Please explain how the answer is found. **Title: Calculating Oxidation Numbers in Organic Molecules**---### Instructions:1. **Overall Oxidation Number Calculation:** - Calculate the overall oxidation number for carbon using the algorithm method, not the structure. - Ensure that you use the correct chemical formula. **Input Field:** - Enter the calculated oxidation number in a decimal form (e.g., if O.N. = \( \frac{5}{2} \), then enter 2.5).2. **Individual Oxidation Numbers:** - Calculate the individual oxidation number for each carbon atom by examining the structure. **Help Button:** - “I’m a little confused still.” (Click for additional guidance)3. **Averaging Oxidation Numbers:** - Average the individual oxidation numbers calculated. - This value should match the one obtained using the algorithm method. - Use this technique to verify your calculations during quizzes. **Help Button:** - “I’m still stuck.” (Click for extra assistance) 4. **Evaluation:** - Click “Evaluate” to check your answers.---**Feedback Section:**- If the calculated average oxidation number is incorrect, a message will prompt: - “The average oxidation number is incorrect. Try only using the chemical formula and ignoring the structure.”---**Diagram:**- **Structure Representation:** - The image includes a structural diagram of a heterocyclic compound with a nitrogen atom in a five-membered ring. This is to aid in visualizing the carbon framework for individual oxidation state calculations.---This exercise is designed to enhance your skills in determining oxidation numbers, ensuring you can switch between viewing a chemical formula and molecular structure seamlessly. Familiarity with this method is crucial for accuracy in organic chemistry assessments.

The structural formula of the molecule is: From the structural formula, the molecule has 5 C, 1 N and 9 H Hence the chemical formula of the given molecule is C5H9N#Part-1: According to the rules of oxidation number, H has +1 oxidation state in its molecules except in metal hydrides. Hence each H has +1 oxidation state. N has an oxidation state of -3. Suppose the oxidation sate of C be "X". In a neutral molecule, the sum of oxidation states of all the atoms is equal to 0. Hence 5*X + 9*(+1) + 1*(-3) = 0 => 5X + 9 - 3 = 0 => 5X = - 6 => X = - 6/5 = - 1.2 Hence the overall oxidation number of C = - 1.2 (Answer)#Part-2: C is more electronegative than H Hence in a C-H bond, H has +1 charge and C has -1 charge N is more electronegative than C. Hence in an N-C bond, C is +1 and N is -1. In a C-C bond, each C has 0 oxidation state.# Calculation for individual oxidation number: For C1, Oxidation number = +1-1 = 0 For C2, Oxidation number = -1 For C3, Oxidation number = -1-1 = -2 For C4, Oxidation number = -1-1+1 = -1 For C5, Oxidation number = -1-1-1+1 = -2 Average of individual oxidation numbers = 0-1-2-1-25= -65= - 1.2 - 1.2, which is the same as the oxidation number calculated in part-1.