R, Y, B and P are the elements belonging to 17^th group of Modern periodic table. The correct trend of SRP values $\left({\mathrm{E}}_{\frac{1}{2}{\mathrm{X}}_2/{\mathrm{X}}^{-}}^0\right)$ of these elements is R₂ > Y₂ > B₂ > P₂. The element ‘B’ in the sequence is

Solution:

Atomic radius….F<Cl<Br<I;

Density ….F<Cl<Br<I;

IP ….F>Cl>Br>I;

EN ….F>Cl>Br>I;

Electron gain enthalpy ….I<Br<F<Cl; Due to small size and inter electronic repulsions in Fluorine ;

IP …..I^-<Br^-<F^-<Cl^-;

IP trend….F>Cl>Br>I>Cl^->F^->Br^->I^-

B.D.E….Cl₂>Br₂>F₂>I₂; Due to small size and repulsions between lone pairs on adjacent Fluorine atoms ;

IP …..I^-<Br^-<F^-<Cl^-;

Reactivity and oxidizing power…..F₂>Cl₂>Br₂>I₂;

Due to high hydration enthalpy of F^- and low B.D.E of F₂, F₂ is the strongest oxidant and highly reactive;

SRP Values of ${\mathrm{E}}_{\frac{1}{2}{\mathrm{F}}_2/{\mathrm{F}}^{-}}^0\left(+2.87 \mathrm{V}\right) > {\mathrm{E}}_{\frac{1}{2}{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}^0\left(+1.36 \mathrm{V}\right) > {\mathrm{E}}_{\frac{1}{2}{\mathrm{Br}}_2/{\mathrm{Br}}^{-}}^0\left(+1.08 \mathrm{V}\right) > {\mathrm{E}}_{\frac{1}{2}{\mathrm{I}}_2/{\mathrm{I}}^{-}}^0\left(+0.54 \mathrm{V}\right)$;

Reducing power….I^- > Br^- > Cl^- > F^- ;

Hydration enthalpy…F^->Cl^->Br^->I^-;

MP and BP ….F₂<Cl₂<Br₂<I₂; Due to increased strength of London dispersion forces;