The number of species having a square planar shape from the following is __
\(XeF_4, SF_4, SiF_4, BF^-_4 BrF^-_4,[Cu(NH_3)_4]^{2+}, [FeCl_4]^{2-}, [PtCl_4]^{2-}\)
\(XeF_4, SF_4, SiF_4, BF^-_4 BrF^-_4,[Cu(NH_3)_4]^{2+}, [FeCl_4]^{2-}, [PtCl_4]^{2-}\)
Correct Answer: 4
Approach Solution - 1
Square Planar Molecular Geometry Analysis
To determine the shapes of these species, we need to consider their electron domain geometry and the effect of lone pairs. Square planar geometry arises from an octahedral electron domain geometry with two lone pairs occupying axial positions. It can also arise from dsp2 hybridisation when the central atom has a d8 configuration in a complex ion.
Let’s analyze each species:
- \( \text{XeF}_4 \): Xe has 8 valence electrons, and 4 are used for bonding with F. This leaves 4 electrons (2 lone pairs). \( \text{XeF}_4 \) has 6 electron domains (4 bonding, 2 lone pairs) adopting an octahedral electron domain geometry. The lone pairs occupy axial positions, resulting in a square planar molecular geometry.
- \( \text{SF}_4 \): S has 6 valence electrons, 4 used for bonding. This leaves 2 electrons (1 lone pair). \( \text{SF}_4 \) has 5 electron domains (4 bonding and 1 lone pair) with a trigonal bipyramidal electron domain geometry. The lone pair occupies an equatorial position leading to a see-saw shape.
- \( \text{SiF}_4 \): Si has 4 valence electrons, all used for bonding. \( \text{SiF}_4 \) has 4 electron domains (4 bonding, 0 lone pairs) and is tetrahedral.
- \( \text{BrF}_4^- \): Br has 7 valence electrons + 1 (from the negative charge) = 8 electrons. 4 are used for bonding. This leaves 4 electrons (2 lone pairs). \( \text{BrF}_4^- \) has 6 electron domains (4 bonding, 2 non-bonding), adopting an octahedral electron geometry. The two lone pairs are in the axial positions, leading to a square planar molecular geometry.
- \( [\text{Cu(NH}_3)_4]^{2+} \): \( \text{Cu}^{2+} \) is a d9 system. Due to Jahn Teller distortion, this complex prefers to be a distorted square planar structure rather than a perfect square planar shape. For this type of question, it’s usually counted as a square planar complex.
- \( [\text{FeCl}_4]^{2-} \): \( \text{Fe}^{2+} \) is a d6 system, which is generally tetrahedral when dealing with weak field ligands.
- \( [\text{PtCl}_4]^{2-} \): \( \text{Pt}^{2+} \) is a d8 system. These often form square planar complexes due to dsp2 hybridization.
Conclusion:
Therefore, \( \text{XeF}_4 \), \( \text{BrF}_4^- \), \( [\text{Cu(NH}_3)_4]^{2+} \), and \( [\text{PtCl}_4]^{2-} \) are square planar. The answer is 4.
Approach Solution -2
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