All s orbitals are spherical in shape, and the smallest principle quantum number will refer to the smallest orbital. As principal quantum number, \(\displaystyle n\), increases, so does the size of the orbital. The correct answer for this question will be the s orbital with the smallest coefficient: \(\displaystyle 1s\).

Answer choices containing p or d orbitals are incorrect because these orbitals are not spherical in shape.

The location of a given electron is described by a set of four quantum numbers. The principle quantum number, \(\displaystyle n\), gives the energy level of the electron. It essentially describes the size of the electron shell, and can be any integer value. As this value increases, the size of the orbital increases and is located farther from the atom nucleus. The azimuthal (angular momentum) quantum number, \(\displaystyle l\), refers to the subshell of the energy level. Each subshell has a different shape, corresponding to spdf configurations. \(\displaystyle l\) can be any value from zero to \(\displaystyle n-1\). The magnetic quantum number, \(\displaystyle m_{l}\), gives the orientation of the subshell given by the azimuthal quantum number. These values correspond to orientations along the x-axis, y-axis, or z-axis. Each orientation constitutes a different orbital and can hold exactly two electrons. The magnetic quantum number can be any integer greater than \(\displaystyle -l\) and less than \(\displaystyle +l\), with the total number of possible orbitals given by \(\displaystyle (2l +1)\). \(\displaystyle m_{s}\) is the electron spin quantum number. This number is used to distinguish between the two electrons in a single orbital.

The \(\displaystyle d\) subshell (\(\displaystyle l=3\)) can carry a total of five orbitals, corresponding to the values of \(\displaystyle m_{l}\). Each orbital will hold two electrons, for a total of ten.

The only false statement concerns the angular momentum quantum number, confusing it with the \(\displaystyle m_{s}\) quantum number.

The correct answer is 50 electrons.

The number of electrons a shell can hold can be found using the formula \(\displaystyle 2n^{^{2}}\). Given the principle quantum number, we can calculate the number of electrons that can fill that energy level.

\(\displaystyle 2n^2\)

\(\displaystyle 2(5)^2\)

\(\displaystyle 2(25)=50\)

The, principal quantum number, \(\displaystyle n\) tells us the shell, which is represented as the leading coefficient in electronic configuration. Since we are given \(\displaystyle n=5\), we know that the leading number will be five.

The angular momentum (azimuthal) quantum number, \(\displaystyle l\), tells us the shape of the orbital. Each shape corresponds with a value of \(\displaystyle l\), and a letter. The values for \(\displaystyle l\) and their corresponding orbital shapes are listed below:

\(\displaystyle 0=s\)

\(\displaystyle 1=p\)

\(\displaystyle 2=d\)

\(\displaystyle 3=f\)

\(\displaystyle 4=g\)

We are given the subshell \(\displaystyle l=3\), which tells us we are looking at an \(\displaystyle f\) orbital.

Together, this gives our answer of \(\displaystyle 5f\).

The location of a given electron is described by a set of four quantum numbers. Of these, the magnetic quantum number (\(\displaystyle m_{l}\)) gives the orientation of the orbital in three-dimensional space.

The principle quantum number, \(\displaystyle n\), gives the energy level of the electron. It essentially describes the size of the electron shell, and can be any integer value.

The azimuthal (angular momentum) quantum number, \(\displaystyle l\), refers to the subshell of the energy level. Each subshell has a different shape, corresponding to spdf configurations. \(\displaystyle l\) can be any value from zero to \(\displaystyle n-1\).

The magnetic quantum number, \(\displaystyle m_{l}\), gives the orientation of the subshell given by the azimuthal quantum number. For example, a p orbital has three different orientations based on three different values for the magnetic quantum number. These values correspond to orientations along the x-axis, y-axis, or z-axis. Each orientation constitutes a different orbital and can hold exactly two electrons. The magnetic quantum number can be any integer greater than \(\displaystyle -l\) and less than \(\displaystyle +l\), with the total number of possible orbitals given by \(\displaystyle (2l +1)\).

\(\displaystyle m_{s}\) is the electron spin quantum number. This number is used to distinguish between the two electrons in a single orbital.

(\(\displaystyle n-1\)) is not a quantum number. This is a formula that allows us to determine the possible values of \(\displaystyle l\) for every principal quantum number.

\(\displaystyle 1s^{2}2s^{2}2p^{^{6}}3s^{1}\) is the correct configuration for a neutral sodium atom.

A neutral sodium atom contains eleven electrons, so you can eliminate \(\displaystyle 1s^{2}2s^{2}3s^{2}3p^{^{6}}\) and \(\displaystyle 1s^{2}2s^{2}2p^{^{6}}\) which contain twelve and ten electrons, respectively.

Next, \(\displaystyle 1s^{2}2s^{2}2p^{^{5}}3s^{2}\) can be eliminated because of the Aufbau principal, which states that electrons are placed into orbitals from the lowest to highest energy. In this answer choice, the \(\displaystyle 2p\) (lower energy) orbital is not completely filled, containing only five out of the possible six possible electron that it can hold. This orbital must be filled completely before electrons can be placed in the \(\displaystyle 3s\) (higher energy) orbital.

\(\displaystyle 1s^{2}2s^{2}2p^{^{7}}\) is incorrect because there are seven electrons in the \(\displaystyle 2p\) orbital. This is impossible, as p orbitals hold a maximum of six electrons.

The noble gas electron configuration for fluorine, \(\displaystyle F\), in its ground state is \(\displaystyle [He]2s^{2}2p^{5}\) . We can see from this configuration that there are seven valence electron in its outer shell of \(\displaystyle n=2.\) 

When an element gains or loses charge, becoming a cation or an anion, it is either gaining or loosing electrons. Anions (negatively charged ions) have gained electrons and cations (positively charged ions) have lost electrons. 

The \(\displaystyle F^{-}\) ion has gained one electron, increasing its total valence electrons by one. The new noble gas electron configuration for the ion will be \(\displaystyle [He]2s^{2}2p^{6}\), with eight valence electrons. The ion is now isoelectronic to the noble gas neon, and satisfies the octet rule.

To write the noble gas configuration of an element, choose the noble gas in the row above the element to put in brackets. For sulfur, that is the noble gas neon, \(\displaystyle [Ne]\).

Then, you write the rest of the configuration beginning on the same row as the given element. Sulfur is in the third row, or period, which means that it must start with the \(\displaystyle 3s\) orbital. Continuing to fill the orbitals, we get the configuration \(\displaystyle [Ne]3s^{2}3p^{4}\).

Writing electron configuration this way can save time, and easily tells us how many valence electrons an element has. Sulfur has six valence electrons, shown by the number of electrons in the third energy level.

In its ground state, phosphorous has five valence electron.

We can determine the number of valence electrons by examining the ground state electron configuration of an element. The ground state electron configuration of phosphorous is:

\(\displaystyle 1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}\)

From this configuration we can see that the outermost shell is \(\displaystyle n=3\). In its outer shell there are two electrons in the \(\displaystyle s\) subshell and three electrons in the \(\displaystyle p\) subshell, giving us a total of five valence electrons. Keep in mind that the different subshells correspond to different orbitals in the same energy level. When determining valence electrons, all electrons in the highest energy level must be included.

A noble gas electron configuration is achieved when an atom has an octet electron configuration, indicating its most stable state. For example, sulfur (S), at its ground state, has 6 valence electrons. When it gains two electrons (-2 charge), it has eight electrons, fulfilling the octet.

All of the answer choices, except \(\displaystyle P^{2-}\), have octet/noble gas electron configurations. Phosphorus (P), at its ground state, has 5 valence electrons. A -2 charge will create an ion with 7, not 8, electrons.