Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Solutions

Calculate the radius of molybdenum atom if the element crystallizes as body- centred cubic crystals. Given: Density of $Mo=10.3\;g\;cm^{-3}$ and molar mass of $Mo= 95.94 \;g\;mol^{-1}$

$\begin{array}{1 1} 135.5\;pm \\ 415\;pm \\ 678\;pm \\ 516\;pm \end{array} $

Download clay6 mobile app

1 Answer

Answer : 135.5 pm
In a body-centered cubic unit cell, there are two atoms per unit cell.
Hence, from the expression $r= \large\frac{N}{a^3} \bigg( \large\frac{M}{N_A}\bigg)$
We get, $a^3 = \large\frac{N}{\rho} \bigg( \large\frac{M}{N_A}\bigg)$
$\qquad= \bigg( \large\frac{2}{10.3 \;g\;cm^{-3}}\bigg) \bigg( \frac{95.04 \;g \;mol^{-1} }{6.023 \times 10^{23}\;mol^{-1}} \bigg)$
$\qquad= 3.09 \times 10^{-23}\;cm^3$
or $a= 3.13 \times 10^{-8}\;cm =313\;pm$
Now since in the body centered cubilc unit cell , atoms touch each other along the cross diagonal of the cube , we have
$4r= \sqrt 3 a$
$r= (\sqrt 3) (313\;pm)/4=135.5\;pm$
answered Aug 8, 2014 by meena.p

Related questions