Chemistry - CHEMICAL BONDING Question with Solution | TestHub

ChemistryCHEMICAL BONDINGVSEPR, Bond Angle, Bond Length, Bond EnergyEasy2 minPYQ_2018

ChemistryEasysingle choice

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

Answer:

Solution:

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 $\frac{P - 1}{P} = \cos 109' 28$ $\frac{P - 1}{P} = 0 .33$ $\frac{P - 1}{P} = \frac{3}{4}$ $% P = 75 %$

Stream:JEESubject:ChemistryTopic:CHEMICAL BONDINGSubtopic:VSEPR, Bond Angle, Bond Length, Bond Energy

⏱ 2mℹ️ Source: PYQ_2018

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