Well the sum of the oxidation numbers amounts to the fee on the salt or molecule....i.e. The sum of the oxidation numbers amounts to ZERO.

We gots calcium, a team 2 alkaline earth, i.e. #Ca^(2+)#...and #2xxHO^-#. Now hydrogen usually takes a #+I# oxidation number, and it does for this reason here. And finally, we provide oxygen a #-II# oxidation number....

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And #2_"Ca"+2xx(-2_"O")+2xx(+1_"H")=0# together required....

And so us gots #Ca(+II)#, #O(-II)#, and also #H(+I)# as the oxidation numbers....

You have to follow numerous rules to determine oxidation numbers.

The important rules for this concern are:

The oxidation number of Group 2 steels in a link is +2.The oxidation number of oxygen in a link is typically -2.The oxidation variety of hydrogen in a compound is generally +1.The amount of all oxidation numbers in one ion have to equal the charge on the ion.The sum of every oxidation number in a neutral compound must equal zero.

Per Rule 1, the oxidation number of #"Ca"# is +2.

Write the oxidation number above the #"Ca"# in the formula:

#stackrelcolor(blue)("+2")("Ca")"(OH)"_2#.

We will write the sum of the oxidation numbers listed below each atom.

There is just one #"Ca"# atom, therefore we compose +2 below the #"Ca"# atom:

#stackrelcolor(blue)("+2")("Ca")"(OH)"_2##color(white)(stackrelcolor(blue)("+2")("Ca")"(OH)"_2)#

The oxidation variety of the calcium ion is +2.

Per Rule 2, the oxidation number of #"O"# is -2.

Write the oxidation number above the #"O"# in the formula:

#stackrelcolor(blue)("+2")("Ca")(stackrelcolor(blue)("-2")("O")"H")_2#

There is just one #"O"# atom inside the parentheses. For this reason we write -2 listed below the atom.

#stackrelcolor(blue)("+2")("Ca")(stackrelcolor(blue)("-2")("O")"H")_2##color(white)(l)stackrelcolor(blue)("+2")("")color(white)(m)stackrelcolor(blue)("-2")("")#

Per Rule 3, the oxidation number of #"H"# is +1.

Write the oxidation number over the #"H"# in the formula:

#stackrelcolor(blue)("+2")("Ca")(stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H"))_2##color(white)(l)stackrelcolor(blue)("+2")("")color(white)(m)stackrelcolor(blue)("-2")("")#

There is just one #"H"# atom within the parentheses. For this reason we compose +1 listed below the atom.

#stackrelcolor(blue)("+2")("Ca")(stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H"))_2##color(white)(l)stackrelcolor(blue)("+2")("")color(white)(m)stackrelcolor(blue)("-2")("")stackrelcolor(blue)("+1")("")#

Per Rule 4, the sum of every the oxidation number in an ion must equal the charge on the ion.

#stackrelcolor(blue)("+2")("Ca")(stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H"))_2##color(white)(l)stackrelcolor(blue)("+2")("")color(white)(m)color(blue)(underbrace(stackrelcolor(blue)("-2")("")stackrelcolor(blue)("+1")(""))_color(blue)("-1")#

Thus, the full oxidation variety of the hydroxide ion is #"-2 + 1 = -1"#

We could likewise have suggested that because the oxidation number of #"Ca"# is +2, the total oxidation number of the #"OH"# ions need to be -2, the charge on every #"OH"# ion mustbe -1.

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Per Rule 5, the sum of every the oxidation numbers in a neutral compound must equal zero.