20 CHAPTER 2 Bonding and Molecular Structure Problem 2.10 Arrange the s,p,and the three sp-type HO's in order of decreasing energy. The morescharacter in the orbital,the lower the energy.Therefore.the order of decreasing energy is p>sp>sp2>sp>s Problem 2.11 What effect does hybridization have on the stability of bonds? Hybrid orbitals can (a)overlap better and(b)provide greater bond angles,thereby minimizing the repulsion between pairs of electrons and making for great stability. HON =(Number of o bonds)+(Number of unshared pairs of electrons) The hybridi rom Table2.3.If more than four HO's e HO's (Fig 29a uded to octahedral HO's [Fig. 909 (a)sp'd HO's (b)sp'd HO's Figure 2.9 TABLE 2.3 HON PREDICTED HYBRID STATE p 45 The p method can also be used for the multicovalent elements in the second period and,with few exceptions.in the higher periods of the periodic table. Problem 2.12 Use the HON method to determine the hybridized state of the underlined elements: (a)CHCl3 (b)H2C=CH2 (c)O=C=O (d)HC=N:(e)H" NUMBER OF BONDS +NUMBER OF UNSHARED ELECTRON PAIRS HON HYBRID STATE 00 (d)c 22 22 (d)N 1 (e) 4 sp Problem 2.10 Arrange the s, p, and the three sp-type HO’s in order of decreasing energy. The more s character in the orbital, the lower the energy. Therefore, the order of decreasing energy is p > sp3 > sp2 > sp > s Problem 2.11 What effect does hybridization have on the stability of bonds? Hybrid orbitals can (a) overlap better and (b) provide greater bond angles, thereby minimizing the repulsion between pairs of electrons and making for great stability. By use of the generalization that each unshared and σ-bonded pair of electrons needs a hybrid orbital, but π bonds do not, the number of hybrid orbitals (HON) needed by C or any other central atom can be obtained as HON (Number of σ bonds) + (Number of unshared pairs of electrons) The hybridized state of the atom can then be predicted from Table 2.3. If more than four HO’s are needed, d orbitals are hybridized with the s and the three p’s. If five HO’s are needed, as in PCl5 , one d orbital is included to give trigonal-bipyramidal sp3 d HO’s [Fig. 2.9(a)]. For six HO’s, as in SF6 , two d orbitals are included to give octahedral sp3d2 HO’s [Fig. 2.9(b)]. 20 CHAPTER 2 Bonding and Molecular Structure Figure 2.9 HON PREDICTED HYBRID STATE 2 sp 3 sp2 4 sp3 5 sp3d 6 sp3d2 TABLE 2.3 The preceding method can also be used for the multicovalent elements in the second period and, with few exceptions, in the higher periods of the periodic table. Problem 2.12 Use the HON method to determine the hybridized state of the underlined elements: NUMBER OF σ BONDS  NUMBER OF UNSHARED ELECTRON PAIRS HON HYBRID STATE 4 04 sp3 3 03 sp2 2 02 sp 2 02 sp 1 12 sp 3 14 sp3 (a) (b) (c) (d) C (d) N (e)