The answer to your question is: letter b Sulfur Explanation: Remember that in neutral molecules, the sum of the oxidation numbers is zero. Thus, the algebraic sum of the oxidation numbers of the elements of the compound NO is equal to zero. Nitrogen is given an oxidation number of +2, while the oxygen in this compound is given an oxidation number of -2. Here we have a compound NO (nitrogen oxide). NO -> N +2 and O -2 Explanation: Oxidation numbers are assigned to the elements of a compound to track the number of electrons each atom has. For the compound to have zero charge, the sum of the oxidation numbers is zero. The oxidation number of oxygen (-2) is then used as a reference. Let’s assign the oxidation number of nitrogen to N. N=+2 O=-2 Explanation: The compound NO is electrically neutral. And, the oxidation state of bromine increases, it undergoes an oxidation reaction and hence is a reducing agent. In the given chemical equation: Reactive side: Hydrogen oxidation state = +1 Bromine oxidation state = -1 Sulfur oxidation state = +6 Oxygen oxidation state = -2 Product side: Hydrogen oxidation state = +1 Bromine oxidation state = 0 Sulfur oxidation state = +4 Oxygen oxidation state = -2 As oxidation state of sulfur decreases, it undergoes a reduction reaction and is therefore an oxidizing agent. Reducing agent is defined as the reagent that helps another substance to be reduced and itself oxidized. Para 2: Oxidizing agent is defined as the reactant that helps another substance to oxidize and reduce. Thus, the oxidation state of Mn is +4 and the oxidation state of O is -2. To calculate the oxidation state of manganese, in a neutral manganese oxide compound, we take the oxidation state of manganese as ‘x’. Explanation: For 1: The oxidation state of oxygen is assumed to be -2. For 2: The oxidizing agent is and the reducing agent is HBr. The reducing agent is HBr, or Br⁻¹, because the oxidation number of H goes from -2 in HBr to 0 in Br2.įor 1: The oxidation state of Mn is +4 and the oxidation state of O is -2. 2) The oxidizing agent is H2SO4, or S⁺⁶, because the oxidation number of S goes from +6 in H2SO4 to +4 in SO2. X = +2 Therefore, the oxidation of N to NO is +2 and the oxidation of O to NO is -2, respectively.ġ) oxidation number Mn +4, oxidation number O -2. The sum of the oxidation number of a neutral compound is zero. +2, -2 Explanation: According to the oxidation number assignment rule, the oxidation number of the O atom is -2. This is a loss of electrons and O has been oxidized. O goes from an oxidation state of -4 to 0. This is a gain of electrons and K has been reduced. The oxidation number of K in KClO2: K + (-1) + 2(-2) = 0 K-5 = 0 K = +5 The oxidation number of K in KCl: K + (-1) = 0 K = + 1 The oxidation number of Cl in KClO2 is -1 For Cl in KCl, the oxidation number is -1 For O in KClO2, the oxidation number is (2 x -2) = -4 For O in O2, the oxidation number is 0 K goes from an oxidation state of +5 to +1. For charged radicals, their oxidation number is the charge. The algebraic sum of all the oxidation numbers of all the atoms of a neutral compound is zero. The charge on single ions signifies their oxidation number. k in kclo2: k in kclo2: k in kcl: k in kcl: cl in kclo2: cl in kclo2: cl in kcl: cl in kcl: o in kclo2: o in kclo2: o in o2: o in o2: which element is it oxidized? kk oo clcl which element is reduced? oooo lolĮxplanation: The formula of the reaction: KClO₂ → KCl + O₂ To assign oxidation numbers, we must obey certain rules: Elements in the uncombined state or whose atoms combine to form molecules have a number d zero oxidation. For the reaction kclo2⟶kcl+o2 kclo2⟶kcl+o2 assign oxidation numbers to each element on each side of the equation.
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