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As shown in the image, we have two free p orbitals in ethene but we need two p orbitals each for the bonding and anti-bonding orbital. That is a total of four. What is the concept behind this? enter image description here

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  • $\begingroup$ No the antibonding and bonding does not happen at the same time. $\endgroup$ – Ali Caglayan Oct 18 '15 at 8:54
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but we need two p orbitals each for the bonding and anti-bonding orbital. That is a total of four.

No, that is not a total of four: if you start from 2 atomic orbitals ($p_{\mathrm{1}}$ and $p_{\mathrm{2}}$), you should get the same number of molecular orbitals at the end ($\pi$ and $\pi^{*}$). The number of orbitals is perfectly conserved here.

In general, out of $n$ atomic orbitals (at most) the same number of (orthogonal) molecular orbitals could be built. In this particular case, two molecular orbitals that are built from $p_{\mathrm{1}}$ and $p_{\mathrm{2}}$ atomic orbitals are $$ \pi^\phantom{*} = 1/\sqrt{2} p_{\mathrm{1}} + 1/\sqrt{2} p_{\mathrm{2}} \, , \\ \pi^{*} = 1/\sqrt{2} p_{\mathrm{1}} - 1/\sqrt{2} p_{\mathrm{2}} \, . \\ $$ $\pi$ and $\pi^{*}$ will be orthogonal by construction if $p_{\mathrm{1}}$ and $p_{\mathrm{2}}$ were chosen to be orthonormal in the first place.

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