Question

2) A first-row transition metal complex of the formula (M(OH2)612+ has a magnetic moment of 3.9 MB. Determine the most likely number of unpaired electrons and the identity of the metal(s).

inorganic chemistry

Answer 1

im Answer: 1st dow it is 30 transition metal indicates that contains 3d elections. [ m(H₂O) - As J20 is a nutoul ligand, then, it is met Also give the MB2 3.9 BM of the metal, We Janow that, Ур ч учC+2) Гя может я не ne number of the unpaired ebedaon of the centrul atom 7 Now 3.g. ſncnta) n(n+2) = 15.21 n+ 2n-15.2120 According to gridrus Acharto sule: - Rule:- 'If the equation ] nz -2 + 2+ 4.15:21 AX+ Bx + C zo Then Xz-B I/B CAC 27 2-2+48.05 a zona 5 2 3.025 00 - 5.025 As the unpaired election can not be negative, then The number of unpair electrona 3 Short trick to calculate the B.M for MCQ or entrance exami- If the * The number before the point is the od MB value is the number of unpaired election Example: If "B & B.9, unpaired election. 3, If M3 2.89 BM, Then uporired electrone 4 If hp. 5 SBM, unpaired electoon en

Then me ras 3 unpaired electoon. I The two electron acrore from 3 d orbe 48 Oobital. So mat has only 3d electons. These are two possibility in i) dº system. The If it d system then the metal is Vanadium CV) Electronic Configаration:- 1525 206 35 3p645 3 2 For v 2+ = 1525 206 35 36 3d If it dr system, then the metal in Cobalt (Co) · Electronic configuration - 15 25 28 35 30° 45 327 For Co2t :- 15*25 206 35 3p% 327