A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of $$2.0$$ and $$2.5$$ Å respectively. The ratio of densities of lattices fcc to bcc for the metal M is ______ (Nearest integer)

For a crystalline metal, the density of a unit cell is given by
$$\rho = \frac{Z \, M}{N_A \, a^{3}}$$
where $$Z$$ = number of atoms per unit cell, $$M$$ = molar mass of the metal, $$N_A$$ = Avogadro constant, and $$a$$ = edge length of the cubic unit cell.

Because the same metal M forms both lattices, $$M$$ and $$N_A$$ are identical for fcc and bcc forms. Hence, the ratio of densities depends only on $$Z$$ and $$a$$:
$$\frac{\rho_{\text{fcc}}}{\rho_{\text{bcc}}} = \frac{Z_{\text{fcc}}}{Z_{\text{bcc}}}\,\frac{a_{\text{bcc}}^{3}}{a_{\text{fcc}}^{3}}$$ $$-(1)$$

For the two lattices:
• Face-centred cubic (fcc): $$Z_{\text{fcc}} = 4$$, edge length $$a_{\text{fcc}} = 2.0 \text{ Å}$$.
• Body-centred cubic (bcc): $$Z_{\text{bcc}} = 2$$, edge length $$a_{\text{bcc}} = 2.5 \text{ Å}$$.

Substituting these values into $$(1)$$:
$$\frac{\rho_{\text{fcc}}}{\rho_{\text{bcc}}} = \frac{4}{2}\,\frac{(2.5)^{3}}{(2.0)^{3}}$$

Simplify step by step:
$$\frac{4}{2} = 2$$
$$\left(\frac{2.5}{2.0}\right)^{3} = (1.25)^{3} = 1.953125$$

Therefore,
$$\frac{\rho_{\text{fcc}}}{\rho_{\text{bcc}}} = 2 \times 1.953125 = 3.90625$$

The ratio, rounded to the nearest integer, is $$\approx 4$$.

Final Answer: 4