Identify which of the following statements are true. check all that apply. check all that apply. ionization energies are always negative quantities. oxygen has a larger first ionization energy than fluorine. the second ionization energy of an atom is always greater than its first ionization energy. the third ionization energy is the energy needed to ionize three electrons from a neutral atom.

The second ionization energy of an atom is always greater than its first ionization energy.

Explanation:

Ionization energies are always negative quantities. FALSE. Ionization energies are always positive because they represent the energy that has to be added to a neutral gaseous atom to remove 1 electron. By convention, energy absorbed goes with the positive sign.

X(g) + 1 I.E. ⇒ X(g)⁺ + 1 e⁻

Oxygen has a larger first ionization energy than fluorine. FALSE. First ionization energy  (1 I.E) is the energy required to remove 1 electron from a neutral gaseous atom. First ionization energy for Oxygen is 13.61 eV and for Fluorine is 17.42 eV, that is, more energy is required to remove an electron from F than from O.

The second ionization energy of an atom is always greater than its first ionization energy. TRUE. The energy required to remove 1 electron from a +1 cation (second ionization energy) is always greater than the first ionization energy because of the attraction between the cation and the electron.

X(g)⁺ + 2 I.E. ⇒ X(g)⁺² +  1 e⁻

The third ionization energy is the energy needed to ionize three electrons from a neutral atom. FALSE. The third ionization energy (3 I.E.) is the energy required to remove 1 electron from a +2 cation.

X(g)⁺² + 3 I.E. ⇒ X(g)⁺³ +  1 e⁻

lithosphere and asthenosphere

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.