Question

Niobium crystallises in body centred cubic structure. Its density is 8.55 g cm⁻³ and its atomic mass is 93 g mol⁻¹. (N_A = 6.023 × 10²³)

What is the atomic radius of niobium?

पर्याय

• 136 pm

• 140 pm

• 143 pm

• 149 pm

Answer 1

149 pm

Explanation:

Given: Body-centred Cubic (bcc) structure ⇒ Z = 2 atoms/unit cell

Density (ρ) = 8.55 g/cm³

Atomic mass (M) = 93 g/mol

Avogadro’s number (N_A) = 6.023 × 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 × 10⁻²³ cm³

a =  `(3.612 xx 10^(−23))^(1/3)`

a  = 3.4043 × 10⁻⁸ cm

a = 340.43 pm

In a bcc structure, the body diagonal

`sqrt 3 * a = 4 r`

∴ `r = (sqrt 3 a)/4`

`r = (1.732 xx 340.43)/4`

`r = 589.62/4`

r = 147.5 pm 

r ≈ 149 pm (as per nearest option)