Write the formula for sulfur (VI) oxide?

Answer 1

SO2
The oxidation number of O is -2. S is 4. Hint (4 = IV).
Molecule should be neutral. So you need 2 O. Hence SO2

Answer 2

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RE:
Write the formula for sulfur (VI) oxide?

Answer 3

VI means that the oxidation number of S is 6. The oxidation number of oxygen is - usually - 2, so the formula of sulfur(VI) oxide is:

SO3

Answer 4

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> For tungsten(VI) phosphate, do the roman numerals indicate the valency of the element? No, they represent the oxidation state of the metal. For a cation which is only a metal, this will be the charge. So, for example, cobalt(II) would be Co^2+, cobalt(III) would be Co^3+, and titanium(IV) would be Ti^4+. But, there are some cases where there is a difference, of which the most commonly encountered early in chemistry studies is: mercury(II) is Hg^2+ but mercury(I) is (Hg2)^2+ - each mercury atom has a +1 charge, but they are bound together to make a dipositive cation. Thus, tungsten(VI) = W^6+ Note that you have mixed up the Roman numerals for four and six, which are IV and VI, respectively. > Phosphate's valency is -3. So, when I use the cross method, why isn't the formula > W subscript 3 (P0 subscript 4) subscript 4? The answer was W(P0 subscript 4) subscript 2. Using the cross method, combining W^6+ with (PO4)^3- gives W3(PO4)6, but the ratio 3:6 reduces to 1:2, and so tungsten(VI) phosphate would be W(PO4)2. Your answer, W3(PO4)4 would be the formula for tungsten(IV) phosphate. > Also, to write sulfur dioxide, both the valencies of these elements are negative. > S is -2 and O subscript 2 is -2. What happens now? With compounds consisting of two non-metals, the oxidation states are frequently different from the usual valencies. Sulfur dioxide, for example, is sulfur(IV) oxide if you want to use that naming convention. However, the typical naming convention tells you how many atoms are in the formula... "di" in dioxide tells you two O atoms, and the lack of any prefix before sulfur means there is one S atom. So, sulfur dioxide = SO2 Some other examples... carbon dioxide = CO2 carbon monoxide = CO sulfur trioxide = SO3 diphosphorus pentoxide = P2O5 > How would you write cesium cyanate? > With elements ending in "ate" is there are certain rule about it? There are a bunch of oxygen-containing ions ending in "ate" whose formulae you need to memorise, such as: sulfate = (SO4)^2- phosphate = (PO4)^3- nitrate = (NO3)^- cyanate = OCN^- chlorate = (ClO3)^- Once you have these memorise, you should learn the rules for figuring out other ions from these. For example, changing the "-ATE" ending to "-ITE" means the ion has one fewer oxygen atoms. sulfite = (SO3)^2- phosphite = (PO3)^3- nitrite = (NO2)^- chlorite = (ClO2)^- The prefix "PER-" before an "-ATE" ion means add an extra oxygen atom. So, as chlorate is (ClO3)^- it follows that perchlorate is (ClO4)^-. The formula (MnO4)^- for permanganate implies the manganate ion is (MnO3)^-. The prefix "HYPO-" before an "-ITE" ion means remove an oxygen atom, producing a series of ions: PERchlorATE = (ClO4)^- chlorATE = (ClO3)^- chlorITE = (ClO2)^- HYPOchlorITE = OCl^- One final prefix to mention is "THIO-", which means change one oxygen atom into a sulfur atom. Since the sulfate ion is (SO4)^2-, thiosulfate is (S2O3)^2-. Similarly, thiocyanate is SCN^- as cyanate is OCN^-. Anyway... as the cesium ion is Cs^+, we have: cesium cyanate = CsOCN and some more examples... cesium hypochlorite = CsOCl cesium chlorate = CsClO3 cesium sulfate = Cs2SO4 cesium thiosulfate = Cs2S2O3 cesium thiocyanate = CsSCN cesium nitrite = CsNO2 Hopefully this has answered your question *and* given some 'take home advice'. =)

Answer 5

Sulfur (IV) Oxide is SO2 since the Oxygen ion has a -2 charge, two are needed to cancel out the Sulfur ion's +4 charge.
http://www.webelements.com/webelements/compounds/text/S/O2S1-7446095.html

EDIT:
Upon closer examination, I initially misread the charge on the Sulfur ion in the compound, mistaking it for S+4 rather than S+6....apparently an easy mistake to make since another user did it as well.

In this case the proper formula for Sulfur (VI) is SO3 since three, +2, Oxygen ions are needed to cancel out the +6 charge on the Sulfur.

http://www.webelements.com/webelements/compounds/text/S/O3S1-7446119.html

Answer 6

SO3

Answer 7

so4