The strongest bond is present in:

B r_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C l_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

F_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $C l_{2}$
As we move down a group, the size of the atom increases and its bond strength decreases.
According to this, $F_{2}$ has the strongest bond, but due to the very small size of Fluorine, there is a strong electronic repulsion between its electrons which makes it weak & unstable.
Therefore, Chlorine outlooks Fluorine in this matter & has highest bond strength.

Suggest Corrections

Similar questions

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

(x^{2} + y^{2})

x

y

F_{2}, C l_{2}, B r_{2}, I_{2}

F_{2}, C l_{2}, B r_{2}, I_{2}

I_{2}, {B r}_{2}, {C l}_{2}, F_{2}

I_{2} < {B r}_{2} < {C l}_{2} < F_{2}

C l_{2} > B r_{2} > F_{2} > I_{2}

Join BYJU'S Learning Program