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Coordination Compounds question

2024 · Shift 2 · Q12
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Coordination Compounds question

2024 · Shift 2 · Q12

JEE AdvancedChemistryCoordination CompoundsNumerical+4 / −1
Among [Co(CN)4]4−,[Co(CO)3(NO)],XeF4,[PCl4]+,[PdCl4]2−,[ICl4]−,[Cu(CN)4]3−\left[\mathrm{Co}(\mathrm{CN})_4\right]^{4-},\left[\mathrm{Co}(\mathrm{CO})_3(\mathrm{NO})\right], \mathrm{XeF}_4,\left[\mathrm{PCl}_4\right]^{+},\left[\mathrm{PdCl}_4\right]^{2-},\left[\mathrm{ICl}_4\right]^{-},\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}[Co(CN)4​]4−,[Co(CO)3​(NO)],XeF4​,[PCl4​]+,[PdCl4​]2−,[ICl4​]−,[Cu(CN)4​]3− and P4\mathrm{P}_4P4​ the total number of species with tetrahedral geometry is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 4OR5

  1. We examine the geometry of each given species one by one.
  1. [Co(CN)4]4−\left[\mathrm{Co}(\mathrm{CN})_4\right]^{4-}[Co(CN)4​]4−
  • Oxidation state of Co: x+4(−1)=−4⇒x=0x+4(-1)=-4 \Rightarrow x=0x+4(−1)=−4⇒x=0
  • So Co is Co0\mathrm{Co}^0Co0, i.e. d9d^9d9.
  • Coordination number =4=4=4.
  • For 4-coordinate complexes, geometry is usually either tetrahedral or square planar.
  • Co0\mathrm{Co}^0Co0 with cyanide in such a case is not the usual square planar d8d^8d8 situation; this is taken as tetrahedral.

So, this counts as tetrahedral.

  1. [Co(CO)3(NO)]\left[\mathrm{Co}(\mathrm{CO})_3(\mathrm{NO})\right][Co(CO)3​(NO)]
  • CO is neutral and linear NO is treated as NO+\mathrm{NO}^+NO+ in nitrosyl complexes.
  • Hence oxidation state of Co: x+0+0+0+(+1)=0⇒x=−1x+0+0+0+(+1)=0 \Rightarrow x=-1x+0+0+0+(+1)=0⇒x=−1
  • Coordination number =4=4=4 (3 CO + 1 NO).
  • Such 4-coordinate carbonyl/nitrosyl complexes are generally tetrahedral.

So, this also counts as tetrahedral.

  1. XeF4\mathrm{XeF}_4XeF4​
  • Steric number around Xe: 8+42=6\frac{8+4}{2}=628+4​=6
  • Thus arrangement is octahedral electron-pair geometry with 2 lone pairs.
  • Molecular geometry is square planar, not tetrahedral.

So, not tetrahedral.

  1. [PCl4]+\left[\mathrm{PCl}_4\right]^+[PCl4​]+
  • Valence electrons on central P: 5−1=45-1=45−1=4
  • Four bond pairs, no lone pair.
  • By VSEPR, geometry is tetrahedral.

So, this counts as tetrahedral.

  1. [PdCl4]2−\left[\mathrm{PdCl}_4\right]^{2-}[PdCl4​]2−
  • Oxidation state of Pd: x+4(−1)=−2⇒x=+2x+4(-1)=-2 \Rightarrow x=+2x+4(−1)=−2⇒x=+2
  • Pd2+\mathrm{Pd}^{2+}Pd2+ is d8d^8d8.
  • 4-coordinate d8d^8d8 complexes of Pd(II) are characteristically square planar.

So, not tetrahedral.

  1. [ICl4]−\left[\mathrm{ICl}_4\right]^-[ICl4​]−
  • Total electron pairs around I correspond to 4 bond pairs + 2 lone pairs.
  • Electron-pair geometry octahedral; molecular geometry square planar.

So, not tetrahedral.

  1. [Cu(CN)4]3−\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}[Cu(CN)4​]3−
  • Oxidation state of Cu: x+4(−1)=−3⇒x=+1x+4(-1)=-3 \Rightarrow x=+1x+4(−1)=−3⇒x=+1
  • So Cu is Cu+\mathrm{Cu}^+Cu+, i.e. d10d^{10}d10.
  • 4-coordinate d10d^{10}d10 complexes are typically tetrahedral.

So, this counts as tetrahedral.

  1. P4\mathrm{P}_4P4​
  • White phosphorus has a tetrahedral cage structure.
  • The four P atoms occupy the corners of a tetrahedron.

So, this counts as tetrahedral.

  1. Total tetrahedral species

Tetrahedral species are:

  • [Co(CN)4]4−\left[\mathrm{Co}(\mathrm{CN})_4\right]^{4-}[Co(CN)4​]4−
  • [Co(CO)3(NO)]\left[\mathrm{Co}(\mathrm{CO})_3(\mathrm{NO})\right][Co(CO)3​(NO)]
  • [PCl4]+\left[\mathrm{PCl}_4\right]^+[PCl4​]+
  • [Cu(CN)4]3−\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}[Cu(CN)4​]3−
  • P4\mathrm{P}_4P4​

Hence total number: 555

  1. Comparison with stored answer

Stored correct answer: 4 OR 54\text{ OR }54 OR 5

Our derived answer is 555, which matches one of the accepted values.

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