Solving part-1 only
#1
KMnO_4
- Transition metal is Manganese (Mn)
#2
Actually it's the oxidation number of Mn
Let's find how?
[tex]\\ \tt\Rrightarrow x+1+4(-2)=0[/tex]
[tex]\\ \tt\Rrightarrow x+1-8=0[/tex]
[tex]\\ \tt\Rrightarrow x-7=0[/tex]
[tex]\\ \tt\Rrightarrow x=+7[/tex]
- x is the oxidation number
#3
- Purple as per the color of potassium permanganate
#4
[tex]\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}[/tex]
