How to calculate oxidation number

Cu + O-2 = 0 (+2) + (-2)=0 = CuO = copper II oxide (+1) + (-2)=0 = 2x(+1) + (-2)=0 = Cu2O = copper I oxide. KNO3 K+N+3O =9 (+1) + X + 3x(-2) = 0 (+1) + X + (-6) = 0 X= 6-1 = 5 N= +5

Oxidation Number is the loss of electron of an atom present in a chemical bonded compound. It may be 0,+1, -1 Hint: Oxidation number is the number of unpair electron (lose or gain according to stability rule ) of an atom known as valency 》Copper (1) oxide have a chemical formula of Cu₂O Suppose the oxidation state of Cu is x Cu₂O of 2(x) + (-2) =0 2x -2 =0 2x = 0+2 x = 2/2 x= +1 So the Oxidation number of Cu₂O is 2(+1)+ (-2) =0 》Potassium trioxonitrate (v) have a chemical formula of KNO3 Suppose the oxidation number of nitrogen is x KNO3 of (+1)+(x)+3(-2) =0 (+1)+ (x) -6 =0 x = +6 - 1 x = +5 So the oxidation number of KNO3 is (+1)+(+5)+3(-2) =0 Note: Sum of oxidation number ( charge ) is always equal to zero in a neutral compound

This is a great question! It looks like you got some answers below but if you would like to do a session so that I could explain the concept of oxidation number further, I would love to help.

Oxidation number of an element in a compound is the charge carried by the element in that compound. Charge could be negative or positive depending wheather it loses or gains electron to attain stability. In Cu2O, the charge of Cu is calculated as follows: 2x+(-2) = 0. Therefore, x=2/2, x=+1. Hence oxidation number of Cu in Cu2O is +1. Similarly in KNO3, charge on N is calculated as +1+x+3(-2) = 0, therefore x= +5.

he oxidation number is the charge of each atom in a compound if the compound is composed of ions. It is calculated as follows: i. Cu2O this compound overall charge is zero so 2Cu+(-2) of O=0 , 2Cu=+2,divide both sides by two then Cu=+1,as the compound states it is copper I(meaning oxidation number is 1 you can contact for more explanation .Regards,Joshua. ii.

1) Cu2O Oxidation state of O is (-2) and that of Cu is x (say) Since, molecule is neutral Therefore, x + (-2) =0 => x=2 Hence, oxidation state of Cu in Cu2O is +2 2) KNO3 Oxidation state of K is +1 That of O is (-2) And that of N is y (say) Since, molecule is neutral Therefore, 1+x+(-2)3 = 0 => 1+x-6=0 => x=+5 Hence, oxidation state of N is (+5)

Oxidation number is calculated just like the usual algebra. The oxidation state of the unknown is assigned an alphabet while the other elements are given their standard oxidation numbers.

You need to specify the particular element that you want to calculate its oxidation number in the given compound. If not, then the oxidation number of the whole compound will be zero because the sum of the oxidation numbers of the elements in a neutral compound (uncharged) is zero. However, from most of the answers already given, the oxidation number of a NAMED element in the compound has been calculated for you: COPPER in copper (I) oxide and NITROGEN in potassium trioxonitrate (V).

The oxidation number of both the compounds copper oxide and potassium trioxonitrate is zero since there is no charge on them

The oxidation number of both the compounds copper oxide and potassium trioxonitrate is zero since there is no net charge on them. For more clarity and easy understanding pls connect with me

LEO=Loss of electrons is oxidation GER= Gain of electrons is reduction The oxidation number of an atom in an element is always zero The sum of oxidation number of all atom in neutral compound is also zero The sum of oxidation number of all the atom in an ion is equal to the charge on ion Oxidation number of oxygen in most compounds is _2 The oxidation number of Hydrogen in most compounds is +1 While fluorine has _1 oxidation number. By following these rules you can determine the oxidation number of any atom in a compound.

There are two forms of Copper Oxide: Cupperous oxide (Cu2O) and Cupperic oxide (CuO). Because Oxygen has a -2 oxidation number in oxides and the total charge of a compound should be zero. Thus, Cu has a +1 oxidation state in Cu2O and a +2 oxidation state in CuO.

Both have zero oxidation number, as positive and negative charges on both compounds are equal. We will have a session in which i will explain you elaborately how to find because first you need to know basics of oxidation number and how it finds.

The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state.

+5

Oxidation number is calculated just like the usual algebra.

The oxidation number of copper(1) oxide is +1 for copper and -2 for oxygen, giving the compound a net oxidation number of 0. The oxidation number of potassium trinitrate (V) is +5 for nitrogen, -2 for oxygen, and +1 for potassium, giving the compound a net oxidation number of 0.

Cu2O is chmeical farmula of copper(1)oxide as the oxidation number of oxugen is 2 so if oxidation number of copper is calculated Cu2O cu2 +(-2) =0 Cu2=0+2 Cu2= 2 Cu=1 so oxidation number of copper is 1. in case of potassium trioxonitrate(v) KNO3 the oxidation number of nitrogen can be calculated as KNO3 (+1) +N+(-2)(3) = 0 (+1) +N+(-6) = 0 +1+N-6 =0 N-5 = 0 N= 5

LEO=Loss of electrons is oxidation GER= Gain of electrons is reduction The oxidation number of an atom in an element is always zero The sum of oxidation number of all atom in neutral compound is also zero The sum of oxidation number of all the atom in an ion is equal to the charge on ion Oxidation number of oxygen in most compounds is _2 The oxidation number of Hydrogen in most compounds is +1 While fluorine has _1 oxidation number. By following these rules you can determine the oxidation number of any atom in a compound

To calculate the oxidation state of an element in a compound, you need to follow a few steps: Determine the overall charge of the compound: This can be done by considering the charge of each ion or atom in the compound. For example, in the compound NaCl, the overall charge is 0 since Na has a +1 charge and Cl has a -1 charge. Assign the known oxidation states: The oxidation states of certain elements in a compound are usually known. For example, in the compound H2O, the oxidation state of oxygen is -2 and the oxidation state of hydrogen is +1. Determine the unknown oxidation state: To do this, you need to use the information you have and the fact that the overall charge of the compound must be zero. For example, in the compound FeCl3, the oxidation state of Fe is unknown. The overall charge of the compound is -1 (3 Cl atoms each with a -1 charge). From this, we know that the sum of the oxidation states of Fe and Cl must be +2 (since -1 + -1 + -1 + x = -1, where x is the oxidation state of Fe). Therefore, the oxidation state of Fe in FeCl3 is +3. Check your work: The sum of the oxidation states of all the elements in a compound should equal the overall charge of the compound. For example, in FeCl3, the sum of the oxidation states of Fe and Cl is +3 + (-1) + (-1) + (-1) = 0, which is equal to the overall charge of the compound. It's important to note that oxidation states are assigned to individual atoms within a compound, and not to the compound as a whole. The same element can have different oxidation states in different compounds, so it's important to consider the specific compound you're working with.

The oxidation number is told in the brackets if it is 1 then it has an oxidation number of 1 a single atom has an oxidation number of zero

Cu(I) oxide Cu2O You know that because Cu(I) - the Roman numeral tells you Cu its 1+. So Cu + Cu = 2+, and O is 2- therefore the oxidation number of Cu2O is (2-) + (2+) = 0 Potassium trioxonitrate (v): Known as KNO3 Oxidation number of K is 1+ Oxidation number of O3 is (2-) x 3 = -6 The entire formula needs to equal 0 and you need to find N So (1+) + (-6) = -5 In order to make -5 into 0 you need +5. Therefore N is -5 the sum of oxidation number always needs to equal 0 unless its an ion

Oxidation number is the net charge on a particle, for atoms or rather elements it is zero by the fact that atoms are electrically neutral. When atoms gain or lose electrons, the form ions. We can get the oxidation numbers from the charge they acquire ie +1,+2, +3 when they lose one, two, three electrons respectively. In compounds or radicals, the sum of oxidation numbers either gives us zero(for compounds) or the net charge(for radicals).

The oxidation number of copper(1) oxide is +1 for copper and -2 for oxygen. The oxidation number of potassium trioxonitrate(V) is +5 for nitrogen, -2 for oxygen, and +1 for potassium.

Dear Janet Bademosi In order to calculate the oxidation number of copper in cuprous oxide We move as CU2O 2Cu-2=0 Cu=2/2 Cu=1 In order to calculate oxidation number of nitrogen in potassium trioxonitrate We move as KNO3 1+N-6=0 N=6-1 N=5 For further assistance don't hesitate to contact me

you can separate pottasium with ion (+) and NO3 (-) K(+) + NO3(-) and number oxidation of K is +1 and O=-2 and N is -1+(-6) = -7 oxidation number CuO is O=-2 so Cu is +2

oxidation number of sulfur in H2SO4 (+1) x 2 + x + (-2) x 4 = 0 x = +6 Therefore, the oxidation number of sulfur in H2SO4 is +6.

CuO (Copper oxide). Oxygen always gets -2 or -1 unless bonded with Fluorine. therefore, the Cu in CuO will have +2 oxidation state. No matter whats the composition or bonding an oxidation state or valency will never be more than 8. KNO3, the oxidation state of K is +1. as two oxygen atoms are in double bond with Nitrogen and the third one is in ionic mode, which gives one bond with K and one bond with N. therefore, the oxidation state of K in KNO3 is +1

Oxidation number of copper (I) oxide, Cu2O and Potassium trioxonitrate (V), KNO3. The sum of oxidation number of all atoms in a compound adds up to zero. Therefore, Cu2O = 0 KNO3 = 0

Oxidation number of copper (I) oxide, Cu2O and Potassium trioxonitrate (V), KNO3. The sum of oxidation number of all atoms in a compound adds up to zero. Therefore, Cu2O = 0 KNO3 = 0

Cuprious oxide Cu2O 2Cu+(-2)=0 2Cu=2 Cu=2/2 Cu=1

The oxidation number of copper(1) oxide (Cu2O) is +1 for copper and -2 for oxygen. To determine the oxidation number of each element in potassium trioxonitrate(V) (KNO3), we can use the following steps: The oxidation number of potassium is always +1 in its compounds. Oxygen has an oxidation number of -2 in most compounds, so we can assign -6 to the three oxygen atoms in KNO3. To determine the oxidation number of nitrogen, we can use the fact that the sum of the oxidation numbers in a compound is equal to zero. In KNO3, we have: (+1) + x + (-6) = 0 Solving for x, we get: x = +5 Therefore, the oxidation number of nitrogen in potassium trioxonitrate(V) (KNO3) is +5.

The oxidation state of elements can be calculated as Number of electrons are less or higher than the elemental state. Ex: KMnO4 Here KMnO4 = k+ + MnO4- Oxidation state of K is +1 Oxidation state of O is -2 Oxidation state of Mn is +7

Calculate the oxidation number by subtracting the charges of the other elements in the compound from the total charge of the compound, or by deducing from the brackets. Copper has an oxidation state of +1 in copper oxide, and nitrogen is +5 in potassium trioxonitrate(v), as can be deduced from the brackets.

The oxidation number of copper oxide is 2 and the the oxidation number of potassium trioxonitrate is 3

The oxidation number of an ion is equal to the charge of the ion. Example :Na+ has an oxidation number of +1, and S2- has an oxidation number -2 The sum of the oxidation number in a netural compound is zero, The sum of the oxidation number of a poly atomic ion is equal to the charge of the ion.

Ability of ions of respective atoms to form a chemical bond which in turn is associated with the number of electron vaccines or abundance in its bonding electronic shells.

The oxidation number of any free element always remains 0. The oxidation number in the case of a monatomic ion is always equal to the value of the charge of the ion. For instance, the oxidation number of Na3- is 3-. The expected oxidation number of hydrogen is plus 1.

By substracting the sum of lone pairs and electron it gains from bond from the number of valence electrons

This is a really good query! It appears that you have received some responses below, but if you would want to schedule a meeting so that I can better explain the idea of the oxidation number, I would be happy to assist.