Niobium crystallises in body centred cubic structure. Its density is 8.55 g cm−3 and its atomic mass is 93 g mol−1. (NA = 6.023 $\times$ 1023) What is the atomic radius of niobium?

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149 pm Explanation: Given: Body-centred Cubic (bcc) structure $\Rightarrow$ Z = 2 atoms/unit cell Density (ρ) = 8.55 g/cm3 Atomic mass (M) = 93 g/mol Avogadro’s number (NA) = 6.023 $\times$ 1023 mol−1 rho = (Z xx M)/(N_A xx a^3) a^3 = (Z xx M)/(rho xx N_A) `a^3 = (2 xx 93)/(6.023 xx 10^23 xx 8.55) a^3 = 186/(51.48765 xx 10^23)` a3 = 3.612 $\times$ 10−23 cm3 a = (3.612 xx 10^(−23))^($\frac{1}{3}$) a = 3.4043 $\times$ 10−8 cm a = 340.43 pm In a bcc structure, the body diagonal sqrt 3 * a = 4 r $\therefore$ r = (sqrt 3 a)/4 r = (1.732 xx 340.43)/4 r = 589.$\frac{62}{4}$ r = 147.5 pm r ≈ 149 pm (as per nearest option)

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