The $PCl_5$ molecule is a result of the hybridisation of

$H - C \equiv C - C \equiv C - CH_3$
Compare the bond lengths $a$ and $b$.

Among $SO_2, NF_3, NH_3, XeF_2, ClF_3$ and $SF_4$,the hybridization of the molecule with non$-zero$ dipole moment and highest number of lone$-pairs$ of electrons on the central atom is

In which hybridisation is an angular shape not possible?

The state of hybrid orbitals of carbon in $CO_{2}$,$CH_{4}$ and $CO_{3}^{2-}$ respectively is

$BF_3$ is a planar and electron-deficient compound. The hybridization and the number of electrons around the central atom,respectively,are: