Question

Molybdenum ($\text{Atomic wt.}=96g/mol$) crystallizes as bcc crystal. If density of crystal is $10.3g/c{m}^3,$ what will be the radius of $Mo$ atom? $[\text{U}se:N_A=6\times {10}^{23}]$

• A. $111pm$
• B. $314pm$
• C. $136pm$
• D. $152pm$

Answer 1

The correct option is C $136pm$

Since, molybdenum crystallizes as bcc crystal.
$\text{No. of atoms per unit cell, Z}=2$
Expression for density is:
$\rho =\frac{Z\times M}{N_A\times a^3}$
$a^3=\frac{Z\times M}{N_A\times \rho }$
$a^3=\frac{2\times 96}{6\times {10}^{23}\times 10.3}$
$a^3=3.1\times {10}^{-23}c{m}^3$
$a=3.14\times {10}^{-8}cm=314pm$

For bcc crystal, the atoms at corner and body centre are in contact.
Thus,
$4r=\sqrt{3}a$
$r=\frac{\sqrt{3}a}{4}$
$r=\frac{\sqrt{3}}{4}\times 314$
$r=135.96pm$
$r\approx 136pm$