An ionic solid AB has bcc structure. If the distance of closest approach between the two ions is $1.73\mathop {\rm A}\limits^\circ $ then density in $(gm\;c{m^{ - 3}})$of the solid is $({N_A} = $Avogadro’s number)
(1) $\dfrac{{M \times {{10}^{24}}}}{{8 \times {N_A}}}$
(2) $\dfrac{{M \times {{10}^{30}}}}{{8 \times {N_A}}}$
(3) $\dfrac{{M \times {{10}^{24}}}}{{{N_A} \times 3\sqrt 3 }}$
(4) $\dfrac{{M \times {{10}^{24}}}}{{4 \times {N_A}}}$

Hint: As we know that a bcc structure of a unit cell possesses $8$ lattice point at $8$ corners and one additional lattice point at the centre of the body. By knowing the dimensions of a unit cell we can calculate the density of the unit cell.

Complete Step by step answer:
As we know that each unit cell is adjacent to another and most of the atoms are shared by neighbouring unit cells due to which only some portion of each atom belongs to a particular unit cell.
So, by knowing the dimensions of a unit cell, we can calculate the volume of the unit cell, the mass and number of atoms in the unit cell and the density of a unit cell as well.

In the given question we have the distance of closest approach which is $1.73\mathop {\rm A}\limits^\circ $ and we need to calculate the density of the solid AB. Let the edge length of a cubic crystal of an element or compound $ = a$ and we know that body diagonal is the closest distance of approach for bcc which is $ = \dfrac{{\sqrt 3 a}}{2}$.
So, $a = \dfrac{{2 \times 1.73}}{{\sqrt 3 }}$
Thus, the volume of the unit cell$ = {a^3}$
${a^3} = {\left( {\dfrac{{2 \times 1.73}}{{\sqrt 3 }}} \right)^3}$
$
\Rightarrow {a^3} = \dfrac{{8 \times 3\sqrt 3 }}{{3\sqrt 3 }} \\
\Rightarrow {a^3} = 8 \times {10^{ - 24}}c{m^3} \\
 $
And density of the cell will be $d = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}$,
So using the above formula we will get:
$d = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}$
$\Rightarrow d = \dfrac{{2 \times M}}{{{N_A} \times 8 \times {{10}^{ - 24}}}}$
$\Rightarrow d = \dfrac{{M \times {{10}^{24}}}}{{4 \times {N_A}}}$

Therefore, the correct answer is (D).

Note: The body centred cubic arrangement contains $8$ lattice point at the corners and each one of these is shared by $8$ cells and another lattice point is completely inside the unit cell or present at the centre of the body thus the number of points or atoms per unit cell of bcc is \[8 \times \dfrac{1}{8} + 1 = 2\] therefore the value of $Z = 2$ similarly value of ccp arrangement can be calculated which comes out to be $Z = 4$.