Copy and complete the following table Element Number of Neutrons Electronic Configuration Group in the periodic Table \(^{23}_{11}Na\) --- \(1s^{2}2s^{2}2p^...
Question 1 Report
Copy and complete the following table
| Element | Number of Neutrons | Electronic Configuration | Group in the periodic Table |
| \(^{23}_{11}Na\) | --- | \(1s^{2}2s^{2}2p^{6}3s^{1}\) | 1 |
| \(^4_2He\) | 2 | ---- | --- |
| ---- | 7 | \(1s^{2}2s^{2}2p^{2}\) | ---- |
Answer Details
Completed table
Reasoning for each entry:
- \( ^{23}_{11}Na \): number of neutrons \( = 23 - 11 = 12 \). The configuration \( 1s^2\,2s^2\,2p^6\,3s^1 \) ends in \( 3s^1 \) (one outer electron), so it is in Group 1.
- \( ^{4}_{2}He \): neutrons \( = 4 - 2 = 2 \); with 2 electrons the configuration is \( 1s^2 \). Its outer shell is full, so it is a noble gas in Group 0.
- Third element: the configuration \( 1s^2\,2s^2\,2p^2 \) has 6 electrons, so the atomic number is 6 (carbon). With 7 neutrons the mass number is \( 6 + 7 = 13 \). The two \( p \) electrons make it a Group IV (Group 4) element.
| Element | Number of neutrons | Electronic configuration | Group in the periodic table |
|---|---|---|---|
| \( ^{23}_{11}Na \) | 12 | \( 1s^2\,2s^2\,2p^6\,3s^1 \) | 1 |
| \( ^{4}_{2}He \) | 2 | \( 1s^2 \) | 0 |
| \( ^{13}_{6}C \) (carbon) | 7 | \( 1s^2\,2s^2\,2p^2 \) | IV (4) |
The group number for a main-group element equals the number of electrons in its outermost shell (helium being the exception, placed in Group 0 with the noble gases).
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