Question

Hybridization of central atom is independent of the phase/state of the compound in case of
A.$Be{{H}_{2}}$
B.${{N}_{2}}{{O}_{5}}$
C.$Xe{{F}_{6}}$
D.$P{{F}_{5}}$

Answer 1

Hint: One element can show different hybridization in different conditions or we can say on different experimental conditions so we need to see each compound hybridization and its state/condition accordingly to find the independent one.

Complete answer:
Let us consider all the options one by one.
First in Beryllium hydride $Be{{H}_{2}}$ it is in gaseous state when it is a monomer having $sp$ hybridization and it is in solid state when it is as a polymer having $s{{p}^{3}}$ hybridization, which means $Be{{H}_{2}}$ is dependent of the phase/state respect to hybridization.

Now going to second option which is Dinitrogen pentoxide ${{N}_{2}}{{O}_{5}}$ , it is present in gaseous state where all nitrogen atoms are $s{{p}^{2}}$ hybridized and when it is present in solid state ${{N}_{2}}{{O}_{5}}$ is present in the form of ${{\left[ N{{O}_{2}} \right]}^{+}}$ having $sp$ hybridization and ${{\left[ N{{O}_{3}} \right]}^{-}}$ having $s{{p}^{2}}$ hybridization, which means ${{N}_{2}}{{O}_{5}}$ is also dependent on hybridization respect to phase/state.

Now continuing to third option which is Xenon hexafluoride $Xe{{F}_{6}}$ in gaseous state the centre atom Xenon is having $s{{p}^{3}}{{d}^{3}}$ hybridization and in solid state it is having $s{{p}^{3}}{{d}^{2}}$ hybridization, from which we can conclude that $Xe{{F}_{6}}$ is also depended on hybridization with respect to phase.

Now the last option which is Phosphorus pentafluoride ($P{{F}_{5}}$) in gaseous state and in solid state it central atom has $s{{p}^{3}}$ hybridization so now we have got our answer as $P{{F}_{5}}$ the compound which is independent of phase/state with respect to hybridization.

Note:
Hybridization is a mixture of orbitals and not electrons so in hybridization there can be fully filled, half-filled or empty orbitals can take part and one more point is that the number of hybrid orbitals on central atom of the molecule is equal to the sum of number of sigma bonds and lone pair of electrons.
And last point is that One element can represent many hybridization states depending on experimental conditions just like we have seen above.