Answer
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Hint: We have to know that the carbides contain the \[{C_4}^ - \] unit. We can look at the options given to identify the species given which has \[{C_4}^ - \] this unit thus it is easy to answer this question. \[{C_2}^{2 - }\] is an acetylide so it is not considered to be a carbide as it has no \[{C_4}^ - \] unit. Carbides are formed from metals or metal oxides.
Complete answer:
We need to remember that the methanides are carbides that have a carbon centre having formula \[{C_4}^ - \]. These species contain carbon atoms and an atom which has lower electronegativity. The carbide species are stable and usually have a high melting point, they are formed at high temperature from different oxides of metals. They form different chemical bonding depending upon its nature which can be classified as covalent, interstitial and salt like carbides.
We will look at all the options now to answer this question, as we know that carbide that contains \[{C_4}^ - \]unit is methanide.
Option A) this is an incorrect option as \[A{l_4}{S_3}\] this is not considered to be carbide as it has no carbon atom, instead of carbon it has sulphur so we can easily neglect this option.
Option B) this is an incorrect option as \[M{g_2}{C_3}\] this is not considered to be a methanide, thus we can neglect this option.
Option C) this is an incorrect option as \[Ca{C_2}\] is not a methanide as it contains \[{C_2}^{2 - }\] unit and not \[{C_4}^ - \]which is the formula for methanide, thus this option can be neglected.
Option D) this is a correct option as both the molecules contain carbon that has \[{C_4}^ - \] unit which is considered to be methanide. Both \[B{e_2}C\] and\[A{l_4}{C_3}\], Beryllium and Aluminum carbides are an example of methanides.
Note:
As we know that the carbides that are given in options are salt like carbides since they have ions in them. They are ionic in nature and have no covalent bonding in them. Beryllium belongs to alkaline earth metals that form carbide whereas Aluminum belongs to group 14 that forms carbide. \[{C_2}^{2 - }\]is an acetylide so it is not considered to be a carbide as it has no \[{C_4}^ - \]unit.
Complete answer:
We need to remember that the methanides are carbides that have a carbon centre having formula \[{C_4}^ - \]. These species contain carbon atoms and an atom which has lower electronegativity. The carbide species are stable and usually have a high melting point, they are formed at high temperature from different oxides of metals. They form different chemical bonding depending upon its nature which can be classified as covalent, interstitial and salt like carbides.
We will look at all the options now to answer this question, as we know that carbide that contains \[{C_4}^ - \]unit is methanide.
Option A) this is an incorrect option as \[A{l_4}{S_3}\] this is not considered to be carbide as it has no carbon atom, instead of carbon it has sulphur so we can easily neglect this option.
Option B) this is an incorrect option as \[M{g_2}{C_3}\] this is not considered to be a methanide, thus we can neglect this option.
Option C) this is an incorrect option as \[Ca{C_2}\] is not a methanide as it contains \[{C_2}^{2 - }\] unit and not \[{C_4}^ - \]which is the formula for methanide, thus this option can be neglected.
Option D) this is a correct option as both the molecules contain carbon that has \[{C_4}^ - \] unit which is considered to be methanide. Both \[B{e_2}C\] and\[A{l_4}{C_3}\], Beryllium and Aluminum carbides are an example of methanides.
Note:
As we know that the carbides that are given in options are salt like carbides since they have ions in them. They are ionic in nature and have no covalent bonding in them. Beryllium belongs to alkaline earth metals that form carbide whereas Aluminum belongs to group 14 that forms carbide. \[{C_2}^{2 - }\]is an acetylide so it is not considered to be a carbide as it has no \[{C_4}^ - \]unit.
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