What is the oxidation number of each atom in sodium hydrogen carbonate nahco3

### Answers

The oxidation state of carbon in NaHCO3 is +4. Sodium has a +1 oxidation state.

Explanation:Hydrogen has a +1 oxidation state. Oxygen has a -2 oxidation state

1. P₂O₅ → oxidation number of phosphorous is +5 and Oxygen is -2.

2. (SO₄)²⁻ → oxidation number of sulfur is +6 and Oxygen is -2.

3. KClO₃ → oxidation number of Potassium is +1, Chlorine is +5, and Oxygen is -2.

4. NH₄Cl → oxidation number of Nitrogen is -3, Hydrogen is +1, and Chlorine is -1

5. (NH₄)₂S → oxidation number of Nitrogen is -3, Hydrogen is +1, and Sulfur is -2

Explanation:

General Rules for assigning oxidation numbers

The oxidation number of a free element is always 0.

The oxidation number of a mono-atomic ion equals the charge of the ion.

The alkali metals (group I) always have an oxidation number of +1.

The alkaline earth metals (group II) are always assigned an oxidation number of +2.

Oxygen almost always has an oxidation number of -2, except in peroxides (H₂O₂) where it is -1 and in compounds with fluorine (OF₂) where it is +2.

Hydrogen has an oxidation number of +1 when combined with non-metals, but it has an oxidation number of -1 when combined with metals.

The algebraic sum of the oxidation numbers of elements in a compound is zero.

The algebraic sum of the oxidation states in an ion is equal to the charge on the ion.

Using the above rules:

1. P₂O₅

∵ it is a neutral compound its total charge is 0.

Also, we know that oxygen has an oxidation number of -2.

Let oxidation number of P be x

∴ 2(x)+5(-2)=0 → 2x=+10 → x=+5

∴oxidation number of phosphorous is +5.

2. SO₄²⁻:

∵ it is a charged ion its total charge is -2.

Also, we know that oxygen has an oxidation number of -2.

Let oxidation number of S be x

∴ (x)+4(-2)= -2 → x=+6

∴oxidation number of sulfur is +6.

3. KClO₃:

∵ it is a neutral compound its total charge is 0.

Also, we know that oxygen has an oxidation number of -2 and the oxidation number of K (group I) is +1

Let oxidation number of Cl be x

∴ (+1) + (x) + 3(-2) = 0 → x=+5

∴oxidation number of Chlorine is +5.

4. NH₄Cl:

∵ it is a neutral compound its total charge is 0.

Also, we know that chloride has an oxidation number of -1

Hydrogen has an oxidation number of +1 when combined with non-metals

Let oxidation number of N be x

∴ (x) + 4(+1) + (-1) = 0 → x=-3

∴oxidation number of Nitrogen is -3.

5. (NH₄)₂S:

∵ it is a neutral compound its total charge is 0.

Also, we know that chloride has an oxidation number of -1

Ammonium ion (NH₄⁺) has an oxidation number of +1

Let oxidation number of N be x

∴ 2(+1) + (x) = 0 → x= -2

∴oxidation number of sulfur is -2.

E

Explanation:

To solve this question, we need to know the oxidation number in the compound in the question. The name of the compound is tetraoxosulphate(vi) acid. We can obviously see that the oxidation number of sulphur in this particular compound is 6.

A. Is wrong

The name of the compound is trioxosulphate(iv) ion

B. Is wrong

S2O32-

2S + 3(-2) = -2

S = +2

C.is wrong

Oxidation state there is -2

D is wrong

The oxidation number here is zero

E is correct

Oxidation number of oxygen is -2 and that of chlorine is -1

Hence:

S + 2(-2) + 2(-1) = 0

S -4 -2 = 0

S = 4+2 = +6

Let the oxidation number of P be x.

We know that oxidation number of Na= +1, O= -2

therefore, 3*1 + x+ 4* -2 = 0 (net charge on the molecule)

3 + x - 8 =0

x = +5

therefore, the answer is C)+5

No element decreases its oxidation number.

Explanation:

Oxidation number of each atom in sodium hydrogen carbonate ( NaHCO3) are

Na = +1

H = +1

C = +4

O = -2

Explanation:

Let's pick the first atom

NaHCO3

find oxidation number of Na

Na + 1 +4 + 3(-2) = 0( because there is no charge)

Na + 5 - 6 = 0

Na = 6 -5

Na = +1

Always put the sign

Find the oxidation number of H in NaHCO3

+1 + H + 4 + 3(-2) = 0.

H + 1 + 4 + 3(-2) = 0

H + 5 -6 = 0

H = 6 -5

H = +1

Find the oxidation number of C In NaHCO3

+1 +1 +C + 3(-2) = 0

C + 1+1-6 = 0

C +2 - 6= 0

C = 6 -2

C = +4

Find the oxidation number of O in NaHCO3

+1 + 1 + 4 + 3(O) = 0

6 + 3 (O) = 0

3(O) = -6

Divide through by 3 to get O

3(O) / 3 = -6 / 3

O = -6/3

O = -2

The answer to your question is: letter b Sulphur

Explanation:

Remember that in neutral molecules the sum of the oxidation numbers equals zero.

Mg⁺²S⁺⁶O⁻²₄

Magnesium = +2

Sulphur = +6

Oxygen = -2 x 4 = -8

You may find bellow the assignment of the oxidation number for each atom.

Explanation:

Li₃PO₄

Li have the oxidation number +1

P have the oxidation number +5

O have the oxidation number -2

SO₃²⁻

S have the oxidation number +4

O have the oxidation number -2

Cr₂S₃

Cr have the oxidation number +3

S have the oxidation number -2

NO₃⁻

N have the oxidation number +5

O have the oxidation number -2

Learn more about:

oxidation number

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+1*3 + (+5) + (-2*4) = 0

BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄

this is a double displacement reaction

the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄

oxidation number of Cl is -1 in both BiCl₂ and NaCl

oxidation number of Na is +1 in both Na₂SO₄ and NaCl

oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .

Answer for this question is no element decreases its oxidation number.