Question

procedure:
for each molecule in the table below, list the number of total valence electrons in the molecule and draw
the lewis structure.
cf₄
nh₃
so₄²⁻
nh₄⁺
h₂s
po₄³⁻
hcl
nf₃
sih₄
bf₃

Explanation:

Step1: Calculate valence electrons for \(NH_3\)

N has 5 valence electrons and each H has 1. So \(5 + 3\times1=8\)

Step2: Draw Lewis structure for \(NH_3\)

N is central atom. It forms 3 single bonds with H and has 1 lone pair.

Step3: Calculate valence electrons for \(HCl\)

H has 1 and Cl has 7. So \(1 + 7=8\)

Step4: Draw Lewis structure for \(HCl\)

H and Cl form a single bond. Cl has 3 lone pairs.

Step5: Calculate valence electrons for \(SO_4^{2 - }\)

S has 6, each O has 6 and add 2 for charge. So \(6+4\times6 + 2=32\)

Step6: Draw Lewis structure for \(SO_4^{2 - }\)

S is central atom. It can form double bonds with some O (resonance structures possible).

Step7: Calculate valence electrons for \(NF_3\)

N has 5, each F has 7. So \(5+3\times7 = 26\)

Step8: Draw Lewis structure for \(NF_3\)

N is central atom. It forms 3 single bonds with F and has 1 lone pair.

Step9: Calculate valence electrons for \(NH_4^{+}\)

N has 5, each H has 1, subtract 1 for charge. So \(5+4\times1-1 = 8\)

Step10: Draw Lewis structure for \(NH_4^{+}\)

N is central atom. It forms 4 single bonds with H (no lone pairs as it's a cation).

Step11: Calculate valence electrons for \(SiH_4\)

Si has 4, each H has 1. So \(4+4\times1=8\)

Step12: Draw Lewis structure for \(SiH_4\)

Si is central atom. It forms 4 single bonds with H.

Step13: Calculate valence electrons for \(H_2S\)

S has 6, each H has 1. So \(6+2\times1=8\)

Step14: Draw Lewis structure for \(H_2S\)

S is central atom. It forms 2 single bonds with H and has 2 lone pairs.

Step15: Calculate valence electrons for \(PO_4^{3 - }\)

P has 5, each O has 6, add 3 for charge. So \(5+4\times6+3 = 32\)

Step16: Draw Lewis structure for \(PO_4^{3 - }\)

P is central atom. It can form double bonds with some O (resonance structures possible).

Step17: Calculate valence electrons for \(BF_3\)

B has 3, each F has 7. So \(3+3\times7=24\)

Step18: Draw Lewis structure for \(BF_3\)

B is central atom. It forms 3 single bonds with F (B has incomplete octet here as it's a common exception).

Answer:

Molecule	Valence Electrons	Lewis Structure
\(HCl\)	\(8\)	\(\ce{H - Cl}\) (with 3 lone pairs on Cl)
\(SO_4^{2 - }\)	\(32\)	Resonance structures (e.g., \(\ce{O = \underset{\overset{\ce{O}}{\mid}}{S} - O^{-}}\) with double - bonded O and single - bonded O with negative charge)
\(NF_3\)	\(26\)	\(\ce{F - \underset{\overset{\ce{F}}{\mid}}{N} - F}\) (with a lone pair on N)
\(NH_4^{+}\)	\(8\)	\(\ce{H - \underset{\overset{\ce{H}}{\mid}}{N^{+}} - H}\) (all single bonds, no lone pairs)
\(SiH_4\)	\(8\)	\(\ce{H - \underset{\overset{\ce{H}}{\mid}}{Si} - H}\) (all single bonds)
\(H_2S\)	\(8\)	\(\ce{H - S - H}\) (with 2 lone pairs on S)
\(PO_4^{3 - }\)	\(32\)	Resonance structures (e.g., \(\ce{O = \underset{\overset{\ce{O}}{\mid}}{P} - O^{-}}\) with double - bonded O and single - bonded O with negative charge)
\(BF_3\)	\(24\)	\(\ce{F - B - F}\) (all single bonds, B with incomplete octet)