Back to Directory
NEET CHEMISTRYMedium

In which of the following pairs, both the species are not isostructural?

A

SiCl4,PCl4+\text{SiCl}_4, \text{PCl}_4^+

B

Diamond, silicon carbide

C

NH3,PH3\text{NH}_3, \text{PH}_3

D

XeF4,XeO4\text{XeF}_4, \text{XeO}_4

Step-by-Step Solution

To determine if species are isostructural, we need to check their geometry or shape based on VSEPR theory and hybridisation:

  • SiCl4\text{SiCl}_4 and PCl4+\text{PCl}_4^+: Silicon in SiCl4\text{SiCl}_4 has 4 valence electrons, forming 4 bond pairs with 0 lone pairs (sp3sp^3, tetrahedral). Phosphorus in PCl4+\text{PCl}_4^+ has 5 valence electrons, but a +1 charge leaves 4 electrons; it forms 4 bond pairs with 0 lone pairs (sp3sp^3, tetrahedral). They are isostructural.
  • Diamond and Silicon carbide (SiC): Both possess a giant three-dimensional network structure where each atom is sp3sp^3 hybridised and tetrahedrally bonded to four other atoms. They are isostructural.
  • NH3\text{NH}_3 and PH3\text{PH}_3: Both central atoms belong to Group 15, having 5 valence electrons. They form 3 bond pairs and have 1 lone pair. Both possess a trigonal pyramidal shape. They are isostructural.
  • XeF4\text{XeF}_4 and XeO4\text{XeO}_4: Xenon in XeF4\text{XeF}_4 has 8 valence electrons, forming 4 single bonds with fluorine and leaving 2 lone pairs. Its steric number is 6 (sp3d2sp^3d^2), and its shape is square planar. Xenon in XeO4\text{XeO}_4 forms 4 double bonds with oxygen using all 8 of its valence electrons, leaving 0 lone pairs. Its steric number is 4 (sp3sp^3), and its shape is tetrahedral. Therefore, they are not isostructural.
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started