JEE Advanced 2018Paper 1
- Exam
- JEE Advanced
- Year
- 2018
- Paper
- Paper 1
- Subject
- Chemistry
- Chapter
- Chemical Bonding and Molecular Structure
JEE Advanced 2018 · Paper 1 · Chemistry · Chemical Bonding and Molecular Structure
Among the species given below, the total number of diamagnetic...
Among the species given below, the total number of diamagnetic species is ___.
Hatom, N{O}_{2}monomer, {O}_{2}^{-}M{n}_{3}{O}_{4}, {\left(N{H}_{4}\right)}_{2} [ FeC{l}_{4} ] , {\left(N{H}_{4}\right)}_{2} [ NiC{l}_{4} ] , {K}_{2}Mn{O}_{4}, {K}_{2}Cr{O}_{4}





Hatom, N{O}_{2}monomer, {O}_{2}^{-}M{n}_{3}{O}_{4}, {\left(N{H}_{4}\right)}_{2} [ FeC{l}_{4} ] , {\left(N{H}_{4}\right)}_{2} [ NiC{l}_{4} ] , {K}_{2}Mn{O}_{4}, {K}_{2}Cr{O}_{4}
Correct answer
0
Step-by-step solution
H - atom = paramagnetic
odd unpaired electron species paramagnetic
(superocide) = One unpaired electrons in M.O. paramagnetic
(in vapour phase) = same as , two unpaired are present in M.O. paramagnetic