hybridization

The trick to this problem is to look for implicit hydrogens- hydrogens (protons) that are not drawn in, but assumed to be there because there are no charges.

Also remember these rules:

• If an atom has no multiple bonds, the total number of "things" (bonds and lone pairs) surrounding the atom is probably four, so it's sp³ hybridized (s¹ + p³ = 4).
• If an atom has one double bond, the total number of "things"surrounding the atom is probably three, so it's sp² hybridized (s¹ + p² = 3).
• If an atom has one triple bond or two double bonds, the total number of "things" surrounding the atom is probably two, so it's sp hybridized (s¹ + p¹ = 2).
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A sp² This oxygen has one double bond and two lone pairs. Three things total = sp².

B sp² This carbon has one double bond and two single bonds, to a carbon and an oxygen (so two bonds). Three things total = sp².

C sp³ This nitrogen has one long pair, two implicit hydrogens (two bonds), and one single bond to a carbo (another bond). Four things total = sp³.

D sp³ This carbon has one implicit hydrogen and three single bonds. Four things total = sp³.

E sp³ This oxygen has two lone pairs and two bonding pairs. Four things total = sp³.

F sp² This carbon has one double bond (one bond), one implicit hydrogen (another bond), and is bonded to one other carbon (a third bond). So three bonds in total. Three things total = sp².

G sp This carbon has a triple bond (one bond) and a single bond (another bond). So two things in total = sp.

H sp This nitrogen has a lone pair and a triple bond. Two things total = sp.

For angles, remember that electrons repel each other, and so will try to be as far apart from each other as possible:

• Best way to separate two things is a straight line (180°).
• Best way to separate three things is a "trifecta" (120°).
• Best way to separate four things in three dimensional space is a tetrahedron (109.5°).

X 120° The carbon between the two oxygens is sp² hybridized, so the shape is trigonal planar and the bond angle is 120°.

Y 180° The carbon bonded to the nitrogen is sp hybridized, so the shape is linear and the bond angle is 180°.

Z 109.5° The carbon is sp³ hybridized (the most common type of carbon). So the shape is tetrahedral and the bond angle is 109.5°.

The alkyne (triple bond) is the most stable and therefore the least basic. This is because it is sp hybridized.

An s-orbital is closer to the nucleus and more electronegative than a p-orbital, and an sp hybridized atom has 50% s-character. 

Alkenes are sp² hybridized and have 33% s-character, and alkanes are sp³ hybridized and have 25% s-character.

Because sp hybridized carbons have the most s-character, they are more electronegative and are better at stabilizing negative formal charges than sp² or sp³ carbons are.

Therefore, the alkane (sp³) is the least stable/strongest base, while the alkyne (sp) is the most stable/weakest base.