Among $ 10^{-10} $ g (each) of the following elements, which one will have the highest number of atoms?
Element : Pb, Po, Pr and Pt
$\text{No. of atoms} = \frac{\text{Mass in g}}{\text{Molar Mass (g/mol)}} \times N_A$
Therefore for the same Mass element having the least Molar mass will have the higher no. of atoms.
$M_{Pb} = 209$
$M_{Pr} = 141$
$M_{Po} = 207$
$M_{Pt} = 195$
Statement-1: \( \text{ClF}_3 \) has 3 possible structures.
Statement-2: \( \text{III} \) is the most stable structure due to least lone pair-bond pair (lp-bp) repulsion.
Which of the following options is correct?