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I am considering $z$-axis as the internuclear axis in all cases.

When we consider overlap of s and p orbital to form a $\sigma$-bond, the chosen orbital must be p$_z$ orbital for a proper overlap.

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Now, when we consider two hybridised sp orbitals, their properties are different from s or p orbitals. And I think they can reorient themselves in any direction to form bonds.

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Is there any particular direction of this hybrid sp-orbital w.r.t. coordinate axes?

(i) First, take the case of $\ce{BeCl2}$. In the excited state of $\ce{Be}$, there is only one $\ce{e-}$ in the p-orbital. Can any p-orbital participate in hybridisation to form two sp hybridised orbitals?

But these two sp hybridised orbitals must overlap with two p$_z$ orbitals of two $\ce{Cl}$ atoms to form two sp-p$_z$ $\sigma$-bonds.

(ii) Secondly, take the case of $\ce{NO2-}$, in which $\ce{N}$ is sp$^2$ hybridised. In this case, the remaining p-orbital with $\ce{N}$ atom must be a p$_x$ or p$_y$ orbital which will form a $\pi$-bond with p$_x$ or p$_y$ orbital (respectively) of $\ce{O}$ atom. As two p$_z$ orbitals cannot form a $\pi$-bond.

Thus, in the three hybridised sp$^2$ orbitals, one must be a p$_z$ orbital. And one of the half-filled p-orbital of $\ce{O}$ must be a p$_z$ orbital to form sp$^2$-p$_z$ $\sigma$-bond?

Apurvium
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    This concept of the “internuclear axis” always being z (which is the basis of your claim that sigma bonds must be formed from pz) is fine for linear molecules, but breaks down completely for anything nonlinear like NO2. The bonds point along different axes; clearly, they can’t all be z at the same time. – orthocresol Jul 20 '21 at 10:54
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    You need $d_{x^2 - y^2}$ $d_{x^2 - y^2}$. And that comment is implicitly assuming that there's only one bond under discussion, i.e. a diatomic molecule, which is necessarily linear. As long as you only want to talk about one bond, it's perfectly fine to call that axis $z$. If you want to talk about two bonds at once, you can't call both of them $z$ at the same time. – orthocresol Jul 20 '21 at 17:20
  • @orthocresol Understood. But then could not understand your comment to this question: {d${xz}$, d${yz}$}→π , and {d${xy}$, d${x^2−y^2}$}→δ; Why these particular orbitals would form these bonds? – Apurvium Jul 20 '21 at 17:28
  • @orthocresol OK. But then one more confusion, why do we choose d$_{z^2}$ orbital for $\ce{sp_3d}$ hybridisation and likewise there are other examples... Does it anything to do with the orientation of lobes, which are along the axis? – Apurvium Jul 21 '21 at 03:37
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    Well, why not? In NO2, it’s perfectly permissible to choose one of the N-O bonds to be the z-axis, so you can indeed say that that bond is formed from s and p_z. You just can’t also claim that the other bond is also made from p_z. With an octahedral transition metal complex, it’s also perfectly permissible to choose one of the three bonding axes to be z, so that the d_z2 orbital points along that axis. The only restriction is that you can’t claim that the other axes are also z. Also, this sp3d thing is dubious at best, but I’ll let that slide for now. – orthocresol Jul 21 '21 at 08:50
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    Or in a square planar complex, it’s also permissible to choose the “out-of-plane” axis to be z (i.e. the axis in which there are no ligands), and we nearly always do so. Hence the d_z2 orbital doesn’t interact (much) with the ligands. That’s all perfectly fine to say, as long as you’re internally consistent. Calling both axes z in NO2 is not internally consistent, hence it’s not okay. – orthocresol Jul 21 '21 at 08:52
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    Note, that there is one more issue. You seem to suggest that an sp orbital is formed from one s and one p orbital. That is true in the simplest of cases, but you picked a difficult case with NO2-. The bonds don’t point at right angles to each other, so you cannot assign one p orbital per bond: the geometry doesn’t work out. Actually the truth is that the sp orbitals are a mixture of s and p orbitals, such that the total contribution of s orbital is equal to the total contribution of p orbitals. In other words, you could have something like s plus half px plus half py. – orthocresol Jul 21 '21 at 09:00
  • @orthocresol N is sp² hybridised. So, there will be contribution from all 3 p orbitals such that p orbitals contribute two-third part. Now the whole notion of selection of p orbitals seem to be collapsed! – Apurvium Jul 21 '21 at 11:50

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