The given ionization energy has to be matched with the given electronic configuration of atoms. Concept Introduction: An atom can be converted into an ion by removing an electron from its outermost shell or adding electron to the outermost shell. An amount of energy is required to remove an electron from the outermost shell of an atom to form a cation. The minimum energy required for removing an electron from an atom in gaseous phase is known as ionization energy . If one electron is removed from the atom to form a cation, the energy required is known as first ionization energy . If another electron is removed from the formed ion, the energy required is known as second ionization energy and this goes on. Second ionization energy is always higher than the first ionization energy because the effective nuclear charge increases. The trend followed in a periodic table for ionization energy is, it increases across the period and decreases down the group. This is because; across the period, the electrons are added to the same orbital and hence decrease in shielding. However, down the group, the electrons are added to a new sub-shell and hence increase in shielding. To compare: the ionization energies with the electronic configuration given

Question

Want to see the full answer?

Answer 1

Question

Chapter 4, Problem 4.55QP

Interpretation Introduction

Interpretation: The given ionization energy has to be matched with the given electronic configuration of atoms.

Concept Introduction:

An atom can be converted into an ion by removing an electron from its outermost shell or adding electron to the outermost shell. An amount of energy is required to remove an electron from the outermost shell of an atom to form a cation. The minimum energy required for removing an electron from an atom in gaseous phase is known as ionization energy.

If one electron is removed from the atom to form a cation, the energy required is known as first ionization energy. If another electron is removed from the formed ion, the energy required is known as second ionization energy and this goes on. Second ionization energy is always higher than the first ionization energy because the effective nuclear charge increases.

The trend followed in a periodic table for ionization energy is, it increases across the period and decreases down the group. This is because; across the period, the electrons are added to the same orbital and hence decrease in shielding. However, down the group, the electrons are added to a new sub-shell and hence increase in shielding.

To compare: the ionization energies with the electronic configuration given

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Using reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NO2 (g) = N2O4(g) AGº = -5.4 kJ Now suppose a reaction vessel is filled with 4.53 atm of dinitrogen tetroxide (N2O4) at 279. °C. Answer the following questions about this system: Under these conditions, will the pressure of N2O4 tend to rise or fall? Is it possible to reverse this tendency by adding NO2? In other words, if you said the pressure of N2O4 will tend to rise, can that be changed to a tendency to fall by adding NO2? Similarly, if you said the pressure of N2O4 will tend to fall, can that be changed to a tendency to '2' rise by adding NO2? If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO 2 needed to reverse it. Round your answer to 2 significant digits. 00 rise ☐ x10 fall yes no ☐ atm G Ar 1

Why do we analyse salt?

Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. H H CH3OH, H+ H Select to Add Arrows H° 0:0 'H + Q HH ■ Select to Add Arrows CH3OH, H* H. H CH3OH, H+ HH ■ Select to Add Arrows i Please select a drawing or reagent from the question area