Question

Molybdenum (At. wt. $=96g/mol$) crystallizes as $bcc$ crystal. If density of crystal is $10.3g/c{m}^3$, then radius of $Mo$ atom is: [Use : $N_A=6\times {10}^{23}$]

• A. $111$pm
• B. $314$pm
• C. $135.96$pm
• D. none of these

Answer 1

The correct option is C $135.96$pm

Since, Molybladenum crystallizes as $bcc$ rystal. So, $N=2$.

Expression for density is:

$\rho =\frac{N\times M}{N_A\times a^3}$

$a^3=\frac{N\times M}{N_A\times \rho }$

$a^3=\frac{2\times 96}{6\times {10}^{23}\times 10.3}$

$a^3=3.1\times {10}^{-23}{cm}^3$

$a=3.14\times {10}^{-8}cm=314pm$.

For, $bcc$ crystal,

$4r=\sqrt{3}a$

$r=\frac{\sqrt{3}}{4}a$

$r=\frac{\sqrt{3}}{4}\times 314$

$r=136pm$.