The first ionization potential of the elements B, C, and N (atomic numbers 5, 6, and 7) steadily increases, but that of O is less than that of N. The best interpretation of the lowervalue for O is that:1. the ionization potential of N is a maximum and the values decrease steadily for the elements O, F, and Ne.2. there is more shielding of the nuclear charge in O than in B, C, or N.3. the electron removed from O is farther from the nucleus and therefore less tightly bound than that in N.4. the electron removed from O corresponds to a different value of the quantum number L than that of the electron removed from B, C, or N.5. the half-filled set of p orbitals in N makes it more difficult to remove an electron from N than from O.

The best interpretation of the lower value for O is that : The half-filled set of p orbitals in N makes it more difficult to remove an electron from N than from O

Explanation:

The potential for ionized electrons is a measure of how quickly an atom is destroyed.

The capacity for ionization is increased around the table from left to right. The actual electronic N atom structure is 2 2  (half-filled  orbital), Which allows it additional stability.

This increases the energy required to remove the first electron from N in comparison with O.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.