Question

what ground state to excited state combination gives sp hybridization with an ab₂ domain classification vsepr structure? multiple choice: ground --> excited. hint: b bonding groups are single arrows. e lone pairs are double arrows.
format: for example, sp3 is typed in as sp3 with no spaces and no caps.
images of electron configurations and options: e to f, s to t, c to d, a to b

Explanation:

Step1: Recall sp hybridization

sp hybridization occurs when one s orbital and one p orbital mix to form two sp hybrid orbitals. For \( AB_2 \) (linear, 2 bonding domains, 0 lone pairs), the central atom should have 2 unpaired electrons (for bonding) in the ground to excited state transition.

Step2: Analyze electron configurations

- **Option e to f? Wait, no, let's check each option:**
- **a to b**: Check electron configs. a: 1s (paired), 2s (paired), 2p (two unpaired? Wait, no, the diagrams: a has 1s with two arrows (paired), 2s with two arrows (paired), 2p with two single arrows? Wait, no, the problem's hint: B (bonding) are single arrows, E (lone pairs) double. Wait, sp hybridization needs one s and one p to hybridize. Let's look at the ground (left) and excited (right) for each option.
- **c to d**: c: 1s (paired), 2s (paired), 2p (empty? No, c has 1s with two, 2s with two, 2p empty. d: 1s (paired), 2s (one), 2p (two single? No, d's 2s is one, 2p three? No, maybe not.
- **s to t**: s: 1s (paired), 2s (paired), 2p (four? No, s has 1s, 2s, 2p with four? No, t: 1s (paired), 2s (paired), 2p (four? No.
- **e to f? Wait, the options are e to f? No, the options are e to f? Wait, the options given: e to f? Wait, the options are:
- e to f? No, the options are:
- e to f? Wait, the options are:
- e to f (no, the options are "e to f", "s to t", "c to d", "a to b". Wait, let's re-express:

Wait, sp hybridization: central atom has 2 bonding domains (so 2 unpaired electrons for bonding). Let's check the electron configurations:

- **a to b**: Ground (a): 1s (paired, double arrow), 2s (paired, double), 2p (two single arrows? Wait, a's 2p: two single? No, a's 2p has two single? Wait, no, the diagram: a: 1s (↑↓), 2s (↑↓), 2p (↑, ↑, _) [wait, no, the 2p has three orbitals. Wait, a: 1s (paired), 2s (paired), 2p: two orbitals with single arrows, one empty? Then b: 1s (paired), 2s (single arrow), 2p: three single arrows? No, that would be sp² or sp³. Wait, no.

Wait, sp hybridization requires that one electron from 2s is promoted to 2p, so that we have one s and one p (total two orbitals) for hybridization. Wait, ground state: 2s² 2p⁰ (but no, for sp, we need 2s¹ 2p¹ (after excitation). Wait, let's look at the option **c to d**:

Wait, c: ground state: 1s² 2s² 2p⁰ (1s: ↑↓, 2s: ↑↓, 2p: empty). d: excited state: 1s² 2s¹ 2p¹ (1s: ↑↓, 2s: ↑, 2p: ↑, _) [wait, no, d's 2p has two? No, d's 2p has three? Wait, maybe not. Wait, the correct option: sp hybridization for \( AB_2 \) (linear, 2 bonding domains, 0 lone pairs) means the central atom has 2 unpaired electrons (for two bonds). Let's check the ground (left) and excited (right) for each option:

- **c to d**: c (ground): 1s² (↑↓), 2s² (↑↓), 2p⁰ (empty). d (excited): 1s² (↑↓), 2s¹ (↑), 2p¹ (↑), and 2p²? No, d's 2p has three? Wait, no, maybe c to d: promoting one electron from 2s to 2p, so 2s¹ 2p¹, which can hybridize to sp (two orbitals, each with one electron for bonding). Wait, but let's check the other options:

- **a to b**: a (ground): 1s², 2s², 2p² (two unpaired in 2p). b (excited): 1s², 2s¹, 2p³ (three unpaired). That's sp³? No, sp²? No.

- **e to f**: e (ground): 1s², 2s², 2p⁴ (four electrons? No, e's 2p has four? No, e: 1s (↑↓), 2s (↑↓), 2p (↑↓, ↑, ↑) – four electrons. f: 1s (↑↓), 2s (↑↓), 2p (↑↓, ↑, ↑, ↑) – five? No.

- **s to t**: s: 1s², 2s², 2p⁴. t: 1s², 2s², 2p⁴? No.

Wait, the correct one should be c to d? Wait, no, let's re-express: sp hybridization needs one s and one p. So ground state: 2s² 2p⁰ (c), excited state: 2s¹ 2p¹ (d), so that we have two orbitals (sp) with one electron each for bonding (AB₂, two bonds). So c to d is the transition where one electron from 2s (paired) is promoted to 2p (empty), making 2s¹ 2p¹, which can hybridize to sp (two orbitals, each with one electron, forming two bonds, AB₂ linear, sp hybridization).

Wait, but let's check the options. The options are:

- e to f: No.

- s to t: No.

- c to d: Yes, because c is ground (2s², 2p⁰), d is excited (2s¹, 2p¹), so promoting one electron from 2s to 2p, allowing sp hybridization (one s and one p orbitals mix, forming two sp orbitals, each with one electron for bonding, giving AB₂ (linear) with two bonding domains, sp hybridization.

Wait, but let's confirm: sp hybridization: number of hybrid orbitals = number of bonding + lone pairs. AB₂ has 2 bonding, 0 lone pairs, so 2 hybrid orbitals, which is sp (1s + 1p = 2 orbitals). So the central atom must have 2 unpaired electrons (for two bonds) after excitation. In c (ground), 2s is paired (no unpaired), 2p is empty. In d (excited), 2s has one (unpaired), 2p has one (unpaired), so total two unpaired electrons, which can form two bonds (AB₂), and the hybridization is sp (since one s and one p are used to make two hybrid orbitals).

So the correct option is c to d.

Answer:

c to d