Calculate the bond order for $NO, NO^+, CN, CN^-$ and $CO$.
(N/A) The bond order is calculated using the formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
| Species | Bond Order |
| $NO$ | $2.5$ |
| $NO^+$ | $3.0$ |
| $CN$ | $2.5$ |
| $CN^-$ | $3.0$ |
| $CO$ | $3.0$ |
Explore More
Similar Questions
Correct decreasing order of bond order of oxygen species in $I$ to $IV$:
$(I)$ $Na_2O_2$
$(II)$ $O_2[AsF_6]$
$(III)$ $CsO_2$
$(IV)$ $O_2$
$(I)$ $Na_2O_2$
$(II)$ $O_2[AsF_6]$
$(III)$ $CsO_2$
$(IV)$ $O_2$
Medium
View SolutionThe number of antibonding electron pairs in $O_2^{2-}$ molecular ion on the basis of molecular orbital theory is
Difficult
View SolutionWhich of the following has the minimum bond length?
MediumAIPMT 2011
View SolutionWhich one of the following is paramagnetic?
MediumWBJEE 2011
View SolutionThe number of electrons that are paired in an oxygen molecule $(O_2)$ is:
MediumIIT 1995
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo