JEE Main 2018 — Chemistry Inorganic Chemistry

2018mcqeasy

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

Official previous-year question

Held on 15 Apr 2018 · Verified 6 Jul 2026.

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Graphite	Diamond
${\mathrm{sp}}^{2}$ hybridisation	${\mathrm{sp}}^{3}$ hybridisation
$%P=\frac{2}{3}\times 100=67%$	$%P=\frac{3}{4}\times 100=75%$
For ${\mathrm{sp}}^{2}$ : $%P$ character $\frac{P-1}{P}=cos\theta$ $\frac{P-1}{P}=\mathrm{cos}120=-0.5$ $P-1=0.5P$ $1.5P=1$ $P=\frac{2}{3}$ $%P=\frac{2}{3}\times 100=66.67%$	For ${\mathrm{sp}}^{3}$ : $%P$ character $\frac{P-1}{P}=\mathrm{cos}109'28$ $\frac{P-1}{P}=0.33$ $\frac{P-1}{P}=\frac{3}{4}$ $%P=75%$

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2018mcqeasy

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

Official previous-year question

Held on 15 Apr 2018 · Verified 6 Jul 2026.

Graphite	Diamond
${\mathrm{sp}}^{2}$ hybridisation	${\mathrm{sp}}^{3}$ hybridisation
$%P=\frac{2}{3}\times 100=67%$	$%P=\frac{3}{4}\times 100=75%$
For ${\mathrm{sp}}^{2}$ : $%P$ character $\frac{P-1}{P}=cos\theta$ $\frac{P-1}{P}=\mathrm{cos}120=-0.5$ $P-1=0.5P$ $1.5P=1$ $P=\frac{2}{3}$ $%P=\frac{2}{3}\times 100=66.67%$	For ${\mathrm{sp}}^{3}$ : $%P$ character $\frac{P-1}{P}=\mathrm{cos}109'28$ $\frac{P-1}{P}=0.33$ $\frac{P-1}{P}=\frac{3}{4}$ $%P=75%$