1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Column-I contains four entries and Column-II contains five entries. Entries of Column-I are to be matched with some entries of Column-II. One or more than one entries of Column-I may have the matching with the same entries of Column-II

Open in App
Solution

## (A) CN(13);σ1s2σ∗1x2σ2s2σ∗2s2π2px2π2py2σ2pz1Bond order =9−42=2.5; paramagnetic, μBM=1.73HOMO=σ2pzLUMO=π∗2px or π∗2pyCN+ bond order =8−42=2;CN− bond order =10−42=3(B) OF17);σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2p2x=π2p2uπ∗2px=π∗2p1yBond order =10−72=1.5; paragmanetic, μBM=1.73HOMO=π∗2p orbital; OF+ bond order =10−62=2;OF− bond order =10−82=1(C) BN(12)=σ1s2σ∗1s2σ2s2σ∗2s2π2p2x=π2p2yBond order =8−42=2;diamagnetic,μBM=0BN+ bond order =7−42=1.5;BN−=9−42=2.5HOMO=π2p(D) NO(15)=σ1s2σ∗2s2σ2s2σ∗2s2σ2p2zπ2p2x=π2p2yπ∗2p1xBond order =10−52=2.5paramagnetic,μBM=1.73HOMO=π∗2pNO+ bond order 10−42=3;NO− bond order =10−62=2O−2 has one unpaired electron, so μBM=1.73

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Enthalpy
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program