Assignment 30. STRUCTURE OF MOLECULES AND MULTI-ATOM IONS - Part II

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Britton Randall
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1 Assignment 30 STRUCTURE OF MOLECULES AND MULTI-ATOM IONS - Part II In Assignment 10, you predicted the three-dimensional shapes of molecules and multi-atom ions based on electron dot diagrams, which explicitly depicted the number of electron pairs surrounding a central atom. In this Assignment, only the chemical formula will be given. From this formula, you will develop an electron dot diagram, which will allow you to predict the electron pair shape and molecular shape as in Assignment 10. The following step-by-step procedure explains how to deduce molecular shape from a given formula. As an example, the right hand column shows how each step would be used to determine the shape of permanganate ion, MnO 4. Deducing the shape of single center molecules and multi-atom ions from a given formula. 1. From the given formula, you should identify a central atom which is surrounded by ligands. (MnO 4 contains the center Mn atom and four surrounding O atoms.) 2. To construct the electron dot structure, an electron counting mechanism is employed: Determine the total number of electrons in the formula by adding the valence shell electrons for each neutral atom (the valence shell electrons are found from the electron dot diagram for the neutral atom). Adjust this total by adding or subtracting electrons to account for the charge on the formula. From this pool of electrons, each ligand takes 8 electrons to complete an octet (except for H, which takes only 2 electrons). Any electrons remaining from the pool constitute lone pairs on the central atom. Note that the central atom in a hypervalent compound may have more than 8 electrons ( see example 5 ). If the pool does not contain enough electrons to provide an octet on each ligand, then additional d or f inner shell electrons (not shown on electron dot diagrams) must be provided from the central atom. 3. Next draw an electron dot diagram for the formula by taking the central atom -- including any lone pairs -- and surrounding it with the correct number of ligands (each having an octet). (MnO 4 contains one Mn atom, four surrounding O atoms, and has an overall 1 charge, giving a total electron pool of 2e + 4x6e + 1e = 27e.) (To octet each oxygen ligand atom requires 4 x 8e = 32e. Subtracting the 32 required electrons from the available pool of 27e leaves a deficiency of 5 electrons (27e 32e = - 5e). Since there are no leftover electrons, there are no lone pairs on the Mn atom.) (For MnO 4 the five additional electrons required to completely octet the ligand atoms must come from the 3d orbitals of Mn.) Note: the lone pairs on the central atom are NEVER used for bonding with the ligands -- they remain completely with the central atom. 4. Once you have the electron dot diagram, analyze it for electron pair shape and for molecular (or complex ion) shape as described in Assignment 10. tetrahedral tetrahedral

2 5. If your predicted electron dot diagram shows the central atom with less than four electron pairs surrounding it, recall the treatment of this situation using π-bonding and/or resonance ideas from Assignment 10. For second row atoms -- especially C, N and O -- formation of π-bonds is a frequent and important phenomenon for completing octet structures for the central atom. For larger atoms there is only a limited or partial π-bonding contribution. EXAMPLES: + 1) NH 4 ammonium ion Center atom? N Ligand atoms? 4 H Charge? +1 (one electron gone) To "octet" ligands? 1xN = 1x 5e = 5e 4xH = 4 x 1e = 4e Ion charge (one less e) = -1e Total pool = 8e 4 x (H) = 4 x 2e = 8e (Of the total pool of 8e, 8e are used to "octet" ligands) Electrons on central atom? N = 8e - 8e = no e (N, with no lone pairs, and 4 "octeted" ligands) Electron pair shape? Tetrahedral Tetrahedral 2) BrF 3 bromine trifluoride Center atom? Br Ligand atoms? 3 F Charge? zero 1 x Br = 1 x 7e = 7e 3 x F = 3 x 7e = 21e Ion charge = no e Total pool = 28e 3x(F) = 3 x 8e = 24e (Of the total pool of 28e, 24e are used to octet ligands) Electrons on central atom? Br = 28e - 24e = 4e, or 2 lone pairs Electron Pair Shape? trigonal bipyramidal T-shaped (note position of lone pairs)

3 HYBRID ORBITALS In lecture you have learned about atomic orbitals (AOs) the regions of space surrounding an atom s nucleus that house that atom s electrons. A carbon atom has four available atomic orbitals--one 2s AO, and three 2p AOs (2p x, 2p y, 2p z ) to house its four valence electrons. Take a look at the shapes and orientations of carbon s valence AOs shown at right. If a bond is formed by the overlap of atomic orbitals, then to form a molecule of methane, CH 4, each of these C atomic orbitals must overlap with an H 1s atomic orbital to form four C-H bonds. The problem is that such an overlap would result in two different types of C-H bond (one C(2s) - H(1s) overlap and three C(2p) H(1s) overlaps) with the three C(2p) H(1s) overlaps producing C-H bonds mutually perpendicular to each other. Neither of these attributes is observed experimentally! Recall that for a CH 4 molecule, the procedures of this Assignment would predict a tetrahedral geometry in which four identical C-H bonds are spaced as far apart from each other as possible (109.5 bond angles). We can t explain CH 4 s known geometry using our localized electron model of bonding unless we allow for the possibility that the central carbon atom is capable of rearranging its individual atomic orbitals as needed to attain the most stable (lowest energy) condition for the molecule. Although the atomic orbitals in an isolated C atom are not the best for bonding, the four C atomic orbitals can be combined and rearranged (hybridized) to form four new hybrid atomic orbitals (or hybrid orbitals, HOs) which do possess the appropriate orientation for bonding (Fig. 1). In this case, the four orbitals produced are named sp 3 hybrid atomic orbitals because they result from mixing one s-type AO and three p-type AOs of the C atom. The number of AOs used to create hybrid atomic orbitals must equal the number of HOs produced! Thus, there are four sp 3 hybrid orbitals. This phenomenon is called orbital conservation. Figure 1 Note that in the CH 4 example, no π-bonds are expected in the molecule because the central atom has a complete octet. The expectation of no π-bonding also makes sense from an atomic orbital standpoint if all of carbon s p-aos are involved in creating the bonding HOs, there are no empty p-atomic orbitals left to participate in π-bonding. The idea that individual atoms can rearrange their atomic orbitals as needed to achieve the minimum energy for the molecule is an assumption of this model of bonding. The five common orbital hybridization names and the corresponding electron pair shapes are listed in the following table:

4 Number of electron pairs (σ-bonds and lone) Electron pair shape (Not appropriate) Linear Trigonal coplanar Tetrahedral Trigonal bipyramidal Octahedral Orbital Hybridization (Not appropriate) sp sp 2 sp 3 sp 3 d sp 3 d 2 NOTE: Each orbital hybridization name listed above carries built-in clues that describe the respective configuration. Merely from a hybridization name you can determine: the atomic orbitals (of the central atom) that are most affected by that hybridization the spatial dimensionality the resulting arrangement will have (is it one-, or two-, or three- dimensional?) the name given to each of the hybrid orbitals in the final arrangement in how many directions (orientations out from the nucleus) the hybrid orbitals are arranged EXAMPLES: 3) - NO 2 nitrite ion Center atom? N Ligand atoms? 2 O Charge? -1 (one electron extra) 1xN = 1 x 5e = 5e 2xO = 2 x 6e = 12e Ion charge (one extra e) = + 1e Total pool = 18e 2 x (O) = 2 x 8e = 16e (Of the total pool of 18e, 16e are used to octet ligands) Electrons on central atom? N = 18e - 16e = 2e, or 1 lone pair (N, with its lone pair, and 2 octeted oxo-ligands) trigonal coplanar angular sp 2 Note: The trigonal-planar electron pair shape indicates that nitrogen s atomic orbitals must account for three bonds in two dimensions. This requires three N AOs (2s, 2p, 2p) to combine and rearrange to form three new sp 2 hybrid atomic orbitals. Thus the hybridization is sp 2 and one π-bond is expected though it is not explicitly shown in the diagram. (Review the resonance treatment of nitrite in Assignment 10 if necessary.)

5 4) SnBr 2 Tin(II) bromide (stannous bromide) Center atom? Sn Ligand atoms? 2 Br Charge? zero 1xSn = 1 x 4e = 4e 2xBr = 2 x 7e = 14e Ion charge = no e Total pool = 18e To octet? Electrons on central atom? 2x(Br) = 2 x 8e = 16e Sn = 18e - 16e = 2e, or one lone pair trigonal coplanar angular sp 2 5) 2- SnCl 6 hexachlorostannate(iv) ion Center atom? Sn Ligand atoms? 6 Cl Charge? -2 (two electrons extra) 4e + 6 x 7e + 2e = 48e To octet? 6x(Cl) = 6 x 8e = 48e Central electrons? Sn = 48e - 48e = no e (no lone pairs) octahedral sp 3 d 2 octahedral Note: No π-bonds expected because no available empty p atomic orbitals on the central atom. All of Sn s valence p AOs, as well as its valence s AO and two of its n-1 d AOs, are involved in a rearrangement to produce six hybrid sp 3 d 2 hybrid orbitals

6 6) - I 3 Triiodide ion Center atom? I Ligand atoms? 2 I Charge? -1 (one electron extra) 3x7e + 1e = 22e 2x(I) = 2 x 8e = 16e Electrons on central atom? I = 22e - 16e = 6e (three lone pairs) trigonal bipyramidal sp 3 d linear (lone pairs MUST be placed equatorially in order to obtain correct shape) 7) +2 UO 2 uranyl(vi) ion (dioxouranium(vi) ion) Center atom? U Ligand atoms? 2 O Charge? +2 (two electrons gone) 2e + 2 x 6e - 2e = 12e 2x(O) = 2 x 8e = 16e Electrons on central atom? U = 12e - 16e = -4e (NO lone pairs) 4 electrons MUST come from n-1 d-sub-shells of the uranium atom linear sp linear Note: two π-bonds predicted (but their effectiveness is limited by the large size of uranium)

7 8) - SbCl 4 tetrachloroantimonate(iii) ion Center atom? Sb Ligand atoms? 4 Cl Charge? -1 5e + 4x7e + 1e = 34e 4x(Cl) = 4 x 8e = 32e Electrons on central atom? Sb = 34e - 32e = 2e (one lone pair) trigonal bipyramidal sp 3 d teeter The lone pair on Sb is equatorial, NOT at a axial (up or down) position. 9) SO 3 sulfur trioxide Center atom? S Ligand atoms? 3 O Charge? zero 6e + 3 x 6e = 24e 3x(O) = 3 x 8e = 24e Electrons on central atom? S = 24e - 24e = no e (no lone pairs) trigonal coplanar sp 2 trigonal coplanar Note: One π-bond is likely. Each O ligand will share more of its electron octet with the S to provide the central atom with an octet without alteration of the shape. The π-bond is delocalized over all four atoms; one-third of the π-bond comes from each oxygen ligand

8 Problem Set 30 Be able to define the following terms: hybrid atomic orbital orbital conservation hybridization name For each of the following formulas: a) construct the electron dot diagram; b) tell what hybridization applies to central atom; c) give diagram and name of both the electron pair shape and the molecular shape; d) tell how many π-bonds, if any, the actual molecule should contain. (Follow the procedures outlined in the examples. Mark electron dots clearly.) 1) NF 3, nitrogen trifluoride 2) VO 2 +, vanadyl(v) ion 3) IF 4, tetrafluoroiodate(iii) ion 4) SnCl 6 2, hexachlorostannate(iv) ion 5) HgCl 3, trichloromercurate(ii) ion 6) TeCl 4, tellurium tetrachloride 7) HOCl (O is central), hypochlorous acid 8) ClNO (N is central), nitrosyl chloride 9) H 2 CO (C is central), formaldehyde Do the same for these multi-centered species: H 10) CH 3 CH 2 OH (bonding sequence is H 3 C C OH), ethanol. H Determine the number of electrons in the electron pool of these molecules & complex ions: CO 2 HCN (C is central) XeF 6 NO 2 VOCl 3 (V is central) NH 4 + PO 4 3 SO 3 2 PCl 4 + NO 3 BH 4 N

9 Partial answers to P.S. 30: Check your accuracy against the following conclusions. Note: These are not complete answers; yours should also show your reasoning process e.g., the calculation of the electron pool, and accurate depictions of the dot-picture and the electron-pair arrangement (in appropriate geometric style) and the geometric rendition of the entire molecule's shape. The diagrams then serve as basis for deriving the conclusions below. Complete answers may be viewed on the posted answer key on the Chemistry 111 web page. 1) NF 3 has total of 26 electrons in the valence pool, resulting in 1 lone pair on the N; electron-pair shape is tetrahedral (corresponding to sp 3 hybridization); molecular shape is pyramidal; no π-bonding in the final molecule (for the central N already has an octet). 2) VO 2 + has a total of only 13 electrons in the valence pool, so no lone pairs can exist on the V; indeed, just to satisfy ligand octets, V must contribute three d electrons (from its inner shell) to the bonding pool; resulting electron-pair shape is linear (sp hybridization); molecular shape is linear; 2 π-bonds are plausible (but of limited effectiveness, since vanadium isn't a "second row" element). 3) IF 4 has a total of 36 electrons in the valence pool, thus 2 lone pairs remain on the I; electron-pair shape is octahedral (sp 3 d 2 hybridization); molecular shape is square co-planar, with the lone pairs of iodine located opposite from each other; no π-bonding in the final molecule (the central I already has more than an octet). 4) SnCl 6 2 has 48 e ; no lone pr on Sn; octahedral e - prs (sp 3 d 2 hybridization); octahedral mol.shape; no π-bonding. 5) HgCl 3 has 24 e ; no lone pr on Hg; trigonal co-planar e - prs (sp 2 ); trigonal co-planar mol.shape; 1 π-bond is plausible (but of limited effect, since neither element is on"second-row" of P.Table). 6) TeCl 4 has 34 e ; 1 lone pr on Te; trigonal bipyramidal e - prs (sp 3 d); teeter mol.shape; no π-bonding; the lone pair on Te must be placed equatorially, not at a polar position. 7) HOCl has 14 e ; 2 lone prs on O; tetrahedral e - prs (sp 3 ); angular mol.shape; no π-bonding. 8) ClNO has 18 e ; 1 lone pr on N; trigonal co-planar e - prs (sp 2 ); angular mol.shape; 1 π-bond is necessary. 9) H 2 CO has 12 e ; no lone pr on C; trigonal co-planar e - prs (sp 2 ); trigonal co-planar mol.shape; no π-bonding. 10) CH 3 CH 2 OH has three central atoms: C 1 (at left), C 2 (at center), O(at right). Around C 1, no lone pr on C 1 ; tetrahedral e - prs (sp 3 ); tetrahedral mol.shape; no π-bonding. Around C 2, no lone pr on C 2 ; tetrahedral e - prs (sp 3 ); tetrahedral mol.shape; no π-bonding. Around O, 2 lone prs on O; tetrahedral e - prs (sp 3 ); angular mol.shape; no π-bonding. (To design an overall molecular diagram, tilt the tetrahedral shapes so that C-C-O "backbone" lies flat.) Electron pools of the molecules: 16e ; 10e ; 50e ; 17e (there is an unpaired e on N here); 32e. Electron pools of complex ions: 8e ; 32e ; 26e ; 32e ; 24e ; 16e ; 8e

8.2 Hybrid Atomic Orbitals

8.2 Hybrid Atomic Orbitals 420 Chapter 8 Advanced Theories of Covalent Bonding Answer: (a) is a π bond with a node along the axis connecting the nuclei while (b) and (c) are σ bonds that overlap along the axis. 8.2 Hybrid Atomic