Question

Assign oxidation states to atoms underlined in each compound i. (NH4)2HPO4 ii. Na2C2O4 iii. Mg2P2O7 iv. KMnO4

Answer 1

I cannot see any atoms underlined, so I'll answer the question to the best of my understanding.
Remember one thing, sum of all oxidation numbers should be equal to the charge on the species.
(NH₄)₂HPO₄ → This is made of 2 NH₄⁺ and 1 HPO₄-²⁻ ions. In HPO₄²⁻, H is in +1  state, O is in -2 state, and hence, P would be in -2+8-1 = +5 oxidation state.
Na₂C₂O₄ → 2 Na⁺ and 1 C₂O₄²⁻ ions. In C₂O₄²⁻, O is in -2 state, so C would be in (-2+8)/2 = +3 state. We've divided by 2 since there are 2 C atoms in oxalate ion.
Mg₂P₂O₇ → 2 Mg²⁺ and 1 P₂O₇⁴⁻ ion. Oxidation number of O would be -2, and hence, the oxidation number of P would be (-4+14)/2 = +5, again dividing by 2 because there are 2 P atoms.
KMnO₄ → MnO₄⁻ ion is present having O in -2 state, which means O.S. of Mn would be -1+8 = +7. This is the highest O.S. of Mn.
Just remember that the sum of O.S. of all atoms is equal to the charge on the species, and use this concept to find unknown O.S.