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Chapter 8: Q20 E (page 448)

Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.)

Short Answer

Expert verified

Two carbon atoms have been \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) hybridised, two carbon atoms have been \({\rm{s}}{{\rm{p}}^{\rm{2}}}\)hybridised, and two carbon atoms have been \({\rm{sp}}\) hybridised.

Step by step solution

01

Define atom

Atom is the fundamental unit of any element. A.It cannot be broken down further by chemical processes.

02

Explanation

This is the structure created by linking the atoms with the proper number of bonds, and each carbon atom's hybridization is indicated below.

Therefore, \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) hybridization of two carbons, \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridization of two carbons, and \({\rm{sp}}\) hybridization of two carbons.

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Most popular questions from this chapter

Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds: (a) \({\rm{ClNO}}\) (\({\rm{N}}\) is the central atom) (b) \({\rm{C}}{{\rm{S}}_{\rm{2}}}\) (c) \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{CO}}\) (\({\rm{C}}\) is the central atom) (d) \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{SO}}\) (\({\rm{S}}\) is the central atom) (e) \({\rm{S}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}\) (\({\rm{S}}\) is the central atom) (f) \({\rm{Xe}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}\) (\({\rm{Xe}}\) is the central atom) (g) \({\rm{ClO}}{{\rm{F}}_{\rm{2}}}^{\rm{ + }}\) (\({\rm{Cl}}\) is the central atom).

Using the MO diagrams, predict the bond order for the stronger bond in each pair:

B2or B2+

F2orF2+

O2 or O22+

C2+or C2-

Draw the orbital diagram for carbon in showing how many carbon atom electrons are in each orbital.

How many unpaired electrons would be present on a Be22- ion? Would it be paramagnetic or diamagnetic?

What charge would be needed on \({{\rm{F}}_{\rm{2}}}\) to generate an ion with a bond order of \({\rm{2}}\)?
See all solutions

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