Using only the periodic table inside the front cover of the text, write the expected ground-state electron configurations for a. the third element in Group \(5 \mathrm{~A}\). b. element number 116 . c. an element with three unpaired \(5 d\) electrons. d. the halogen with electrons in the \(6 p\) atomic orbitals.

First, we need to find the third element in Group 5A using the periodic table. Group 5A corresponds to elements with an outer electron in the p-block and 5 valence electrons. Counting down three elements in the group gives us nitrogen (N). The electron configuration of nitrogen can be written as follows: \(1s^2 \ 2s^2 \ 2p^3\) So, the expected ground-state electron configuration for the third element in Group 5A is: \(1s^2 \ 2s^2 \ 2p^3.\)

We need to find the electron configuration for element number 116. This element is located in the p-block, specifically in the 7th period and has the symbol Lv (Livermorium). The electron configuration of Livermorium can be written as follows: \(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 \ 3d^{10} \ 4p^6 \ 5s^2 \ 4d^{10} \ 5p^6 \ 6s^2 \ 4f^{14} \ 5d^{10} \ 6p^5\) So, the expected ground-state electron configuration for element number 116 is: \(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 \ 3d^{10} \ 4p^6 \ 5s^2 \ 4d^{10} \ 5p^6 \ 6s^2 \ 4f^{14} \ 5d^{10} \ 6p^5.\)

In this part, we need to find an element with three unpaired 5d electrons. We look for an element whose electron configuration contains the subshell 5d with 3 unpaired electrons. Among the transition metals, the 5d subshell starts filling with Hafnium (Hf, atomic number 72). Counting three elements from Hafnium in the periodic table, we find Tantalum (Ta, atomic number 73). The electron configuration of Tantalum is: \(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 \ 3d^{10} \ 4p^6 \ 5s^2 \ 4d^{10} \ 5p^6 \ 6s^2 \ 4f^{14} \ 5d^3\) So, the element with three unpaired 5d electrons is Tantalum (\(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 \ 3d^{10} \ 4p^6 \ 5s^2 \ 4d^{10} \ 5p^6 \ 6s^2 \ 4f^{14} \ 5d^3\)).

Finally, we need to find the halogen with electrons in the 6p atomic orbitals. Halogens are found in Group 7A, and their electron configuration ends with p^5 in their outermost shell. Since we are looking for a halogen with electrons in the 6p atomic orbitals, we look for an element in Group 7A that has a 6p^5 configuration at the end. This element is Astatine (At, atomic number 85). The electron configuration of Astatine is: \(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 \ 3d^{10} \ 4p^6 \ 5s^2 \ 4d^{10} \ 5p^6 \ 6s^2 \ 4f^{14} \ 5d^{10} \ 6p^5\) So, the halogen with electrons in the 6p atomic orbitals is Astatine (\(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 \ 3d^{10} \ 4p^6 \ 5s^2 \ 4d^{10} \ 5p^6 \ 6s^2 \ 4f^{14} \ 5d^{10} \ 6p^5\)).