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Chemical Bonding and Molecular Structure question

2023 · 6 Apr · Shift 1 · Q16
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Chemical Bonding and Molecular Structure question

2023 · 6 Apr · Shift 1 · Q16

JEE MainChemistryChemical Bonding and Molecular StructureNumerical+4 / −1
The number of species from the following which have square pyramidal structure is ‾\underline{\hspace{2cm}}​ PF5,BrF4−,IF5,BrF5,XeOF4,ICl4−\mathrm{PF}_{5}, \mathrm{BrF}_{4}^{-}, \mathrm{IF}_{5}, \mathrm{BrF}_{5}, \mathrm{XeOF}_{4}, \mathrm{ICl}_{4}^{-}PF5​,BrF4−​,IF5​,BrF5​,XeOF4​,ICl4−​
Numerical answer
View written solutionFree

Correct answer: 3

  1. Use VSEPR to determine shape of each species
    Square pyramidal geometry generally arises for steric number 666 with arrangement AX5EAX_5EAX5​E: 6 electron pairs=5 bond pairs+1 lone pair\text{6 electron pairs} = 5 \text{ bond pairs} + 1 \text{ lone pair}6 electron pairs=5 bond pairs+1 lone pair This gives an octahedral electron-pair geometry and a square pyramidal molecular shape.

We now check each species.


  1. PF5\mathrm{PF_5}PF5​
    Central atom: PPP
    Valence electrons on P=5P = 5P=5
    It forms 555 bonds and has no lone pair.

So it is: AX5AX_5AX5​ Geometry = trigonal bipyramidal, not square pyramidal.

❌ Not counted.


  1. BrF4−\mathrm{BrF_4^-}BrF4−​
    Central atom: BrBrBr
    Valence electrons on Br=7Br = 7Br=7, plus 111 extra for negative charge ⇒8\Rightarrow 8⇒8 electrons.

It forms 444 bonds, leaving 222 lone pairs on Br. So: AX4E2AX_4E_2AX4​E2​ Electron pair geometry = octahedral
Molecular shape = square planar.

❌ Not counted.


  1. IF5\mathrm{IF_5}IF5​
    Central atom: III
    Valence electrons on I=7I = 7I=7
    Forms 555 bonds, leaving 111 lone pair. So: AX5EAX_5EAX5​E Electron pair geometry = octahedral
    Molecular shape = square pyramidal.

✅ Counted.


  1. BrF5\mathrm{BrF_5}BrF5​
    Central atom: BrBrBr
    Valence electrons on Br=7Br = 7Br=7
    Forms 555 bonds, leaving 111 lone pair. So: AX5EAX_5EAX5​E Molecular shape = square pyramidal.

✅ Counted.


  1. XeOF4\mathrm{XeOF_4}XeOF4​
    Central atom: XeXeXe
    Valence electrons on Xe=8Xe = 8Xe=8.

Around Xe: one Xe=OXe=OXe=O double bond, four Xe−FXe-FXe−F single bonds, and one lone pair.
In VSEPR, the double bond counts as one electron domain. Thus total electron domains around Xe: 1(O)+4(F)+1(lone pair)=61(O) + 4(F) + 1(\text{lone pair}) = 61(O)+4(F)+1(lone pair)=6 So: AX5EAX_5EAX5​E Molecular shape = square pyramidal.

✅ Counted.


  1. ICl4−\mathrm{ICl_4^-}ICl4−​
    Central atom: III
    Valence electrons on I=7I = 7I=7, plus 111 extra due to charge ⇒8\Rightarrow 8⇒8.

Forms 444 bonds, leaving 222 lone pairs. So: AX4E2AX_4E_2AX4​E2​ Molecular shape = square planar.

❌ Not counted.


  1. Total number of square pyramidal species

Square pyramidal species are:

  • IF5\mathrm{IF_5}IF5​
  • BrF5\mathrm{BrF_5}BrF5​
  • XeOF4\mathrm{XeOF_4}XeOF4​

Hence, total number is: 3\boxed{3}3​


  1. Comparison with stored correct answer
    Stored correct answer = 333
    Derived answer = 333

So the answer agrees with the stored correct answer.

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