Question

which choice correctly identifies the oxidation number (o.n.) for the given species? answer: a mn²⁺, o.n. = 0 b ca, o.n. = +2 c f₂, o.n. = -1 d se²⁻, o.n. -2 e p₄, o.n. = -1

Explanation:

Step1: Recall oxidation - number rules

The oxidation number of a monatomic ion is equal to its charge. The oxidation number of an element in its elemental form is 0.

Step2: Analyze option A

For $Mn^{2 + }$, the oxidation number (O.N.) is + 2 (not 0) because it is a 2 - charged cation.

Step3: Analyze option B

For $Ca$, it is an element in its elemental form, so the oxidation number is 0 (not + 2).

Step4: Analyze option C

For $F_2$, it is an element in its elemental form, so the oxidation number is 0 (not - 1).

Step5: Analyze option D

For $Se^{2 - }$, the oxidation number is - 2 as it is a 2 - charged anion. This is correct.

Step6: Analyze option E

For $P_4$, it is an element in its elemental form, so the oxidation number is 0 (not - 1).

Answer:

D. $Se^{2 - }, O.N.=-2$