March 14, 2012

Periodic Classification - II Three Marks

1. Calculate the effective nuclear charge experienced by the 4s electron in potassium atom. (S =16.8).
Atomic number / Actual charge of K atom (Z)    = 19
Screening constant (S)                                       = 16.8
Effective nuclear charge (Z*)                         = Z – S
                                                                          = 19 – 16.8
                          Z*  = 2.2
2. Calculate the electronegativity value of fluorine on Mullikan's scale from the following data: Ionisation potential of F = 17.4 eV/atom, Electron affinity of F = 3.62 eV/atom.
         Electronegativity of F   = (IP)F - (EA)F / 2 x 2.8
= 17.4 + 3.62 / 5.6
= 21.02 / 5.6
         Electronegativity of F   = 3.75
3. Compare the first ionisation energy of Aluminium (Z = 13) with that of Magnesium (Z = 12). Justify your answer.
The first ionization energy of Aluminium is lower than that of Magnesium.
The electronic configuration of Magnesium is [Ne]3s2 and that of Aluminium is [Ne] 3s2 3p1.
Thus, one has to remove 3p-electron in case of Aluminium and 3s-electron in the case of Magnesium. But it is easier to remove the p-electron than the s-electron.
Thus, the first ionization energy of Aluminium is lower than that of Magnesium.
4. Compare the ionisation energy of nitrogen with that of oxygen.
1.Ionisation energy of N > O (or) Ionisation energy of O < N
2.Nitrogen atom (Z = 7; 1s2 2s2 2px1 2py12pz1) is having stable half filled 2p orbital.
3.Oxygen atom (Z = 8; 1s2 2s2 2px2 2py12pz1) loses an electron easily to acquire the stable half filled 2p orbital.
5. Define electron affinity.
            Amount of energy which is released when an extra electron enters the valence orbital of an isolated neutral atom to form a negative ion.
      Atom(g) + Electron(g)   Negative ion(g) + Energy
6. Explain why the first ionisation energy of Be is greater than that of B. Or The ionisation energy of boron is less than beryllium. Why?
Boron atom (Z = 5; 1s2 2s2 2px1 2py0 2pz0) is having one unpaired electron in the 2p-subshell. Be-atom (Z = 4; 1s2 2s2) is having paired electrons in the 2s subshell.
As the completely filled 2s-subshell in Be-atom is more stable than B-atom due to symmetry, more energy would be needed to remove an electron from Be-atom. Hence, Be has high ionisation energy.
7. If the d(C – Cl) is 1.76 Å and r(Cl) is 0.99 Å, find the radius of carbon atom.
d(C – Cl)   = 1.76 Å
r(Cl)    = 0.99 Å
    Thus, d(C–Cl)    = r(C) + r(Cl)
    r(C)  = d(C – Cl) – r(Cl)
     = 1.76 – 0.99
Radius of carbon atom r(C) = 0.77 Å
8. Ionisation energy of Neon is greater than that of Fluorine. Give reason. Or Neon has more ionisation energy than fluorine. Why?
1. The nuclear charge of Ne (Z = 10) is more than that of F (Z = 9). Ne > F
2. Greater the nuclear charge, greater would be the force of attraction between nucleus and outermost electron.
3. Neon has the stablest / completely filled / fully filled electronic configuration 1s22s2 2p6
9. Larger the size of atom lesser is the ionisation energy. Explain.
      The electrons are tightly held in smaller atoms whereas in large atoms, electrons are held quite loose, i.e., lesser energy is required for removal of electrons from larger atoms than the smaller one. Hence ionization energy is lower for larger atoms.
10. Mention the disadvantage of Pauling’s and Mulliken’s electronegativity scale.
Disadvantage of Pauling’s scale
Bond energies are not known with any degree of accuracy for many solid elements.
Disadvantage of Mulliken’s scale
Electron affinities with the exception of a few elements are not reliably known.
11. The electron affinities of beryllium and nitrogen are almost zero. Why?
It is due to the Extra stability of the completely filled 2s-orbital in beryllium and of the exactly half-filled 2p-orbital in nitrogen. As these are stable electronic configurations, they do not have tendency to accept electrons.
12. Why is electron affinity of fluorine less than that of chlorine?
1.Due to small size of fluorine atom, there occurs repulsion among electrons of the valence shell and also with electron to be added.
2.The addition of an extra electron produces high electron density which increases strong electron-electron repulsion. The repulsive forces between electrons results in low electron affinity.
13. Why is ionisation energy of fluorine greater than that of oxygen?
F (Z = 9; 1s2 2s2 2px2 2py2 2pz1) is having more nuclear charge than oxygen (Z = 8; 1s2 2s2 2px2 2py12pz1). In both the cases, the electron has to be removed from the same 2p-subshell. As fluorine is having more nuclear charge than oxygen, it means that the nucleus of fluorine will attract the outer 2p-electrons more firmly than oxygen. Hence, first I.E. of fluorine would be more than that of oxygen.
14. Why is the first ionisation energy of Beryllium greater than that of Lithium?
      Because the nuclear charge of Be (Z = 4) is greater than Li (Z = 3). Higher the nuclear charge, greater would be the force of attraction between nucleus and outermost electron. Hence, the first ionisation energy of Be is greater than that of Li.
15. Calculate the effective nuclear charge of the last electron in an atom whose configuration is
1s2 2s2 2p6 3s2 3p5
  Z = 17
Z* = Z – S
      = 17 – [(0.35 x No. of other electrons in nth shell)
              + (0.85 x No. of electrons in (n –1)th shell)
              + (1.00 x total number of electrons in the inner shells)]
      = 17 – [(0.35 x 6) + (0.85 x 8) + (1 x 2)]
      = 17 – 10.9
Z* = 6.1
16. Compare the ionisation energies of Carbon and Boron
Carbon (Z = 6; 1s2 2s2 2px1 2py1 2pz0) is having more nuclear charge than boron (Z = 5; 1s2 2s2 2px1 2py02pz0). In both the cases, one has to remove electron from same 2p-subshell. Carbon is having more nuclear charge than boron. Therefore the nucleus of carbon, attracts the outer 2p-electron more firmly than does boron. Thus, first ionisation energy of carbon would be more than that of boron.

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