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# Orbitals participating in hybridization in ${[Xe{F_8}]^{2 - }}$ is:(A) $s,{p_x},{p_y},{p_z},{p_{xy}},{p_{yz}},{p_{zx}},{d_{{x^2} - {y^2}}},{d_{{z^2}}}$ (B) $s,{p_x},{p_y},{p_z},{p_{xy}},{p_{yz}},{p_{zx}},{d_{{x^2} - {y^2}}}$ (C) $s,{p_x},{p_y},{p_z},{p_{xy}},{p_{yz}},{p_{zx}},{d_{{z^2}}}$ (D) ${p_x},{p_y},{p_z},{p_{xy}},{p_{yz}},{p_{zx}},{d_{{x^2} - {y^2}}},{d_{{z^2}}}$

Last updated date: 09th Sep 2024
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In ${[Xe{F_8}]^{2 - }}$ the lone pair does not participate in hybridization and while calculating the steric number. The steric number $= \dfrac{1}{2}[V + M - C + A]$ the negative charge is not taken into consideration here. So, the steric number will be equal to $\dfrac{1}{2}[8 + 8] = 8$ . As the lone pair does not participate in hybridization, there is only one way for hybridization for steric number eight i.e. when $3p$ orbitals combine with the $5d$ orbitals. So, the hybridization will be ${p^3}{d^5}$ . The orbitals participating in hybridization in ${[Xe{F_8}]^{2 - }}$ are $s,{p_x},{p_y},{p_z},{p_{xy}},{p_{yz}},{p_{zx}},{d_{{x^2} - {y^2}}},{d_{{z^2}}}$
The hybridization occurs when an atom bonds using electrons to form both the $s$ and $p$ orbitals, creating an imbalance in the energy levels of electrons. To find the hybridization, first look at the atom and count the number of atoms connected to it. Count the number of atoms connected, not the bonds. Then count the number of lone pairs attached to it. Add these two numbers. If the sum is four then the atom is $s{p^3}$ , if it is three: $s{p^2}$ and if it is two: $sp$