X(3+) and Z(2-) will give X2Z3.
X(3+) and Z(3-) will give XZ.
X(5+) and Z(2-) will give X2Z5.
X(5+) and Z(3-) will give X3Z5.
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X(3)
Possible Compounds = X2Z3, X2Z5, XZ, X3Z5
Carbon is a pblock element. It shows -4 to +4 oxidation numbers.
Oxidation numbers are very important in much of chemistry because many times atoms do lose or gain electrons. When this happens, they become ions.However, oxidation numbers are especially important when writing chemical formulas for ionic compounds.
hydrogen +1 in most of the compounds -1 in metal hydrides
No, they can't only have positive oxidation numbers in compounds.
Possible Compounds = X2Z3, X2Z5, XZ, X3Z5
Formulas for compounds
Carbon is a pblock element. It shows -4 to +4 oxidation numbers.
Oxidation numbers are very important in much of chemistry because many times atoms do lose or gain electrons. When this happens, they become ions.However, oxidation numbers are especially important when writing chemical formulas for ionic compounds.
hydrogen +1 in most of the compounds -1 in metal hydrides
No, they can't only have positive oxidation numbers in compounds.
there is only one atom of each element
Hydrogen. +1 in most of the compounds -1 in metal hydrides and hydrocarbons
+2 and +3 in its compounds.
-1 is most common, but Cl can exhibit oxidation numbers from -1 to +7 in its compounds.
X(3+) and Z(2-) will give X2Z3. X(3+) and Z(3-) will give XZ. X(5+) and Z(2-) will give X2Z5. X(5+) and Z(3-) will give X3Z5.
One rationalization is the "inert pair effect" - lead and tin have oxidation numbers of +2 and +4 . The inert pair effect also rationaliss the two oxidation numbers of +1 and +3 exhibited by gallium, indium and thallium. In compounds with the lower oxidation numebrs the s electrons are not removed.