Question

Fluorine reacts with dilute NaOH and forms a gaseous product \(A\). The bond angle in molecule of \(A\) is

• A. \(104^\circ 40'\)
• B. \(103^\circ\)
• C. \(107^\circ\)
• D. \(109^\circ 28'\)

Hint:

Greater electronegativity of bonded atoms reduces bond-pair repulsion near central atom and decreases bond angle (as in \(OF_2\)).

Answer 1

The Correct Option is B

Step 1: Write reaction of \(F_2\) with dilute NaOH.
\[ 2F_2 + 2NaOH \rightarrow 2NaF + OF_2 + H_2O \]
So gaseous product \(A = OF_2\) (oxygen difluoride).
Step 2: Determine shape of \(OF_2\).
Central atom is oxygen.
O has 2 bond pairs and 2 lone pairs \(\Rightarrow\) bent structure like \(H_2O\).
Step 3: Compare bond angle.
Bond angle in \(H_2O\) is \(104.5^\circ\).
In \(OF_2\), fluorine is more electronegative, pulling bonding pairs away, reducing repulsion.
So bond angle is less than water: around \(103^\circ\).
Final Answer:
\[ \boxed{103^\circ} \]