Solution Found!
Write the Lewis structure for each molecule(a) N2O(oxygen is terminal)(b) SiH4(c) CI4(d)
Chapter 10, Problem 50P(choose chapter or problem)
Write the Lewis structure for each molecule.
(a) \(\mathrm{N}_{2} \mathrm{O}\) (oxygen is terminal)
(b) \(\mathrm{SiH}_{4}\)
(c) \(\mathrm{CI}_{4}\)
(d) \(\mathrm{Cl}_{2} \mathrm{CO}\) (carbon is central)
Equation Transcription:
Text Transcription:
N_2 O
SiH_4
CI_4
Cl_2 CO
Questions & Answers
QUESTION:
Write the Lewis structure for each molecule.
(a) \(\mathrm{N}_{2} \mathrm{O}\) (oxygen is terminal)
(b) \(\mathrm{SiH}_{4}\)
(c) \(\mathrm{CI}_{4}\)
(d) \(\mathrm{Cl}_{2} \mathrm{CO}\) (carbon is central)
Equation Transcription:
Text Transcription:
N_2 O
SiH_4
CI_4
Cl_2 CO
ANSWER:Solution 50P
Here, we are going to write the Lewis structure for each molecule.
Lewis structure of an element is the representation of the chemical symbol surrounded by dots which represent the valence electrons.
Rules for construct a Lewis structure.
- Calculation of valence electrons are in the molecule that needs to be shown on the Lewis Structure. For a molecule having charge, add an electron for every negative charge and remove an electron for every positive charge.
- The single bonds are drawn in initial framework, called the skeleton, of the molecule.
- The non-central atoms i.e. the terminal atoms should be fulfilled octet by using the lone-pairs of electrons.
- Distribute the electrons among atoms to fulfill the octet or duet for Hydrogen atom.
- If there are extra lone-pair electrons and the octet rule is not filled for the central atom, use the extra electrons to form double or triple bonds around the central atom.
- the formal charge of each atom should be known.
- N2O(oxygen is terminal)
Step 1: Calculation of total number of valence electrons for the molecule to draw Lewis structure.
Valence electron of N = 5
Valence electron of O =6
Total number of valence electrons in N2O molecule = 5 x 2 + 6 = 16 e