Answer to Question #264024 in Inorganic Chemistry for krishen

Question

(a)

Choose and explain which element has the greater value for the following properties:

i. electron affinity.

    Na or F. 

ii. electronegativity.

        Cl or Ca.        

iii. ionization energy.

       S or Se.

iv. atomic radius.

    C or Sn.

v. metallic property.

      Si or Br.

 

(b)

Write the electron configurations for the following atoms/ions:

     i.       Ca.

    ii.       S.

  iii.       Mg²⁺.

  iv.       O^2-.

    v.       Fe³⁺.

Answer 1

a) (i) In the given question,we have been asked about the electron affinity.

Electron affinity is defined as the amount of energy released when an electron is attached to nuetral isolated atom.

Therefore, F will have higher electron affinity as F being a non- metal will release more energy as compared to Na.

Value of Electron Affinity of Na: -53 KJ/mol and F:-328 KJ/mol

Generally, Non-metals have higher electron affinity.

(ii) Electronegativity is defined as the tendency of an atom to attract a shared pair of electrons.

Attraction of electron are generally done by non-metals and therefore, Cl has more electronegativity.

( iii) Ionixation energy is defined as the minimum amount of energy required to remove a electron from nuetral isolated atom.

Removal of electrons are generally happens in metals as they donate electron to form cations.

As we move down a group, the atomic radius of elements gets increased due to which the valence electrons are more away from the nucleus. As the distance between valence electron and nucleus is increased, it becomes easy to remove the electron.Therfore, Se being bigger in size as lower ionization Energy.

Therefore, S being smaller in size has greater ionization energy.

(iv) As we move down a group, the atomic size of the element gets increased.Therefore, Sn being more lower in group has greater atomic size as compared to C.

Therefore, Correct Answer is:Sn

(v) Si has more mettalic property as compared to Br since Si is more in the left part of the periodic table as compared to bromine.

Elements being in the left part have more metallic property as compared to the elements in right side of the periodic table.

Therefore, Correct Answer is : Si

(b)

(i) Ca- 1s ² 2s ² 2p ⁶ 3s ² 3p ⁶ 4s ²

(ii) S- 1s ² 2s ² 2p ⁶ 3s ² 3p ⁴

(iii) Mg ²⁺ - 1s ² 2s ² 2p ⁶

2 electron has been removed from the outermost shell since it has form cation.

(iv) O ^2- - 1s ² 2s ² 2p ⁶

Two electron has been added to the outermost shell since it has formed anion.

(v) Fe ³⁺ - 1s ² 2s ² 2p ⁶ 3s ² 3p ⁶ 3d ⁵

2 electron from 4s orbital and 1 electron from 3d orbital has been removed since it has formed cation.