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Hint: The d orbitals within the transition elements don’t have longer equal power in their complexes. It is a transition where an electron jumps from one d orbital to another orbital. Generally those are degenerate, that is, the d orbitals have the equal energy, but under rare conditions, in which some includes the presence of ligands or the degeneracy can be removed in order that there can be a specific power and hence wavelengths are associated with these transitions.
Complete step by step solution:
Transition metal ions are colored because of the $d - d$ transition. Unpaired electrons in d orbitals are agitated from lower strength stage to better power strength stage. For this, the strength from visible places is absorbed.
$S{c^{3 + }}$ has outer electronic configuration of $3{d^0}\;4{s^0}$ with zero unpaired electrons. Hence, it is colorless as no electron is unpaired.
$T{i^{3 + }}$ has outer electronic configuration with one unpaired electron. Hence, it is purple colored because there’s at least one electron free.
Additional information:
Most of the chemical compounds are colored because the electron shells have different energies. Electron shells can be dealt with as strength stages with the ones which could be far away from the nucleus which is being higher in power. All atoms are capable of soaked up strength and redeeming this energy again. This happens when power is absorbed through an atom or compound and an electron inside the shells is excited or high. When its miles are excited it actions from a lower strength level/shell to a higher power level/shell. When it comes back off from the higher energy stage/shell it has to release the power that it absorbed from there. This occurs when the electron comes back track to its unique strength level releasing the greater energy as a photon.
Note:
Most compounds of metals and non metals now do not show up coloration. For example, sodium chloride is a white crystalline strong, which is similar to magnesium oxide. But complexes generally of Copper oxide and Iron oxide are colored compounds. To understand this we should recognize how coloration is made from compounds.
Complete step by step solution:
Transition metal ions are colored because of the $d - d$ transition. Unpaired electrons in d orbitals are agitated from lower strength stage to better power strength stage. For this, the strength from visible places is absorbed.
$S{c^{3 + }}$ has outer electronic configuration of $3{d^0}\;4{s^0}$ with zero unpaired electrons. Hence, it is colorless as no electron is unpaired.
$T{i^{3 + }}$ has outer electronic configuration with one unpaired electron. Hence, it is purple colored because there’s at least one electron free.
Additional information:
Most of the chemical compounds are colored because the electron shells have different energies. Electron shells can be dealt with as strength stages with the ones which could be far away from the nucleus which is being higher in power. All atoms are capable of soaked up strength and redeeming this energy again. This happens when power is absorbed through an atom or compound and an electron inside the shells is excited or high. When its miles are excited it actions from a lower strength level/shell to a higher power level/shell. When it comes back off from the higher energy stage/shell it has to release the power that it absorbed from there. This occurs when the electron comes back track to its unique strength level releasing the greater energy as a photon.
Note:
Most compounds of metals and non metals now do not show up coloration. For example, sodium chloride is a white crystalline strong, which is similar to magnesium oxide. But complexes generally of Copper oxide and Iron oxide are colored compounds. To understand this we should recognize how coloration is made from compounds.
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