Coloured complexes - class-XI

The colour of the following compounds are:

Since $Zn^{+2}$ have fully filled $d$ orbital having no unpaired electrons,it does not undergo electronic transitions,hence colourless.However,rest of the elements given as $Ni^{+2}$ , $Cu^{+2}$ and $Cr^{+2}$ has unpaired electrons which can undergo electronic transitions and hence show colours.

$Cu^{+}$ and $Ti^{+4}$ have fully filled and empty d orbital respectively. Hence due to absence of unpaired electrons,they will be colourless. But $Fe^{+2}$ and $Co^{+2}$chave unpaired electrons. Hence they will be coloured.

TiF$ _6^{2-}$ -

Colour of transition metal ions are due to absorption of the same wavelength. This results in $d-d$ transition.

In Na3VO4, vanadium is in the +5 oxidation state (d0 configuration). The color arises from ligand-to-metal charge transfer (LMCT) rather than d-d transitions.

Both potassium dichromate and potassium permanganate derive their intense colors from ligand-to-metal charge transfer (LMCT) transitions, where electrons are transferred from oxygen orbitals to empty metal d-orbitals. Copper sulfate color is due to d-d transitions, making options A and B the correct choices.

Solution:- (B) ${Sc}^{+3}$

coloured compound of transition elements is assosiated with partially filled (n-1)d orbitals. the transition metal ions containing unpaired d-electrons undergoes electronic transition from one d-orbital to another. during this d-d transition process the electrons absorb certain energy from the radiation and emit the remainder of energy as colored light. the color of ion is complementary of the color absorbed by it. hence, colored ion is formed due to d-d transition which falls in visible region for all transition elements.

Transition metals are often coloured.

$Cu\longrightarrow $ Blue colour

Colour of $Fe^{2+}$ is light green.

The silver salts like$AgBr$ and $AgI$ are colored, the color of transition metal ions is due  to the presence of unpaired electron . The electrons which absorbs radiations of color that from the visible or UV spectrum and undergo transition from ground state to excited state within d-sub shells,so it is d-d transitions of electrons.

The outer shell electronic configuration of $Cu^{2+}$ has $3d^{9}$ configuration and contain 1  unpaired electron, thus it is coloured.

The color in $KMnO _4$ arises from an electronic transition.

In the permanganate ion $MnO _4^{-}$, manganese is in the +7 oxidation state, so it has no d electrons. Photons promote an electron from the highest energy molecular orbital to an empty d orbital on the manganese. When a photon of light is absorbed, this charge transfer takes place from L to M.

During this d-d transition process, the electrons absorb certain energy from the radiation and emit the remainder of energy as colored light. The color of ion is complementary of the color absorbed by it. hence, colored ion is formed due to d-d transition which falls in the visible region for all transition elements.

TiF6^2- is colorless (Ti^4+ is d^0, no d-d transitions possible). Cu2Cl2 is also colorless/slightly yellowish as it's copper(I) with full d^10 configuration. The other complexes have partially filled d-orbitals giving color: CoF6^3- (green, Co^3+ d^6) and NiCl4^2- (blue, Ni^2+ d^8). Note: Option D has a typo 'CiF' should be 'CoF'.

Zinc oxide is white at room temperature, but upon heating, it loses oxygen reversibly and forms zinc excess defects. The trapped electrons in interstitial sites occupy specific energy levels, absorbing blue light and reflecting a yellow color. Thus, heated ZnO appears yellow.

The intensely purple color of the permanganate ion is not due to d-d transitions because manganese is in the +7 oxidation state with a d0 configuration. Instead, it arises from a ligand-to-metal charge transfer (LMCT), where electrons move from the oxygen ligand orbitals to the empty metal orbitals.

V(4+) (as vanadyl ion) and Cu(2+) (as hydrated copper ion) are well-known for their blue colors in aqueous solutions.

Both K2Cr2O7 and KMnO4 exhibit color due to charge transfer transitions (LMCT) because the metal centers are in high oxidation states with d0 configurations.

Vanadium in K3[VF6] is in the +3 oxidation state with a d2 configuration and undergoes d-d transitions, making it colored. However, the statement claims it is a colourless compound, which is incorrect since many V(III) complexes show color. Therefore, option C describes the false statement.

${Ni}^{2+}$ is light green, ${Fe}^{2+}$ is pale green, 

In case of transition elements, where the d-level is not completely filled, it is possible to promote an electron from a lower energy d-orbital to a higher energy d-orbital by absorption of a photon of electromagnetic radiation having an appropriate energy. Electromagnetic radiations in the visible region of the spectrum often possess the appropriate energy for such transitions. 

$ VCl _3$ is green.

$ Co^{2+}$ = $3d^7$  

Sc(3+) has a d0 electronic configuration, meaning there are no d-electrons available for d-d transitions, resulting in a colorless compound.

As both $Zn^{2+}  = d^{10}, Sc^{3+} = d^0$ have zero unpaired d-electron so they both are colourless.

compound with unpaired d-electron will show color.

In copper (I) ion there are no vacant d orbitals as it is diamagnetic. Copper(I) ion being less charged has small ligand field effect and the transition is in the infrared region in which no color is perceived by human eye.

Transition metal ion                     Colour

$(A)$ $Cr\quad [Ar]3d^54s^1$

The reaction between MnO$ _2$ and KOH in presence of air is given as:

Solution:- (B) ${d} _{xy}, {d} _{yz}$