**Question 24 of 28****Determine the number of valence electrons in phosphoric acid (H₃PO₄) and then draw the corresponding Lewis structure (with minimized formal charges).**- A) 32- B) 29- C) 31- D) 35- E) 34**Additional Information:**You can access more details by visiting OpenStax or exploring the Periodic Table using the resources provided. **Explanation:**To solve this problem, calculate the total number of valence electrons in H₃PO₄ by summing the valence electrons of each atom:- Hydrogen (H): 1 electron each, 3 atoms total = 3 electrons- Phosphorus (P): 5 electrons, 1 atom total = 5 electrons- Oxygen (O): 6 electrons each, 4 atoms total = 24 electronsThus, the total number of valence electrons is 3 + 5 + 24 = 32. Therefore, the correct answer is A) 32. To complete the problem, sketch the Lewis structure based on these valence electrons, ensuring formal charges are minimized.

Given compound: H3PO4 We have to find the number of valence electrons in H3PO4. Also, we have to draw the Lewis structure of H3PO4.The atomic numbers of H, P and O are 1, 15, and 8 respectively. Hence, the electron configurations of H, P, and O are: H: 1s1 P: [Ne] 3s2 3p3 O: [He] 2s2 2p4 Hence, the valence electrons of H, P, and O are 1, 5, and 6 respectively. Hence, the total number of valence electrons in H3PO4 = 3×1+1×5+4×6 = 32 So, the correct option is A. Now, the Lewis structure of H3PO4 will be: NOTE: P will be the central atom because it is the largest in size among H, P, and O.