In graphite and diamond, the percentage of p-characters of the hybrid orbitals in hybridization respectively, are

Question

In graphite and diamond, the percentage of p-characters of the hybrid orbitals in hybridization respectively, are, 33 and 25, 67 and 75, 50 and 75, 33 and 75

MARKS App · Accepted Answer

Graphite Diamond sp 2 hybridisation sp 3 hybridisation % P = 2 3 × 100 = 67 % % P = 3 4 × 100 = 75 % For sp 2 : % P character P − 1 P = cosθ P − 1 P = cos 120 = − 0 .5 P − 1 = 0 .5 P 1 .5 P = 1 P = 2 3 % P = 2 3 × 100 = 66 .67 % For sp 3 : % P character P − 1 P = cos 109 ' 28 P − 1 P = 0 .33 P − 1 P = 3 4 % P = 75 %

Graphite	Diamond
${sp}^{2}$ hybridisation	${sp}^{3}$ hybridisation
$% P = \frac{2}{3} \times 100 = 67 %$	$% P = \frac{3}{4} \times 100 = 75 %$
For ${sp}^{2}$ : $% P$ character $\frac{P - 1}{P} = cosθ$ $\frac{P - 1}{P} = \cos 120 = - 0 .5$ $P - 1 = 0 .5 P$ $1 .5 P = 1$ $P = \frac{2}{3}$ $% P = \frac{2}{3} \times 100 = 66 .67 %$	For ${sp}^{3}$ : $% P$ character $\frac{P - 1}{P} = \cos 109' 28$ $\frac{P - 1}{P} = 0 .33$ $\frac{P - 1}{P} = \frac{3}{4}$ $% P = 75 %$

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