Question

which of the following compounds are colourless a) fe3+ b) ti4+ C) CO2+ D) ni2+​

Answer 1

option b) ti4

Ions which do not have unpaired electrons do not show colour in aqueous solution. These are known as diamagnetic ions.

Answer 2

Answer:

Out of the given compounds, the colourless compound is $Ti^{4+}$.

Therefore, option b) $Ti^{4+}$ is correct.

Explanation:

Out of the given compounds, the colourless compound is =?

Case 1) $Fe^{3+}$

As we know, the atomic number of iron ( $Fe$ ) is $26$.

Therefore,

• The electronic configuration of $Fe$ = $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{6}$
• The electronic configuration of $Fe^{3+}$ = $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{5}$

It has unpaired d-electrons. Therefore, the d-d transition occurred.

• The unpaired electrons present in the lower energy d-orbital get promoted to higher energy d-orbitals ( d-d transition ), by the absorption of visible light.

Hence, $Fe^{3+}$ is coloured because of the presence of unpaired electrons.

Case 2) $Ti^{4+}$

As we know, the atomic number of titanium ( $Ti$ ) is $22$.

Therefore,

• The electronic configuration of $Ti$ = $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{2}$
• The electronic configuration of $Ti^{4+}$ = $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}$

It has no unpaired d-electrons. Therefore, no d-d transition occurred.

Hence, $Ti^{4+}$ is colourless because of the absence of unpaired electrons.

Case 3) $Co^{2+}$

As we know, the atomic number of cobalt ( $Co$ ) is $27$.

Therefore,

• The electronic configuration of $Co$ = $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{7}$
• The electronic configuration of $Co^{2+}$ = $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{7}$

It has unpaired d-electrons. Therefore, the d-d transition occurred.

• The unpaired electrons present in the lower energy d-orbital get promoted to higher energy d-orbitals ( d-d transition ), by the absorption of visible light.

Hence, $Co^{2+}$ is coloured because of the presence of unpaired electrons.

Case 4) $Ni^{2+}$

As we know, the atomic number of nickel ( $Ni$ ) is $28$.

Therefore,

• The electronic configuration of $Ni$ = $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{8}$
• The electronic configuration of $Ni^{2+}$ = $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{8}$

It has unpaired d-electrons. Therefore, the d-d transition occurred.

• The unpaired electrons present in the lower energy d-orbital get promoted to higher energy d-orbitals ( d-d transition ), by the absorption of visible light.

Hence, $Ni^{2+}$ is coloured because of the presence of unpaired electrons.  

Therefore, the colourless compound is $Ti^{4+}$.

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