See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Oxidation state is determined by assigning electrons in bonds to the more electronegative atom and summing the formal charges. Step 2 - Oxidation state of Os in 7B (OsO4): OsO4 has osmium bonded to four oxygen atoms via double bonds. Each oxygen has oxidation state -2. The molecule is neutral, so: Os + 4(-2) = 0, giving Os = +8. Step 3 - Oxidation state of Os in 7C (osmate ester): In 7C, the osmium is chelated by two oxygen atoms from the diol (single-bond Os-O linkages, each O carries -2) and retains two oxo (Os=O) ligands (each O carries -2). The total oxygen count bonded to Os is four, each with -2. However, in the osmate ester the two O atoms that were previously double-bonded oxygens in OsO4 are now single-bonded (ester-type) oxygens still assigned -2, but Os has been reduced by two electrons (one from each new C-O bond formed). The complex is an Os(VI) species: Os + 4(-2) + overall charge considerations for the chelate ring (the two alkoxide oxygens are each -2, and two oxo oxygens are each -2, total -8 from O; for a neutral complex Os = +8 would apply, but the actual product is an Os(VI) osmate ester with formal charge on Os = +6 because the two new O-C bonds donate electron density and the product is the cyclic osmate(VI) ester). The standard result from organometallic/organic chemistry: OsO4 is Os(+8), and the osmate ester intermediate is Os(+6). This is consistent with osmium being reduced from +8 to +6 during the dihydroxylation reaction (a [3+2] cycloaddition), gaining two electrons from the alkene pi bond. Step 4 - Matching to options: 7B: Os = +8, 7C: Os = +6 → answer is (b) 8, 6. Step 5 - Why other options fail: (a) 6, 8 reverses the values incorrectly. (c) 6, 6 gives wrong value for OsO4. (d) 8, 8 ignores the reduction of Os during osmate ester formation. Therefore, the correct answer is B.