In a reaction one mole of ${{N}_{2}}{{H}_{4}}$ loses ten moles of electrons to form a new compound Y. Assuming that all nitrogen appears in the new compound. What is the oxidation number of N in Y?
Answer
602.4k+ views
Hint: We need to know the number of electrons which is needed to make the bond, in order to find the oxidation state of the nitrogen in ${{N}_{2}}{{H}_{4}}$ or hydrazine. Then by determining whether the nitrogen gains or loses electrons we can determine the oxidation state of nitrogen in Y.
Complete step by step answer:
-Let’s begin with the concept of oxidation number. It can be defined in various terms such as addition of oxygen, in terms of hydrogen removal, addition of electronegative elements or in term of change of electrons in an oxidation process.
-In the context of the given question we need to consider oxidation number in terms of the electron transfer. As we know in chemical reactions it’s important to know whether an atom loses or gain an electron. Oxidation number helps us to keep a track of these electron transfers.
- The given question can be represented as follows
\[{{N}_{2}}{{H}_{4}}\to Y+10{{e}^{-}}\]
- As we can see one mole of hydrazine loses ten moles of electron to form the new compound Y. It’s given that all nitrogen appears in the new compound, hence the oxidation state of hydrogen remains unchanged and the electrons are lost from nitrogen.
- One mole of ${{N}_{2}}{{H}_{4}}$ contains two moles of nitrogen. That is two moles of nitrogen loses ten moles of electron. Therefore, one mole of nitrogen will lose five moles of electron.
-Thus, the initial oxidation state of nitrogen in hydrazine is −2. Then it loses five moles of electron and the final oxidation state of nitrogen can be calculated as follows
The final oxidation state of N =−2 + 5
=+3
Since electrons are getting removed, the charge should be positive.
Therefore the oxidation state of N in Y is +3.
Note: Keep in mind that for an uncombined element the oxidation number will be zero. Generally, oxygen has an oxidation number of $+2$ and Hydrogen has an oxidation number of $+1$. For the neutral compounds, the sum of oxidation numbers of its constituent atoms will be zero and in the case of polyatomic ions, the sum of oxidation numbers of atoms will be equal to the net charge of the polyatomic ion.
Complete step by step answer:
-Let’s begin with the concept of oxidation number. It can be defined in various terms such as addition of oxygen, in terms of hydrogen removal, addition of electronegative elements or in term of change of electrons in an oxidation process.
-In the context of the given question we need to consider oxidation number in terms of the electron transfer. As we know in chemical reactions it’s important to know whether an atom loses or gain an electron. Oxidation number helps us to keep a track of these electron transfers.
- The given question can be represented as follows
\[{{N}_{2}}{{H}_{4}}\to Y+10{{e}^{-}}\]
- As we can see one mole of hydrazine loses ten moles of electron to form the new compound Y. It’s given that all nitrogen appears in the new compound, hence the oxidation state of hydrogen remains unchanged and the electrons are lost from nitrogen.
- One mole of ${{N}_{2}}{{H}_{4}}$ contains two moles of nitrogen. That is two moles of nitrogen loses ten moles of electron. Therefore, one mole of nitrogen will lose five moles of electron.
-Thus, the initial oxidation state of nitrogen in hydrazine is −2. Then it loses five moles of electron and the final oxidation state of nitrogen can be calculated as follows
The final oxidation state of N =−2 + 5
=+3
Since electrons are getting removed, the charge should be positive.
Therefore the oxidation state of N in Y is +3.
Note: Keep in mind that for an uncombined element the oxidation number will be zero. Generally, oxygen has an oxidation number of $+2$ and Hydrogen has an oxidation number of $+1$. For the neutral compounds, the sum of oxidation numbers of its constituent atoms will be zero and in the case of polyatomic ions, the sum of oxidation numbers of atoms will be equal to the net charge of the polyatomic ion.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

