Ammonium chloride crystallizes in a body centred cubic lattice with edge length of unit cell of $$390$$ pm. If the size of chloride ion is $$180$$pm, the size of ammonium ion would be

Ammonium chloride crystallizes in the CsCl-type structure (a body-centred cubic arrangement of two different ions).
In this lattice:

• The chloride ions $$Cl^-$$ occupy the eight corners of the cube.
• The ammonium ion $$NH_4^+$$ sits at the body centre.
• Along the body diagonal, a corner $$Cl^-$$ ion touches the central $$NH_4^+$$ ion, which in turn touches the opposite corner $$Cl^-$$ ion.

Hence, along the body diagonal we have the contact sequence
$$Cl^- \;|\; NH_4^+ \;|\; Cl^-$$

Let
$$r_- =$$ radius of $$Cl^- = 180\text{ pm}$$
$$r_+ =$$ radius of $$NH_4^+ = ?$$
$$a =$$ edge length of the cubic unit cell $$= 390\text{ pm}$$

The length of the body diagonal of a cube is given by the geometry formula
$$\text{body diagonal} = \sqrt{3}\,a$$

The same body diagonal is also equal to the sum of the three touching radii in sequence:
$$\text{body diagonal} = r_- + r_+ + r_- + r_+$$
$$\qquad\qquad\;\; = 2\,(r_- + r_+)$$

Equating the two expressions:

$$\sqrt{3}\,a = 2\,(r_- + r_+)$$

Substitute the numerical values:

$$\sqrt{3}\,(390) = 2\,(180 + r_+)$$

$$1.732 \times 390 = 2\,(180 + r_+)$$

$$675.48 = 360 + 2\,r_+$$

$$2\,r_+ = 675.48 - 360 = 315.48$$

$$r_+ = \frac{315.48}{2} \approx 157.7\text{ pm}$$

Rounded to the nearest whole number, the ionic radius of $$NH_4^+$$ is $$158\text{ pm}$$.

Option B which is: $$158\text{ pm}$$