Unit 2 · Compound Structure and PropertiesPractice · 化合物结构与性质

Bond polarity depends on the electronegativity difference (ΔEN) between the two atoms: the larger ΔEN, the more polar the bond. Electronegativities: C 2.55, S 2.58, N 3.04, O 3.44, F 3.98. So ΔEN is C–S 0.03, C–N 0.49, C–O 0.89, and C–F 1.43. C–S joins two different elements, yet it is almost nonpolar.

Ionic bonds hold cations and anions together, while the atoms inside a polyatomic ion are held by covalent bonds. $\ce{K2SO4}$ ($\ce{SO4^2-}$), $\ce{NH4NO3}$ ($\ce{NH4+}$ and $\ce{NO3-}$), and NaCN ($\ce{CN-}$) all contain polyatomic ions, so they have both kinds of bonds. $\ce{CaCl2}$ is made of monatomic ions only, so it has only ionic bonds; $\ce{CH3OH}$ is molecular, with only covalent bonds.

Force is read from the slope of the curve: where the slope is zero, the net force is zero, and that is the minimum, point Y. Its distance is the bond length, and its depth is the bond energy. At W and X the curve falls steeply to the right, so the atoms are too close and the net force is repulsive. Zero potential energy at X only means the energy equals that of two separated atoms; it doesn’t mean there is no force. At Z the curve rises to the right, so the net force is attractive and pulls the atoms together.

An I atom is larger than a Cl atom, so the H–I bond is longer and the minimum moves to the right. The nuclei are farther apart, so the attraction is weaker, the bond energy is smaller, and the well is shallower. Data: H–Cl 127 pm and 431 kJ/mol; H–I 161 pm and 299 kJ/mol.

By Coulomb’s law, the attraction between ions depends on their charges and the distance between them. Both compounds have the same charges (+2 and −2), so distance decides: $\ce{Mg^2+}$ has a radius of 72 pm and $\ce{Ca^2+}$ 100 pm, so the ions in MgO are closer together and its lattice energy is greater. Measured lattice energies: MgO 3791 kJ/mol, CaO 3401 kJ/mol. The electronegativity fact in B is true, but electronegativity isn’t part of Coulomb’s law, so it isn’t a valid reason.

A C atom’s radius is only 0.50 times that of Fe, small enough to fit in the holes between Fe atoms, so carbon steel is an interstitial alloy. The radius ratio of Zn to Cu is 1.00: the atoms are about the same size, so Zn atoms take the places of some Cu atoms in the lattice, and brass is a substitutional alloy.

Total valence electrons: 6 + 3 × 6 + 2 = 26 (the 2− charge adds 2 electrons). A single bond from S to each O uses 6 electrons, leaving 20. Each O gets 3 lone pairs (18 electrons in all), and the last 2 electrons stay on S as 1 lone pair, for 10 lone pairs in total. The lone pair on S makes $\ce{SO3^2-}$ trigonal pyramidal, not trigonal planar like $\ce{SO3}$.

The higher the bond order, the shorter the bond. $\ce{NO+}$ has a single structure with an N≡O triple bond: bond order 3. $\ce{NO2-}$ has 2 equivalent resonance structures, so one double bond is shared over 2 N–O bonds: average bond order 1.5. $\ce{NO3-}$ has 3, so one double bond is shared over 3 bonds: average bond order 4/3. So the bond lengths are $\ce{NO+}$ < $\ce{NO2-}$ < $\ce{NO3-}$.

Formal charge = valence electrons − lone-pair electrons − bonds. C: 4 − 2 − 3 = −1; O: 6 − 2 − 3 = +1. It’s unusual for the more electronegative O to carry the positive formal charge, but only the triple bond gives both C and O an octet (with a double or single bond, C has fewer than 8 electrons), and the octet rule comes first.

Total valence electrons: 7 + 4 × 7 + 1 = 36 (the 1− charge adds an electron). Four I–Cl bonds use 8, three lone pairs on each Cl use 24, and the remaining 4 electrons are 2 lone pairs on I. That gives I 6 electron domains in an octahedral arrangement. The two lone pairs sit opposite each other, leaving the 4 Cl atoms in one plane, so the ion is square planar. Octahedral is the arrangement of the electron domains, not the shape, and counting only the 4 bonds while ignoring the lone pairs on I suggests tetrahedral.

First work out each shape with VSEPR: $\ce{PCl5}$ trigonal bipyramidal (0 lone pairs); $\ce{IF5}$ square pyramidal (1 lone pair); $\ce{SiCl4}$ tetrahedral (0 lone pairs); $\ce{BrF3}$ T-shaped (2 lone pairs); $\ce{XeF2}$ linear (3 lone pairs). When all the outer atoms are the same, whether the bond dipoles cancel depends on the shape. Trigonal bipyramidal and tetrahedral molecules are fully symmetric. In $\ce{XeF2}$ the 3 lone pairs lie around the equator and the two F atoms sit in a straight line, so its bond dipoles cancel too: these three are nonpolar. In $\ce{IF5}$ (square pyramidal) and $\ce{BrF3}$ (T-shaped), the lone pairs on the central atom make the molecule unsymmetric, so they are polar.

Every pair of bonded atoms shares exactly one σ bond: 3 C–H, C=C, C–C, and C≡N, 6 in all. The double bond adds 1 π bond and the triple bond 2 more, 3 in all. The carbon bonded to N is attached to 2 atoms and has no lone pairs, so it has 2 electron domains and is sp hybridized (linear); the other two carbons have 3 domains each and are sp². Counting every line as a σ bond gives 9.