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# Here,$A{{l}_{4}}{{C}_{3}}$ is an ionic carbide, named as:(a)- Acetylide(b)- Methanide(c)- Allylide(d)- Alloy

Last updated date: 25th Jul 2024
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Hint: $A{{l}_{4}}{{C}_{3}}$ is an ionic carbide because there is one metal element and the other is non-metal. By finding the oxidation number of the carbon we can find the type of the compound and the oxidation number of aluminium is +3.

Carbides are the group of compounds in which the carbon atom is bonded with highly electropositive elements like alkali metal elements, alkaline earth metal elements, and group III elements. So, based on the oxidation number of carbon or type of carbon we can classify them as methanide, acetylides, and sesqui carbides.
If the carbon is in ${{C}^{4-}}$ form then they are methanides, if the carbon is in $C_{2}^{2-}$ form then they are acetylides, and if the carbon is in $C_{3}^{4-}$ form then they are sesqui carbides.
So, the given compound is $A{{l}_{4}}{{C}_{3}}$, we can find the oxidation of carbon, as the oxidation number of aluminium is +3.
$(+3)4+3x=0$
$12+3x=0$
$3x=-12$
$x=-4$
So, the oxidation number of carbon is -4 which means the carbon is in ${{C}^{4-}}$ form. Therefore, it is a methanide.

So Option (b) is the correct answer.

So, $A{{l}_{4}}{{C}_{3}}$ is known as aluminium carbide and it is a crystal of pale yellow to brown in color. We can make aluminium carbide in an electric arc furnace by the direct reaction of aluminium and carbon. They are called ionic carbides because aluminium is a metal and carbon is a non-metal.
Another example of a compound that belongs to the methanide group is beryllium carbide whose formula is $B{{e}_{2}}C$ as in this the oxidation number of Be is +2 and oxidation number of carbon is -4.