Question

Consider the following electron configurations to answer the next two questions: (i) 1s22s22p@3s! (ii) 1s22s22p3s2 (iii) 1s22s22p3s23p1 (iv) 1s22s22p3s23p (v) 1s22s22p@3s23p5 39) The electron configuration belonging to the atom with the highest second ionization energy is A) (0) B) () C) (iii) D) (iv) E) (v) 40) The electron configuration that belongs to the atom with the lowest second ionization energy is A) (i) B)(ii) C) (iii) D) (iv) E) (v)

Answer 1

(i) no of electrons are = 11 => sodium

(ii) no of electrons are = 12 => magnesium

(iii) no of electrons are = 13 => aluminium

(iv) no of electrons are = 16 => sulphur

(v) no of electrons are = 15 => phosphorous

39) the sodium having the 11 electrons and when one electron is removed then it forms the Na+ ion i.e having the 10 electrons and gets the noble gas configuration of the Neon and it is more stable and requires more energy to remove the second electron so more ionization energy is required & remaining all contains the lowest second ionization energy than the sodium

answer => second highest ionization energy is A) (i) => sodium

40) magnesium has the valency of 2 i.e it readily forms the Mg2+ ion and so it is easy to remove electron from Mg+ ( which is formed by the removal of first electron ) and requires less ionization energy when compare with other elements

answer => lowest is => B) (ii) => magnesium