Out of $Sc^{3+}$, $Cr^{3+}$, $Ti^{4+}$ and $Fe^{2+}$ which pair contains both colourless ions:

Question

Accepted Answer

The correct option is (1):

Ions of transition metals are typically colored due to d-d electronic transitions. For an ion to be colorless, it must either have a completely empty d-orbital ($d^0$ configuration) or a completely filled d-orbital ($d^{10}$ configuration). This is because the absence of unpaired d-electrons or available d-orbitals for transition prevents the absorption of visible light, thus making the ion appear colorless. Let's analyze the electronic configuration of each ion:  * **$Sc^{3+}$:** Scandium (Sc) has an atomic number of 21, and its electronic configuration is $$$Ar$$ 3d^1 4s^2$. When it forms a $Sc^{3+}$ ion, it loses all three valence electrons (two from $4s$ and one from $3d$). Therefore, the electronic configuration of $Sc^{3+}$ is $$$Ar$$ 3d^0$. Since the d-orbital is completely empty, $Sc^{3+}$ is a colorless ion.

* **$Ti^{4+}$:** Titanium (Ti) has an atomic number of 22, and its electronic configuration is $$$Ar$$ 3d^2 4s^2$. When it forms a $Ti^{4+}$ ion, it loses all four valence electrons (two from $4s$ and two from $3d$). Therefore, the electronic configuration of $Ti^{4+}$ is $$$Ar$$ 3d^0$. Since the d-orbital is completely empty, $Ti^{4+}$ is also a colorless ion.Thus, the pair $Sc^{3+}$ and $Ti^{4+}$ both contain colorless ions because both have a $d^0$ electronic configuration.

The incorrect option is (2): $Sc^{3+}$ and $Cr^{3+}$
While $Sc^{3+}$ is colorless ($3d^0$), $Cr^{3+}$ (Chromium, atomic number 24) has an electronic configuration of $$$Ar$$ 3d^3$. The presence of three unpaired d-electrons allows for d-d transitions, making $Cr^{3+}$ colored (typically violet).

The incorrect option is (3): $Cr^{3+}$ and $Ti^{4+}$
As explained above, $Cr^{3+}$ is a colored ion due to its $3d^3$ configuration, while $Ti^{4+}$ is colorless. Therefore, this pair does not contain both colorless ions.

The incorrect option is (4): $Sc^{3+}$ and $Fe^{2+}$
While $Sc^{3+}$ is colorless ($3d^0$), $Fe^{2+}$ (Iron, atomic number 26) has an electronic configuration of $$$Ar$$ 3d^6$. The presence of four unpaired d-electrons allows for d-d transitions, making $Fe^{2+}$ colored (typically pale green).