QQuestionChemistry
QuestionChemistry
| atom | subshell from which electron removed to form + 1 cation | subshell to which electron added to form - 1 anion |
| --- | --- | --- |
| H | 1s | 1s |
| Ne | 2p | 3s |
| Cl | 11 | 1 |
| Sc | 4s | 3d |
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Answer
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Step 1:: Identify the subshell from which an electron is removed to form a cation and the subshell to which an electron is added to form an anion for each atom.
The given table provides the information needed. We will analyze each row and find the corresponding subshells.
Step 2:: Analyze the first row (Hydrogen atom).
For a Hydrogen atom (H) to form a + 1 cation, an electron is removed from the 1s subshell. To form a - 1 anion, an electron is added to the 1s subshell.
Step 3:: Analyze the second row (Neon atom).
For a Neon atom (Ne) to form a + 1 cation, an electron is removed from the 2p subshell. To form a - 1 anion, an electron is added to the 3s subshell.
Step 4:: Analyze the third row (Chlorine atom).
For a Chlorine atom (Cl) to form a + 1 cation, an electron is removed from the 3p subshell (since the 11 subshell does not exist). To form a - 1 anion, an electron is added to the 1s subshell (since the 1 subshell does not exist).
Step 5:: Analyze the fourth row (Scandium atom).
For a Scandium atom (Sc) to form a + 1 cation, an electron is removed from the 4s subshell. To form a - 1 anion, an electron is added to the 3d subshell.
Final Answer
| atom | subshell from which electron removed to form + 1 cation | subshell to which electron added to form - 1 anion | | --- | --- | --- | | H | 1s | 1s | | Ne | 2p | 3s | | Cl | 3p | 1s (does not exist, so 2s, 2p, or 3s instead) | | Sc | 4s | 3d |
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