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Let’s go step by step for each molecule. We’ll use VSEPR theory to find:

1. Electron Dot Structure – Show all valence electrons (dots) around atoms.
2. Ball Diagram – Simple sketch showing central atom and bonded atoms (we’ll describe it in words since we can’t draw).
3. Molecular Shape (Drawing) – The actual 3D shape based on electron pairs.
4. Molecular Shape (Name) – Name of the shape (like linear, trigonal planar, etc.).

We’ll do this for each compound one at a time.

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1. SH₄

Wait — sulfur (S) has 6 valence electrons. Hydrogen (H) has 1 each → 4 H = 4 electrons. Total = 10 electrons? But that doesn’t make sense because S can’t have 5 bonds normally. Actually, SH₄ does not exist as a stable molecule. Sulfur usually forms H₂S. But if we force it for homework, maybe they mean SF₄ or something else? Let’s check common molecules.

Actually, looking again — probably typo? Maybe it’s SF₄? Or perhaps SiH₄? SiH₄ is silane, very common.

But let’s assume it’s SiH₄ (silicon tetrahydride), which is real and follows octet rule.

If it’s really SH₄ — sulfur would have 4 bonds + 1 lone pair → 5 electron domains → seesaw shape. But SH₄ isn’t stable. I think it’s likely a typo and should be SiH₄.

Let me confirm with standard curriculum: In many worksheets, #1 is often CH₄ or SiH₄. Since C is not listed, maybe SiH₄.

Alternatively, maybe it’s PH₄⁺? No, charge not shown.

I’ll proceed assuming it’s SiH₄, because SH₄ is not a real neutral molecule under normal conditions.

But wait — let’s double-check: Sulfur *can* form SH₄ in extreme conditions, but it’s rare. For school level, it’s probably a mistake.

Looking at other compounds: BH₃, BCl₃, AsH₅ — these are all valid except AsH₅ (arsine is AsH₃; AsH₅ is unstable). So maybe this worksheet includes some theoretical or less common ones.

To avoid confusion, I’ll treat them as written, even if unstable, and apply VSEPR rules strictly.

So for SH₄:

- Sulfur (Group 16) → 6 valence e⁻
- 4 H → 4 × 1 = 4 e⁻
- Total valence e⁻ = 10
- Each bond uses 2 e⁻ → 4 bonds = 8 e⁻ used
- Remaining 2 e⁻ → 1 lone pair on S
- Electron domains: 4 bonding + 1 lone pair = 5 → trigonal bipyramidal electron geometry
- Molecular shape: seesaw (because one position is lone pair)

But again — SH₄ is not typical. However, for academic exercise, we’ll go with it.

Alternatively, maybe it’s CH₄? But carbon isn't listed. Hmm.

Wait — looking at list: 1. SH₄, 2. H₂Se, 3. BeCl₂, 4. BCl₃, 5. AsH₅, 6. CF₄, 7. OF₂, 8. SiCl₄, 9. NCl₃, 10. CO₃F₂

AsH₅ is also unusual — arsenic typically forms AsH₃. But again, for VSEPR practice, we may consider hypotheticals.

I think best approach: Solve each as written using VSEPR rules, regardless of stability.

So let’s begin properly.

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## Step-by-step for each:

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1. SH₄

- Central atom: S
- Valence e⁻: S=6, H×4=4 → total 10
- Bonds: 4 single bonds → use 8 e⁻
- Leftover: 2 e⁻ → 1 lone pair on S
- Electron domains: 5 (4 bonding + 1 lone pair)
- Electron geometry: trigonal bipyramidal
- Molecular shape: seesaw (lone pair occupies equatorial position)
- Ball diagram: S in center, 4 H around — two axial, two equatorial, one equatorial spot empty (lone pair)
- Drawing name: seesaw
- Shape name: seesaw

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2. H₂Se

- Se (Group 16) → 6 valence e⁻
- H×2 = 2 e⁻
- Total = 8 e⁻
- Two bonds → 4 e⁻ used
- Leftover: 4 e⁻ → 2 lone pairs on Se
- Electron domains: 4 (2 bonding + 2 lone pairs)
- Electron geometry: tetrahedral
- Molecular shape: bent (or angular)
- Ball diagram: Se center, two H attached, two lone pairs
- Drawing: bent
- Name: bent

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3. BeCl₂

- Be (Group 2) → 2 valence e⁻
- Cl×2 = 14 e⁻? Wait — better: total valence e⁻
- Be: 2, Cl: 7 each → 2 + 14 = 16 e⁻
- But Be is exception — only needs 4 e⁻ (duet rule for period 2)
- Forms two single bonds → 4 e⁻ used in bonds
- Each Cl gets 3 lone pairs (6 e⁻ each) → 12 e⁻
- Total: 4 + 12 = 16 — good.
- Electron domains on Be: 2 bonding, 0 lone pairs
- Electron geometry: linear
- Molecular shape: linear
- Ball diagram: Cl—Be—Cl straight line
- Drawing: linear
- Name: linear

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4. BCl₃

- B (Group 13) → 3 valence e⁻
- Cl×3 = 21 e⁻
- Total = 24 e⁻
- Three single bonds → 6 e⁻ used
- Each Cl has 3 lone pairs → 3×6=18 e⁻
- Total: 6+18=24 — good.
- B has no lone pairs, 3 bonding domains
- Electron geometry: trigonal planar
- Molecular shape: trigonal planar
- Ball diagram: B center, three Cl at 120° in plane
- Drawing: flat triangle
- Name: trigonal planar

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5. AsH₅

- As (Group 15) → 5 valence e⁻
- H×5 = 5 e⁻
- Total = 10 e⁻
- Five bonds → 10 e⁻ used → no lone pairs
- Electron domains: 5 bonding
- Electron geometry: trigonal bipyramidal
- Molecular shape: trigonal bipyramidal (all positions occupied by H)
- Ball diagram: As center, 3 H in equatorial plane, 2 H axial up/down
- Drawing: trigonal bipyramid
- Name: trigonal bipyramidal

(Note: AsH₅ is not stable, but for VSEPR, we accept it.)

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6. CF₄

- C (Group 14) → 4 valence e⁻
- F×4 = 28 e⁻
- Total = 32 e⁻
- Four single bonds → 8 e⁻ used
- Each F has 3 lone pairs → 4×6=24 e⁻
- Total: 8+24=32 — good.
- C has 4 bonding domains, 0 lone pairs
- Electron geometry: tetrahedral
- Molecular shape: tetrahedral
- Ball diagram: C center, four F at corners of tetrahedron
- Drawing: tetrahedron
- Name: tetrahedral

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7. OF₂

- O (Group 16) → 6 valence e⁻
- F×2 = 14 e⁻
- Total = 20 e⁻
- Two single bonds → 4 e⁻ used
- O has 2 lone pairs (4 e⁻) → total on O: 4 (bonds) + 4 (lone) = 8
- Each F has 3 lone pairs → 2×6=12 e⁻
- Total: 4 (bonds) + 4 (O lone) + 12 (F lone) = 20 — good.
- Electron domains on O: 2 bonding + 2 lone pairs = 4
- Electron geometry: tetrahedral
- Molecular shape: bent
- Ball diagram: O center, two F attached, two lone pairs
- Drawing: bent
- Name: bent

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8. SiCl₄

- Si (Group 14) → 4 valence e⁻
- Cl×4 = 28 e⁻
- Total = 32 e⁻
- Four single bonds → 8 e⁻ used
- Each Cl has 3 lone pairs → 4×6=24 e⁻
- Total: 8+24=32 — good.
- Si has 4 bonding domains, 0 lone pairs
- Electron geometry: tetrahedral
- Molecular shape: tetrahedral
- Ball diagram: Si center, four Cl at tetrahedral angles
- Drawing: tetrahedron
- Name: tetrahedral

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9. NCl₃

- N (Group 15) → 5 valence e⁻
- Cl×3 = 21 e⁻
- Total = 26 e⁻
- Three single bonds → 6 e⁻ used
- N has 1 lone pair (2 e⁻) → total on N: 6 (bonds) + 2 (lone) = 8
- Each Cl has 3 lone pairs → 3×6=18 e⁻
- Total: 6 + 2 + 18 = 26 — good.
- Electron domains on N: 3 bonding + 1 lone pair = 4
- Electron geometry: tetrahedral
- Molecular shape: trigonal pyramidal
- Ball diagram: N center, three Cl, one lone pair pushing down
- Drawing: pyramid with triangular base
- Name: trigonal pyramidal

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10. CO₃F₂

This is tricky. Carbon is central? Probably.

Carbonate-like but with fluorines? Likely structure: C bonded to 3 O and 2 F? That would be 5 bonds — too many for carbon.

Wait — maybe it’s [CO₃]²⁻ with two F? Not clear.

Perhaps it’s carbonyl fluoride derivative? Common molecule is COF₂ (carbonyl fluoride), but here it’s CO₃F₂.

Maybe it’s a typo? Perhaps COF₂?

Or maybe it’s C with double bond to O, and single bonds to two O⁻ and two F? Too messy.

Another possibility: It might be trifluoroacetate ion or something, but formula doesn’t match.

Wait — perhaps it’s C(O)F₂ with an extra O? Unlikely.

I recall there is a molecule called “carbonic difluoride” but it’s COF₂.

Perhaps it’s CO₃F₂ meaning O=C(OF)₂? Like ester? But then carbon has 4 bonds: double bond O, and two -OF groups.

That could work.

Assume: Central C, double bond to one O, single bond to two O atoms, each of those O bonded to F.

So structure: O=C(–O–F)₂

Then:

- C: 4 valence e⁻
- O (double bond): contributes 4 e⁻? Better calculate total valence.

Total valence electrons:

- C: 4
- O×3: 18
- F×2: 14
- Total: 4+18+14=36 e⁻

Structure: C double bond O (uses 4 e⁻), C single bond O (2 e⁻), that O single bond F (2 e⁻), similarly another C–O–F.

So bonds:

- C=O : 4 e⁻
- C–O : 2 e⁻
- O–F : 2 e⁻
- C–O : 2 e⁻
- O–F : 2 e⁻
Total bonding e⁻: 4+2+2+2+2=12 e⁻

Remaining e⁻: 36-12=24 e⁻ → distributed as lone pairs.

Double-bonded O: 2 lone pairs (4 e⁻)
Each single-bonded O: 2 lone pairs (4 e⁻ each) → 8 e⁻
Each F: 3 lone pairs (6 e⁻ each) → 12 e⁻
Total lone pairs e⁻: 4+8+12=24 — perfect.

Now, electron domains on C:

- Double bond counts as 1 domain
- Two single bonds to O → 2 domains
Total: 3 electron domains on C

No lone pairs on C.

Electron geometry: trigonal planar

Molecular shape: trigonal planar (since all domains are bonding)

Ball diagram: C center, double bond O, and two –O–F groups, all in same plane at ~120°

Drawing: flat triangle with C in center, O (double), O-F, O-F

Name: trigonal planar

Note: The oxygens bonded to F will have their own shapes, but the question asks for molecular shape — probably referring to the overall shape around central atom C.

So yes, trigonal planar.

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Now, compiling answers in table format as requested.

Since we can't draw diagrams, I'll describe ball diagram briefly and give shape names.

Final Answer Table:

| MOLECULE | ELECTRON DOT | BALL DIAGRAM (description) | MOLECULAR SHAPE (Drawing description) | MOLECULAR SHAPE (Name) |
|----------|--------------|-----------------------------|----------------------------------------|-------------------------|
| 1. SH₄ | S with 4 H bonds and 1 lone pair | S center, 4 H: 2 axial, 2 equatorial, 1 equatorial lone pair | Seesaw shape | seesaw |
| 2. H₂Se | Se with 2 H bonds and 2 lone pairs | Se center, 2 H, 2 lone pairs | Bent | bent |
| 3. BeCl₂ | Be with 2 Cl bonds, no lone pairs | Cl—Be—Cl straight line | Linear | linear |
| 4. BCl₃ | B with 3 Cl bonds, no lone pairs | B center, 3 Cl at 120° | Flat triangle | trigonal planar |
| 5. AsH₅ | As with 5 H bonds, no lone pairs | As center, 3 H equatorial, 2 H axial | Trigonal bipyramid | trigonal bipyramidal |
| 6. CF₄ | C with 4 F bonds, no lone pairs | C center, 4 F at tetrahedral angles | Tetrahedron | tetrahedral |
| 7. OF₂ | O with 2 F bonds and 2 lone pairs | O center, 2 F, 2 lone pairs | Bent | bent |
| 8. SiCl₄ | Si with 4 Cl bonds, no lone pairs | Si center, 4 Cl at tetrahedral angles | Tetrahedron | tetrahedral |
| 9. NCl₃ | N with 3 Cl bonds and 1 lone pair | N center, 3 Cl, 1 lone pair | Pyramid with triangular base | trigonal pyramidal |
| 10. CO₃F₂ | C double bond O, single bonds to two O (each bonded to F) | C center, double bond O, two –O–F groups in plane | Flat triangle | trigonal planar |

For Electron Dot column, since we can't draw dots, in actual worksheet you'd draw Lewis structures. Here, I described key features.

In real assignment, student should draw:

- For SH₄: S surrounded by 4 H, and one pair of dots on S.
- Etc.

But since text-based, descriptions suffice.

Final Answer:

The completed table with molecular shapes is as above. For each molecule, the molecular shape name is:

1. SH₄: seesaw
2. H₂Se: bent
3. BeCl₂: linear
4. BCl₃: trigonal planar
5. AsH₅: trigonal bipyramidal
6. CF₄: tetrahedral
7. OF₂: bent
8. SiCl₄: tetrahedral
9. NCl₃: trigonal pyramidal
10. CO₃F₂: trigonal planar

Note: Some molecules like SH₄ and AsH₅ are not commonly stable, but for VSEPR theory exercises, we apply the rules as taught.