Question

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Subject: Inorganic Chemistry

Question: COUNT the number of valence electron per metal of the reaction products. Are they obeying the 18 valence-electron rule? Give comments.

Note:
bipy: Bipyridine
dppe: bis(diphenylphosphino)ethane

(a) (PR3)2 Ir (CO) Cl + dppe — Ir (dppe)2 CI + 2PR3 + CO (b) (PR3)2 Ir (CO) CI + CO - (PR3)2 Ir (CO)2 CI (C) V(CO)6 + Na [V(CO)6] + Nat (d) (15 - C5H5)2 Ni + NO — ņ5-C5H5 Ni (NO) + C5H5 (e) (C5H5)2 Ni Na/Hg 95 - C5H5 Ni (n3 - C5H7) EtOH

Answer 1

Yes, these all 5 products follows the 18 electron - valance count rule. The explanation of contribution of electrons by every metal, ligands and total valence electron count of complex is in the following 3 pages. The page number is marked on the paper.
| Page No. 1 - - product Part (i) (ii) Question No. of valonce election per metal of the reaction Are they obeying the 18 valence election rule? Ans :- (0) (Pro). Inco + dppe Irlappe). Ce + 2.PR.3 + co Solution : (i) Ir a transition mutal of 9th group @Aler electronic (onliaunchen 2 (Xe] + 5d 6 =) metal have total valonce é =/9 Aug ci contribution by her element in complex UO * dppe Izle LG PA | Luhau nhº nhung gia || 'Ph contribute = 4 - - Complex = = [ Ir (appe). W] 9+ (4) X2 +1 9+8tt it follows 18 é sule.

Pace Page No. 2) (PR3) Ir (colace (6) (PR3) e Irlo ut co Solution :- (i) Ir a 9 valance é explained in part (a). - PR₂ = 2 è contribution. " O PR - have a pair of e to contribute = 2 e contribution. complen Total valance é [ (PR3)2 Iy (cola el 2x2)+9+ 2x2) + 1 4 +9+4+1 = 18- valence-electrons on complex (c) UCCO) + Na [uccodo] + Nat Solution - - ci) V 5th group transition metal e electronic configuration Rell [Ar]3d , 45² - tatal valonce é in V 5 - [Uccolo 5+ 6x0) += 18 tatal valonce count: on complex (d) (45_Cs Hs ) a Ni + No n s - Cats Nic No) +CHS c) Ni => [Ar] 3. 4g? - 10 total valence on sitel cii [n5 CG Hs] = 5 é contribution bcoz metal will bind all 5 Carbon atoms.

NO - [Page No - 3 No contributes by a types ca If no is bind with metal in linear shape - N=0 = 3 é contribution bind with metal in bent shape O M = only é contributios. Here in this complex Ni metal form 18 e complex always (it is known) - [ Ni Cys-CsHs) No] 10 + 5+ 3 = 10 total valansee ei oo complex - CCS Hs) a Ni Nang (y5 – Cs Hs) Nil n3EC5 H₂) Etoh co Ni metal contribution lo é 12 = 3e contribution. cin IT In shows that metal is binding only with 3 Carbon atoms that's why é - contribution by this => Ⓡ. [ (ns. GHS) Ni Cy3 - CsH7)] 5 + 10 + 3 =118 total valance e