molecule or ion,number of valence electrons,total number of atoms and sets of lone pairs surrounding central atom(s),structure geometry (name)
ccl4
h2o
co2
h2o2
ch4
c2h4
ch2o
ch3oh
ch3nh2
nf3
nh3
nh4+
no2-
so2
of2
h2s
sf6
icl5

Question

Accepted Answer

# Explanation:
## Step1: Determine valence - electrons for CCl₄
Carbon has 4 valence electrons and each chlorine has 7. So, $4+(4	imes7)=32$ valence electrons. The central carbon has 4 bonding pairs and 0 lone - pairs, total of 4 electron - groups, and the structure geometry is tetrahedral.
## Step2: Determine valence - electrons for H₂O
Oxygen has 6 valence electrons and each hydrogen has 1. So, $6+(2	imes1)=8$ valence electrons. The central oxygen has 2 bonding pairs and 2 lone - pairs, total of 4 electron - groups, and the structure geometry is bent.
## Step3: Determine valence - electrons for CO₂
Carbon has 4 valence electrons and each oxygen has 6. So, $4+(2	imes6)=16$ valence electrons. The central carbon has 2 bonding pairs and 0 lone - pairs, total of 2 electron - groups, and the structure geometry is linear.
## Step4: Determine valence - electrons for H₂O₂
Each oxygen has 6 valence electrons and each hydrogen has 1. So, $(2	imes6)+(2	imes1)=14$ valence electrons. Each oxygen has 2 bonding pairs and 2 lone - pairs, but considering the central atoms (the two oxygen atoms), the structure is non - linear.
## Step5: Determine valence - electrons for CH₄
Carbon has 4 valence electrons and each hydrogen has 1. So, $4+(4	imes1)=8$ valence electrons. The central carbon has 4 bonding pairs and 0 lone - pairs, total of 4 electron - groups, and the structure geometry is tetrahedral.
## Step6: Determine valence - electrons for C₂H₄
Each carbon has 4 valence electrons and each hydrogen has 1. So, $(2	imes4)+(4	imes1)=12$ valence electrons. Each carbon has 3 bonding pairs and 0 lone - pairs, and the structure geometry is trigonal planar around each carbon.
## Step7: Determine valence - electrons for CH₂O
Carbon has 4 valence electrons, oxygen has 6, and each hydrogen has 1. So, $4 + 6+(2	imes1)=12$ valence electrons. The central carbon has 3 bonding pairs and 0 lone - pairs, and the structure geometry is trigonal planar.
## Step8: Determine valence - electrons for CH₃OH
Carbon has 4 valence electrons, oxygen has 6, and each hydrogen has 1. So, $4+6+(4	imes1)=14$ valence electrons. The central carbon in the methyl part has 4 bonding pairs and 0 lone - pairs (tetrahedral), and the oxygen has 2 bonding pairs and 2 lone - pairs (bent).
## Step9: Determine valence - electrons for CH₃NH₂
Carbon has 4 valence electrons, nitrogen has 5, and each hydrogen has 1. So, $4 + 5+(5	imes1)=14$ valence electrons. The central carbon has 4 bonding pairs and 0 lone - pairs (tetrahedral), and the nitrogen has 3 bonding pairs and 1 lone - pair (trigonal pyramidal).
## Step10: Determine valence - electrons for NF₃
Nitrogen has 5 valence electrons and each fluorine has 7. So, $5+(3	imes7)=26$ valence electrons. The central nitrogen has 3 bonding pairs and 1 lone - pair, total of 4 electron - groups, and the structure geometry is trigonal pyramidal.
## Step11: Determine valence - electrons for NH₃
Nitrogen has 5 valence electrons and each hydrogen has 1. So, $5+(3	imes1)=8$ valence electrons. The central nitrogen has 3 bonding pairs and 1 lone - pair, total of 4 electron - groups, and the structure geometry is trigonal pyramidal.
## Step12: Determine valence - electrons for [NH₄]⁺
Nitrogen has 5 valence electrons and each hydrogen has 1, and we subtract 1 for the positive charge. So, $5+(4	imes1)-1 = 8$ valence electrons. The central nitrogen has 4 bonding pairs and 0 lone - pairs, total of 4 electron - groups, and the structure geometry is tetrahedral.
## Step13: Determine valence - electrons for [NO₂]⁻
Nitrogen has 5 valence electrons, each oxygen has 6, and we add 1 for the negative charge. So, $5+(2	imes6)+1 = 18$ valence electrons. The central nitrogen has 2 bonding pairs and 1 lone - pair, total of 3 electron - groups, and the structure geometry is bent.
## Step14: Determine valence - electrons for SO₂
Sulfur has 6 valence electrons and each oxygen has 6. So, $6+(2	imes6)=18$ valence electrons. The central sulfur has 2 bonding pairs and 1 lone - pair, total of 3 electron - groups, and the structure geometry is bent.
## Step15: Determine valence - electrons for OF₂
Oxygen has 6 valence electrons and each fluorine has 7. So, $6+(2	imes7)=20$ valence electrons. The central oxygen has 2 bonding pairs and 2 lone - pairs, total of 4 electron - groups, and the structure geometry is bent.
## Step16: Determine valence - electrons for H₂S
Sulfur has 6 valence electrons and each hydrogen has 1. So, $6+(2	imes1)=8$ valence electrons. The central sulfur has 2 bonding pairs and 2 lone - pairs, total of 4 electron - groups, and the structure geometry is bent.
## Step17: Determine valence - electrons for SF₆
Sulfur has 6 valence electrons and each fluorine has 7. So, $6+(6	imes7)=48$ valence electrons. The central sulfur has 6 bonding pairs and 0 lone - pairs, total of 6 electron - groups, and the structure geometry is octahedral.
## Step18: Determine valence - electrons for ICl₅
Iodine has 7 valence electrons and each chlorine has 7. So, $7+(5	imes7)=42$ valence electrons. The central iodine has 5 bonding pairs and 1 lone - pair, total of 6 electron - groups, and the structure geometry is square pyramidal.

# Answer:
| Molecule or ion | Number of valence electrons | Total number of atoms and sets of lone pairs surrounding central atom(s) | Structure geometry (name) |
|---|---|---|---|
| CCl₄ | 32 | 4 (4 bonding, 0 lone) | Tetrahedral |
| H₂O | 8 | 4 (2 bonding, 2 lone) | Bent |
| CO₂ | 16 | 2 (2 bonding, 0 lone) | Linear |
| H₂O₂ | 14 | Non - linear structure | Non - linear |
| CH₄ | 8 | 4 (4 bonding, 0 lone) | Tetrahedral |
| C₂H₄ | 12 | 3 (3 bonding, 0 lone) around each C | Trigonal planar around each C |
| CH₂O | 12 | 3 (3 bonding, 0 lone) | Trigonal planar |
| CH₃OH | 14 | Tetrahedral around C in methyl, bent around O | - |
| CH₃NH₂ | 14 | Tetrahedral around C in methyl, trigonal pyramidal around N | - |
| NF₃ | 26 | 4 (3 bonding, 1 lone) | Trigonal pyramidal |
| NH₃ | 8 | 4 (3 bonding, 1 lone) | Trigonal pyramidal |
| [NH₄]⁺ | 8 | 4 (4 bonding, 0 lone) | Tetrahedral |
| [NO₂]⁻ | 18 | 3 (2 bonding, 1 lone) | Bent |
| SO₂ | 18 | 3 (2 bonding, 1 lone) | Bent |
| OF₂ | 20 | 4 (2 bonding, 2 lone) | Bent |
| H₂S | 8 | 4 (2 bonding, 2 lone) | Bent |
| SF₆ | 48 | 6 (6 bonding, 0 lone) | Octahedral |
| ICl₅ | 42 | 6 (5 bonding, 1 lone) | Square pyramidal |

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QUESTION IMAGE

Question

Explanation:

Step1: Determine valence - electrons for CCl₄

Step2: Determine valence - electrons for H₂O

Step3: Determine valence - electrons for CO₂

Step4: Determine valence - electrons for H₂O₂

Step5: Determine valence - electrons for CH₄

Step6: Determine valence - electrons for C₂H₄

Step7: Determine valence - electrons for CH₂O

Step8: Determine valence - electrons for CH₃OH

Step9: Determine valence - electrons for CH₃NH₂

Step10: Determine valence - electrons for NF₃

Step11: Determine valence - electrons for NH₃

Step12: Determine valence - electrons for [NH₄]⁺

Step13: Determine valence - electrons for [NO₂]⁻

Answer:

Molecule or ion	Number of valence electrons	Total number of atoms and sets of lone pairs surrounding central atom(s)	Structure geometry (name)
H₂O	8	4 (2 bonding, 2 lone)	Bent
CO₂	16	2 (2 bonding, 0 lone)	Linear
H₂O₂	14	Non - linear structure	Non - linear
CH₄	8	4 (4 bonding, 0 lone)	Tetrahedral
C₂H₄	12	3 (3 bonding, 0 lone) around each C	Trigonal planar around each C
CH₂O	12	3 (3 bonding, 0 lone)	Trigonal planar
CH₃OH	14	Tetrahedral around C in methyl, bent around O	-
CH₃NH₂	14	Tetrahedral around C in methyl, trigonal pyramidal around N	-
NF₃	26	4 (3 bonding, 1 lone)	Trigonal pyramidal
NH₃	8	4 (3 bonding, 1 lone)	Trigonal pyramidal
[NH₄]⁺	8	4 (4 bonding, 0 lone)	Tetrahedral
[NO₂]⁻	18	3 (2 bonding, 1 lone)	Bent
SO₂	18	3 (2 bonding, 1 lone)	Bent
OF₂	20	4 (2 bonding, 2 lone)	Bent
H₂S	8	4 (2 bonding, 2 lone)	Bent
SF₆	48	6 (6 bonding, 0 lone)	Octahedral
ICl₅	42	6 (5 bonding, 1 lone)	Square pyramidal