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State True or False
The bond angle in ${\text{C}}{{\text{l}}_{\text{2}}}{\text{O}}$ is equal to that ${\text{O}}{{\text{F}}_{\text{2}}}$.
 A True
 B False

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Last updated date: 15th Jun 2024
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Answer
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Hint: Halogens readily react with oxygen to form halogen oxides. Two such compounds are dichlorine monoxide \[{\text{C}}{{\text{l}}_{\text{2}}}{\text{O}}\] and oxygen difluoride ${\text{O}}{{\text{F}}_{\text{2}}}$. Both the compounds have a bent structure according to VSEPR.

Complete step by step answer:
\[{\text{C}}{{\text{l}}_{\text{2}}}{\text{O}}\] is and organic compound, synthesized by Antoine Jérôme Balard. It is a brownish yellow gas. It is a strong oxidising agent and chlorinating agent.
It was earlier prepared by the reaction of mercuric oxide and chlorine gas, as shown in the reaction below:
${\text{2C}}{{\text{l}}_{\text{2}}}{\text{ + HgO}} \rightarrow {\text{HgC}}{{\text{l}}_{\text{2}}}{\text{ + C}}{{\text{l}}_{\text{2}}}$
But this reaction is no more used due to its toxicity and high expenses.
Now-a-days, a safer method is used for its preparation, the reaction is shown below:
${\text{2C}}{{\text{l}}_{\text{2}}}{\text{ + 2N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{0}} \rightarrow {\text{C}}{{\text{l}}_{\text{2}}}{\text{O + 2NaHC}}{{\text{O}}_{\text{3}}} + 2{\text{NaCl}}$ ${\text{2NaCl + 2C}}{{\text{l}}_{\text{2}}}{\text{ + 2NaHC}}{{\text{O}}_{\text{3}}} \rightarrow {\text{C}}{{\text{l}}_{\text{2}}}{\text{O + 2C}}{{\text{O}}_{\text{2}}}{\text{ + 2NaCl + }}{{\text{H}}_{\text{2}}}{\text{O}}$
The structure of ${\text{O}}{{\text{F}}_{\text{2}}}$ is similar to that of water. The structure is bent in shape with bond angle of 110.9° and bond length of 170 pm.
On the other hand, if we talk about ${\text{2C}}{{\text{l}}_{\text{2}}}{\text{ + HgO}} \to {\text{HgC}}{{\text{l}}_{\text{2}}}{\text{ + C}}{{\text{l}}_{\text{2}}}$ it was first synthesised in 1929. The reaction used for the production of oxygen difluoride involves fluorine and sodium hydroxide, as shown in the reaction below:
${\text{2}}{{\text{F}}_{\text{2}}}{\text{ + 2NaOH}} \rightarrow {\text{O}}{{\text{F}}_{\text{2}}}{\text{ + 2NaF + }}{{\text{H}}_{\text{2}}}{\text{O}}$
Sodium fluoride is a side product of this reaction.
The structure of ${\text{O}}{{\text{F}}_{\text{2}}}$ is similar as water i.e. bent in shape. Its bond angle is \[103^\circ \] and bond length is \[{\text{140}}{\text{.5 pm}}\].
Thus, \[{\text{C}}{{\text{l}}_{\text{2}}}{\text{O}}\] and ${\text{O}}{{\text{F}}_{\text{2}}}$ doesn’t have same bond angle. Bond angle of \[{\text{C}}{{\text{l}}_{\text{2}}}{\text{O}}\] is \[110.9^\circ \] and for it is ${\text{O}}{{\text{F}}_{\text{2}}}$ is \[103^\circ \]
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Note:
Halogens and oxygen react to form halogen oxides, but all the halogen oxides are not identical. The bind length and bond angle of each halogen oxide is different. The bond angle of ${\text{O}}{{\text{F}}_{\text{2}}}$ is smaller than \[{\text{C}}{{\text{l}}_{\text{2}}}{\text{O}}\] because \[{\text{C}}{{\text{l}}_{\text{2}}}{\text{O}}\] in oxygen is more electronegative than chlorine. Due to which the shared electron pair is close to oxygen. The shared pair of electrons show repulsion with each other and thus increases the bond angle of \[{\text{C}}{{\text{l}}_{\text{2}}}{\text{O}}\]. Whereas, in ${\text{O}}{{\text{F}}_{\text{2}}}$, fluorine is more electronegative than oxygen and keeps the shared pair of electrons towards itself, thus its bond angle is less