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Chemical Bonding and Molecular Structure question

2004 · Shift 0 · Q65
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Chemical Bonding and Molecular Structure question

2004 · Shift 0 · Q65

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
Which one of the following has the regular tetrahedral structure? (Atomic nos : B = 5, S = 16, Ni = 28, Xe = 54)
  1. A
    XeF4XeF_4XeF4​
  2. B
    [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2−
  3. C
    BF4−BF_4^-BF4−​
  4. D
    SF4SF_4SF4​
View written solutionFree

Correct answer: C

  1. Determine the geometry of each species using VSEPR / known coordination geometry

We need the species with regular tetrahedral structure.

  1. Option A: XeF4XeF_4XeF4​
  • Xenon has 8 valence electrons.
  • In XeF4XeF_4XeF4​, xenon forms 4 bonds with fluorine and has 2 lone pairs.
  • Total electron pairs around Xe = 6.
  • Electron pair geometry = octahedral.
  • With 2 lone pairs opposite each other, molecular shape = square planar.

So, XeF4XeF_4XeF4​ is not tetrahedral.

  1. Option B: [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2−
  • Oxidation state of Ni: x+4(−1)=−2⇒x=+2x + 4(-1) = -2 \Rightarrow x=+2x+4(−1)=−2⇒x=+2
  • So Ni is Ni2+Ni^{2+}Ni2+, which is 3d83d^83d8.
  • CN−CN^-CN− is a strong field ligand, causing pairing of electrons.
  • Hence the complex is dsp2dsp^2dsp2 hybridised and has square planar geometry.

So, [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2− is not tetrahedral.

  1. Option C: BF4−BF_4^-BF4−​
  • Boron has 3 valence electrons.
  • In BF4−BF_4^-BF4−​, boron forms 4 sigma bonds with fluorine.
  • There is no lone pair on boron.
  • Steric number = 4.
  • Therefore, geometry = tetrahedral.

Since all four bond pairs are identical and there are no lone pairs, the structure is regular tetrahedral.

So, BF4−BF_4^-BF4−​ has regular tetrahedral structure.

  1. Option D: SF4SF_4SF4​
  • Sulfur has 6 valence electrons.
  • In SF4SF_4SF4​, sulfur forms 4 bonds and has 1 lone pair.
  • Total electron pairs = 5.
  • Electron pair geometry = trigonal bipyramidal.
  • Molecular shape = see-saw.

So, SF4SF_4SF4​ is not tetrahedral.

  1. Final conclusion

Only BF4−BF_4^-BF4−​ has a regular tetrahedral structure.

Therefore, the correct option is: C\boxed{C}C​

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