Question

determine the oxidation number (o.n.) of each element in these compounds:
a) cao
b) kno₃
c) hso₄⁻
d) caco₃
e) n₂
f) h₂o
g) no₃⁻

Explanation:

Step1: Oxidation number rules

• Oxidation number of free element is \(0\).
• For monatomic ion, oxidation number equals its charge.
• In compounds:
• Oxidation number of \(H\) is \(+1\) (except in metal hydrides).
• Oxidation number of \(O\) is \(-2\) (except in peroxides and \(O_2F_2\)).
• Sum of oxidation numbers of all atoms in a neutral compound is \(0\), and in a polyatomic ion is equal to its charge.

Step2: For \(CaO\)

Let oxidation number of \(Ca\) be \(x\) and \(O\) is \(-2\).
Since \(CaO\) is neutral: \(x+(-2)=0\)
\(x = + 2\)
Oxidation number of \(Ca\) is \(+2\), \(O\) is \(-2\)

Step3: For \(KNO_3\)

Oxidation number of \(K\) is \(+1\) (group 1 metal). Let oxidation number of \(N\) be \(y\), \(O\) is \(-2\).
Since \(KNO_3\) is neutral: \(+1 + y+3\times(-2)=0\)
\(+1 + y - 6=0\)
\(y=+5\)
Oxidation number of \(K\) is \(+1\), \(N\) is \(+5\), \(O\) is \(-2\)

Step4: For \(HSO_4^-\)

Oxidation number of \(H\) is \(+1\), let oxidation number of \(S\) be \(z\), \(O\) is \(-2\).
Since charge on \(HSO_4^-\) is \(-1\): \(+1+z + 4\times(-2)=-1\)
\(+1+z-8=-1\)
\(z=+6\)
Oxidation number of \(H\) is \(+1\), \(S\) is \(+6\), \(O\) is \(-2\)

Step5: For \(CaCO_3\)

Oxidation number of \(Ca\) is \(+2\). Let oxidation number of \(C\) be \(w\), \(O\) is \(-2\).
Since \(CaCO_3\) is neutral: \(+2+w + 3\times(-2)=0\)
\(+2+w-6=0\)
\(w = + 4\)
Oxidation number of \(Ca\) is \(+2\), \(C\) is \(+4\), \(O\) is \(-2\)

Step6: For \(N_2\)

Since it is a free element, oxidation number of \(N\) is \(0\)

Step7: For \(H_2O\)

Oxidation number of \(H\) is \(+1\), let oxidation number of \(O\) be \(m\).
Since \(H_2O\) is neutral: \(2\times(+1)+m=0\)
\(m=-2\)
Oxidation number of \(H\) is \(+1\), \(O\) is \(-2\)

Step8: For \(NO_3^-\)

Let oxidation number of \(N\) be \(n\), \(O\) is \(-2\).
Since charge on \(NO_3^-\) is \(-1\): \(n+3\times(-2)=-1\)
\(n - 6=-1\)
\(n=+5\)
Oxidation number of \(N\) is \(+5\), \(O\) is \(-2\)

Answer:

a) \(Ca:+2\), \(O:-2\)
b) \(K:+1\), \(N:+5\), \(O:-2\)
c) \(H:+1\), \(S:+6\), \(O:-2\)
d) \(Ca:+2\), \(C:+4\), \(O:-2\)
e) \(N:0\)
f) \(H:+1\), \(O:-2\)
g) \(N:+5\), \(O:-2\)