A) \[C{{O}_{2}}<CO_{3}^{2-}<CO\]
B) \[CO<CO_{3}^{2-}<C{{O}_{2}}\]
C) \[CO_{3}^{2-}<C{{O}_{2}}<CO~\]
D) \[CO<C{{O}_{2}}<CO_{3}^{2-}\]
Correct Answer: D
Solution :
| A bond length is the average distance between the centres of nuclei of two bonded atoms. A multiple bond (double or triple bond) is always shorter than die corresponding single bond. |
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| The C-atom in \[CO_{3}^{2-}\] is \[s{{p}^{2}}\] hybridized as shown |
| The C-atom in \[C{{O}_{2}}\] is sp-hybridized with bond distance of carbon-oxygen is 122 pm. |
| \[\begin{align} & O==C==O\overset{{}}{\longleftrightarrow}{{\,}^{+}}O\,\equiv \equiv \,C\equiv \equiv C-\bar{O}\overset{{}}{\longleftrightarrow} \\ & \bar{O}--C\equiv \equiv \overset{+}{\mathop{O}}\, \\ \end{align}\] |
| The C-atom in CO is sp-hybridized with \[C-O\] bond distance is 110 pm : \[C\equiv {{O}^{+}}\]: So the correct order is \[CO<C{{O}_{2}}<CO_{3}^{2-}\]. |
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