Question

compute the theoretical density of diamond, given that c--c and bond angle are 0.154nm and 109.5 deg, respectively how does this value compare with the measured density?

Answer 1

Answer: According to the given informations ,

Here, one-half of the bond angle = 109.5/2 = 54.75, which means that ,angle = 90-?54.75= 35.25⁰

Forthermore , X = a/4 = y sin[Q]

a= 4ysin?= (4)(0.154 nm)(sin 35.25?) = 0.356 nm

Now , The unit cell density is = a3=(3.56* 10^-8cm)³ = 4.51* 10?^-23cm3

And we know that there are 8 equivalent atoms in diamond unit cell , Hence by applying density formula we get

density = Z M / a³ Na

here Z = 8 , M is molar mass of C = 12 , a³ = 4.51* 10^-23 , and Na is avogadro constant , by putting all these values we get

density = 3.54 g / cm^{3 .}

^{[Note the measured density of diamond is generally 3.51 g/cm3 ]}

Hence it is all about the given question . Thank you :)