrons. The sodium atom, with eleven electrons, is the only one listed that could not have this configuration. Ionized sodium, however, symbolized as Na+, does apply. (Be careful to distinguish neutral atoms and ions).

Potassium (K) is orignially in the electron configuration of \[Ar\]4s1. To obtain a +1 charge it loses an electron, resulting in a configuration of \[Ar\].

Potassium (K) is orignially in the electron configuration of \[Ar\]4s1. To obtain a +1 charge it loses an electron, resulting in a configuration of \[Ar\].

When an element loses an electron it is generally taken away from the highest electron shell. The electron configuration of iron (Fe) is . The 4s orbital is farther away from the nucleus than the 3d orbital, therefore the electron configuration of Fe+ will be .

When an element loses an electron it is generally taken away from the highest electron shell. The electron configuration of iron (Fe) is . The 4s orbital is farther away from the nucleus than the 3d orbital, therefore the electron configuration of Fe+ will be .

When determining electronic configuration, the answer is made much easier by starting with the next smallest noble gas in brackets. As a result, \[Ar\] is an appropriate way to incorporate every previous electron before argon.

After argon, vanadium has five other electrons to distribute, and because vanadium is a transitional element, it will fill its 3d subshells before filling the 4p subshells. The 4s subshell is filled first, and the last three electrons are placed into the 3d subshells.

\[Ar\]3d34s2

When determining electronic configuration, the answer is made much easier by starting with the next smallest noble gas in brackets. As a result, \[Ar\] is an appropriate way to incorporate every previous electron before argon.

After argon, vanadium has five other electrons to distribute, and because vanadium is a transitional element, it will fill its 3d subshells before filling the 4p subshells. The 4s subshell is filled first, and the last three electrons are placed into the 3d subshells.

\[Ar\]3d34s2

Nuclear orbitals will always fill from the innermost to the outermost subshells. Using the rule, we can approximate the order in which these orbitals will fill. Because electrons fill starting with the centermost orbitals, the electron that is closest to the nucleus will belong to the orbital that fills first.

corresponds to the principle quantum number, the first number in the given orbital location. is the azimuthal quantum number, and dictates the shape of the orbital.

5s and 3d produce the same number from the equation, but in the event of a tie we always pick the orbital with the lowest letter (s < p < d < f). The innermost orbital will be 5s.

Nuclear orbitals will always fill from the innermost to the outermost subshells. Using the rule, we can approximate the order in which these orbitals will fill. Because electrons fill starting with the centermost orbitals, the electron that is closest to the nucleus will belong to the orbital that fills first.

corresponds to the principle quantum number, the first number in the given orbital location. is the azimuthal quantum number, and dictates the shape of the orbital.

5s and 3d produce the same number from the equation, but in the event of a tie we always pick the orbital with the lowest letter (s < p < d < f). The innermost orbital will be 5s.