are all tetrahedral complexes high spin?
313382
Unlike octahedral complexes, the ligands of tetrahedral complexes come in direct contact with the dxz, dxy, and dyz orbitals. ... Usually, electrons will move up to the higher energy orbitals rather than pair. Because of this, most tetrahedral complexes are high spin.
1390235
The electronic configuration of silver is [Kr] 4d105s1. The diamagnetic contribution of the full inner shells is much greater than the paramagnetism of the single electron in the 5s orbital. Several other sources also characterize all the noble metals (copper, silver and gold) are being diamagnetic.
902701
Due to the presence of unpaired d electrons, transition metals can form paramagnetic compounds. Diamagnetic compounds have d-electrons that are all paired up.
1266896
In tetrahedral molecular geometry, a central atom is located at the center of four substituents, which form the corners of a tetrahedron. ... In square planar molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane.
1385640
There are no internal mirror planes, so every carbon atom is different. The four chiral centres (with their attached groups) are: C2 (OH,C1, C3, H)
669368
For TeF4 and ICl3, the arrangement of these molecules is such that the individual bond dipoles do not all cancel, so each has an overall dipole moment (is polar).
1290125
(Crystal Field Theory) How many unpaired electrons are there in a strong field iron(II) octahedral complex? 18. (Crystal Field Theory) Consider the complex ion [Mn(OH2)6]2+ with 5 unpaired electrons.
1125678
Since there are 4 ligands, the compound must be tetrahedral or square planar. The tetrahedral complex would yield 2 unpaired electrons.
153229
In CHCl3, the molecular shape is tetrahedral, meaning that the H and the three Cl atoms will occupy the vertices of a triangular based pyramid around the central C atom. All of these bonds are polar (C-H only very slightly so). ... Therefore, the molecule is polar.
1314891
The shape of molecules depends on hybridization, structural geometry and presence of lone pairs on the central atom. ... Also due to the presence of lone pair would be distorted due to the electron dense cloud of the lone pairs. So the shape of XeF4 would be square planar rather than tetrahedral.
1373409
The symmetric shapes are linear, trigonal planar, and tetrahedral. ... If the shape is symmetric, look to see whether all of the atoms attached to the central atom are the same. If so, the molecule is nonpolar. If there is more than one kind of atom attached to the central atom, the molecule is polar.
305222
Explanation: Mn has [ Ar ] 3d(5) 4s(2) CONFIGURATION . NOW THESE FOUR ELECTRONS CAN BE ARRANGED IN THE FOLLOWING WAYS : tg(3) eg(1) OCTAHEDRAL weak field 4.
208786
Trans [CO (en)2Cl2] is able to exhibit the phenomena of optical isomerism because it can form a superimposable mirror image. Which one of the following oxides is expected to exhibit paramagnetic behaviour?
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The bonds between C and the four Cl atoms are covalent with sp3 hybridization to the central atom. The geometrical structure of CCl4 is tetrahedral. Each covalent bond in this molecule is polar because of the difference in electronegativity between carbon and chloride atom.
1198983
Since, Zn, Cd, and Hg have completely filled shells and there is no unpaired electrons. Due to this there is no d-d transition as a result, the complexes formed by Zn, Cd, and Hg are colourless.
1395622
Due to the presence of symmetry elements, Ni(NH3)2Cl2 does not exhibit isomerism. The Geometry of Hexaamminenickel chloride is square planar. The reason for this is that the nickel is giving away four electrons and has no extra electron pairs floating around. Option (c) is the answer.
860115
Hence, [NiCl4]2- is paramagnetic. In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2. But CO is a strong field ligand. ... Since no unpaired electrons are present in this case, [Ni(CO)4] is diamagnetic.