Question

part c
enter the correct ground - state (or lowest energy) configuration based on the number of electrons: $1s^2 2s^2 2p^6 2d^4$.
express your answer in complete form in the order of orbital filling as a string without blank space between orbitals. for example, $1s^2 2s^2$ should be entered as 1s^22s^2.

Explanation:

Step1: Recall orbital filling order

Orbitals fill in the order \(1s\), \(2s\), \(2p\), \(3s\), \(3p\), \(4s\), \(3d\), etc. The \(d\)-orbitals start at \(n = 3\) (i.e., \(3d\)), so \(2d\) is not a valid orbital (since for \(n = 2\), the orbitals are \(2s\) and \(2p\); \(d\)-orbitals require \(n\geq3\)). The given incorrect configuration has \(2d^4\), which should be corrected by moving those electrons to the next valid orbitals. After \(2p^6\), the next orbital is \(3s\), then \(3p\), then \(4s\), then \(3d\). But let's count the electrons: \(1s^22s^22p^6\) has \(2 + 2+ 6=10\) electrons, plus \(4\) more (from \(2d^4\)) gives \(14\) electrons. Wait, no: the original incorrect configuration is \(1s^22s^22p^62d^4\), but \(2d\) doesn't exist. The correct filling after \(2p^6\) (which is Ne, 10 electrons) is \(3s^23p^2\)? Wait, no, wait: the electron count: \(1s^2\) (2) + \(2s^2\) (2) + \(2p^6\) (6) + \(2d^4\) (4) – but \(2d\) is invalid. The correct orbitals for \(n = 2\) are \(s\) and \(p\); \(d\) starts at \(n = 3\). So the \(4\) electrons in \(2d^4\) should be placed in the next available orbitals. The order after \(2p\) is \(3s\), \(3p\), \(4s\), \(3d\). Wait, maybe the intended element is Si? Wait, no. Wait, let's re-express. The incorrect \(2d^4\) should be replaced with the correct orbitals. The correct ground - state configuration for the number of electrons here: \(1s^22s^22p^6\) is 10 electrons, plus 4 electrons. The next orbitals after \(2p\) are \(3s\) (can hold 2) and then \(3p\) (can hold 6). So 10 + 2 (3s) + 2 (3p) = 14? Wait, no, the electron count from \(1s^22s^22p^62d^4\) is \(2 + 2+6 + 4=14\) electrons. The correct configuration for 14 electrons: \(1s^22s^22p^63s^23p^2\)? Wait, no, 14 electrons: \(1s^2\) (2) + \(2s^2\) (2) + \(2p^6\) (6) + \(3s^2\) (2) + \(3p^2\) (2) – total 14. But wait, the problem is to correct the \(2d^4\) to valid orbitals. Wait, maybe the original mistake is using \(2d\) instead of \(3d\)? No, \(3d\) comes after \(4s\). Wait, no, the correct orbital filling order is \(1s\), \(2s\), \(2p\), \(3s\), \(3p\), \(4s\), \(3d\), \(4p\), etc. So the \(2d\) is invalid. So we need to replace \(2d^4\) with the correct orbitals. The electrons in \(2d^4\) should be placed in \(3s^23p^2\)? Wait, no, let's count again. The given configuration is \(1s^22s^22p^62d^4\). The total number of electrons is \(2 + 2+6 + 4 = 14\). The correct ground - state electron configuration for 14 electrons (which is silicon, Si) is \(1s^22s^22p^63s^23p^2\). But wait, the problem says "Enter the correct ground - state (or lowest energy) configuration based on the number of electrons: \(1s^22s^22p^62d^4\)". So we need to correct the \(2d^4\) to the appropriate orbitals. Since \(2d\) is not a valid orbital, we move those 4 electrons to the next available orbitals. The order is \(3s\) (2 electrons), then \(3p\) (2 electrons, since we have 4 electrons to place: 2 in \(3s\) and 2 in \(3p\)). So the correct configuration is \(1s^22s^22p^63s^23p^2\), which in the required format is \(1s^22s^22p^63s^23p^2\) (written as \(1s^22s^22p^63s^23p^2\) without spaces, so \(1s^22s^22p^63s^23p^2\)). Wait, but let's check the electron count: \(2+2 + 6+2+2=14\), which matches the incorrect configuration's electron count (2 + 2+6 + 4 = 14, since we moved the 4 electrons from \(2d^4\) to \(3s^23p^2\)).

Step2: Write the correct configuration

The correct orbital filling order means we replace the invalid \(2d^4\) with the correct orbitals. The correct configuration is \(1s^22s^22p^63s^23p^2\), which is written as \(1s^22s^22p^63s^23p^2\) (no spaces between orbitals as per the example).

Answer:

\(1s^22s^22p^63s^23p^2\)