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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91 Problem 92 Problem 93 Problem 94 Problem 95 Problem 96 Problem 97 Problem 98 Problem 99 Problem 100 Problem 101 Problem 102 Problem 103 Problem 104 Problem 105 Problem 106 Problem 107 Problem 108 Problem 109 Problem 110 Problem 111 Problem 112 Problem 113 Problem 114 Problem 115 Problem 116

Problem 37 Easy Difficulty

Name each of the following compounds:
(a) $\mathrm{TeO}_{2}$
(b) $\mathrm{Sb}_{2} \mathrm{S}_{3}$
(c) $\mathrm{GeF}_{4}$
(d) $\mathrm{SiH}_{4}$
(e) $\mathrm{GeH}_{4}$

Answer

a. $\mathrm{TeO}_{2}$ In this compound, the cation is tellerium and anygen. As the compound is neutral in nature and the oxidation number of oxygen is $-2,$ the oxidation number of tellerium will be $+4$ , the name of the compound will be Tellurium (IV) oxide
b. In this compound, the cation is antimony and anion is sulfur. As the compound is neutral in nature
and the oxidation number of sulfur is $-2,$ the oxidation number of antimony will be $+3,$ the
name of the compound will be Antimony (III) sulfide
c. In this compound, the cation is germanium and anion is fluorine. As the compound is neutral in
nature and the oxidation number of flourine is $-1$ , the oxidation number of germanium will be
$+4,$ the name of the compound will be Germanium (N) fluoride
d. In this compound, the cation is silicon and anion is hydrogen. As the compound is neutral in
nature and the oxidation number of hydrogen is $-1,$ the oxidation number of silicon will be $+4$ ,
the name of the compound will be Silicon (IV) Hydraide
e. In this compound, the cation is germanium and anion is hydrogen. As the compound is neutral in
nature and the oxidation number of hydrogen is $-1,$ the oxidation number of germanium will be
$+4,$ the name of the compound will be Germanium (IV) Hydride.

Related Courses

Chemistry 101
Chemistry

Chapter 18

Representative Metals, Metalloids, and Nonmetals

Related Topics

Nonmetals Chemistry

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Video Transcript

in this compound. The Kodai in is delirium on a nyeon is oxygen. The compound is in neutral state oxidation. Number off excision is negative. Toe by lofty Librium is positive for so the name will be delirium for oxide income bound to be anti money is a nine. Anti money is contained, while sulfur is a nine oxidation number of sun for his negative. Too vile off anti money is plus three. The name will be anti money tree still fight often. See, Gordon is Jemaine Iam A nine is fury oxidation. Number off Lorena's negative one on oxidation number off Romania Miss plus four. The name will be germanium for fluoride option D. Satine is silicon. A nine is hydrogen oxidation. Number off. Contain is blissful. The name will be silicon for hydride option E Catania, Satine, Germanium and I and hydrogen oxidation. Number off germanium is plus for the name will be Jermaine Iam four. Hide right

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