Question

enter the correct ground - state (or lowest energy) configuration based on the number of electrons: $1s^22s^22p^9$.
express your answer in complete form in the order of orbital filling as a string without blank space between orbitals. for example, $1s^22s^2$ should be entered as 1s^22s^2.

part c

enter the correct ground - state (or lowest energy) configuration based on the number of electrons: $1s^22s^22p^82d^4$.
express your answer in complete form in the order of orbital filling as a string without blank space between orbitals. for example, $1s^22s^2$ should be entered as 1s^22s^2.

part d

enter the correct ground - state (or lowest energy) configuration based on the number of electrons: $1s^21p^4$.
express your answer in complete form in the order of orbital filling as a string without blank space between orbitals. for example, $1s^22s^2$ should be entered as 1s^22s^2.

part e

enter the correct ground - state (or lowest energy) configuration based on the number of electrons: $1s^22s^22p^83s^23p^1$.
express your answer in complete form in the order of orbital filling as a string without blank space between orbitals. for example, $1s^22s^2$ should be entered as 1s^22s^2.

Explanation:

Part (Let's assume we start with the first part, say the initial one before Part C, but since the user's image has multiple parts, let's take one, e.g., the first problem: "Enter the correct ground - state... \(1s^2 2s^2 2p^9\)")

Step 1: Recall orbital filling rules

The \(p\) orbital can hold a maximum of 6 electrons (since \(p\) subshell has 3 orbitals, each with 2 electrons: \(2\times3 = 6\)). The given configuration has \(2p^9\), which is incorrect. We need to fill orbitals in the correct order: \(1s\), \(2s\), \(2p\), \(3s\), \(3p\), etc. After filling \(1s^2\) and \(2s^2\), the \(2p\) can take at most 6 electrons. So the remaining electrons after filling \(1s^22s^22p^6\) will go to the next orbital, which is \(3s\) and then \(3p\). The total number of electrons in \(1s^22s^22p^9\) is \(2 + 2+9=13\). After filling \(1s^22s^22p^6\) (10 electrons), we have \(13 - 10 = 3\) electrons left. These will fill \(3s\) (2 electrons) and then \(3p\) (1 electron).

Step 2: Construct the correct configuration

So the correct configuration is \(1s^22s^22p^63s^23p^1\) (written as \(1s^22s^22p^63s^23p^1\) without spaces as per the example, so \(1s^22s^22p^63s^23p^1\)).

Part C: Problem \(1s^22s^22p^62d^4\)

Step 1: Recall orbital filling order

The \(d\) orbitals start filling after \(4s\) (for the 3d series) or in the correct order, the orbitals fill as \(1s\), \(2s\), \(2p\), \(3s\), \(3p\), \(4s\), \(3d\), etc. There is no \(2d\) orbital because the second energy level (\(n = 2\)) has only \(s\) and \(p\) orbitals (\(l = 0\) for \(s\), \(l = 1\) for \(p\); \(l\) can be at most \(n - 1\), so for \(n=2\), \(l\) can be 0 or 1, no \(l = 2\) (which is \(d\) orbital)). So after \(1s^22s^22p^6\), the next orbital is \(3s\), then \(3p\), then \(4s\), then \(3d\). The total number of electrons in \(1s^22s^22p^62d^4\) is \(2 + 2+6 + 4=14\). After filling \(1s^22s^22p^6\) (10 electrons), we have \(14 - 10 = 4\) electrons. These will fill \(3s\) (2), \(3p\) (6, but we don't need that many), wait, no: the correct order is \(1s\), \(2s\), \(2p\), \(3s\), \(3p\), \(4s\), \(3d\). Wait, for \(n = 3\), \(l\) can be 0,1,2 (so \(3s\), \(3p\), \(3d\)), but the filling order is \(1s\to2s\to2p\to3s\to3p\to4s\to3d\). So after \(2p^6\) (n=2), we go to \(3s\) (n=3, l=0), then \(3p\) (n=3, l=1), then \(4s\) (n=4, l=0), then \(3d\) (n=3, l=2). So the electrons in \(2d^4\) are incorrect. The total electrons: \(2+2 + 6+4=14\). The correct filling: \(1s^22s^22p^63s^23p^4\) (since \(1s^22s^22p^6=10\), \(14 - 10 = 4\), so \(3s^23p^2\)? Wait, no: \(3s\) takes 2, then \(3p\) takes 2 (total 4). Wait, \(3s^23p^2\)? Wait, no, \(1s^22s^22p^6\) is 10 electrons. 14 electrons: \(10+4\). So \(3s^2\) (2) and \(3p^2\) (2)? Wait, no, \(3s\) is filled first (2 electrons), then \(3p\) (up to 6). So \(1s^22s^22p^63s^23p^2\) is wrong? Wait, no, the original configuration has \(2d^4\), which is invalid. The correct order is that after \(n = 2\) (s and p), we go to \(n = 3\) (s, p, then d after 4s). So the correct configuration for 14 electrons is \(1s^22s^22p^63s^23p^2\)? Wait, no, 14 electrons: \(1s^2(2)+2s^2(2)+2p^6(6)+3s^2(2)+3p^2(2)\)? Wait, 2 + 2+6 + 2+2 = 14. But actually, the correct electron configuration for silicon (14 electrons) is \(1s^22s^22p^63s^23p^2\), which is correct. So the error is \(2d^4\), we replace it with \(3s^23p^2\) (but wait, the original is \(1s^22s^22p^62d^4\), so total electrons 14. So correct is \(1s^22s^22p^63s^23p^2\)? Wait, no, \(3s\) is filled before \(3p\), so after \(2p^6\), we fill \(3s^2\), then \(3p^2\) (since 2 + 2+6+2+2=14).

Part D: Problem \(1s^21p^5\)

Step 1: Recall orbital filling

The \(p\) orbitals start at \(n = 2\) (i.e., \(2p\)). The first energy level (\(n = 1\)) has only an \(s\) orbital (since \(l\) can be at most \(n - 1\), so for \(n = 1\), \(l = 0\), which is \(s\) orbital). So there is no \(1p\) orbital. The correct filling order is \(1s\), then \(2s\), then \(2p\). The total number of electrons in \(1s^21p^5\) is \(2 + 5=7\). After filling \(1s^2\) (2 electrons), we fill \(2s^2\) (2 electrons), then we have \(7-(2 + 2)=3\) electrons left, which go to \(2p^3\).

Step 2: Construct the correct configuration

So the correct configuration is \(1s^22s^22p^3\) (written as \(1s^22s^22p^3\) without spaces).

Part E: Problem \(1s^22s^22p^63s^23p^1\) (wait, the original is \(1s^22s^22p^63s^23p^1\)? Wait, no, the user's Part E is \(1s^22s^22p^63s^23p^1\)? Wait, no, the user's Part E says "Enter the correct ground - state... \(1s^22s^22p^63s^23p^1\)"? Wait, no, the original in the image for Part E is \(1s^22s^22p^63s^23p^1\)? Wait, no, the user's Part E is "Enter the correct ground - state (or lowest energy) configuration based on the number of electrons: \(1s^22s^22p^63s^23p^1\)". Wait, but maybe there was a typo, but if the given is \(1s^22s^22p^63s^23p^1\), let's check. The total number of electrons: \(2+2 + 6+2+1=13\), which is aluminum. The electron configuration for aluminum is \(1s^22s^22p^63s^23p^1\), which is correct. Wait, but maybe the original had a typo, but if we assume the given is to correct, but if it's already correct? Wait, no, maybe the user's image has a typo, but let's proceed.

(Note: Since the user's problem has multiple parts, we can provide solutions for each part as above. For the purpose of this answer, let's take the first part (the one with \(1s^22s^22p^9\)) as an example of the solution format.)

Answer:

(for the first part: \(1s^22s^22p^9\) correction):
\(1s^22s^22p^63s^23p^1\)

(For Part C: \(1s^22s^22p^62d^4\) correction)