Tetrahedral Carbon, Bonding, and Hybridization Practice Questions — JAMB UTME

Question 1Question →

Consider the unsaturated hydrocarbon 2-methylbut-1-en-3-yne, which has the condensed structural formula $CH_2=C(CH_3)C\equiv CH$ . What is the total number of sigma ( $\sigma$ ) bonds present in one molecule of this compound?

Show answer & explanation

Answer: 10; 10 sigma bonds; 10 bonds

Answer

The total number of sigma (

\sigma

) bonds present in one molecule of 2-methylbut-1-en-3-yne is 10.

In 2-methylbut-1-en-3-yne (

CH_2=C(CH_3)C\equiv CH

), there are 6 carbon-hydrogen single bonds (2 from C1, 3 from the methyl group, and 1 from C4) and 4 carbon-carbon sigma bonds (1 from the C1=C2 double bond, 1 connecting C2 to the methyl carbon, 1 connecting C2 to C3, and 1 from the C3\equiv C4 triple bond). Summing these gives 10 sigma bonds in total.

Step-by-Step Solution

1

Expand the condensed structural formula to identify all individual carbon-hydrogen (C-H) bonds.

The

=CH_2

group contains 2 C-H

\sigma

bonds, the methyl group (

-CH_3

) contains 3 C-H

\sigma

bonds, and the terminal alkynyl group (

\equiv CH

) contains 1 C-H

\sigma

bond, giving a total of 6 C-H

\sigma

bonds.

Every single bond between carbon and hydrogen is a single sigma bond.

2

Identify all carbon-carbon (C-C) sigma bonds in the backbone and side chain.

The

C=C

double bond contributes 1 C-C

\sigma

bond, the single bond to the methyl branch contributes 1 C-C

\sigma

bond, the C2-C3 single bond contributes 1 C-C

\sigma

bond, and the

C\equiv C

triple bond contributes 1 C-C

\sigma

bond, giving a total of 4 C-C

\sigma

bonds.

Multiple bonds (double or triple) contain exactly one sigma bond each, with the remaining bonds being pi (

\pi

) bonds.

3

Sum the total number of C-H and C-C sigma bonds.

6 \text{ (C-H } \sigma\text{ bonds)} + 4 \text{ (C-C } \sigma\text{ bonds)} = 10 \text{ total } \sigma \text{ bonds}

.

Adding all localized

\sigma

bonds yields the total count for the molecule.

Key Concept

Determination of sigma (\sigma) and pi (\pi) bond counts in complex open-chain hydrocarbons

Question 2Question →

Which of the following describes the type of orbital overlap that forms the carbon-carbon ( $\text{C}-\text{C}$ ) single bond in an ethane ( $C_2H_6$ ) molecule?

Head-on overlap of two

sp^3

hybrid orbitals

Sideways overlap of two unhybridized

p

orbitals

Head-on overlap of two

sp^2

hybrid orbitals

Sideways overlap of two

sp^3

hybrid orbitals

Show answer & explanation

Answer: Head-on overlap of two $sp^3$ hybrid orbitals

Answer

Head-on overlap of two

sp^3

hybrid orbitals

The carbon atoms in ethane (

C_2H_6

) are each bonded to four other atoms, giving them a tetrahedral arrangement and

sp^3

hybridization. The carbon-carbon single bond is a sigma (

\sigma

) bond formed by the direct head-on overlap of one

sp^3

hybrid orbital from each carbon atom.

Step-by-Step Solution

1

Determine the hybridization state of carbon atoms in ethane (

C_2H_6

)

Each carbon atom forms 4 single sigma (

\sigma

) bonds, corresponding to

sp^3

hybridization with tetrahedral geometry.

Saturated hydrocarbons (alkanes) undergo

sp^3

hybridization to accommodate four equivalent single bonds.

2

Identify the mode of overlap for the carbon-carbon single bond

The single bond between the two carbon atoms is a sigma (

\sigma

) bond.

Sigma bonds are always formed by direct end-to-end (head-on) overlap of atomic or hybrid orbitals along the bond axis.

Key Concept

Orbital overlap and hybridization in tetrahedral carbon (

sp^3

sigma bonding)

Question 3Question →

Consider the organic compound 3-methylbut-1-yne, which has the condensed structural formula $HC\equiv C-CH(CH_3)_2$ . How many carbon atoms in a single molecule of this compound are $sp^3$ hybridized?

Show answer & explanation

Answer: 3; three; 3 carbon atoms; 3 carbons

Answer

There are 3

sp^3

hybridized carbon atoms in one molecule of 3-methylbut-1-yne.

In 3-methylbut-1-yne (

HC\equiv C-CH(CH_3)_2

), there are 5 total carbon atoms. The two terminal/alkyne carbons (

C_1

and

C_2

) participate in a triple bond, giving them 2

\sigma

bonds each and an

sp

hybridization state. The central methine carbon (

C_3

) is bonded to four distinct atoms (

C_2

,

H

, and two methyl carbons) via single

\sigma

bonds, making it

sp^3

hybridized. The two methyl group carbons (

C_4

and

C_5

) are each single-bonded to three hydrogen atoms and

C_3

, making them

sp^3

hybridized as well. Thus, exactly 3 carbon atoms are

sp^3

hybridized.

Step-by-Step Solution

1

Expand the condensed structural formula to identify every carbon atom and its bonding environment.

The expanded structure is

H-C_1 \equiv C_2 - C_3H(C_4H_3)(C_5H_3)

, containing a total of 5 carbon atoms.

Expanding the formula clarifies the number of single (

\sigma

) and multiple bonds connected to each carbon atom.

2

Determine the hybridization state of the triply bonded carbon atoms (

C_1

and

C_2

).

C_1

and

C_2

are each involved in one triple bond and one single bond, forming 2

\sigma

bonds and 2

\pi

bonds. Thus, both

C_1

and

C_2

are

sp

hybridized.

A carbon atom with 2 steric domains (linear geometry) uses

sp

hybrid orbitals.

3

Determine the hybridization state of the methine carbon atom (

C_3

) and the two methyl carbon atoms (

C_4

and

C_5

).

C_3

forms four single

\sigma

bonds (one to

C_2

, one to

H

, and two to methyl carbons).

C_4

and

C_5

each form four single

\sigma

bonds (one to

C_3

and three to

H

). Therefore,

C_3

,

C_4

, and

C_5

are all

sp^3

hybridized.

A carbon atom bonded to 4 separate atoms via single

\sigma

bonds has 4 steric domains (tetrahedral geometry) and undergoes

sp^3

hybridization.

4

Count the total number of

sp^3

hybridized carbon atoms.

3 carbon atoms (

C_3

,

C_4

, and

C_5

) are

sp^3

hybridized.

Combining the results from steps 2 and 3 gives 2

sp

carbons and 3

sp^3

carbons.

Key Concept

Identification of carbon hybridization states (

sp^3, sp^2, sp

) in aliphatic molecules

Estimated Time:1m 30s

Question 4Question →

What is the hybridization state and geometry of the central carbon atom in a molecule of methane ( $CH_4$ )?

sp^3

hybridization with tetrahedral geometry and bond angles of

109.5^\circ

sp^2

hybridization with trigonal planar geometry and bond angles of

120^\circ

sp

hybridization with linear geometry and bond angles of

180^\circ

sp^3

hybridization with square planar geometry and bond angles of

90^\circ

Show answer & explanation

Answer: $sp^3$ hybridization with tetrahedral geometry and bond angles of $109.5^\circ$

Answer

sp^3

hybridization with tetrahedral geometry and bond angles of

109.5^\circ

In methane (

CH_4

), the central carbon atom forms four single covalent

\sigma

bonds with four hydrogen atoms. To achieve equivalent bonding, one

2s

and three

2p

orbitals hybridize to form four

sp^3

hybrid orbitals directed toward the corners of a regular tetrahedron, giving bond angles of

109.5^\circ

.

Step-by-Step Solution

1

Determine the valence electron configuration and bonding of the central carbon atom.

Carbon has 4 valence electrons and forms 4 single

\sigma

(sigma) bonds with four hydrogen atoms in methane (

CH_4

).

The number of single bonds and lone pairs determines the steric number of the central atom.

2

Determine the hybridization state based on the steric number.

Steric number = 4 (four

\sigma

bonds, zero lone pairs), requiring four equivalent hybrid orbitals formed by combining one

2s

and three

2p

atomic orbitals (

sp^3

).

Mixing one

s

orbital and three

p

orbitals yields four equivalent

sp^3

hybrid orbitals.

3

Determine the spatial geometry and ideal bond angle.

According to VSEPR theory, four bonding pairs position themselves as far apart as possible in 3D space, forming a tetrahedral geometry with bond angles of

109.5^\circ

.

The tetrahedral arrangement minimizes electrostatic repulsion among the four bonding electron pairs.

Key Concept

Tetrahedral Carbon, Bonding, and Hybridization

Estimated Time:45s

Question 5Question →

In a propyne molecule ( $CH_3-C\equiv CH$ ), which hybrid orbitals overlap head-on to form the carbon-carbon single bond between the methyl carbon atom and the adjacent triple-bonded carbon atom?

sp^3

and

sp

hybrid orbitals

sp^3

and

sp^2

hybrid orbitals

sp^2

and

sp

hybrid orbitals

sp^3

and

sp^3

hybrid orbitals

Show answer & explanation

Answer: $sp^3$ and $sp$ hybrid orbitals

Answer

The carbon-carbon single bond in propyne is formed by the head-on overlap of an

sp^3

hybrid orbital from the methyl carbon atom and an

sp

hybrid orbital from the adjacent acetylenic carbon atom.

In propyne (

CH_3-C\equiv CH

), the methyl carbon atom is attached to four atoms via single bonds, giving it a tetrahedral arrangement and

sp^3

hybridization. The central carbon atom is involved in a triple bond and one single bond, giving it a linear arrangement and

sp

hybridization. Therefore, the single bond connecting these two carbon atoms is a

\sigma

bond formed by the head-on overlap of an

sp^3

hybrid orbital from the methyl carbon and an

sp

hybrid orbital from the central acetylenic carbon.

Step-by-Step Solution

1

Determine the hybridization state of the methyl carbon atom (

CH_3-

).

The methyl carbon atom forms 4

\sigma

bonds (3 with H atoms, 1 with C), giving it 4 electron domains and a tetrahedral geometry with

sp^3

hybridization.

Four equivalent bonding electron domains around a carbon atom require four

sp^3

hybrid orbitals.

2

Determine the hybridization state of the adjacent triple-bonded carbon atom (

-C\equiv

).

This carbon atom forms 2

\sigma

bonds (1 with methyl C, 1 with terminal C) and 2

\pi

bonds, giving it 2 electron domains and a linear geometry with

sp

hybridization.

Two linear bonding domains around a carbon atom require two

sp

hybrid orbitals.

3

Identify the orbitals participating in the

\sigma

single bond between these two carbon atoms.

The

\sigma

single bond is formed by the end-to-end (head-on) overlap of one

sp^3

hybrid orbital from the methyl carbon and one

sp

hybrid orbital from the acetylenic carbon.

Single bonds (

\sigma

bonds) between hybridized carbon atoms result from direct axial overlap of their respective hybrid orbitals.

Key Concept

Orbital Overlap and Carbon Hybridization in Alkynes

Question 6Question →

What is the bond angle between adjacent single covalent bonds formed by a tetrahedral $sp^3$ hybridized carbon atom in a saturated organic compound?

109.5^\circ

120^\circ

180^\circ

90^\circ

Show answer & explanation

Answer: $109.5^\circ$

Answer

The bond angle of a tetrahedral sp³ hybridized carbon atom is 109.5°.

In saturated organic compounds such as alkanes, the central carbon atom forms four single sigma bonds using four equivalent

sp^3

hybrid orbitals. According to VSEPR theory, four bonding electron pairs surrounding a central atom arrange themselves in a regular tetrahedral shape to minimize electrostatic repulsion, yielding a standard bond angle of

109.5^\circ

.

Step-by-Step Solution

1

Determine the hybridization state of a carbon atom forming four single sigma bonds.

Mixing one 2s orbital and three 2p orbitals gives four equivalent

sp^3

hybrid orbitals.

Saturated carbon forms four single bonds by directed valence orbital mixing.

2

Apply VSEPR theory to determine spatial orientation for four electron pairs around the central carbon.

To minimize electron pair repulsion, the four orbitals point toward the vertices of a regular tetrahedron.

Symmetrical four-coordinate electron pair repulsion produces tetrahedral spatial orientation.

3

Identify the characteristic inter-bond angle of a regular tetrahedron.

The angle between any two adjacent bonds is

109.5^\circ

(or

109^\circ 28'

).

This angle maximizes the distance between the four bonding electron pairs in three-dimensional space.

Key Concept

Tetrahedral Geometry and sp³ Hybridization Bond Angle

Estimated Time:45s

Question 7Question →

Consider the cumulated diene compound penta-1,2-diene, represented by the condensed structure $\text{CH}_2=\text{C}=\text{CH}-\text{CH}_2-\text{CH}_3$ . What are the hybridization states of the central carbon atom (C2) and the methylene carbon atom (C4), respectively?

sp

and

sp^3

sp^2

and

sp^3

sp

and

sp^2

sp^2

and

sp^2

Show answer & explanation

Answer: $sp$ and $sp^3$

Answer

The central carbon atom (C2) is

sp

hybridized and the methylene carbon atom (C4) is

sp^3

hybridized.

In penta-1,2-diene, the central allene carbon atom at position 2 (C2) forms two double bonds (

2\ \sigma

bonds and

2\ \pi

bonds), which necessitates

sp

hybridization. The methylene carbon atom at position 4 (C4) forms four single

\sigma

bonds with two hydrogen atoms and two carbon atoms, which corresponds to

sp^3

tetrahedral hybridization.

Step-by-Step Solution

1

Determine the number of

\sigma

and

\pi

bonds on carbon-2 (C2).

C2 forms two double bonds, which consists of

2\ \sigma

bonds and

2\ \pi

bonds.

Carbon atoms involved in two double bonds (cumulated dienes/allenes) use two

sp

hybrid orbitals to form

\sigma

bonds at an angle of

180^\circ

.

2

Determine the hybridization state of C2.

C2 is

sp

hybridized with linear geometry.

Two

\sigma

bonding domains correlate to

sp

hybridization.

3

Determine the bonding domains and hybridization of carbon-4 (C4).

C4 is bonded to two hydrogen atoms, C3, and C5 via single covalent bonds, giving

4\ \sigma

bonds.

Four single

\sigma

bonding domains require

sp^3

hybridization with tetrahedral geometry.

Key Concept

Hybridization in Cumulated Dienes and Saturated Carbons

Estimated Time:1m 30s

Question 8Question →

In a saturated acyclic alkane such as propane ( $\text{C}_3\text{H}_8$ ), each carbon atom exhibits tetrahedral geometry through $sp^3$ hybridization. What is the percentage of $s$ -orbital character present in each of these hybrid orbitals?

25\%

33.3\%

50\%

75\%

Show answer & explanation

Answer: $25\%$

Answer

The percentage of

s

-orbital character in each

sp^3

hybrid orbital of a tetrahedral carbon atom is

25\%

.

In tetrahedral carbon compounds,

sp^3

hybridization involves mixing one

s

orbital and three

p

orbitals to produce four degenerate orbitals. Therefore, the proportion of

s

-character in each hybrid orbital is

\frac{1}{4}

, which equals

25\%

.

Step-by-Step Solution

1

Determine the composition of orbitals in

sp^3

hybridization.

An

sp^3

hybrid orbital is produced by combining one

s

atomic orbital and three

p

atomic orbitals, yielding a total of 4 equivalent hybrid orbitals.

Hybridization mixes pure atomic orbitals to form equivalent hybrid orbitals around a central tetrahedral carbon atom.

2

Calculate the fractional contribution of the

s

orbital.

The fraction of

s

-character is

\frac{1}{1 + 3} = \frac{1}{4} = 0.25

.

One out of the four constituent atomic orbitals is an

s

orbital.

3

Convert the fraction into a percentage.

0.25 \times 100\% = 25\%

.

Multiplying the fractional contribution by 100 gives the percentage of

s

-character.

Key Concept

Orbital Hybridization and s/p Character in Tetrahedral Carbon

Question 9Question →

Match each structural feature or carbon center of 2-methylbut-1-en-3-yne ( $HC\equiv C-C(CH_3)=CH_2$ ) on the left with its corresponding hybridization state, geometric descriptor, or orbital overlap description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: (1) The

C_2-C_3

single bond matches with the

\sigma

-bond formed by head-on

sp-sp^2

hybrid orbital overlap. (2) The methyl carbon center matches with

sp^3

hybridization, tetrahedral geometry, and

\approx 109.5^\circ

bond angles. (3) The alkene double bond matches with one

\sigma

-bond (

sp^2-sp^2

) and one

\pi

-bond (

2p-2p

). (4) The terminal acetylenic carbon matches with

sp

hybridization, linear geometry, and

180^\circ

bond angle.

Each carbon atom in 2-methylbut-1-en-3-yne adopts a hybridization state determined by its steric number (number of attached atoms and lone pairs).

C_1

and

C_2

are

sp

-hybridized (linear,

180^\circ

),

C_3

and

C_4

are

sp^2

-hybridized (trigonal planar,

120^\circ

), and the methyl group carbon is

sp^3

-hybridized (tetrahedral,

109.5^\circ

). Consequently, single bonds between differently hybridized carbons utilize hybrid orbitals corresponding to each carbon (

sp-sp^2

for

C_2-C_3

), and double bonds consist of one

\sigma

bond (

sp^2-sp^2

) plus one

\pi

bond (

2p-2p

).

Step-by-Step Solution

1

Analyze the expanded structural formula of 2-methylbut-1-en-3-yne

H-C(1)\equiv C(2)-C(3)(CH_3)=C(4)H_2

Determining the bonding domains around each carbon atom establishes its hybridization state and structural role.

2

Determine hybridization state for each carbon center

C_1

(

sp

),

C_2

(

sp

),

C_3

(

sp^2

),

C_4

(

sp^2

), and methyl carbon (

sp^3

)

Carbons with 2 electron domains are

sp

(linear), 3 domains are

sp^2

(trigonal planar), and 4 domains are

sp^3

(tetrahedral).

3

Map orbital overlap types to specific bonds

C_2-C_3

is an

sp-sp^2

\sigma

-bond; double bond

C_3=C_4

comprises an

sp^2-sp^2

\sigma

-bond and a

2p-2p

\pi

-bond.

Single bonds are formed by head-on overlap of hybrid orbitals, while double bonds consist of one coaxial

\sigma

bond and one collateral

\pi

bond.

4

Match left items with their corresponding right item descriptions based on hybridization and geometry principles

All 4 items are accurately matched to their structural characteristics.

Ensures complete alignment between structural features and underlying orbital hybridization properties.

Key Concept

Orbital Hybridization and Overlap Types in Hydrocarbon Frameworks

Estimated Time:2m 0s

Question 10Question →

What is the total number of sigma ( $\sigma$ ) bonds in a single molecule of prop-2-enal (acrolein, $\text{CH}_2=\text{CH}-\text{CHO}$ )?

Show answer & explanation

Answer: 7; seven; 7 sigma bonds; 7 sigma

Answer

7

Expanding prop-2-enal (

\text{CH}_2=\text{CH}-\text{CHO}

) reveals four

\text{C}-\text{H}

single bonds, one

\text{C}-\text{C}

single bond, one

\text{C}=\text{C}

double bond (composed of one

\sigma

and one

\pi

bond), and one

\text{C}=\text{O}

double bond (composed of one

\sigma

and one

\pi

bond). Summing all head-on orbital overlaps yields a total of 7 sigma (

\sigma

) bonds.

Step-by-Step Solution

1

Draw the expanded structural formula of prop-2-enal.

The expanded formula showing all individual atoms and bonds is

\text{H}_2\text{C}=\text{CH}-\text{C}(=\text{O})\text{H}

.

Expanding the structural formula ensures that all implicit single bonds, double bonds, and hydrogen attachments are explicitly visible for counting.

2

Count all carbon-hydrogen (

\text{C}-\text{H}

) single sigma bonds.

There are 2

\text{C}-\text{H}

bonds on the terminal alkene carbon, 1

\text{C}-\text{H}

bond on the central alkene carbon, and 1

\text{C}-\text{H}

bond on the aldehyde carbon, giving a total of 4

\text{C}-\text{H}

\sigma

bonds.

Every single covalent bond formed with hydrogen involves head-on

sp^2-s

orbital overlap and constitutes one

\sigma

bond.

3

Count the sigma bonds among the carbon-carbon and carbon-oxygen links.

The

\text{C}=\text{C}

double bond contains 1

\sigma

bond, the

\text{C}-\text{C}

single bond contains 1

\sigma

bond, and the

\text{C}=\text{O}

double bond contains 1

\sigma

bond, giving 3 heavy-atom

\sigma

bonds.

Every covalent bond—whether single, double, or triple—contains exactly one

\sigma

bond resulting from axial head-on orbital overlap.

4

Sum the total number of sigma bonds.

4\text{ (C-H \sigma\ bonds)} + 3\text{ (C-C and C-O \sigma\ bonds)} = 7\text{ \sigma\ bonds}

.

Adding all localized head-on orbital overlaps gives the total count of sigma bonds in the molecule.

Key Concept

Determination of sigma (

\sigma

) and pi (

\pi

) bond counts in organic structures

Question 11Question →

Match each carbon species or bond descriptor on the left with its corresponding hybridization state, geometric configuration, or orbital overlap mode on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The central carbon in methane matches

sp^3

hybridization with tetrahedral geometry (

109.5^\circ

); the

\pi

bond component matches sideways overlap of unhybridized

p

orbitals; the central carbon in carbon dioxide matches

sp

hybridization with linear geometry (

180^\circ

); and the carbon in ethene matches

sp^2

hybridization with trigonal planar geometry (

120^\circ

).

Each carbon atom's hybridization and spatial arrangement depend directly on its steric number (number of

\sigma

bonds). Methane features four

\sigma

bonds (

sp^3

,

109.5^\circ

tetrahedral). Ethene features three

\sigma

bonds per carbon (

sp^2

,

120^\circ

trigonal planar). Carbon dioxide features two

\sigma

bonds (

sp

,

180^\circ

linear).

\Pi

bonds are characterized by the sideways overlap of unhybridized

2p

atomic orbitals.

Step-by-Step Solution

1

Determine the steric number and geometry of the carbon in methane (

\text{CH}_4

).

Four single

\sigma

bonds give a steric number of 4, which dictates

sp^3

hybridization and a tetrahedral angle of

109.5^\circ

.

Mixing one

s

and three

p

orbitals forms four equivalent

sp^3

hybrid orbitals pointing to tetrahedral corners.

2

Identify how a carbon-carbon

\pi

bond is formed.

It forms via lateral/sideways overlap of parallel, unhybridized

p

orbitals above and below the internuclear axis.

Head-on overlap forms

\sigma

bonds, whereas parallel side-by-side overlap creates

\pi

electron clouds.

3

Analyze the steric environment around the carbon in carbon dioxide (

\text{CO}_2

).

The carbon forms two

\sigma

bonds (one to each oxygen atom) and two

\pi

bonds, yielding a steric number of 2, corresponding to

sp

hybridization and

180^\circ

linear geometry.

Two hybrid orbitals position themselves as far apart as possible at

180^\circ

.

4

Determine the hybridization and bond angles of carbon in ethene (

\text{C}_2\text{H}_4

).

Each carbon atom forms three

\sigma

bonds (steric number 3), requiring

sp^2

hybridization with a trigonal planar shape and

120^\circ

bond angles.

Three hybrid orbitals lie in a single plane separated by

120^\circ

.

Key Concept

Carbon Hybridization, Geometry, and Orbital Overlap

Tetrahedral Carbon, Bonding, and Hybridization

Answer

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