Abstract Reaction of HMo(CO) 3C 5H 5 and Sn(C 5H 5) 2 produces the tin hydride HSn[Mo(CO) 3C 5H 5] 3 (I). Reaction of I with CCl 4, CHCl 3, or CH 2Cl 2 gives ClSn[Mo(CO) 3C 5H 5] 3 (II). With hydrogen chloride the hydride I reacts to produce the dichloride Cl 2Sn[Mo(CO) 3C 5H 5] 2. The first step in this reaction is cleavage of the SnH bond to produce the chloride II. The hydride I reacts with acetic acid to produce the diacetate (CH 3COO) 2Sn[Mo(CO) 3C 5H 5] 2.