A) \[N{{H}_{3}},P{{H}_{3}}\]
B) \[Xe{{F}_{4}},Xe{{O}_{4}}\]
C) \[SiC{{l}_{4}},PC{{l}_{4}}\]
D) Diamond, silicon carbide
Correct Answer: B
Solution :
Hybridization of \[N{{H}_{3}}[\sigma =3,lp=1]\] \[s{{p}^{3}}\]geometry: tetrahedral (ii) Structure of\[Xe{{F}_{4}}\]is square planar. Structure of\[Xe{{O}_{4}}\]is tetrahedral So \[Xe{{F}_{4}}\]and \[Xe{{O}_{4}}\]are not isostructural (iii) Structure of \[SiC{{l}_{4}}\]is tetrahedral Structure of \[PCl_{4}^{+}\]is tetrahedral (iv) Both diamond and SiC are isostructural because both have tetrahedral arrangement and central atom is \[s{{p}^{3}}\]hybridised.You need to login to perform this action.
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