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

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

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.