
Which one of the following has a square planar geometry?
a) ${\left[ {CoC{l_4}} \right]^{2 - }}$
b) ${\left[ {FeC{l_4}} \right]^{2 - }}$
c) ${\left[ {NiC{l_4}} \right]^{2 - }}$
d) ${\left[ {PtC{l_4}} \right]^{2 - }}$
Answer
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Hint: We know that the square planar sub-atomic geometry in science portrays the stereochemistry (spatial course of action of atoms) that is received by certain substance compounds. As the name recommends, particles of this geometry have their atoms situated at the corners.
Complete answer:
A square planar complex additionally has a coordination number of\[4\]. The design of the perplexing varies from tetrahedral on the grounds that the ligands structure a basic square on the x and y tomahawks. Along these lines, the gem field parting is likewise unique. Since there are no ligands along the z-pivot in a square planar unpredictable, the repugnance of electrons in the\[{d_{xz}}\], \[{d_{yz}}\] , and the \[{d_z}^2\] orbitals are extensively lower than that of the octahedral complex. The \[{d_{x2 - y2}}\] orbital has the most energy, trailed by the \[{d_{xy}}\] orbital, which is trailed by the excess orbitals (in spite of the fact that \[{d_z}^2\] has marginally more energy than the \[{d_{xz}}\] and \[{d_{yz}}\] orbital). This example of orbital parting stays steady all through all geometries.
We know that \[P{t^{ + 2}}\] has a \[{d^8}\] configuration. \[C{l^ - }\] is a feeble field ligand. In this way, ${\left[ {PtC{l_4}} \right]^{2 - }}$has square planar geometry.
Note:
Tetrahedral geometry is somewhat harder to picture than square planar math. Tetrahedral math is undifferentiated from a pyramid, where every one of corners of the pyramid compares to a ligand, and the focal atom is in the pyramid. This geometry likewise has a coordination number of \[4\] since it has \[4\] ligands bound to it. At long last, the bond point between the ligands is\[{109.5^o}\]. An illustration of the tetrahedral particle methane.
In a tetrahedral complex, \[\Delta t\] is moderately small even with solid field ligands as there are less ligands to security with. It is uncommon for the \[\Delta t\] of tetrahedral edifices to surpass the blending energy.
Complete answer:
A square planar complex additionally has a coordination number of\[4\]. The design of the perplexing varies from tetrahedral on the grounds that the ligands structure a basic square on the x and y tomahawks. Along these lines, the gem field parting is likewise unique. Since there are no ligands along the z-pivot in a square planar unpredictable, the repugnance of electrons in the\[{d_{xz}}\], \[{d_{yz}}\] , and the \[{d_z}^2\] orbitals are extensively lower than that of the octahedral complex. The \[{d_{x2 - y2}}\] orbital has the most energy, trailed by the \[{d_{xy}}\] orbital, which is trailed by the excess orbitals (in spite of the fact that \[{d_z}^2\] has marginally more energy than the \[{d_{xz}}\] and \[{d_{yz}}\] orbital). This example of orbital parting stays steady all through all geometries.
We know that \[P{t^{ + 2}}\] has a \[{d^8}\] configuration. \[C{l^ - }\] is a feeble field ligand. In this way, ${\left[ {PtC{l_4}} \right]^{2 - }}$has square planar geometry.
Note:
Tetrahedral geometry is somewhat harder to picture than square planar math. Tetrahedral math is undifferentiated from a pyramid, where every one of corners of the pyramid compares to a ligand, and the focal atom is in the pyramid. This geometry likewise has a coordination number of \[4\] since it has \[4\] ligands bound to it. At long last, the bond point between the ligands is\[{109.5^o}\]. An illustration of the tetrahedral particle methane.
In a tetrahedral complex, \[\Delta t\] is moderately small even with solid field ligands as there are less ligands to security with. It is uncommon for the \[\Delta t\] of tetrahedral edifices to surpass the blending energy.
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