Select from following gropus, the one which has the largest radius `Li,Na,Rb`

Explanation

Co (The others have the same nuclear charge, but less electrons)
(ii) `S^(2-)` (All have the same electronic configuration, but the `S` nucleus has the smallest positive charge)
(iii) Rb (It is in the largest period)
(iv) `C` (It is the farthest left in the periodic table)
(v) na(It is the first element of a new period)
(vi) La(Lu is smaller because of lanthaniod contraction)
(vii) Au (Ag is almost the same size, because of the effect of the lanthanoid contraction on Au)
(viii) Ba(`HF` is much smaller because of the effect of the lanthanoid contraction)
(ix) na(The size decreases along the period `(rarr)` and also as electrons are removed)
(b) Less (`IE` decreases down the group `(darr)` due to increasing atomic size, but other factors also play an important role) The observed value is 8.1 eV
(c) It is difficult to remove `e^(-)` from `P` than from `S` because of the added stability of the half- filled p-subshell The observed value is `10.9 eV`
(d) The `IE` of Na should be intermediate between that of `Li` and `K` The `IE` of Na should be close to the arithmatic average of the two, or `4.9 eV` The observed is `5.1 eV`
(e) Generally `IE` increases along the period `(rarr)` However there is a larger increase in `IE` from `Li(Z=3)` to Be `(Z=4)` than from Be to `C(Z=6)` `IE` of `B` actually is less than that of Be due to penetration effect in `B` The observed `IE` of `B` is `8.3 eV` N has a half -filled `2p` subshell and should have extra stability for this reason The increase in going from `N(Z=5)` to `(Z=6)` is `3.0 eV` and the additional increase should be greater than this, thus the `IE` of `N` should exceed `14.3 eV` The experimental value is `14.5 eV`
(f) Sr (It is a metal)
(g) (i) `CI` (All of them belong to 3rd period and `CI` is farthest to the right of the three elements)
(ii) He (iii) Ne
(h) (i) `K gt Rb gt Cs` (Generally `IE` decreases down the group `(darr)`)
(ii) `C gt Be gtB` (The `2se^(-)` is more difficult to remove than 2p (penetration effect)
(iii) `Au gt Cu gtAg` (The lanthanoid contraction makes Ag and Au about equal in size but Au has a much greater nuclear charge)
(iv) `N gt O gt C` (Due to stability of half- filledorbitals)
(v) `F gtN gt O` (The half- filed p-subshell of `N` imparts enough extra stability to make its `IE gt IE` of O)
(vi) `Sc gt Ca gt K` (In each of these cases, a `4s e^(-)` is being removed and the order is as shown)
(vii) `Mg gtAI gt Na` (It is more difficult to remove an `e^(-)` from Mg because the `e^(-)` being removed is a 3s `e^(-)` from a filled subshell The `3pe^(-)` of AI is easily removed It is easiest to remove an `e^(-)` from na because of its large size)
(viii) `Fe^(3+) gt Fe^(2+) gt Fe` (All have same nuclear charge and the number of `e^(-')s` increases in the order listed) (ix) `K^(o+) gt Ar gt CI^(Θ)` (All have the same electronic configuration and the nuclear chared decreases in the order listed)
(i) `Cu(Z =29) implies 3d^(10) 4s^(1) , Cu^(o+) = 3d^(10) 4s^(@)`
`Cu^(+2) = 3d^(9) 4s^(@)`
(ii) `K(Z =19) implies 4s^(1), K^(o+) = 4s^(@), k^(+2) = 3p^(5)`
Cu has 10 more protons and more `e^(-')s` than does `K` but due to imperfect screening effect of `de^(-')s, IE_(1)` of `Cu gt IE_(1)` of `K` and `IE_(2)` for `K` involves the removal of an `e^(-)` from an octet (`3s^(2) 3p^(6))` whereas that of `Cu` involves the more easily ionised `d^(10)` configuration Hence `IE_(2)` of `K gt IE_(2)` of `Cu`
(j) `K^(o+)` loses an `e^(-)` from its `3p^(6)` subhell, `Ca^(o+)` from its 4s subshhell, which requires less energey
Hence `IE(2)` of `K^(o+) gt IE_(2)` of `Ca^(o+)`
(ii) `Zn (Z =30) implies 3d^(10) 4s^(2) , Zn^(o+) implies 3d^(10) 4s^(1)`
`Cu (Z =29) implies 3d^(10) 4s^(1) ,Cu^(o+) implies 3d^(10) 4s^(@)`
`Zn^(o+)` loses an `e^(-)` from `4s^(1)` more easily than `Cu^(o+)` loses an `e^(-)` from `3d^(10)`
Hence `IE_(2)` of `Cu^(o+) gt IE_(2)` of `Zn^(o+)`
(k) The `IE` of an `e^(-)` in `eV` is numerically equal to `IP` in volts
Note If a particle being accelerated by a potential has a charge equal in magnitude to the charge on an `e^(-)` the number of `eV` of energy is numerically equal to the potential in valts `(V)` .