上海交通大学：《大学化学》教学资源_加州理工课件学习_Series14The18ElectronRule

The 18 Fule Rule References:Gray:chapter 5 OGC:chapter 8 XIV-1

The 18 Electron Rule References: Gray: chapter 5 OGC: chapter 8 XIV-1

Element Groups Alkali metals Inert or Noble gases Alkali earths Halogens I He g Transition metals B C N F Ne Al Si P S Ci Ar Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr b Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te Xe Ba La Hf Ta W Re Os Ir Pt Au Hg TI Pb Bi Po At Rn Fr Ra Ac Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Lanthanides Actinides XIV-2

H L i N a K R b C s F r B e M g C a S r B a R a S c Y L a A c T i Z r H f V N b T a C r M o W M n T c R e F e R u O s C o R h I r N i P d P t C u A g A u Z n C d H g B A l G a I n T l C S i G e S n P b N P A s S b B i O S S e T e P o F C l B r I A t H e N e A r K r X e R n C e P r N d P m S m E u G d T b D y H o E r T m Y b L u T h P a U N p P u A m C m B k C f E s F m M d N o L r Element Groups Alkali metals Transition metals Alkali earths Halogens Inert or Noble gases Lanthanides Actinides XIV-2

The Chemists Geoffrey Wilkinson E.O.Fischer Nobel Prize,chemistry,1973 "For their pioneering work,performed independently,on the chemistry of the organometallic,so called sandwich compounds" XIV-3

Geoffrey Wilkinson E. O. Fischer Nobel Prize, chemistry, 1973 “For their pioneering work, performed independently, on the chemistry of the organometallic, so called sandwich compounds” The Chemists XIV-3

18 e Rule for Transition Elements Remember the dot structure for CO? -1 +1 C∷O:> :C≡O:> ∷C=O: Consider Cr,a transition element: Chromium:[Ar](4s)2(3d)4-6 valence e Like most transition elements,Cr needs 18 e in its shell. O 12 e from CO's +6 e from Cr =18 total e;formula is Cr(CO)e Dot structures can predict C≡O: molecules;if we are given :O蕊C8 an exception,we can explain its existence,and figure out some of its chemical properties. 4

Cr 18 e- Rule for Transition Elements • Remember the dot structure for CO? C O C O • Consider Cr, a transition element: Chromium: [Ar] (4s)2 (3d)4 → 6 valence e- • Like most transition elements, Cr needs 18 e- in its shell. O C C O C O C O 12 e- from COʼs + 6 e- from Cr = 18 total e- ; formula is Cr(CO)6 Dot structures can predict molecules; if we are given an exception, we can explain its existence, and figure out some of its chemical properties. C O -1 +1 XIV-4

Using the 18e rule Given that H2Fe(CO)x exists,what does x equal? Iron:[Ar](4s)2(3d)6 -8 valence e Hydrogen:(1s)1 →1 valence e Fe wants to have 18 e,because it's a transition element,but it only has 8.The H's give 2 e,but we still need 8 electrons. Since each CO supplies 2 e,there must be 4 CO's: O H2Fe(CO)4 Ⅲ O CO: R. H -H OC: XIV-5

Using the 18 e- rule • Given that H2Fe(CO)x exists, what does x equal? Iron: [Ar] (4s)2 (3d)6 → 8 valence e￾Hydrogen: (1s)1 → 1 valence e- • Fe wants to have 18 e- , because itʼs a transition element, but it only has 8. The Hʼs give 2 e- , but we still need 8 electrons. Since each CO supplies 2 e- , there must be 4 COʼs: C O C O H H H2Fe(CO)4 Fe C O H H C O Fe XIV-5

Dimer-Forming Transition Elements Given that Mn(CO)5 exists,find its chemical properties: Manganese:[Ar](4s)2(3d)5-7 valence e 5 CO's provide 10 electrons to Mn,leaving Mn with 17 total e; but Mn wants 18 electrons.So,Mn forms a dimer: CO CO co co Mn. Mn. C0二 Mn co co CO co CO CO CO CO CO Mn2(CO)10 Halides,like fluorine,also act this way,because they also need only one electron to fill their shell.There are other similarities between transition elements with 7 valence e and halides... XIV-6

Mn Mn Dimer-Forming Transition Elements • Given that Mn(CO)5 exists, find its chemical properties: Manganese: [Ar] (4s)2 (3d)5 → 7 valence e- • 5 COʼs provide 10 electrons to Mn, leaving Mn with 17 total e- ; but Mn wants 18 electrons. So, Mn forms a dimer: Mn Mn • Halides, like fluorine, also act this way, because they also need only one electron to fill their shell. There are other similarities between transition elements with 7 valence e- and halides... Mn2(CO)10 CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO XIV-6

Transition Metals That Are Like Halides As we have seen,transition elements with 7 valence e (like Mn and Re)which are bonded to 2e donors (like CO) form dimers,because they need only one extra e. Another similarity is reactivity with light: Br2 hv→2Br Mnz(CO)o-hv2Mn(CO) Another similarity is a phenomenon called "coupling": Br+ Br-l co cO cO CO- Re, cO co co Mn(CO)5 Re(CO)5 CO XIV-7

Transition Metals That Are Like Halides • As we have seen, transition elements with 7 valence e- (like Mn and Re) which are bonded to 2 e- donors (like CO) form dimers, because they need only one extra e- . • Another similarity is reactivity with light: Br2 2 Br Mn2(CO)10 2 Mn(CO)5 • Another similarity is a phenomenon called “coupling”: Br + I Br—I Mn(CO)5 + Re(CO)5 Mn Re CO CO CO CO CO CO CO CO CO CO hν hν XIV-7

Another Transition Element Structure Given that Osa(CO)12 exists,what is its structure? Osmium:[Xe](6s)2(5d)6-8 valence e By symmetry,there must be 4 CO's attached to every Os. That would give us 8+4(2)=16 e for each Os.But each Os needs 2 more e to make 18.So the Os's can form a triangle, with each Os contributing 2 valence e's to the single bonds: COCO CO CO co 。。 :Os: :Os: CO Os: Os .Os: CO CO CO CO CO CO CO CO CO CO COCO XIV-8

Another Transition Element Structure • Given that Os3(CO)12 exists, what is its structure? Osmium: [Xe] (6s)2 (5d)6 → 8 valence e- • By symmetry, there must be 4 COʼs attached to every Os. That would give us 8 + 4(2) = 16 e- for each Os. But each Os needs 2 more e- to make 18. So the Osʼs can form a triangle, with each Os contributing 2 valence e- ʼs to the single bonds: Os Os Os CO CO CO CO CO CO CO CO CO CO CO CO Os Os Os CO CO CO CO CO CO CO CO CO CO CO CO XIV-8

What's the Best Way to Count Things? 1 e donor:Anything that has one e that is not in a bond.Examples: H :F ●● 2 e donor:Anything that has two e's that are not in a bond (called a "lone pair").Examples: H-N-H CO (the one we have been using) H 3 e donor:Anything having three e's to spare.Example: The allyl radical: This bond can donate H two e to a metal:So this radical is either a 1 or a 3e donor Radical:atom or molecule with an incomplete valence shell,making it very reactiveiv-9

Whatʼs the Best Way to Count Things? • 1 e- donor: Anything that has one e- that is not in a bond. Examples: H F Cl CH3 • 2 e- donor: Anything that has two e- ʼs that are not in a bond (called a “lone pair”). Examples: CO H N H H (the one we have been using) • 3 e- donor: Anything having three e- ʼs to spare. Example: The allyl radical: C C H C H H H H This bond can donate two e- to a metal; So this radical is either a 1 or a 3 e- donor Radical: atom or molecule with an incomplete valence shell, making it very reactiveXIV-9

How Many e Do We Want Donated? It depends on the specific case.For example, allyl-Mn(CO)s-Mn(CO)5 has 17 e,so we want 1 e donated ( → Mn: H H On the other hand,if we have... allyl-Mn(CO)4 we now need 3 e from the allyl radical Co H木/H co H co Mn' + co H-C Mn' co H CO XIV-10

How Many e- Do We Want Donated? • It depends on the specific case. For example, allyl—Mn(CO)5 → Mn(CO)5 has 17 e- , so we want 1 e- donated • On the other hand, if we have... allyl—Mn(CO)4 Mn C C H C H H H H Mn C H C H H C H H we now need 3 e- from the allyl radical Mn C C H C H H H H Mn C C H C H H H H CO CO CO CO CO CO CO CO + + XIV-10