QJEEM23Ch708f: JEE Main 2023 Chemistry - Class 12 JEEM23Ch708f Solution: Sodium metal crystallizes in a body centred cubic lattice with…

In a body-centered cubic (BCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by :

$\sqrt{3}a = 4r$

Given the unit cell edge length (a) of sodium metal as 4 Å :

$a = 4 ~\mathop A\limits^o$

We can now solve for the radius (r) of the sodium atom :

$4r = \sqrt{3}a$

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

$r = \frac{\sqrt{3} \times 4 ~\mathop A\limits^o}{4}$

$r = \sqrt{3} ~\mathop A\limits^o$

Now, we can approximate the numerical value :

$r \approx 1.732 ~\mathop A\limits^o$

To express the radius as a multiple of 10⁻¹ $\mathop A\limits^o$ :

$r = 1.732 \times 10^{1} \times 10^{-1} ~\mathop A\limits^o = 17.32 \times 10^{-1} ~\mathop A\limits^o$

So, the radius of the sodium atom is 17.32 × 10⁻¹ $\mathop A\limits^o$ .