Use these answers for questions 4 - 7.

(A) 1s2 2s22p5 3s23p5
(B) 1s2 2s22p6 3s23p6
(C) 1s2 2s22p62d10 3s23p6
(D) 1s2 2s22p6 3s23p63d5
(E) 1s2 2s22p6 3s23p63d3 4s2
4. An impossible electronic configuration

5. The ground-state configuration for the atoms of a transition element

6. The ground-state configuration of a negative ion of a halogen

7. The ground-state configuration of a common ion of an alkaline earth element

Electron configurations really confuse me. How come E goes from 3d3 to 4s2? I need help with #6 & 7. I don't understand what they're asking or how to figure them out. This is a very weak area for me!

For #6, all of the halogens are in group 17 (VIIA depending upon the system you are using). Therefore, they have a ns2, np5 configuration (note 7 electrons in the outside shell) where n is 2, 3, 4, 5 etc depending upon the particular halogen we are discussing. They become the -1 ion by adding an electron to the p level to make the outside shell ns2, np6.

For #7, all alkaline earths are in group 2 (or IIA) and have a ns2 configuration. If they lose those two electrons, their electronic configuration becomes that of the noble gas two elements below it; for example, 12Mg is 1s2 2s2 2p6 3s2 (12 total). Losing those two electrons makes it 1s2 2s2 2p6 (total 10) which makes it like Ne (12-10=2 elements below Mg).

Therefore, the answers to #6 and #7 are (B) and (E) respectively.

6. The ground-state configuration of a negative ion of a halogen is obtained by adding an electron to the p level to make the outside shell ns2, np6.

For example, let's consider the halogen chlorine (Cl). The ground-state configuration of chlorine is 1s2 2s2 2p6 3s2 3p5. By adding one more electron to the 3p level, chlorine becomes a negative ion (Cl-) with the electron configuration 1s2 2s2 2p6 3s2 3p6.

7. The ground-state configuration of a common ion of an alkaline earth element is obtained by losing two electrons from the ns2 configuration. After losing those two electrons, the electronic configuration of the ion becomes that of the noble gas two elements below it.

For example, let's consider the alkaline earth element magnesium (Mg). The ground-state configuration of magnesium is 1s2 2s2 2p6 3s2. By losing two electrons from the 3s level, magnesium becomes a common ion (Mg2+) with the electron configuration 1s2 2s2 2p6, which is the same as the noble gas neon (Ne) configuration.

To figure out the answers to questions #6 and #7, we need to understand the electron configurations of halogens and alkaline earth elements.

For #6, we're asked about the ground-state configuration of a negative ion of a halogen. Halogens are found in group 17 (or VIIA) of the periodic table. They have a general electron configuration of ns2 np5, where n represents the energy level (2, 3, 4, etc.) depending on the specific halogen. Halogens become negative ions by gaining an electron, which adds to the p orbital, resulting in a configuration of ns2 np6. This means that the outermost shell of the negative ion of a halogen will have 8 valence electrons (2 from the s orbital and 6 from the p orbital).

For #7, we're asked about the ground-state configuration of a common ion of an alkaline earth element. Alkaline earth elements are found in group 2 (or IIA) of the periodic table. They have a general electron configuration of ns2. When these elements lose their two valence electrons, their electronic configuration becomes similar to the noble gas located two elements below them in the periodic table. For example, let's take Magnesium (Mg) as an alkaline earth element. Its ground-state configuration is 1s2 2s2 2p6 3s2. By losing its two valence electrons (in the 3s orbital), it becomes 1s2 2s2 2p6, which has the same configuration as Neon (Ne). Essentially, the alkaline earth element becomes isoelectronic with the noble gas.

Regarding your confusion about answer choice (E) in question #5, where the electron configuration jumps from 3d3 to 4s2, it's important to note that electron configurations can vary due to different energy levels and subshells. In some cases, energy levels can fill up before a subshell is completely filled. This is known as electron configuration anomalous order. In the specific case of (E), the 3d subshell is not yet completely filled, but the 4s subshell can fill up. This is because the 4s orbital has slightly lower energy than the 3d orbital. This anomaly occurs in the transition metals, which have partially filled d orbitals.

I hope this explanation helps clarify the electron configurations and how to approach questions #6 and #7.