|
| |
(E) The Lewis structure of ClF3 is shown on the right. There are three atoms
|
| |
9.
| F Cl F|
and two lone pairs attached to the central atom, its hybridization is sp 3d , which
Clhyb [Ne] dsp 3
is achieved as follows. Cl unhyb [Ne]
3s 3p
3d
|
| F|
3d
The orbital diagram of F is [He]2s 2p
Each of the three sigma bonds are formed by
the overlap of a 2 p orbital on F with one of the half-filled dsp3 orbitals on Cl. Since the
dsp3 hybridization has a trigonal bipyramidal shape, and the two lone pairs occupy the
equatorial positions in the molecule, ClF3 is T-shaped.
13.
(M)
(a) This is a planar molecule. The hybridization on C is sp2
(one bond to each of the three attached atoms).
(b)
| N≡≡ C —C ≡≡ N| is a linear molecule. The hybridization for each C is sp
(one bond to each of the two ligands).
(c)
Trifluoroacetonitrile is neither linear nor planar. The shape around the
left-hand C is tetrahedral and that C has sp 3 hybridization. The shape
around the right-hand carbon is linear and that C has sp
hybridization.(N atom is sp hybridized).
(d)
34.
S
C
F
F
C
F
−
N is a linear molecule. The hybridization for C is sp.
(M)
(a) In each case, the number of valence electrons in the species is determined first; this
is followed by the molecular orbital diagram for each species.
ΚΚ σ2s σ2s* π2p σ2p π2p* σ2p*
+
C2 no. valence e = ( 2 × 4 ) − 1 = 7 ΚΚ
ΚΚ
O no. valence e = ( 2 × 6 ) +1 = 13 ΚΚ
no. valence e- = ( 2×7 ) -1=13 ΚΚ
NO+ no. valence e- = 5 + 6 – 1 = 10 ΚΚ
–
2
F2+
(b)
σ
2s
σ
2s*
σ
2p
π
2p
π
2 p*
σ
2 p*
-
Bond order = (no. bonding electrons – no. antibonding electrons) ÷ 2
This species is stable.
C +2 bond order = (5 − 2 ) ÷ 2 = 1.5
O 2– bond order = (8 − 5) ÷ 2 = 1.5
This species is stable.
453
C
N
Chapter 11: Bonding II: Additional Aspects
F2+ bond order = (8 − 5) ÷ 2 = 1.5
NO+ bond order = (8 –2) ÷ 2 = 3.0
(c)
36.
This species is stable.
This species is stable.
+
C2 has an odd number of electrons and is paramagnetic, with one unpaired electron.
O2- has an odd number of electrons and is paramagnetic, with one unpaired electron.
F2+ has an odd number of electrons and is paramagnetic, with one unpaired electron.
NO+ has an even number of electrons and is diamagnetic.
In Exercise 33, the species that are isoelectronic are those with 8 electrons: CN + , BN;
with 10 electrons: NO+, CO, CN-
43.
(M) We expect to find delocalized orbitals in those species for which bonding cannot be
represented thoroughly by one Lewis structure, that is, for compounds that require several
resonance forms.
(a)
(b)
In C2H4 , there are a total of 2X4+4X1=12 valence electrons, or 6 pairs.
C atoms are the central atoms. Thus, the bonding description of C2H4
does not involve the use of delocalized orbitals.
H
C
C
H
H
In SO2 , there are a total of 6+2X6= 18 valence electrons, or 9 pairs. N is the central
atom. A plausible Lewis structure has two resonance forms. The bonding description
of SO 2 will require the use of delocalized molecular orbitals.
O
(c)
H
S
O
O
S
O
454
H C
H
| |
In H2 CO , there are a total of 2X1+4+6=12 valence electrons, or 6 pairs.
C is the central atom. A plausible Lewis structure is shown on the right.
Because one Lewis structure adequately represents the bonding in the
molecule, the bonding description of H2CO does not involve the use of
delocalized molecular orbitals.
O