question_answer

                The correct order of C ?O bond length among CO, \[CO_{3}^{2-}\], CO2 is:                                                         

A)                 CO2 < \[CO_{3}^{2-}\] < CO  

B)                 CO < \[CO_{3}^{2-}\]< CO2               

C)                 \[CO_{3}^{2-}\]< CO2 < CO          

D)                 CO < CI2 < \[CO_{3}^{2-}\]

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.                                 The C-atom in \[CO_{3}^{2-}\]is sp2 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                 \[CO<C{{O}_{2}}<CO_{3}^{2-}\]