Answer
1) Percentage of mass of Re in the following Oxides are as:
$ReO_{2}$ = $85.3358% Re$
$ReO_{3}$ = $79.5063% Re$
$Re_{2}O_{3}$ = $88.5833% Re$
$Re_{2}O_{7}$ = $76.8804% Re$
2) The charge on Re in $ReO_{2}$ is +4
The charge on Re in $ReO_{3}$ is +6
The charge on Re in $Re_{2}O_{3}$ is +3
The charge on Re in $Re_{2}O_{7}$ is +7
3) The increasing Order of charge is : $Re_{2}O_{3}$ $\lt$$ReO_{2}$$\lt$$ReO_{3}$$\lt$ $Re_{2}O_{7}$
4) As the charge on Re increased, the % of Re in Rhenium oxide decreased.
Work Step by Step
1) Percentage of Re in each Compound:
$ReO_{2}$ :
% of Re =[ (Atomic Weight of Re)$\div$( Formula Weight of $ReO_{2}$ )]$\times$ 100
% of Re = [ (186.207 g) $\div$(218.205g)]$\times$100
% of Re = 85.3358
$ReO_{3}$:
% of Re =[ (Atomic Weight of Re)$\div$( Formula Weight of $ReO_{3}$ )]$\times$ 100
% of Re = [ (186.207 g) $\div$(234.204g)]$\times$100
% of Re = 79.5063
$Re_{2}O_{3}$:
% of Re =[ (2$\times$Atomic Weight of Re)$\div$( Formula Weight of $Re_{2}O_{3}$ )]$\times$ 100
% of Re = [ (372.414 g) $\div$(420.411g)]$\times$100
% of Re = 88.5833
$Re_{2}O_{7}$:
% of Re =[ (2$\times$Atomic Weight of Re)$\div$( Formula Weight of $Re_{2}O_{7}$ )]$\times$ 100
% of Re = [ (372.414 g) $\div$(484.407g)]$\times$100
% of Re = 76.8804
2) Each compound is neutral and charge on oxygen is taken as -2.
$ReO_{2}$ :
0 = 1(Charge on Re) + 2(Charge on O)
0= x + 2(-2)
x = +4
$ReO_{3}$ :
0 = 1(Charge on Re) + 3(Charge on O)
0= x + 3(-2)
x = +6
$Re_{2}O_{3}$ :
0 = 2(Charge on Re) + 3(Charge on O)
0= 2x + 3(-2)
x = +3
$Re_{2}O_{7}$ :
0 = 2(Charge on Re) + 7(Charge on O)
0= 2x + 7(-2)
x = +7
3) The increasing Order of charge is: $Re_{2}O_{3}$ (Re=+3)$\lt$$ReO_{2}$(Re=+4)$\lt$$ReO_{3}$(Re=+6)$\lt$ $Re_{2}O_{7}$(Re=+7)
4) As the charge on Re increased, the % of Re in Rhenium oxide decreased.
