To find the hybridisation of nitrogen in different species, we use this rule:
| Number of regions | Hybridisation | Geometry |
|---|---|---|
| 2 | sp | Linear |
| 3 | sp2 | Trigonal planar |
| 4 | sp3 | Tetrahedral |
1) NO3− (Nitrate ion):
2) NO2 (Nitrogen dioxide):
3) NO2− (Nitrite ion):
4) NO2+ (Nitronium ion):
Conclusion: Only NO2+ has nitrogen in sp hybridisation.
Correct answer: (C) NO2+

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:
A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of:

Such a group of atoms is called a molecule. Obviously, there must be some force that holds these constituent atoms together in the molecules. The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond.
There are 4 types of chemical bonds which are formed by atoms or molecules to yield compounds.