Question

many choice question
why can the p atom have more than four bonds, and therefore more than eight electrons?
(choose all that apply)
a.
it’s in n=3, so it can use \p\ orbitals
b.
it’s in n=2, so it can use \d\ orbitals
c.
it’s in n=3, so it can use \d\ orbitals
d.
it’s in n=3, so it can use \s\ orbitals
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Explanation:

1. Analyze the principal quantum number (n) of P: Phosphorus (P) has an atomic number of 15, and its electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^3\). The principal quantum number \(n = 3\) for the valence shell (3s, 3p) and also, for \(n=3\), there are empty 3d orbitals available.
2. Evaluate Option A: Using "p" orbitals alone doesn't explain exceeding the octet (since 3p has 3 orbitals, which with 3s gives a total of 4 orbitals for octet). So A is incorrect.
3. Evaluate Option B: P is in \(n = 3\), not \(n=2\), and \(n=2\) has no d - orbitals (\(l\) (azimuthal quantum number) for d is 2, and for \(n = 2\), \(l\) can be 0 (s) or 1 (p)). So B is incorrect.
4. Evaluate Option C: Since P is in \(n=3\), it has access to 3d orbitals. These empty d - orbitals can be used for hybridization (e.g., \(sp^3d\) hybridization) which allows P to form more than four bonds (and thus have more than 8 electrons around it). So C is correct.
5. Evaluate Option D: Using "s" orbitals alone (3s) doesn't explain the ability to form more than four bonds. The key is the availability of d - orbitals for expansion of the octet. So D is incorrect.

Answer:

C. It’s in N = 3, so it can use "d" orbitals