Answer:
NaClO₄
Solution:
Let us calculate the oxidation state of Chlorine in each compound one by one;
1) NaClO
O.N of Na = +1
O.N of O = -2
So,
(+1) + (Cl) + (-2) = 0
1 + Cl - 2 = 0
Cl = -1 + 2
Cl = +1
2) NaClO₃
(+1) + (Cl) + (-2)3 = 0
(+1) + (Cl) - 6 = 0
1 + Cl - 6 = 0
Cl = -1 + 6
Cl = +5
3) NaClO₂
(+1) + (Cl) + (-2)2 = 0
(+1) + (Cl) + (-4) = 0
1 + Cl - 4 = 0
Cl = -1 + 4
Cl = +3
4) NaClO₄
(+1) + (Cl) + (-2)4 = 0
(+1) + (Cl) + (-8) = 0
1 + Cl - 8 = 0
Cl = -1 + 8
Cl = +7
