Question

KH2PO4 степень окисления. Подскажите пожалуйста!

Answer 1

(K⁺¹)(H⁺¹)₂(P⁺⁵)(O⁻²)₄

Answer 2

KH2PO4 Oxidation State

The oxidation state of an element in a compound is the charge it would have if all the shared electrons were assigned to the more electronegative atom. In the case of KH2PO4, let's determine the oxidation state of each element.

- K (Potassium): Potassium is an alkali metal and typically has an oxidation state of +1 in compounds. - H (Hydrogen): Hydrogen usually has an oxidation state of +1 when combined with nonmetals. - P (Phosphorus): The oxidation state of phosphorus in KH2PO4 can be calculated using the formula: \[ 2x + 1 + 4(-2) = 0 \] where \(x\) is the oxidation state of phosphorus. Solving for \(x\), we get: \[ 2x - 7 = 0 \] \[ 2x = 7 \] \[ x = \frac{7}{2} = 3.5 \]

Since oxidation states are typically whole numbers, it's important to note that phosphorus can exhibit multiple oxidation states in different compounds due to its ability to form various bonds.

Therefore, the oxidation state of phosphorus in KH2PO4 is +3.5. [[1]]

It's worth mentioning that the concept of fractional oxidation states is not common in simple ionic compounds, and in this case, it indicates the covalent nature of the P-H bonds in the phosphate ion.

If you have any further questions or need additional information, feel free to ask!