When calculating oxidation numbers, use '-2' (oxygen) as the 'yardstick'(standard).

Since SO3 has three oxygens, then the oxygen moiety is 3 x -2 = -6

Since SO3 is a neutrally charged molecule, overall oxidation is '0'.

Hence

S + -6 = 0

Hence S = 6 is the oxudation number for sulphur.

NB Sulphur exhibits different oxidation states in different molecules.

S = 0

SO2 = 4

SO3 = 6 ( as above)

H2SO4) = 6

H2S2O7 (Oleum) = 6

Elemental sulphur, like all other elements, has an oxidation number of zero(0).

However, when combined with other elements into molecules, it displays different oxidation number.s

H2S = -2

SO2 = +4

SO3 = +6

H2SO4 = + 6

HCN ( hydrogen cyanide) has the structurte H-C///N ( /// = triple bond).

The oxidation number of an atom is the number of outer most electrons used in forming a bond. Not all the outer most electrons are used, some can remain as 'lone pairs'.

Carbon ALWAYS has an oxidation no. of '+4' . It never varies. Hydrogen has an oxidaton number of '+1'. It is the nitrogen that has the variable oxidation. number.

HCN forms the ions , H^(+) & CN^(-)

Since carbon is always '4' and the overall charge of the ion is '-1' , then we can form a sum .

4 + N = -1

N = -1 -4

N = -5 is the oxid'n no/ of nitrogen in HCN. ( All five of nitrgens outer most electrons are involved in the bonding process. ). Compare with ammonia NH3 , Here nitrogen's oxid'n No. is '3' leaving a 'lone pair' of electrons not used in the outer most electron shell.

'4' .

Carbon atoms have four electrons in their outer energy shell. This leaves space to accept four more electrons to formvfour covalent bonds.

The oxidation number refers to the number of electrons in the outer energy shell that are used in combination with other atoms. Since carbon uses all four electrons , then its oxidation number is '4'.

Other elements, notably nitrogen have variable oxidation number.

Nitrgen when combined with hydrogen to form ammonia has an oxidation number of '3'. Because only three electrons out of the five electrons , in the outer energy shell, are used in combination with hydrogen.

In MnO2, the oxidation number of oxygen is typically -2. Since there are two oxygen atoms in MnO2, the total oxidation number contributed by oxygen is -4. The overall charge of the compound is neutral, so the oxidation number of manganese (Mn) can be calculated by setting the total oxidation number equal to zero. Therefore, the oxidation number of Mn in 2MnO2 is +4.

When calculating oxidation numbers, use '-2' for oxygen as a standard. It does vary in some oxygen molecules, but '-2' is a good starting point.

So in KNO3

The oxygen moiety is 3 x -2 = -6

Potassium(K) is always '+1' because it only loses one electron .

Hence the nitrgen os

+1 + N - 6 = 0

N- 5 = 0

N = +5 (oxidation state)

NB For neutrally charged molecule, the sum of the oxidation number is always 'zero'(0).

When calculating oxidation states of compounds containing PPh3 (triphenylphosphine), each phenyl group (Ph) is considered electrically neutral and contributes zero charge. Therefore, the charge of PPh3 is equal to the charge of the phosphorus atom in the compound. Phosphorus typically has an oxidation state of +3 in PPh3, making the overall charge of PPh3 neutral.

Well, honey, the oxidation number of strontium in the hydride of strontium is +2. It's like asking if Betty White is a national treasure - the answer is obviously yes. So, in this case, strontium is happily rocking that +2 oxidation number like a boss.

-1

In the compound NH3, hydrogen has an oxidation number of +1. This is because hydrogen typically has an oxidation number of +1 when bonded to nonmetals like nitrogen in ammonia. The sum of the oxidation numbers in a neutral compound is zero, so the nitrogen in NH3 has an oxidation number of -3 to balance the +1 oxidation number of each hydrogen atom.

Well, isn't that a happy little question! In Na3PO3, the oxidation number for Na is +1 and for O is -2. Since the overall charge of Na3PO3 is neutral, we can calculate the oxidation number for P to be +3. Just remember, there are no mistakes in oxidation numbers, only happy little accidents!

The answer is +7 because the oxygen has a charge of -8 and the charge on the whole compound is -1 therefore the charge on Cl must be +7.

In the compound O2PtF6, oxygen has an oxidation number of -1. This is because fluorine is more electronegative than oxygen and is assigned an oxidation number of -1. Since there are two fluorine atoms in the compound, the total oxidation number contributed by fluorine is -2. Therefore, the oxidation number of oxygen must be -1 to balance the overall charge of the compound, which is 0.

In K2SiO3, potassium (K) has an oxidation number of +1 since it belongs to Group 1 of the periodic table. Oxygen (O) typically has an oxidation number of -2, and there are three oxygen atoms in the compound, giving a total of -6. Therefore, the oxidation number of silicon (Si) can be calculated as +4 to balance the overall charge of the compound, which is 0.

In CaSO4 (calcium sulfate), the oxidation number of calcium (Ca) is +2, as it typically forms ionic compounds with a 2+ charge. Oxygen (O) usually has an oxidation number of -2 in compounds, so in CaSO4, the total oxidation number of the oxygen atoms is -8. To find the oxidation number of sulfur (S), we can set up the equation: 2(+2) + x + 4(-2) = 0, where x represents the oxidation number of sulfur. Solving this equation gives us the oxidation number of sulfur as +6.

1. NH₄NO₃ (ammonium nitrate), the oxidation number of nitrogen in the NH₄⁺ ion is -3, and in the NO₃⁻ ion is +5. The overall charges balance with the ammonium ion being +1 and the nitrate ion being -1.

In H2SiOF6, the oxidation number of hydrogen is +1 and the oxidation number of oxygen is -2. Since the overall charge of the compound is 0, the sum of the oxidation numbers must equal 0. Therefore, the oxidation number of silicon (Si) can be calculated as follows:

2(+1) + x + (-2) + 6(-1) = 0 2 + x - 2 - 6 = 0 x - 6 = 0 x = +6

Therefore, the oxidation number of silicon (Si) in H2SiOF6 is +6.

It can be, but not always so!!!!

For Groupos (I) & (II) the oxidation number is '1' & '2' respectively, the same as the ionic charge.

However, The Oxidation number can vary particularly in the Transition metals. It refers to the number of electrons in the outer (valence) energy shell , that acutally take part in the combination of a chemcial reaction.. NOT all electrons in the outer valence shell take part in a chemcial reaction.

For nitrogen in HNO3 it is '5'.

To calculate oxidation numbers.

1. Use oxygen as the stabndard at '-2'

2. -2 x 3 = -6 is the oxygen moiety.

3. Since hydrogen only oxides to '+1' as H^(+).

4. Create a little sum +1 +N - 6 = 0

To balance then N must be '+5'.

The phosphate anion is PO4^(3-)

Using oxygen as the standard at '-2' .

The nthe oxygen moiety is 4 X -2 = -8

Since the overall charge on the anion is '-3'

Then we can make a sum

P + - 8 = -3

Add '8' to bothe sides

P = +5 The oxidation state of phosphorus.

Alkali metals have a single electron in their outermost shell that is easily lost, resulting in a positive oxidation number. When they lose this electron to form ions, they attain a stable electron configuration, which typically involves having a full valence shell with a noble gas electron configuration.

The oxidation number for carbonate is -2, as it is composed of one carbon atom with a formal charge of +4 and three oxygen atoms each with a formal charge of -2. The oxidation number for barium is +2, as it is a group 2 element in the periodic table, meaning it typically forms 2+ cations.

MnO2: oxidation number +4
KMnO4: oxidation number +7

The oxidation number of Br in BrO3- is +5. This is because oxygen typically has an oxidation number of -2, and since there are three oxygen atoms in the BrO3- ion, the total negative charge is -6. Since the overall charge of the ion is -1, the oxidation number of Br must be +5 to balance the charges.

In CaN2, nitrogen (N) has an oxidation number of -3. To find this, we know that the overall charge of CaN2 is zero (Ca is +2), so the two nitrogens must have a total charge of -6, making the oxidation number of each nitrogen -3.