For the reaction kclo2⟶kcl+o2 kclo2⟶kcl+o2 assign oxidation numbers to each element on each side of the equation. k in kclo2: k in kclo2: k in kcl: k in kcl: cl in kclo2: cl in kclo2: cl in kcl: cl in kcl: o in kclo2: o in kclo2: o in o2: o in o2: which element is oxidized? kk oo clcl which element is reduced? oo kk cl

K in KClO2 = +1

Cl in KClO2 = +3

O in KClO2 = -2

K in KCl = +1

Cl in KCl = -1

O in O2 = 0

Chlorine is going from +3 to -1 so it is being reduced

Oxygen is going from -2 to 0 so it is being oxidized

Explanation:

Potassium is a constant +1

Chlorine could be -1, +1, +3, +5, or +7

Oxygen can be either -2 or +2

Every reactant has to equal zero if there is not a given charge.

Each element can only have certain charges, for example in the previous answer K = +5. Potassium can only be the charge of +1

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.

the awnser is group 12