My text book just referred to this term. I have absolutely no idea what it is. Articles on the internet are too difficult for me to comprehend.

In some alkenes, steric crowding in the planar state may be so severe that the latter is no longer an energy minimum but becomes an energy barrier with the ground state becoming twisted. In this context the compounds in figure are of interest. The drop in rotational barrier from stilbene ($\pu{42.8 kcal mol-1}$, $\pu{179 kJ mol-1}$)to compound $\ce{P}$ ($\pu{21.1 kcal mol-1}$, $\pu{88.3 kJ mol-1}$) to compound $\ce{Q}$ (“small barrier”, cf.) may be explained by increasing stabilization of the triplet biradical transition state. However, the “negative” barrier in $\ce{R}$, that is, the fact that the stable ground state of $\ce{R}$ is non-planar and that the planar conformation represents the transition state for rotation requires a different explanation; the difference between $\ce{Q}$ and $\ce{R}$ is presumably due to the steric interaction of the four chlorine substituents in the planar conformation of $\ce{R}$. This situation resembles that in o,o'- tetrasubstituted biphenyls to alkenes with low or “negative” barriers.

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This is the question given: for which the above is the solution how is it applicable to this question?

Source: "Advanced Problems in Organic Chemistry by MS. Chouhan 9th Edition", Chapter GOC, page 53

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    $\begingroup$ For example, when the overlap of p orbitals are not sufficient to form a double bond, therefore the two electrons behave like radical electrons weakly coupled to each other instead of an electron pair in bonds. $\endgroup$ – Greg Dec 9 '17 at 6:30

A triplet biradical state simply implies that the bonding pair will be cleaved homolytically . For example the triplet biradical transition state for
enter image description here Will be
enter image description here

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