Just a second...2

electronegativity of F >Cl so (B) is true

so easier to remove an electron from Cl atom than F

we have an added electron to 2p orbital so more repulsions as size of 2p orbital is smaller than 3p

so Cl^- >F^- and Cl >F^-

so (C) and (D) are true

having doubt over (A)

when we have Cl^- and F then added electron ll b repelled and also due to electronegativity reason F >Cl^- so (A) is true

i think as the size of fluorine is very very small nd due to this size when an electron is added to it....electron repulsion takes place due to wich it is easier to pul out the electron

i think in the case of fluorine repulsion factor is more dominating factor than noble gas configuration...................................mujhe aisa lagta haii

Waise, overall...i can figure out this at a first sight-

Firstly....F^- gains an e- to attain a stable config. lyke Ne, moreover the e- is closer 2 the nucleus.

n Secondly.... There will be repulsion no doubt; but also the 1st factor can't also be ignored tht easily.
('coz this dosen't look very nice - Cl>Ne)

So, if 2 dominates C & D are very well explained, n if 1 dominates A can be explained (owing to the size factor only)

@uttara... is this an aakash aits Qn?

@Avik : This qs is not from aakash aits . It's from a book

Basically this question needs no concept if one has good visualization. Even with 10^th grade science one can solve this.

First we take option B because both elements are without any charge to irritate us.

Electronegativity of F is more than Cl, so ionization energy of F > Cl.

Next we take option D because both are having charges.

1. Cl^- has more nuclear charge than F^-
2. F^- has more repulsion of electrons since the electrons are crowded together.

From these observations we can conclude that in Chlorine ion, due to more nuclear charge and less repulsion between electrons (since size is more so electron density is less in outermost shell) the electrons are held firm. But in Fluorine ion, the electrons repel each other, and there is a lot of energy. Hence removing an electron from fluorine ion is easier.

Hence ionisation energy of Cl^- is more than that of F^- or rather, I must say that ionisation energy of F^- is less than Cl^- (More appropriate reason)

Next we choose option B. When such options come, they are a bonus because they are surely a part of the answer.

Use the reasoning for option D in this option also.

It is generally true that when we compare a larger element and a smaller ion from the same group, the ionisation energy of the element is more than that of the ion. Fluorine ion will still have the same reactivity, and Chlorine will only lose an electron in presence of more energy to become an ion.

(Very effective short cut for S and P blocks mostly but not for D and F block because sizes are somewhat same)

Now finally we go to option A.

Fluorine is more electronegative than Chlorine. To remove an electron from F element, we require more energy, but to remove an electron from negatively charged ion, we require less energy. This is because in Fluorine element, now there are no excess electrons for repulsion. But in chlorine, there is shielding effect, electron repulsion as well as less electronegativity. For these reasons, we can say that option A is correct.

Hence options A,B,C,D are all correct. :)