Q77E - Chemistry (OpenStax)

Question

Assign an oxidation state to phosphorus in each of the following:

(a) \({\rm{Na}}{{\rm{H}}_2}{\rm{P}}{{\rm{O}}_3}\)

(b) \({\rm{P}}{{\rm{F}}_5}\)

(c) \({{\rm{P}}_4}{{\rm{O}}_6}\)

(d) \({{\rm{K}}_3}{\rm{P}}{{\rm{O}}_4}\)

(e) \({\rm{N}}{{\rm{a}}_3}{\rm{P}}\)

(f) \({\rm{N}}{{\rm{a}}_4}{{\rm{P}}_2}{{\rm{O}}_7}\)

 

Step-by-Step Solution

Answer

(a)  The phosphorus oxidation state is \( + 3\) in\({\rm{Na}}{{\rm{H}}_2}{\rm{P}}{{\rm{O}}_3}\)

(b)  The phosphorus oxidation state is \( + 5\) in\({\rm{P}}{{\rm{F}}_5}\)

(c)  The phosphorus oxidation state is \( + 3\) in\({{\rm{P}}_4}{{\rm{O}}_6}\)

(d)  The phosphorus oxidation state is \( + 5\) in\({{\rm{K}}_3}{\rm{P}}{{\rm{O}}_4}\)

(e)  The phosphorus oxidation state is \( - 3\) in\({\rm{N}}{{\rm{a}}_3}{\rm{P}}\)

(f)  The phosphorus oxidation state is \( - 3\) in\({\rm{N}}{{\rm{a}}_4}{{\rm{P}}_2}{{\rm{O}}_7}\)

 

Step 1: Defition of oxidation state

• The number of electrons lost or acquired by an atom as it establishes a bond is referred to as the oxidation number.
• A Group I A element's oxidation number is \( + 1\), a Group II element's is \(A =  + 2\), and a Group VII element's is \(A =  - 1\).
• The oxidation number of H is normally \( + 1\), and the oxidation number of O is usually \( - 2\).
• The total of oxidation numbers in a neutral substance is zero.

 

Step 2: Find the phosphorus oxidation state of \({\rm{Na}}{{\rm{H}}_2}{\rm{P}}{{\rm{O}}_3}\)

\({\rm{Na}}{{\rm{H}}_2}{\rm{P}}{{\rm{O}}_3}\)is the given compound.

 

Oxidation state (OS) of \({\rm{Na}} =  + 1\)

 

\({\rm{OS}}\;{\rm{of H}} =  +  + 1\)

 

OS of \(O =  - 2\)

 

Let ‘x‘ be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

 

\(\begin{array}{l}1 + 2(1) + x + 3( - 2) = 0\\x = {\rm{OS of P}} =  + 3\end{array}\)

Step 3: Find the phosphorus oxidation state of \({\rm{P}}{{\rm{F}}_5}\) \({\rm{P}}{{\rm{F}}_5}\)is the given compound.

Oxidation state \(({\rm{OS}})\) of \(F =   - 1\)

 

Let ‘x‘ be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

 

\(\begin{array}{l}x + 5( - 1) = 0\\x = OS{\rm{ of P}} =  + 5\end{array}\)

S tep 4: Find the phosphorus oxidation state of \({{\rm{P}}_4}{{\rm{O}}_6}\) \({{\rm{P}}_4}{{\rm{O}}_6}\) is the given compound.

Oxidation state \(({\rm{OS}})\)of \({\rm{O}} =  - 2\)

 

Let ‘x‘ be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

 

\(\begin{array}{l}4x + 6( - 2) = 0\\x = {\rm{ OS of P}} =  + 3\end{array}\)

 

Step 5: Find the phosphorus oxidation state of \({{\rm{K}}_3}{\rm{P}}{{\rm{O}}_4}\) \({{\rm{K}}_3}{\rm{P}}{{\rm{O}}_4}\)is the given compound.

Oxidation state \(({\rm{OS}})\)of \({\rm{O}} =  - 2\)

 

\({\rm{OS}}\) of \({\rm{K}} =  + 1\)

 

Let ‘x‘ be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

 

\(\begin{array}{l}3 + x + 4( - 2) = 0\\x = {\rm{OS of P}} =  + 5\end{array}\)

 

Step 6: Find the phosphorus oxidation state of \({\rm{N}}{{\rm{a}}_3}{\rm{P}}\) Oxidation state \(({\rm{OS}})\)of \({\rm{O}} = - 2\) \({\rm{OS}}\) of \({\rm{K}} = + 1\)

Let ‘x‘ be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

 

\(\begin{array}{l}3 + x + 4( - 2) = 0\\x = {\rm{OS of P}} =  + 5\end{array}\)

 

 

\({\rm{N}}{{\rm{a}}_3}{\rm{P}}\)is the given compound.

 

Oxidation state \(({\rm{OS}})\) of \({\rm{Na}} =  + 1\)

 

\({\rm{OS}}\) of \({\rm{K}} =  + 1\)

 

Let ‘x‘ be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

 

\(\begin{array}{l}3 + x = 0\\x = {\rm{OS of P}} =  - 3\end{array}\) 

 

Step 7: Find the phosphorus oxidation state of \({\rm{N}}{{\rm{a}}_4}{{\rm{P}}_2}{{\rm{O}}_7}\) \({\rm{N}}{{\rm{a}}_4}{{\rm{P}}_2}{{\rm{O}}_7}\) is the given compound.

Oxidation state \(({\rm{OS}})\)of \({\rm{Na}} =   + 1\)

 

\({\rm{OS}}\) of \({\rm{O}} =  - 2\)

 

Let ‘x‘ be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

 

\(\begin{array}{l}4 + 2x + 7( - 2) = 0\\x = OS{\rm{ of P}} =  + 5\end{array}\)