SolnsChap14EVENZum7e

# N22 and n2 both have 14 protons in the two nuclei

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: -energy antibonding orbitals ( ! * p ) , so they should have electrons that are easier to 2 remove as compared to N2, which has no ! * p electrons. To differentiate which has the easiest 2 ! * p to remove, concentrate on the number of electrons in the orbitals attracted to the number 2 of protons in the nucleus. N22− and N2− both have 14 protons in the two nuclei combined. Because N22− has more electrons, one would expect N22− to have more electron repulsions, which translates into having an easier electron to remove. Between O2 and O2+, the electron in O2 should be easier CHAP. 14 595 COVALENT BONDING: ORBITALS - EVEN-NUMBER PROBLEMS to remove. O2 has one more electron than O2+, and one would expect the fewer electrons in O2+ to be better attracted to the nuclei (and harder to remove). Between N22− and O2, both have 16 electrons; the difference is the number of protons in the nucleus. Because N22− has 78 78. two fewer protons than O2, one would expect the N22− to have the easiest electron to remove which translates into the smallest ionization energy. 80. 2p 2s O2 O2− O2 + O The order from lowest IE to highest IE is: O2− < O2 < O2+ < O. The electrons for O2−, O2 , and O2 + that are highest in energy are in the ! * p MOs. But for O2−, 2 these electrons are paired. O2− should have the lowest ionization energy (its paired ! * p 2 electron is easiest to remove). The species O2+ has an overall positive charge, making it harder to remove an electron from O2+ than from O2. The highest energy electrons for O (in the 2p atomic orbitals) are lower in energy than the ! * p electrons for the other species; O will 2 have the highest ionization energy because it requires a larger quantity of energy to remove an electron from O as compared to the other species. CHAP. 14 596 82. COVALENT BONDING: ORBITALS - EVEN-NUMBER PROBLEMS The molecular orbitals for BeH2 are formed from the two hydrogen 1s orbitals and the 2s and one of the 2p orbitals from beryllium. One of the sigma bonding orbitals forms from overlap of the hydrogen 1s o...
View Full Document

## This note was uploaded on 01/26/2014 for the course CHEM 001 taught by Professor Giancoli during the Fall '12 term at UPenn.

Ask a homework question - tutors are online