The correct options are
B wx=0.25
D yz=1
Tl2O3(s)+NH2OH(aq)→TlOH(s)+N2(g)
Oxidation state of Tl in Tl2O3=+3
Oxidation state of Tl in TlOH=+1
Oxidation state of N in NH2OH=−1
Oxidation state of N in N2=0
N is undergoing oxidation.
Tl is undergoing reduction.
Finding nf:
nf of Tl2O3=4
nf of NH2OH=1
Cross multiplying these with nf of each other.
we get,
Tl2O3(s)+4NH2OH(aq)→TlOH(s)+N2(g)
Balancing the elements other than oxygen and hydrogen on both sides,
Tl2O3(s)+4NH2OH(aq)→2TlOH(s)+2N2(g)
Adding the H2O to balance the oxygen,
Tl2O3(s)+4NH2OH(aq)→2TlOH(s)+2N2(g)+5H2O
Hydrogen is already balanced.
Charges on both sides is also balanced.
So, it is the final balanced equation.
So, the value of w,x,y,z is 1,4,2,2 respectively.
wx=14=0.25yz=22=1