Can someone help me with hybridization?

(a) The rule for hybridization: All sigma bonds (single bonds and only one bond in double or triple bonds) and lone pairs (nonbonding pairs) are found in hybrid orbitals.

Since it is impossible to draw a structural formula in the answer part, it will be difficult to explain, but I’ll try.

Ethyl acetate: CH3 — C = O

………………………… I

………………………… O — CH2 –CH3

I think you understand the structure.

First C atom has 4 single (sigma) bonds. therefore its type of hybridization is sp3 (1 s and 3 p = 4)

Second C atom has 2 single(sigma) and 1 double bond (one of them is sigma , the other is pi bond). Total sigma bonds : 3. Since only sigma bonds are hybridized, its type of hybridization is sp2 (1 s and 2 p = 4)

Third C atom has 4 single (sigma) bonds. (one with oxygen, one with 4th C atom and 2 with H atoms). Total sigma bonds: 4 and therefore the type of hybridization is sp3.

Finaly the fourth C atom is like the first one and the type of the hybridization is sp3.

(b) Each C atom has 4 valence electrons (Group IV A)

4 C atoms have 16 valence electrons.

Each O atom has 6 valence electrons (Group VI A)

2 O atoms have 12 valence electrons.

Each H atom has 1 valence electron and total 8 H atoms have 8 valence electrons.

Hence, the total number of valence electrons in ethylacetate molecule is 16 + 12 + 8 = 36

(c) All valence electrons of C and H atoms are used to make bonds (16 + 8 = 24). In addition 2 electrons of each O atom are used to make bonds (each O atom has 2 bonds) Valence electrons used by O to make bonds : 4

Total valence electrons used for making bonds are : 28

(d) I think c and d ask the same question.

(e) 28 valence electrons are used to make bonds out of 36, and 8 valence electrons (4 of each O atom) remain in nonbonding pair.

Note: in the molecule there are 14 bonding pair and 4 nonbonding pair (lone pair).

Ethyl Acetate Lewis Structure

your answers are correct expect for letter c. c: 26. There are 8 nonbonding e- and 2 e- used to make the pi bond so the answer is 36-10 = 26

sp2 hybridisation will reult in a bonding pair being used for a sigma bond between the two O atoms, and two lone pairs at approximately 120 degrees to each other (all planar). The unhybridised p orbital, containing one electron and perpendicular to the plane just mentioned, can overlap sideways with its neighbour on the other oxygen atom, forming a pi bond.

See ethyl acetate valence electron at http://www.curvedarrow.com/

PW

sp3, sp2, sp3, sp3

36

26

2

8