Question

Bonding and Hybridization Worksheet 1. For each of the following compounds a. b. c. Give the hybridization for each atom except hydrogen Give the approximate bond angles for each atom except hydrogen Draw an orbital diagram using lines, wedges and dashed lines for sigma bonds; draw the p orbital interaction for pi bonds 1. ?,?. Il. (CH,)N

Answer 1

I. $H_3O^+$

a.

Here, O is the central atom. It has 3 bonds and a +ve charge.

Each H atom contribute +1 electron.

Oxygen has 6 atom in its outer shell if its nuetral. When it has a positive charge it has 6-1 = 5 electrons in its outer shell.

Hence, total number of electrons outside the central atom O is 3+6-1=8.

Hence, the total number of electron pairs around oxygen is 8/2=4 (3 sigma bond pair and 1 lone pair) .

The only possible hybridization for 4 electron pairs on O atom is sp3.

3 of the hybridizzed sp3 orbital from bonds with H and another carries the lone pair.

b.

The electronic geometry of $H_3O^+$ is tetrahedral while the molecular geometry is pyramidal. Hence, the bond angle H-O-H is around $109 ^\circ$ .

c.

II.  $(CH_3)_4N^+$

a.

Here, N is the central atom with 4 methyl groups connected to it.

Each methyl group gives 1e.

N has 5 outermost electrons when neutral. Hence, it has 4 electron when it has +1 charge.

Hence, total number of electrons around N is 4+5-1=8.

Number of electron pairs = 8/2=4.

Only possible hybridization for N with 4 electron pair(all singma bond pairs with C) is sp3.

Similarly, for each C atom in CH3 groups:

3 H contribute 1e each.

1 N contribute 1e.

C has 4 electron in its outermost shell.

Hence, total number of elctrons around C is 3+1+4=8.

Number of electron pairs = 8/2=4.

Hence, only possible hybridization for 4 electron pairs(all sigma bond pairs) is sp3.

b.

The C-N-C bond angle in the tetrahedra geometry is around $109 ^\circ$ .

The H-C-H bond angle or N-C-H bond angle in tetrahedral geometry is also around $109 ^\circ$ .

c.