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

Solution for problem 15Q Chapter 10

Solved: Give the bond angles for each of the geometries in the preceding question

Introductory Chemistry | 5th Edition


Problem 15Q

Give the bond angles for each of the geometries in the preceding question.

Accepted Solution
Step-by-Step Solution:
Step 1 of 3

Solution 15Q

Here, we are going to determine the bond angles for each geometry.

The molecular geometry of a molecule depends on (i)  number of electron group around the central atom (ii) Number of bonding groups and lone pairs.

(a) two electron groups

Both the electron group is bonding pair electron.

The geometry of the molecule having two electron groups is Linear.

For example: BeH2, HCN and CO2

Therefore, the bond angle for Linear geometry is 180o.

(b) three electron groups

Three  electron groups are bonding pair of electrons.

The geometry of the molecule having three electron groups is Trigonal Planar.

For example: COCl2, CO32-

Therefore, the bond angle for Trigonal planar geometry is 120o.

(c) four electron groups

Here, Four electron groups are bonding pair of electrons.

The geometry of the molecule having four electron groups is Tetrahedral.

For example: CH4,

Therefore, the bond angle for Tetrahedral geometry is 109.5o.

Chapter 10, Problem 15Q is Solved

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Solved: Give the bond angles for each of the geometries in the preceding question