Test: Group-17 Elements - NEET MCQ

Electron affinity decreases down the group as atomic size increases.

Because electronegativity decreases down the group thereby e⁻ attracting tendency decreases down the group. F with high electronegativity will have strongest bond and I with least electronegativity will have least bond strength.

F being highly electronegative has the ability to form hydrogen bonds due to which boiling point of HF is high.

F is having low dissociation energy as F being highly electronegative and small thereby repel each other, hence have low dissociation energy.

Shows −1 oxidation state as it can accept one e⁻ in order to complete its octet.

It has lower e⁻ affinity than chlorine due to its high electronegativity and small size it repels the new incoming e⁻.

HF has highest boiling point due to the strong bond formation.

Electron affinity of Cl is maximum. The correct trend is Cl > F > Br > I.

F₂ has lower bond dissociation energy than Cl₂ due to its small size which results in interelectronic repulsion.

More reactive halogen can displace less reactive halogen from its salt solution. F₂ can displace Cl₂, Br₂, I₂ from their salt solutions while Cl₂ can displace Br₂ and I₂ from NaBr and Nal. Br₂ can displace only I₂ from Nal.

2NaI + Br₂ → 2NaBr + I₂

F_2​ is easily reduced as F⁻¹ is the strongest oxidising agent. Oxidising power can be explained by the fact that the ease with which a halogen [X] forms the halide ion H⁻ decreases down the group.

Fluorine is an element that has high electronegativity. Fluorine has low bond enthalpy and highest reduction potential so more ability to loose the electrons. Fluorine can also be easily reduced. Fluorine is thus one of the best oxidizing agents. 

Oxygen florides forms HF, O₂​ on hydrolysis.

OF₂​ + H₂​O → O₂ ​+ 2HF

2O₂​F₂​ + 2H₂O → O_​2 + 4HF

so oxygen florides do not form forms hypohalous acids.

So the given statement oxygen florides forms oxoacids is wrong.