
In \[M{n_2}{(CO)_{10}}\] no. of metal-metal bonds and no. of \[CO\] bridges are \['x'\] and \[F{e_2}{(CO)_9}\] no. of metal -metal bonds and no. of \[CO\] bridges are \[y\]. Then \[x + y\]
Answer
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Hint: Metal–metal bonds form due to the attractive interactions between metal centers. Bridging bonds are formed when a ligand connects two or more central atoms together. The simplest method to solve this question is to draw the structure of both the compounds and count the metal–metal bonds and bridge bonds.
Complete step-by-step answer: Let’s draw the structure of \[M{n_2}{(CO)_{10}}\] and \[F{e_2}{(CO)_9}\].
Complete step by step solution:
As you can see, in \[M{n_2}{(CO)_{10}}\], there is only one metal–metal bond between manganese atoms. There is no bridge bond in \[M{n_2}{(CO)_{10}}\] as no ligand is connecting the two manganese atoms. While, in \[F{e_2}{(CO)_9}\], there is one metal–metal bond between ferrous atoms and three bridge bonds between \[CO\] and \[Fe\].
To calculate the value of \[x + y\], there are two cases:
\[x\] represents the no of metal – metal bonds and \[y\] represents the no of \[CO\] bridge bonds:
Total number of metal – metal bonds in \[M{n_2}{(CO)_{10}} = 1\]
Total number of metal – metal bonds in \[F{e_2}{(CO)_9} = 1\]
\[x = 1 + 1 = 2\]
Total number of \[CO\] bridge bonds in \[M{n_2}{(CO)_{10}} = 0\]
Total number of \[CO\] bridge bonds in \[F{e_2}{(CO)_9} = 3\]
\[y = 0 + 3 = 3\]
Hence,
\[x + y = 2 + 3\]
\[x + y = 5\]
\[x\] represents the total no of metal -metal bonds and bridge bonds in \[M{n_2}{(CO)_{10}}\] and \[y\] represents the total no of metal -metal bonds and bridge bonds in \[F{e_2}{(CO)_9}\]:
Total number of bonds in \[M{n_2}{(CO)_{10}}{\text{, }}x = 1 + 0 = 1\]
Total number of bonds in \[F{e_2}{(CO)_9},{\text{ }}y = 1 + 3 = 4\]
Hence,
\[x + y = 1 + 4\]
\[x + y = 5\]
Hence, in both the cases, the value of \[x + y\] is \[5\].
Note:
Be careful while drawing the structure of compounds. Any error in the structure will lead to an incorrect answer. While counting the \[CO\] bridge bonds, do not count any metal - \[CO\] only bonds, count metal - \[CO\]- metal bonds. Remember that the ligands that connect two metal atoms form the bridge bonds.
Complete step-by-step answer: Let’s draw the structure of \[M{n_2}{(CO)_{10}}\] and \[F{e_2}{(CO)_9}\].
Complete step by step solution:
As you can see, in \[M{n_2}{(CO)_{10}}\], there is only one metal–metal bond between manganese atoms. There is no bridge bond in \[M{n_2}{(CO)_{10}}\] as no ligand is connecting the two manganese atoms. While, in \[F{e_2}{(CO)_9}\], there is one metal–metal bond between ferrous atoms and three bridge bonds between \[CO\] and \[Fe\].
To calculate the value of \[x + y\], there are two cases:
\[x\] represents the no of metal – metal bonds and \[y\] represents the no of \[CO\] bridge bonds:
Total number of metal – metal bonds in \[M{n_2}{(CO)_{10}} = 1\]
Total number of metal – metal bonds in \[F{e_2}{(CO)_9} = 1\]
\[x = 1 + 1 = 2\]
Total number of \[CO\] bridge bonds in \[M{n_2}{(CO)_{10}} = 0\]
Total number of \[CO\] bridge bonds in \[F{e_2}{(CO)_9} = 3\]
\[y = 0 + 3 = 3\]
Hence,
\[x + y = 2 + 3\]
\[x + y = 5\]
\[x\] represents the total no of metal -metal bonds and bridge bonds in \[M{n_2}{(CO)_{10}}\] and \[y\] represents the total no of metal -metal bonds and bridge bonds in \[F{e_2}{(CO)_9}\]:
Total number of bonds in \[M{n_2}{(CO)_{10}}{\text{, }}x = 1 + 0 = 1\]
Total number of bonds in \[F{e_2}{(CO)_9},{\text{ }}y = 1 + 3 = 4\]
Hence,
\[x + y = 1 + 4\]
\[x + y = 5\]
Hence, in both the cases, the value of \[x + y\] is \[5\].
Note:
Be careful while drawing the structure of compounds. Any error in the structure will lead to an incorrect answer. While counting the \[CO\] bridge bonds, do not count any metal - \[CO\] only bonds, count metal - \[CO\]- metal bonds. Remember that the ligands that connect two metal atoms form the bridge bonds.
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