The density of \(\beta\)-Fe is 7.6 g/cm\(^3\). It crystallizes in a cubic lattice with \( a = 290 \) pm.
What is the value of \( Z \)? (\( Fe = 56 \) g/mol, \( N_A = 6.022 \times 10^{23} \) mol\(^{-1}\))
Step 1: Use the density formula for a unit cell \[ {Density} = \frac{Z \cdot M}{N_A \cdot a^3} \] Where: - \( Z \) = number of atoms per unit cell - \( M \) = molar mass of Fe = 56 g/mol - \( N_A \) = Avogadro’s number = \( 6.022 \times 10^{23} \) - \( a = 290 \) pm = \( 290 \times 10^{-10} \) cm - Given density \( \rho = 7.6 \) g/cm³
Step 2: Calculate the unit cell volume \[ a^3 = (290 \times 10^{-10})^3 = 2.44 \times 10^{-23} { cm}^3 \] Step 3: Solve for \( Z \) \[ Z = \frac{\rho \cdot N_A \cdot a^3}{M} \] \[ Z = \frac{(7.6) \times (6.022 \times 10^{23}) \times (2.44 \times 10^{-23})}{56} \] \[ Z = \frac{1.116 \times 10^1}{56} = 2 \] Thus, \( Z = 2 \), indicating a body-centered cubic (BCC) structure.
Arrange the following in the increasing order of number of unpaired electrons present in the central metal ion:
I. \([MnCl_6]^{4-}\)
II. \([FeF_6]^{3-}\)
III. \([Mn(CN)_6]^{3-}\)
IV. \([Fe(CN)_6]^{3-}\)
The number of \( d \) electrons in Fe is equal to which of the following?
(i) Total number of \( s \)-electrons of Mg
(ii) Total number of \( p \)-electrons of Cl
(iii) Total number of \( p \)-electrons of Ne