If an electron is removed from the anti-bonding orbital, then it will tend to increase the bond order.
The HOMO in NO and O2 is antibonding molecular orbital .
Hence, in NO and O2 bond order will increase on loss of electron.
So, the correct option is (C).
Identify the correct orders against the property mentioned:
A. H$_2$O $>$ NH$_3$ $>$ CHCl$_3$ - dipole moment
B. XeF$_4$ $>$ XeO$_3$ $>$ XeF$_2$ - number of lone pairs on central atom
C. O–H $>$ C–H $>$ N–O - bond length
D. N$_2$>O$_2$>H$_2$ - bond enthalpy
Choose the correct answer from the options given below:
Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
The magnitude of heat exchanged by a system for the given cyclic process ABC (as shown in the figure) is (in SI units):

Covalent bonds can be characterized on the basis of several bond parameters such as bond length, bond angle, bond order, and bond energy (also known as bond enthalpy). These bond parameters offer insight into the stability of a chemical compound and the strength of the chemical bonds holding its atoms together.
For example, The H—H bond enthalpy in hydrogen is 435.8 kJ mol-1. \
Bond order of H2 (H —H) =1
Bond order of 02 (O = O) =2
Bond order of N2 (N = N) =3
Read More: Chemical Bonding and Molecular Structure