Question

16. Assign oxidation numbers to all the atoms in each of the following: a. HNO3 b. CuClb С. 02 d. H2O2 e. C,H120 f. Ag g. PbSO4 h. PbO2 i. Na2C204 j. CO2 k. (NH42Ce(SO4)3 . Cr2O3

Answer 1

a) HNO_3 H = +1 O = -3 (there are 3 so -3*2 = -6) O here = -6 So N+1 - 6 = 0 N = +5 b) Cucl_2 Cu = +2 Cl = -1 c) O_2 Oxidation state of O_2 is zero(o) \r\n d) H_2O_2 H = +1 O = -1 \r\n e) C_6H_12O-6 H= +1 C = n O = -2 6*n +12(+1)+6(-2) = 0 n = 0 C = 0 Well this is general method but if we go deep we find C in Glucose have different oxidation sates for that you can see the below explanation. \r\n e) bullet C_1 atom has one O one C & one H atoms attached. Oxidation State of C_1 i.e n Implies n + (-2) + (o) + (+1) = 0 n = =1 bullet C_2: C_3: C_4: C_5 all have one OH, two C & one H attached. Therefore each one's Oxidation state n + (-1) + 2(0) + (+1) = 0 n = 0 bullet C_6 atom has one OH, one C & two H atoms attached. Therefore its oxidation State n + (-1) + (0) + 2(+1) = +1 Therefore n = -1 Oxidation States of all carbon atoms are +1, 0,0,0,0,-1 respectively. Hence addition of these is O, the overall Oxidation State of Glucose is (zero). \r\n F_2 Ag Oxidation State of Ag is (+1) g) PbSO_4 Pb = n S = +6 O= -2 n + 6 + 4(-2) = 0 n = +2 Pb = +2 h) PbO_2 O = -2 Pb = n n + 2(-2) = 0 n = 4 Pb = +4 i) Na_2C_2O_4 Na = +1 O = -2 C = n 2(+1) + 2(n) + 4(-2) = 0 n = +3 C = +3 j) CO_2 O = -2 C = n n + 2(-2) = 0 n = +4 C = +4 \r\n k) (NH_4)_2 Ce(SO_4)_3 NH_4 is (+1) (Oxidation number of N in NH_4+ is -3 Oxidation number of each H in NH_4+ is +1) Ce is (+4) SO_4 is (-2) (Oxidation number of S is +6 & O is -2 in SO_4) There are two polyatomic ions NH_4 & SO_4 NH_4 is (+1): SO_4 is (-2) & there are 3 that gives than a total of (-6). The compound should be balanced so 6 - 2 = 4: hence Ce has to be (+4). L. Cr_2O_3 O = -2 Cr = n 2(n) + 3(-2) = 0 n = +3 hence C = +3