Explain the following, give appropriate reasons.
1)Ionization potential of N is greater than that of O
2) First ionization potential of C-atom is greater than that of B-atom, where as the reverse is true for second ionization potential.
3) The electron affinity values of Be, Mg and noble gases are zero and those of N (0.02 eV) and P (0.80 eV) are very low
4) The formation of F– (g) from F(g) is exothermic while that of O2-(g) from O (g) is endothermic
Answers
Answer:
Answer:
1. N (Z = 7) 1s2 2s2 2px1 12py1 2pz1. It has exactly half filled electronic configuration and it is more stable. Due to stability, ionization energy of nitrogen is high.
O (Z = 8) 1s2 2s2 2px1 2py1 2pz1. It has incomplete electronic configuration and it requires less ionization energy.
I.E1 N > I.E1O
2. C (Z = 6) 1s2 2s2 2px1 2py1. The electron removal from p orbital is very difficult. So carbon has highest first ionization potential. B (Z = 5) 1s2 2s2 2p1. In boron nuclear charge is less than that of carbon, so boron has lowest first ionization potential.
I.E1 C > I.E1 B
But it is reverse in the case of second ionization energy. Because in case of B+ the electronic configuration is 1s2 2s2, which is completely filled and it has high ionization energy. But in C+ the electronic configuration is 1s2 2s2 2p1, one electron removal is easy so it has low ionization energy.
I.E2 B > I.E2 C
3. Be (Z = 4) 1s2 2s2
Mg (Z = 12) 1s2 2s2 2p6 3s2
Noble gases has the electronic configuration of ns2 np6. All these are completely filled and are more stable. For all these elements Be, Mg and noble gases, addition of electron is unfavorable and so they have zero electron affinity.
Nitrogen (Z = 7) 1s2 2s2 2px1 2py1 2pz1. It has half filled electronic configuration. So addition of electron is unfavorable and it has very low electron affinity value of 0.02 eV. Phosphorus (Z = 15) 1s2 2s2 2p6 3s2 3px1 3py1 3pz1. It also has half filled electronic configuration. Due to the symmetry and more stability, it has very low electron affinity value of 0.80 eV.
4. F(g) + e– → F(g)– exothermic
F (Z = 9) 1s2 2s2 2p5. It is ready to gain one electron to attain the nearest inert gas configuration. By gaining one electron, energy is released, so it is an exothermic reaction.
O(g) + 2e– → O2-(g) endothermic
O (Z = 8) 1s2 2s2 2px1 2py1 2pz1. It is the small atom with high electron density. The first electron affinity is negative because energy is released in the process of adding one electron to the neutral oxygen atom. Second electron affinity is always endothermic (positive) because the electron is added to an ion which is already negative, therefore it must overcome the repulsion