InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The correct order of bond order values among the following (i) `NO^(-)` (ii) `NO^(+)` (iii) `NO` (iv) `NO^(2+)` (v) `NO^(2-)`A. (i) lt (iv) lt (iii) lt (ii) lt (v)B. (iv) = (ii) lt (i) lt (v) lt (iii)C. (v) lt (i) lt (iv) = (iii) lt (ii)D. (ii) lt (iii) lt (iv) lt (i) lt (v) |
|
Answer» Correct Answer - C `NO(7+8=15e^(-)s)`. Its electron configuration is `KK(sigma2s)^2(sigma^**2s)^2(sigma2p_z)^2(pi2p_x^2=pi2p_y^2)(pi^**2p_x^1=pi^**2p_y^0)` `NO: bo=(10-5)/(2)=2.5` In `NO^(-)`, there is one more `e^(-)` which enters `pi^**2p_y` orbital, thus, `NO^(-): bo=(10-6)/(2)=2` In `NO^(2-)`, there are 2 more `e^(-)s` which enter into antibonding orbitals, thus, `NO^(2-): bo=(10-7)/(2)=1.5` In `NO^(+)`, one `e^(-)` is removed from antibonding orbital: thus, `NO^(+): bo =(10-4)/(2)=3` In `NO^(2+)` one `e^(-)` is lost from bonding `MO` while another `e^(-)` is lost from antibonding `MO`. Thus, `NO^(2+): bo=(9-4)/(2)=2.5` Thus, the correct order of bond order is `NO^(2-) lt NO^(-) lt NO = NO^(2+) lt NO^(+)` i.e., `(v) lt (i) lt (iii) = (iv) lt (ii)` |
|