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(B) By convention,the internuclear axis is taken as the $z$-axis.$(i) \ 2p_y - 2p_y$ overlap: Since the $p_y$ orbitals are perpendicular to the $z$-ax

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$(i)$ and $(iv)$

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(B) By convention,the internuclear axis is taken as the $z$-axis.$(i) \ 2p_y - 2p_y$ overlap: Since the $p_y$ orbitals are perpendicular to the $z$-axis,they undergo lateral (sideways) overlap,forming a $\pi$ bond,not a $\sigma$ bond.$(ii) \ 2p_z - 2p_z$ overlap: These orbitals are oriented along the internuclear axis,resulting in head-on (axial) overlap,which forms a $\sigma$ bond.$(iii) \ 2s - 2p_x$ overlap: The $2s$ orbital is spherical and the $2p_x$ orbital is perpendicular to the $z$-axis. Their overlap is zero (non-bonding) because of symmetry.$(iv) \ 2p_x - 2p_x$ overlap: Similar to $p_y$,these orbitals are perpendicular to the $z$-axis and undergo lateral overlap,forming a $\pi$ bond,not a $\sigma$ bond.Therefore,combinations $(i)$,$(iii)$,and $(iv)$ do not form a $\sigma$ bond. Given the options,$(i)$ and $(iv)$ are the most appropriate choices for lateral overlap.

Which of the following combinations does not form a $\sigma$ bond?
$(i) \ 2p_y - 2p_y$
$(ii) \ 2p_z - 2p_z$
$(iii) \ 2s - 2p_x$
$(iv) \ 2p_x - 2p_x$

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Which of the following combinations does not form a $\sigma$ bond?$(i) \ 2p_y - 2p_y$$(ii) \ 2p_z - 2p_z$$(iii) \ 2s - 2p_x$$(iv) \ 2p_x - 2p_x$