7. A weaker penetration of p orbital results in a. Stronger shielding of electro

Question

7. A weaker penetration of p orbital results in a. Stronger shielding of electrons b. Weaker shielding of electrons c. Stronger attraction of electrons d. Weaker repulsion of electrons 8. Energy ordering of orbitals in multi electron atoms is regulated by the a. Energy of the orbitals b. Energy of the principal quantum number c. Energy of the angular quantum number d. All of the above 9. One mole of mercury has a. 200.59 gms of mercury b. 200.59 gms of carbon C. 80 gams of Mercury d. 80 gms of carbon 10. When l-2, the values of r mi will be b. s, p, d, f. c. 1, 2, 3, 4 d. None of the above 11.0.5 moles of CO2 has a. 12 grams of Carbon b. 1 mole of carbon c. 16 grams of Oxygen d. 2 moles of Oxygen 12. The condensed electron configuration of Arsenic, element 33, is a. [Kr] 3d10 4s2 4p4 b. [Ar] 3d10 4s2 4p3 b. [Kr] 3d10 4s2 4p3 d. [Ar] 3d10 4s2 4p4 13. In the modern periodic table a. column is a group and rows are periods b. There are 18 groups and 7 periods c. Traversing down the group increases the atomic radius d. All of the above 14. Across a period a. Metallic Character increases b. Atomic radius increases c. lonization energy increases d. All of the above

Explanation / Answer

7.

Pentration power depends upon the shape of the orbitals.

b.weaker shielding of electrons

8. principal quatum number indicate the energy of lectrons and size of atom

b. Energy of the principal quantum number

9.gram atomic mass of an atom is equal to mole

1 mole of Hg = 200.59 of Hg

a. 200.59gms of mercury

10. l =2      ml = -2,-1,0, +1,+2

a. -2,-1,0,+1,+2

11.1 mole of Co2 contains 2 moles of oxygen

      0.5 moles of CO2 contains 1 mole of oxygen

           = no of moles * gram atomic mass of oxygen

                      = 1*16 = 16g

c. 16grams of oxygen

12. arsenic = 33 [Ar]3d^10 4s^2 4p^3

b. [Ar]3d^10 4s^2 4p^3

13. d all of he above

14.c ionization energy increases

    

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