Which of the following diagrams or statements | vedclass

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(D) For $sp$ hybridization: The mixing of one $s$ and one $p$ orbital (e.g.,$p_y$) results in two $sp$ hybrid orbitals oriented along the axis of the

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All of these

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(D) For $sp$ hybridization: The mixing of one $s$ and one $p$ orbital (e.g.,$p_y$) results in two $sp$ hybrid orbitals oriented along the axis of the $p$ orbital used. Thus,$s + p_y 	o 2sp$ along the $y$-axis is correct.For $sp^2$ hybridization: The mixing of one $s$ and two $p$ orbitals (e.g.,$p_y$ and $p_z$) results in three $sp^2$ hybrid orbitals lying in the plane defined by those two $p$ orbitals (the $yz$ plane). Thus,$s + p_y + p_z 	o sp^2$ orbitals in the $yz$ plane is correct.For $sp^3d^2$ hybridization: The mixing of one $s$,three $p$,and two $d$ orbitals results in six $sp^3d^2$ hybrid orbitals,which are directed towards the corners of an octahedron. Thus,this statement is also correct.Since all statements are correct,the correct option is $D$.

List-$I$ (Molecule / Species)	List-$II$
$(P)$ $NO_2$ (Unpaired electron in)	$(1)$ Vacant $p$-orbital involved in hybridization
$(Q)$ $B_2H_6$	$(2)$ $sp^2$-orbital
$(R)$ $ClO_3$ (Unpaired electron in)	$(3)$ $sp^3$-orbital
$(S)$ $CH_3^+$	$(4)$ $120^\circ$ Bond angle

Which of the following diagrams or statements is correct for the hybrid orbitals?

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