Question

MgO has NaCl type structure and its unit cell edge length is $421$ pm. Its density (in g cm${}^{-3}$) will be:

• A. $3.54$
• B. $1.46$
• C. $2.14$
• D. None of the above

Answer 1

The correct option is A $3.54$

The expression for density is as follows:

$\rho =\frac{ZM}{N_0{a}^3}$

As MgO has the same structure as NaCl, we can say that MgO has FCC structure.

So, putting values accordingly, we get

$\rho =\frac{4\times 40}{(6.022\times {10}^{23})\times (4.21\times {10}^{-8}{)}^3}=3.54$ g cm${}^{-3}$