Actual Degree of Freedom of Diatomic Molecule Ok, I have 2 very different values for degree of freedom(DOF) of diatomic molecules arising due to the difference in the vibrational DOF of the diatomic molecules.
According to this DOF wiki page:-
Under Gas molecules section, one can see the Vibrational DOF of linear molecules, which is $3N-5$.
So, putting $N=2$, Vibrational DOF(Diatomic Molecule)=1 and so,
Total DOF(Diatomic Molecule)=6
In other wiki page namely, Molecular Vibration:-
It says

A diatomic molecule has one normal mode of vibration.

But my textbook says otherwise. It's says the vibration contributes 2 to total DOF of diatomic molecule and so, Total DOF(Diatomic Molecule)=7
Here's the image:-

A VIDEO also says as per my textbook.
Look under the Syed Sumaid comment. There he explained why vibrational DOF of diatomic molecule should be 2 not 1(According to him).
I am not able to understand which one is delivering the right information.
Is it 7 DOF as per my book/that video or 6 DOF as per the wiki links I have given? 
 A: The term 'degrees of freedom' is ambiguous.
In dynamics, and actually in most areas, it means the number of independent parameters needed to describe the system. These can be expressed in different ways (such as cartesian or polar co-ordinates) but the number is always the same.  So a diatomic molecule has 6. (If the bond were rigid it would have 5.)
In thermodynamics a 'degree of freedom' is a quadratic term in the energy. Each of these 'degrees of freedom' contributes ${1 \over 2} kT$ to the energy (classically). 
Often these match. A particle described by a co-ordinate $x$ (dynamic d.o.f.)  has a kinetic energy contribution ${1\over 2} m \dot x^2$ (thermal d.o.f.), likewise $y$ and $z$, likewise angles and inertia. But for a spring, which corresponds to 1 (dynamic) degree of freedom there are two (thermal) degrees of freedom from two quadratic terms ${1 \over 2 } \mu \dot \xi^2+{1\over 2} k \xi^2$.
Same story for polyatomic molecules but even more complicated.
So  diatomic molecule has 6 dynamic degrees of freedom and 7 thermal degrees of freedom.  If a test paper asks 'How many d.o.f. has a diatomic molecule?' check the title of the paper before answering.      
A: The textbook is correct, there are 7 degrees of freedom if the molecules are vibrating. The statement "a diatomic molecule has one normal mode of vibration" is true, however this vibrational energy has a kinetic energy $K=\frac{1}{2}m\dot\eta^2$ and a potential energy $U=\frac{1}{2}k\eta^2$ such that $\epsilon_v=K+U$. Two degrees of freedom from the vibration plus three degrees of freedom $\epsilon_t=\frac{1}{2}m(v_x^2+v_y^2+v_z^2)$ for the $(x,y,z)$ translational kinetic energy plus two degrees of freedom $\epsilon_r=\frac{1}{2}I_y\omega_y^2+\frac{1}{2}I_z\omega_z^2$ for the rotational kinetic energy adds up to seven total degrees of freedom. The rotational kinetic energy $\frac{1}{2}I_x\omega_x^2$ along the axis of the bond is zero because for diatomic molecules $I_x\approx0$.
