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Write the Lewis structure for each molecule(a) N2O(oxygen

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Problem 50P Chapter 10

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 50P

PROBLEM 50P

Write the Lewis structure for each molecule

(a) N2O(oxygen is terminal)

(b) SiH4

(c) CI4

(d) Cl2CO (carbon is central)

Step-by-Step Solution:

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

Step 2: Distribution of electrons around the atoms to fulfil the octet or duet.

Step 3:  Each atom should fulfil the octet.

Here, each atom in structure in (I) and (II) satisfy the octet .

Therefore, the Lewis structure for N2O(oxygen is terminal) is

(b) SiH4

Step 1:  Calculation of total number of valence electrons for the molecule to draw Lewis structure.

Valence electron of Si = 4

Valence electron of H =1

Total number of valence electrons in SiH4  molecule = 4 + 4x1 = 8 e

Step 2: Distribution of electrons around the atoms to fulfil the octet or duet.

Step 3:  Each atom should fulfil the octet.

Here, each atom H satisfies the duet and Si satisfies the octet. .

Therefore, the Lewis structure for SiH4 is

(c) CI4

Step 1:  Calculation of total number of valence electrons for the molecule to draw Lewis structure.

Valence electron of C = 4

Valence electron of I =7

Total number of valence electrons in SiH4  molecule = 4 + 4x7 = 32 e

Step 2: Distribution of electrons around the atoms to fulfil the octet or duet.

Step 3:  Each atom should fulfil the octet.

Here, each atom I and C satisfy the octet. .

Therefore, the Lewis structure for CI4 is

(d) Cl2CO (carbon is central)

Step 1:  Calculation of total number of valence electrons for the molecule to draw Lewis structure.

Valence electron of C = 4

Valence electron of Cl =7

Valence electron of O =6

Total number of valence electrons in Cl2CO  molecule = 2 x7 + 4+6 = 24 e

Step 2: Distribution of electrons around the atoms to fulfil the octet or duet.

Step 3 of 3

Chapter 10, Problem 50P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Write the Lewis structure for each molecule(a) N2O(oxygen

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