Question

Compare size of Be+,C+,O+,F+

Answer 1

The order of size is: Be+ > C+ > O+ > F+

Answer 2

To compare the size of Be+, C+, O+, and F+, we need to consider their positions in the periodic table and the general trends in ionic radii.

1. Identify the elements and their positions: All these elements (Be, C, O, F) belong to the second period of the periodic table.

   • Be is in Group 2.
   • C is in Group 14.
   • O is in Group 16.
   • F is in Group 17.

2. Identify the charge of the ions: All the given species are cations with a +1 charge.

3. Recall the trend for ionic radii across a period: For ions of the same charge, as we move from left to right across a period, the ionic radius generally decreases. This is because:

   • The nuclear charge (number of protons, Z) increases steadily.
   • The number of electron shells remains the same.
   • The effective nuclear charge (Zeff) experienced by the outermost electrons increases significantly.
   • This increased attraction between the nucleus and the valence electrons pulls the electron cloud closer to the nucleus, resulting in a smaller ionic size.

4. Apply the trend to the given ions:

   • Be+ has 4 protons.
   • C+ has 6 protons.
   • O+ has 8 protons.
   • F+ has 9 protons.

   As we move from Be+ to F+ across Period 2, the nuclear charge increases (4 < 6 < 8 < 9). This increasing nuclear charge exerts a stronger pull on the electrons, leading to a decrease in ionic size.

Therefore, the order of decreasing size is: Be+ > C+ > O+ > F+

Explanation: The ions Be+, C+, O+, and F+ are all +1 cations derived from elements in the second period. As we move from left to right across a period, the nuclear charge increases while the number of electron shells remains constant. This leads to an increase in the effective nuclear charge, pulling the electrons more strongly towards the nucleus and consequently decreasing the ionic radius. Hence, Be+ (Z=4) is the largest, and F+ (Z=9) is the smallest among them.