VSEPR Theory Test

According to VSEPR theory, BrF₄^- and XeF₄ both have four bond pairs of electrons and two lone pairs of electrons with square planar shape. TeCl₄ has four bond pairs of electrons and one lone pair of electrons with distorted trigonal bipyramidal shape.
Hence, 1 and 2 have square planar shape.

SF₄ and XeO₂F₂ have sp3d hybridisation and one lone pair of electrons with seasaw geometry. XeOF₂ has sp³d hybridisation and two lone pair of electrons with T-shaped geometry. XeO₃F₂ has sp³d hybridisation and zero lone pair of electrons with trigonal bipyramidal geometry.

This option is correct because the geometry of I₃⁺ is bent while XeF₆ is distorted octahedral.

XeOF₅^- has six bond pairs and one lone pair with pentagonal pyramidal shape. WCl₂F₄ has six bond pairs and a zero lone pair with octahedral geometry. TeCl₄ has four bond pairs and one lone pair with seasaw geometry. XeO₄ has four bond pairs and zero lone pair with tetrahedral shape.
Hence, the representation of codes A - 2, B - 1, C - 3, D - 4 is correct.

The capped square antiprismatic molecular geometry describes the shape of compounds where nine atoms, groups of atoms, or ligands are arranged around a central atom, defining the vertices of a gyroelongated square pyramid. Example is ReH₉^2-.

1. NH₃ and PCl₃ both molecules have one lone pair of electrons with trigonal pyramidal geometry.
2. OF₂ has two lone pair of electrons with bent geometry.
3. SO₃ has zero lone pair of electrons with trigonal planar geometry.
4. CO₂ has zero lone pair of electrons with linear geometry. XeCl₂ has three lone pair of electrons with linear geometry.
   Hence, 1 and 4 statements are incorrect.

ReCl₈^- has square antiprismatic molecular geometry (the shape of compounds where eight atoms, groups of atoms, or ligands are arranged around a central atom, defining the vertices of a square antiprism). W(CH₃)₆ has trigonal prismatic geometry (the shape of compounds where six atoms, groups of atoms, or ligands are arranged around a central atom, defining the vertices of a triangular prism). Na[CoCl₂(NH₃)₄] has octahedral geometry. SeF₄ has one lone pair of electrons with distorted trigonal bipyramidal shape.
Hence, the representation of codes A - 2, B - 1, C - 4, D - 3 is correct.

I₃^- and XeF₂ both have three lone pair of electrons with linear geometry. HgCl₂ has zero lone pair of electrons with linear geometry.
Hence, 1, 2, 3 all molecules have linear geometry.

This option is correct because the lone pairs are present in XeF₂, I₃^- are ICl₂^- maximum, i.e. three.

SnCl₂ has one lone pair of electrons with V-shaped geometry. SeH₂ has two lone pairs of electrons with V-shaped geometry.
SO₂ has one lone pair of electrons with V-shaped geometry while CO₂ has zero lone pair of electrons with linear shape.

XeOF₅^- has one lone pair of electrons with a pentagonal pyramidal shape. PI₃ has one lone pair of electrons with a trigonal pyramidal shape. SF₆ has zero lone pair of electrons with an octahedral geometry. BrCl₃ has two lone pair of electrons and a bent geometry.
Hence, the correct answer using the codes is P - 3, Q - 4, R - 1, S - 2.

NH₄⁺ and ClO₄^- both ions have four bond pairs of electrons and zero lone pair of electrons with tetrahedral geometry. IF₄⁺ has four bond pairs of electrons and one lone pair of electrons with seasaw geometry. ClO₂^- has three bond pairs of electrons and two lone pairs of electrons with V-shaped geometry.
Hence, i and iv are identical in geometry.

SbCl₅ has zero lone pair of electrons with trigonal bipyramidal geometry. SnCl₂ has one lone pair of electrons with V-shaped geometry. ICl has three lone pairs of electrons with linear shape. ICl₄^- has two lone pairs of electrons with square planar shape.
Hence, the correct answer using the codes is P - 2 - i, Q - 3 - iv, R - 4 - iii, S - 1 - ii.

1. XeCl₂ has three lone pairs of electrons with linear geometry.
2. XeF₆ has six bond pairs of electrons and one lone pair of electrons with distorted pentagonal bipyramidal shape.
3. ICl₄^- ion and XeF₄ molecules have four bond pairs of electrons and two lone pairs of electrons with square planar geometry.
   Hence, only statement P is false.

BH₂⁺ has sp hybridization and zero number of lone pair of electrons with linear shape. SnBr₂ has sp² hybridisation and one lone pair of electrons with V-shaped geometry. XeF₃⁺ has sp³d hybridisation and two lone pairs of electrons with T-shaped geometry. IF₅ has sp³d² hybridisation and one lone pair of electrons with square pyramidal geometry.
Hence, the representation of codes A - 2, B - 1, C - 4, D - 3 is correct.