Unformatted text preview: )2(π2p)4(σ2p)2, B.O. = 3, diamagnetic (0 unpaired e−)
N2 (1st excited state): (σ2s)2(σ2s*)2(π2p)4(σ2p)1(π2p*)1
B.O. = (7 − 3)/2 = 2, paramagnetic (2 unpaired e-)
The first excited state of N2 should have a weaker bond and should be paramagnetic. 68. a. Yes, both have four sets of electrons about the P. We would predict a tetrahedral
structure for both. See part d for the Lewis structures.
b. The hybridization is sp3 for P in each structure since both structures exhibit a tetrahedral
arrangement of electron pairs.
c. P has to use one of its d orbitals to form the π bond since the p orbitals are all used to
form the hybrid orbitals.
d. Formal charge = number of valence electrons of an atom – [(number of lone pair
electrons) + 1/2 (number of shared electrons)]. The formal charges calculated for the O
and P atoms are next to the atoms in the following Lewis structures.
Cl +1 P Cl Cl 0 O
Cl P 0 Cl Cl In both structures, the formal charges of the Cl atoms are all zeros. The structure with the
P=O bond is favored on the basis of formal charge since it has a zero formal charge for
all atoms. CHAP. 14
592 70. COVALENT BONDING: ORBITALS - EVEN-NUMBER PROBLEMS For carbon, nitrogen, and oxygen atoms to have formal charge values of zero, each C atom
will form four bonds to other atoms and have no lone pairs of electrons, each N atom will
form three bonds to other atoms and have one lone pair of electrons, and each O atom will
form two bonds to other atoms and have two lone pairs of electrons. Following these
bonding requirements gives the following two resonance structures for vitamin B6:
b a H
f O C HH c C
e C O N C
C C H O
H H C C C HH C H C H C
N O H H
H a. 21 σ bonds; 4 π bonds (The electrons in the three π bonds in the ring are delocalized.)
b. Angles a), c), and g): ≈109.5°; angles b), d), e), and f): ≈120°
c. 6 sp2 carbons; the five carbon atoms in the ring are sp 2 hybridized, as is the carbon with
the double bond to oxygen.
d. 4 sp3 a...
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