Question

use the rules (in order) to assign oxidation numbers to each of the elements in the compounds below.

odic acid h i o
hio₃

hydrazine n h
n₂h₄

metaboric acid h b o
hbo₂

Explanation:

Step1: Assign oxidation number to H in \(HIO_3\)

Hydrogen in most compounds has an oxidation number of \(+ 1\). So for \(HIO_3\), \(H = +1\)

Step2: Assign oxidation number to O in \(HIO_3\)

Oxygen in most compounds has an oxidation number of \(-2\). In \(HIO_3\), there are 3 O atoms. So total oxidation number from O is \(3\times(-2)=-6\)

Step3: Calculate oxidation number of I in \(HIO_3\)

Let oxidation number of I be \(x\). Since the compound is neutral (\(HIO_3\) has overall charge \(0\)), we have \(+1+x+(-6)=0\). Solving for \(x\): \(x - 5=0\), so \(x = +5\)

Step4: Assign oxidation number to H in \(N_2H_4\)

Hydrogen in most compounds has an oxidation number of \(+1\). In \(N_2H_4\), there are 4 H atoms. So total oxidation number from H is \(4\times(+1) = +4\)

Step5: Calculate oxidation number of N in \(N_2H_4\)

Let oxidation number of N be \(y\). Since the compound is neutral (\(N_2H_4\) has overall charge \(0\)), we have \(2y+4 = 0\). Solving for \(y\): \(2y=-4\), so \(y=-2\)

Step6: Assign oxidation number to H in \(HBO_2\)

Hydrogen in most compounds has an oxidation number of \(+1\)

Step7: Assign oxidation number to O in \(HBO_2\)

Oxygen in most compounds has an oxidation number of \(-2\). In \(HBO_2\), there are 2 O atoms. So total oxidation number from O is \(2\times(-2)=-4\)

Step8: Calculate oxidation number of B in \(HBO_2\)

Let oxidation number of B be \(z\). Since the compound is neutral (\(HBO_2\) has overall charge \(0\)), we have \(+1+z+(-4)=0\). Solving for \(z\): \(z - 3=0\), so \(z = +3\)

Answer:

For \(HIO_3\): \(H = +1\), \(I=+5\), \(O=-2\)
For \(N_2H_4\): \(N=-2\), \(H = +1\)
For \(HBO_2\): \(H = +1\), \(B=+3\), \(O=-2\)