Calculate the oxidation number of each atom in the following: CrO₄²⁻, NO₃⁻, CaO, Zn, H₂O, Sr²⁺

• A. CrO₄²⁻: Cr = +6, O = −2
  NO₃⁻: N = +3, O = -2
  CaO; Ca = +1, O = −1
  Zn: Zn = 0
  H₂O; H = +1, O = −2
  Sr²⁺: Sr = +2
• B. CrO₄²⁻: Cr = +6, O = −2
  NO₃⁻: N = +5, O = -1
  CaO; Ca = +1, O = −1
  Zn: Zn = 0
  H₂O; H = +1, O = −1
  Sr²⁺: Sr = +2
• C. CrO₄²⁻: Cr = +4, O = −1
  NO₃⁻: N = +5, O = -2
  CaO; Ca = +2, O = −2
  Zn: Zn = 0
  H₂O; H = +1, O = −1
  Sr²⁺: Sr = +2
• D. CrO₄²⁻: Cr = +6, O = −2
  NO₃⁻: N = +5, O = -2
  CaO; Ca = +2, O = −2
  Zn: Zn = 0
  H₂O; H = +1, O = −2
  Sr²⁺: Sr = +2

Answer: (D)