Why are the two bonds in sulphur dioxide identical? Triatomic molecules are either linear or bent. If we analyse the $\ce{SO2}$ molecule,  it turns out that it is bent.
I know that $\pi$ bonds do not alter shape, but merely the bond lengths.
Now since one of the bonds between S and O is a p$\pi$-p$\pi$ bond and the other bond is a d$\pi$-p$\pi$ bond, the two $\pi$ bonds are not same, so the bond length should not be the same, but my book says the bond lengths are the same which is quite puzzling. I thought maybe its due to resonance but there's no way resonance takes place here.  I read this question,but it don't answer many questions as:
Why d-orbitals of sulphur cannot be used  ?
And  in answer by @Kenny Lau ,
what's specify reason for 'Sulfur sharing two electrons among itself and the other two oxygen atoms.'
So, that makes me think then, what could  be the reason for same bond length in $\ce{SO2}$?
It is not clear as to why d-orbitals of sulphur cannot be used and in the answer to this question @Kenny Lau does not specify reason for 'Sulfur sharing two electrons among itself and the other two oxygen atoms.'
There are two ways to look at the $\ce{SO2}$ molecule. The valence bond way is this:

$\ce{SO2}$ is a bent molecule with C2v symmetry point
group. A valence bond theory approach considering just s and p
orbitals describes the bonding in terms of resonance between two
resonance structures:

The sulfur–oxygen bond has a bond order of 1.5 and does not invoke d
orbital participation. In terms of electron-counting formalism, the
sulfur atom has an oxidation state of +4 and a formal charge of +1.
(Wikipedia)

However, there is a molecular orbital approach which seems to give a deeper appreciation of the bonding. The picture is this:

(Picture source with detail explanation)
This is my simple explanation:

The sulfur is hybridized sp$^2$-p. The bond angle is 119$^o$ - you
can hardly get closer to 120$^o$!

We will assume that the oxygens are also hybridized sp$^2$-p.


Then the sigma bonding is designated in the diagram by the numbers 1 and 2 for the sigma bonds between sulfur and oxygen. Lone pairs are designated by numbers 3 (for S), and 4, 5, 6 and 7 (for O).
The pi system has three molecular orbitals, one from each atom and four electrons: one from each oxygen, two from sulfur. The lobes in these orbitals are drawn at the lower right side of the picture. Orbital 8 has the lowest energy, with all lobes bonding. The non-bonding pi orbital (#8) has reversed signs for the lobes of oxygens and therefore a node at sulfur. The antibonding pi orbital (#10) has two nodes and is unoccupied.
So the final count is two filled sigma bonds and one filled pi bond; five lone pairs and one filled non-bonding pi orbital. The formal bond order is 1.5, same as in the valence bond explanation with resonance.
The only thing I might change in the diagram is the relative energies of the sulfur orbitals and the oxygen orbitals: I would think they might be closed to the same energy. The diagram might be drawn as it is in order to make it clearer and give more room for all the numbers.
We just don't need d orbitals to explain anything, but their involvement cannot be entirely eliminated.