To answer this question, we need to find the electron configuration of both elemental sodium and sodium cation. If we look at the periodic table we can see that sodium is found on the first column. Since it is found in the first column, sodium has one valence electron. To complete octet, sodium will readily lose an electron and become a positively charged sodium ion. The electron configuration for sodium is . The electron configuration for sodium ion is  (because it lost its electron in the  orbital). This means that elemental sodium has an unpaired electron in its  orbital; the sodium ion has no unpaired electrons. Recall that an unpaired electron can generate its own magnetic field and is called paramagnetic; therefore, solid sodium is paramagnetic. The number of electrons in the  orbitals for both sodium and sodium ion is the same (6 electrons total in the  orbital). The outermost shell of sodium is the third shell (because sodium is located on the third row of periodic table). Elemental sodium contains one electron in the  orbital in the outermost shell whereas the sodium ion contains 6 electrons in its outermost shell.

Hybridization is a process involving the fusion, or hybridization, of  and  orbitals to form a unique orbital. It is possible for various combinations of  and  hybridization. Recall that there is one  orbital and three  orbitals in each shell. This means that the one  orbital can hybridize with 1, 2, or all 3  orbitals. Since there are three total combinations, there are three types of hybridized orbitals. These are , , and .  orbital has one  and one  orbital hybridized. This means that the  orbital and the first  orbital become a new  orbital. A molecule with  hybridization will have two  orbitals and two  orbitals. Similarly, an  orbital is made from the hybridization of one  and two  orbitals. In  hybridization, there are three  orbitals and one  orbital. Finally, an  orbital has one  and all three  orbitals; therefore, an  hybridized molecule will have four  orbitals and no  orbitals. The question states that there are three hybridized orbitals in this molecule; therefore, this molecule must be  hybridized. The single  orbital is unhybridized because the molecule probably has a double bond. Electrons in  bonds in double and triple bonds cannot be found in hybridized orbitals; therefore, they need their own  orbital. If a molecule has one  bond (double bond), then it will need one  orbital and will be  hybridized (because this will give three  hybridized orbitals and one  orbital). If it has two  bonds (triple bond), then it will need two  orbitals and will be  hybridized. If a molecule has all  bonds (single bonds), then the molecule will require no empty  orbitals for the delocalized electrons, and will be  hybridized.

Carbon tetrachloride, , has a central carbon atom attached to four chlorine atoms. The bonds between the carbon atom and chlorine atoms are single covalent bonds. The electrons in a single bond ( bond) can be found in hybridized orbitals. Since carbon tetrachloride only has single bonds, the carbon atom can hybridize all of its orbitals (one  and three ) in the outermost shell and form a  hybridization; therefore, three  orbitals and one  orbital participate in hybridization leading us to the correct answer. Carbon dioxide, , has a central carbon atom bonded to two oxygen atoms. To complete octet, carbon and oxygen atoms have double bonds. This means that carbon dioxide has two  bonds (two double bonds). Recall that electrons in  bonds cannot reside in hybridized orbitals; therefore, to accommodate the two  bonds we need two empty, unhybridized  orbitals. This means that carbon dioxide will have hybridization of one  and one  orbital, giving it an  hybridization.

When determining the number of valence electrons for an atom, simply count the number of electrons present in the outermost shell's s and p orbitals. Phosphorus has two electrons in the 3s subshell, and three more in the 3p subshell, making a total of five valence electrons.

The types of subshells, from smallest to largest, are as follows: s, p, d, and f. These four subshells correspond respectively to the following quantum numbers: 0, 1, 2, and 3. From the periodic table, it is known that sulfur has 16 electrons. Additionally, the maximum number of electrons the s sublevel can hold is 2. The maximum number of electrons that the p subshell can hold is 6, and electrons fill orbitals from lowest to highest energy. 1s²2s²2p⁶3s²3p⁴ is the only choice that meets the criteria. 

Notice that all the ions have the same electron configuration as neon: 1s²2s²2p^6. Substances with the same number of electrons and the same electron configuration are isoelectronic, meaning the number of electrons is the same but the number of protons is not. In an isoelectronic series, the ion with the most protons is smallest because the nucleus exerts a stronger force of attraction and the electrons are pulled closer to the nucleus. Consequently, the ion with the fewest protons is largest.  has the most protons, therefore it has the smallest ionic radius. Continuing with this inverse relationship, the correct answer is: .

Although it may seem counterintuitive, atomic radius does decrease from left to right on the periodic table. The reason for this is because the added positive charge in the nucleus causes the elctrons to be pulled more strongly towards the center, which decreases the atomic radius.

Electronegativity values become greater as you move up and to the right on the periodic table. Of the four atoms listed, sulfur is the highest up and farthest to the right, giving it the greatest electronegativity.

Ionization energy is the amount of energy that an atom in the ground state must absorb to emit an electron. Upon ionization, a cation is formed. Ionization energy increases from bottom to top within a group, and from left to right within a row of the periodic table which is the opposite trend that atomic radius follows. Referring to the periodic table, we can see that of these group VII elements, fluorine has the highest ionization energy.

All of these atoms are in group VII. Within a group, atomic radii increase from top to bottom due to the increased number of electron shells.