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Select from each of the following gropus, the one which has the largest radius (i) Co,Co^(2+),Co^(3+) (ii) S^(2-), Ar, K^(o+) (iii) Li,Na,Rb (iv) C,N,O (v) Ne,Na,Mg (vi) La,Lu (vii) Cu, Ag, Au (viii) Ba, H_(r) (ix) Mg, na Na^(o+) , Mg^(+2) ,Al (b) IE_(1) of C is 11.2 eV What would be the value of iE_(1) of Si to be greater or less than this amount ? (c) IE_(1) of Li and K are 5.4 and 4.3 eV respectively What would be the value of IE_(1) of P ? (d) IE_(1) of Na? . (e) The IE' s of Li, Be and C are 5.4, 9.3 and 11.3 eV What would be the value of IE s of B and N ? (f) Which of these elements have the lowest IE_(1) Sr, As, Xe, S,F ? (g) Select from each of the following group the element which has the largest IE (i) Na,P,CI (ii) He,Ne,Ar (iii) O,F,Na (h) Arrange the species in each group in order of decreasing IE' s and in each case explain the reason for the sequence (i) K,Rb,Cs (ii) Be,B,C (iii) Cu,Ag,Au ? (iv) C,N,O (v) N,O,F (vi) K, Ca, Sc (vii) Na,Mg,AI (viii) Fe,Fe^(+2),Fe^(+3) (ix) K^(o+),Ar,CI^(Θ) (i) Explain why IE_(1) of Cu is higher than that of K whereas IE_(2) are in reverse order (j) Account for the difference in IE (i) between K^(o+) and Ca^(o+) (ii) between Cu^(o+) and Zn^(o+) (k) Ionisation potential is an old term for ionisation energy. Explain why yhe two are synonymous ? . |
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Answer» Solution :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)` . |
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