Determining 3d Electrons in Transition Metal Ions

The question asks us to arrange the ions $Cr^{2+}$, $Cu^{+}$, $Ti^{3+}$, and $Mn^{+}$ based on the increasing number of electrons in their 3d subshell. To do this, we need to determine the electron configuration for each ion. We'll start by finding the electron configuration of the neutral atom and then remove electrons according to the ion's charge, remembering that electrons are removed from the outermost shell (highest principal quantum number, n) first. For transition metals, this usually means removing electrons from the ns subshell before the (n-1)d subshell.

Electron Configurations of Neutral Atoms

• Titanium (Ti): Atomic number (Z) = 22. Electron configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2$, which can be shortened to $[Ar] 3d^2 4s^2$.
• Chromium (Cr): Atomic number (Z) = 24. Electron configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$, which is $[Ar] 3d^5 4s^1$. This is an exception to the usual filling order.
• Manganese (Mn): Atomic number (Z) = 25. Electron configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2$, which is $[Ar] 3d^5 4s^2$.
• Copper (Cu): Atomic number (Z) = 29. Electron configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$, which is $[Ar] 3d^{10} 4s^1$. This is also an exception.

Electron Configurations and 3d Electrons of the Ions

Now, let's find the configurations for the given ions:

• (A) $Cr^{2+}$: Neutral Cr is $[Ar] 3d^5 4s^1$. To form $Cr^{2+}$, we remove two electrons. The first electron is removed from the 4s orbital, and the second is removed from the 3d orbital. Configuration: $[Ar] 3d^4$. Number of 3d electrons = 4.
• (B) $Cu^{+}$: Neutral Cu is $[Ar] 3d^{10} 4s^1$. To form $Cu^{+}$, we remove one electron from the 4s orbital. Configuration: $[Ar] 3d^{10}$. Number of 3d electrons = 10.
• (C) $Ti^{3+}$: Neutral Ti is $[Ar] 3d^2 4s^2$. To form $Ti^{3+}$, we remove three electrons. The two electrons are removed from the 4s orbital, and one electron is removed from the 3d orbital. Configuration: $[Ar] 3d^1$. Number of 3d electrons = 1.
• (D) $Mn^{+}$: Neutral Mn is $[Ar] 3d^5 4s^2$. To form $Mn^{+}$, we remove one electron from the 4s orbital. Configuration: $[Ar] 3d^5 4s^1$. Number of 3d electrons = 5.

Ordering the Ions

We need to arrange the ions in increasing order of their 3d electrons:

• $Ti^{3+}$ (C): 1 electron
• $Cr^{2+}$ (A): 4 electrons
• $Mn^{+}$ (D): 5 electrons
• $Cu^{+}$ (B): 10 electrons

Therefore, the increasing order is $Ti^{3+} < Cr^{2+} < Mn^{+} < Cu^{+}$. This corresponds to the sequence (C), (A), (D), (B).