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Chapter 18: Q77E (page 1045)

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

Short Answer

Expert verified

(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 by step solution

01

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.
02

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 鈥榵鈥� be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

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

03

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 鈥榵鈥� be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

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

04

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 鈥榵鈥� be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

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

05

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 鈥榵鈥� be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

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

06

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 鈥榵鈥� 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 鈥榵鈥� be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

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

07

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 鈥榵鈥� be the OS of P. The sum of the \(OS = 0\) since the molecule is neutral.

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

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