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Chemistry

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  • $\begingroup$ I think it is true. $\endgroup$
    – Soumik Das
    Commented May 19, 2018 at 10:41
  • $\begingroup$ may i know why? $\endgroup$
    – Avnish Singh
    Commented May 19, 2018 at 10:41
  • $\begingroup$ Because there is very little chance of overlap. In $d_{x^2 - y^2}$ the lobe oriented along x axis will overall make no overlap with $p_y$ as those are orthogonal, and also the lobe oriented along y axis is much far from $p_y$. If you draw the diagrams it will be more clear. Then you can also see $p_y$ and $d_{xy}$ can form stong $\pi$ -bonds. $\endgroup$
    – Soumik Das
    Commented May 19, 2018 at 10:52
  • 2
    $\begingroup$ When $y$ changes sign, $p_y$ changes sign, but $d_{x^2-y^2}$ doesn't, hence their overlap equals $0$ by parity. $\endgroup$
    – Ivan Neretin
    Commented May 19, 2018 at 13:20