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Chemical Bonding and Molecular Structure question

2014 · Shift 1 · Q5
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  5. /2014 · Shift 1 · Q5

Chemical Bonding and Molecular Structure question

2014 · Shift 1 · Q5

JEE AdvancedChemistryChemical Bonding and Molecular StructureNumerical+3 / −1
A list of species having the formula XZ4XZ_4XZ4​ is given below. XeF4XeF_4XeF4​, SF4SF_4SF4​ ,SiF4SiF_4SiF4​, BF4−BF_4^-BF4−​, BrF4−BrF_4^-BrF4−​, [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+, [FeCl4]2−[FeCl_4]^{2-}[FeCl4​]2−, [CoCl4]2−[CoCl_4]^{2-}[CoCl4​]2− and [PtCl4]2−[PtCl_4]^{2-}[PtCl4​]2− Defining shape on the basis of the location of X and Z atoms, the total number of species having a square planar shape is
Numerical answer
View written solutionFree

Correct answer: 4

We need to count how many of the given XZ4XZ_4XZ4​ species are square planar, considering only the positions of XXX and the four ZZZ atoms.

1. List of species

The species are:

  1. XeF4XeF_4XeF4​
  2. SF4SF_4SF4​
  3. SiF4SiF_4SiF4​
  4. BF4−BF_4^-BF4−​
  5. BrF4−BrF_4^-BrF4−​
  6. [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+
  7. [FeCl4]2−[FeCl_4]^{2-}[FeCl4​]2−
  8. [CoCl4]2−[CoCl_4]^{2-}[CoCl4​]2−
  9. [PtCl4]2−[PtCl_4]^{2-}[PtCl4​]2−

We now determine the geometry of each.

2. Geometry of each species

(1) XeF4XeF_4XeF4​

Using VSEPR:

  • Xe has 8 valence electrons.
  • It forms 4 bonds with F and has 2 lone pairs.
  • Steric number =6= 6=6.
  • Electron pair geometry: octahedral.
  • With two lone pairs opposite each other, molecular shape is square planar.

So, XeF4XeF_4XeF4​ is square planar.

(2) SF4SF_4SF4​

  • S has 6 valence electrons.
  • 4 bonds + 1 lone pair.
  • Steric number =5= 5=5.
  • Electron pair geometry: trigonal bipyramidal.
  • Molecular shape: see-saw.

So, SF4SF_4SF4​ is not square planar.

(3) SiF4SiF_4SiF4​

  • 4 bond pairs, no lone pair on Si.
  • Steric number =4= 4=4.
  • Shape: tetrahedral.

So, SiF4SiF_4SiF4​ is not square planar.

(4) BF4−BF_4^-BF4−​

  • B forms 4 bonds, no lone pair.
  • Shape: tetrahedral.

So, BF4−BF_4^-BF4−​ is not square planar.

(5) BrF4−BrF_4^-BrF4−​

Count electrons on central atom:

  • Br has 7 valence electrons, one extra due to negative charge ightarrow8 ightarrow 8ightarrow8 electrons.
  • 4 bonds + 2 lone pairs.
  • Steric number =6= 6=6.
  • Electron pair geometry: octahedral.
  • Two lone pairs opposite each other give square planar shape.

So, BrF4−BrF_4^-BrF4−​ is square planar.

(6) [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+

For Cu2+Cu^{2+}Cu2+:

  • Electronic configuration: 3d93d^93d9.
  • Four-coordinate Cu(II)Cu(II)Cu(II) complexes are commonly square planar or distorted square planar in such questions.
  • In standard coordination-chemistry classification used in JEE, [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+ is taken as square planar.

So, [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+ is square planar.

(7) [FeCl4]2−[FeCl_4]^{2-}[FeCl4​]2−

  • Oxidation state of Fe: x+4(−1)=−2⇒x=+2x + 4(-1) = -2 \Rightarrow x=+2x+4(−1)=−2⇒x=+2
  • So Fe is Fe2+Fe^{2+}Fe2+, i.e. 3d63d^63d6.
  • Cl−Cl^-Cl− is a weak-field ligand.
  • Four-coordinate Fe2+Fe^{2+}Fe2+ with weak-field ligand is generally tetrahedral.

So, [FeCl4]2−[FeCl_4]^{2-}[FeCl4​]2− is not square planar.

(8) [CoCl4]2−[CoCl_4]^{2-}[CoCl4​]2−

  • Oxidation state of Co: x+4(−1)=−2⇒x=+2x + 4(-1) = -2 \Rightarrow x=+2x+4(−1)=−2⇒x=+2
  • So Co is Co2+Co^{2+}Co2+, i.e. 3d73d^73d7.
  • With weak-field ligand Cl−Cl^-Cl−, four-coordinate complex is generally tetrahedral.

So, [CoCl4]2−[CoCl_4]^{2-}[CoCl4​]2− is not square planar.

(9) [PtCl4]2−[PtCl_4]^{2-}[PtCl4​]2−

  • Oxidation state of Pt: x+4(−1)=−2⇒x=+2x + 4(-1) = -2 \Rightarrow x=+2x+4(−1)=−2⇒x=+2
  • So Pt is Pt2+Pt^{2+}Pt2+, a 5d85d^85d8 metal ion.
  • d8d^8d8 ions of 5d series commonly form square planar complexes.

So, [PtCl4]2−[PtCl_4]^{2-}[PtCl4​]2− is square planar.

3. Count square planar species

Square planar species are:

  1. XeF4XeF_4XeF4​
  2. BrF4−BrF_4^-BrF4−​
  3. [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+
  4. [PtCl4]2−[PtCl_4]^{2-}[PtCl4​]2−

Total number: 444

4. Comparison with stored answer

Stored correct answer = 444

Our derived answer = 444

Hence, the answer agrees with the stored correct answer.

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