Q.Draw the Lewis structures for the following molecules and ions: $H_2S$, $SiCl_4$, $BeF_2$, $CO_3^{2-}$, HCOOH.

Concept: Lewis Dot Structures

Reasoning steps:

1. $H_2S$ – S has 6 valence electrons, each H has 1. Total = 8. S is central, forms single bonds with two H atoms, leaving 2 lone pairs on S.

   Structure: H–S–H with two lone pairs on S.

2. $SiCl_4$ – Si has 4 valence electrons, each Cl has 7. Total = 32. Si is central, forms four single bonds with Cl atoms. No lone pairs on Si; each Cl has 3 lone pairs.

   Structure: tetrahedral, Si bonded to four Cl atoms.

3. $BeF_2$ – Be has 2 valence electrons, each F has 7. Total = 16. Be is central, forms two single bonds with F atoms. Be has no lone pairs; each F has 3 lone pairs.

   Structure: linear, F–Be–F.

4. $CO_3^{2-}$ – C has 4, each O has 6, plus 2 from charge = 24 total. C is central, forms one double bond with one O and single bonds with the other two O atoms. The two singly bonded O atoms each carry a negative charge. All O atoms have 3 lone pairs.

   Structure: resonance hybrid with one C=O and two C–O⁻. …

Lewis dot structures are a visual shorthand for how atoms bond in a molecule. The core idea is simple: atoms share or transfer electrons to achieve a stable octet (or duet for hydrogen). You count all valence electrons, place the least electronegative atom in the center (except hydrogen, which is always terminal), connect atoms with single bonds, then distribute remaining electrons to satisfy octets. If electrons run short, form multiple bonds.

Let’s work through each molecule step by step.

1. $H_2S$ (Hydrogen sulfide)

Step 1: Count valence electrons.

Sulfur (Group 16) has 6 valence electrons. Each hydrogen (Group 1) has 1. Total = $6 + 2 \times 1 = 8$ electrons.

Step 2: Arrange atoms.

Sulfur is the central atom (less electronegative than H, and H is always terminal). Place S in the center, with two H atoms bonded to it.

Step 3: Draw single bonds.

Each S–H bond uses 2 electrons. Two bonds use $2 \times 2 = 4$ electrons. Remaining electrons: $8 - 4 = 4$ electrons (2 lone pairs).

Step 4: Complete octets.

Sulfur already has 2 bonds (4 electrons from bonds). Add the 2 lone pairs (4 electrons) to give sulfur a full octet. Each hydrogen has 2 electrons (duet) from its bond — satisfied.

Final structure:

With two lone pairs on S (not shown in ASCII, but imagine two dots above and below S). This is a bent shape (VSEPR), but the Lewis structure is straightforward.

2. $SiCl_4$ (Silicon tetrachloride)

Step 1: Count valence electrons.

Silicon (Group 14) has 4 valence electrons. Each chlorine (Group 17) has 7. Total = $4 + 4 \times 7 = 32$ electrons.

Step 2: Arrange atoms.

Silicon is less electronegative than chlorine, so Si is central. Four Cl atoms surround it.

Step 3: Draw single bonds.

Four Si–Cl bonds use $4 \times 2 = 8$ electrons. Remaining: $32 - 8 = 24$ electrons.

Step 4: Complete octets.

Each chlorine needs 6 more electrons (3 lone pairs) to complete its octet. Four chlorines require $4 \times 6 = 24$ electrons — exactly what remains. Silicon already has 8 electrons from its four bonds (octet satisfied).

Final structure:

Each Cl has three lone pairs (not drawn). No multiple bonds needed.

3. $BeF_2$ (Beryllium fluoride)

Step 1: Count valence electrons.

Beryllium (Group 2) has 2 valence electrons. Each fluorine (Group 17) has 7. Total = $2 + 2 \times 7 = 16$ electrons.

Step 2: Arrange atoms.

Beryllium is central (less electronegative), with two F atoms on either side.

Step 3: Draw single bonds.

Two Be–F bonds use $2 \times 2 = 4$ electrons. Remaining: $16 - 4 = 12$ electrons.

Step 4: Complete octets.

Each fluorine needs 6 more electrons (3 lone pairs). Two fluorines need $2 \times 6 = 12$ electrons — exactly what’s left. Beryllium, however, only has 4 electrons from its two bonds. It does not have an octet.

Final structure:

Each F has three lone pairs. Be has no lone pairs. The molecule is linear.

4. $CO_3^{2-}$ (Carbonate ion)

Step 1: Count valence electrons.

Carbon (Group 14) has 4. Each oxygen (Group 16) has 6. The 2− charge adds 2 more electrons. Total = $4 + 3 \times 6 + 2 = 24$ electrons.

Step 2: Arrange atoms.

Carbon is central (less electronegative than oxygen). Three oxygen atoms surround it.

Step 3: Draw single bonds.

Three C–O single bonds use $3 \times 2 = 6$ electrons. Remaining: $24 - 6 = 18$ electrons.

Step 4: Complete octets (tentative).

Give each oxygen 6 more electrons (3 lone pairs) to complete its octet. That uses $3 \times 6 = 18$ electrons — exactly what’s left. But now carbon has only 6 electrons (from three bonds) — it needs 2 more.

Step 5: Form a double bond.

To give carbon an octet, convert one lone pair from an oxygen into a second bond with carbon. This creates a C=O double bond. Now carbon has 8 electrons. The oxygen that lost a lone pair now has 2 bonds and 2 lone pairs (still 8 electrons). The other two oxygens each have 1 bond and 3 lone pairs (8 electrons each).

Step 6: Resonance. …