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


We gots calcium, a team 2 alkaline earth, i.e. #Ca^(2+)#...and #2xxHO^-#. Currently hydrogen normally takes a #+I# oxidation number, and also it does therefore here. And finally, we give oxygen a #-II# oxidation number....

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

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


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You must follow number of rules to identify oxidation numbers.

The essential rules because that this concern are:

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

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

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

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

We will write the amount of the oxidation numbers below each atom.

There is just one #"Ca"# atom, so we create +2 listed 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 variety 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 only one #"O"# atom inside the parentheses. So 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 variety of #"H"# is +1.

Write the oxidation number above 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 inside the parentheses. For this reason we write +1 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 all the oxidation numbers in one ion have to 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 additionally have suggested that because the oxidation number of #"Ca"# is +2, the complete oxidation numbers of the #"OH"# ions have to be -2, the fee on every #"OH"# ion mustbe -1.

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