In a saturated acyclic alkane such as propane (\text{C}_3\text{H}_8), each carbon atom exh… — JAMB UTME

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?

$25\%$ Answer
B
$33.3\%$
C
$50\%$
D
$75\%$

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

Determine the composition of orbitals in

sp^3

hybridization.

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.

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.

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

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