Question

15. Account for the following
(i) Fluoride ion can not be formed as easily as compared to chloride ion from their respective elements.
(ii) The increasing order of reactivity among group I elements is Li < Na <K< Rb < Cs
(iii) Mg+2 is smaller than O-2 in size, though both have same electronic configuration ​

Answer 1

Explanation:

• For (i):

Fluorine is the 9th element of the periodic table having electronic configuration $1s^22s^22p^5$. It requires only 1 electron to attain stable electronic configuration and forms a negative ion. The size of this element is very small due to the attraction of electrons and protons.

Chlorine is the 17th element of the periodic table having electronic configuration $1s^22s^22p^63s^22p^7$. It requires only 1 electron to attain stable electronic configuration and forms a negative ion. The size of this element is larger than the size of the fluorine element.

Chlorine atom can easily form a negative ion because the size is large and fluorine cannot form due to interelectronic repulsion in such small atom.

• For (ii):

Reactivity is defined as the tendency of an element or gain or to loose electrons. Group 1 elements are metals and hence, their reactivity will be defined as the tendency to loose electrons.

As we know, that the size of an element increases as we move down the group. So, the valence electron in the cesium will be less attracted by the nucleus and will be easily lost.

Hence, the reactivity will increase as we move down the group, which is Li < Na <K< Rb < Cs.

• For (iii):

Magnesium is the 12th element having 12 electrons and 12 protons and $Mg^{2+}$ has 10 electrons and 12 protons

Oxygen is the 8th element having 8 protons and 8 electrons and $O^{2-}$ has 10 electrons and 8 protons.

The 10 electrons of $Mg^{2+}$ ion is easily attracted to the 12 protons, while the 10 electrons of $O^{2-}$ ions are less attracted to the 8 protons.

Hence, the size will shrink for $Mg^{2-+}$ ion and not for $O^{2-}$ ion although having same electronic configuration.

Answer 2

Explanation:

For (i):

Fluorine is the 9th element of the periodic table having electronic configuration

1

�

2

2

�

2

2

�

5

1s

2

2s

2

2p

5

. It requires only 1 electron to attain stable electronic configuration and forms a negative ion. The size of this element is very small due to the attraction of electrons and protons.

Chlorine is the 17th element of the periodic table having electronic configuration

1

�

2

2

�

2

2

�

6

3

�

2

2

�

7

1s

2

2s

2

2p

6

3s

2

2p

7

. It requires only 1 electron to attain stable electronic configuration and forms a negative ion. The size of this element is larger than the size of the fluorine element.

Chlorine atom can easily form a negative ion because the size is large and fluorine cannot form due to interelectronic repulsion in such small atom.

For (ii):

Reactivity is defined as the tendency of an element or gain or to loose electrons. Group 1 elements are metals and hence, their reactivity will be defined as the tendency to loose electrons.

As we know, that the size of an element increases as we move down the group. So, the valence electron in the cesium will be less attracted by the nucleus and will be easily lost.

Hence, the reactivity will increase as we move down the group, which is Li < Na <K< Rb < Cs.

For (iii):

Magnesium is the 12th element having 12 electrons and 12 protons and

�

�

2

+

Mg

2+

has 10 electrons and 12 protons

Oxygen is the 8th element having 8 protons and 8 electrons and

�

2

−

O

2−

has 10 electrons and 8 protons.

The 10 electrons of

�

�

2

+

Mg

2+

ion is easily attracted to the 12 protons, while the 10 electrons of

�

2

−

O

2−

ions are less attracted to the 8 protons.

Hence, the size will shrink for

�

�

2

−

+

Mg

2−+

ion and not for

�

2

−

O

2−

ion although having same electronic configuration.