$8.4 g$ of $Fe X^{2 +}$ ions (take molar mass $56 g mo l^{- 1}$ and $N_{A} = 6.022 \times 1 0^{23}$ ) is considered. The total number of unpaired electrons in the sample, in units of $1 0^{23}$ , to two decimal places, is

Numerical answer type, answer in x 10^23, tolerance 0.02.

Answer: 3.61 x 10^23

Moles of

Fe X^{2 +} = 8.4/56 = 0.15

. Each

Fe X^{2 +}

(

3 d^{6}

) has 4 unpaired electrons. Total

= 0.15 \times 4 \times 6.022 \times 1 0^{23} = 3.61 \times 1 0^{23}