Question

in the series shown below, which of the following has the most unpaired electrons? select the single best answer. ar⁺, ar²⁺, ar³⁺, ar

Explanation:

Step1: Determine electron configuration of Ar

Argon (Ar) has an atomic number of 18, so its electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^6$. All orbitals are fully filled, so unpaired electrons = 0.

Step2: Electron configuration of $\boldsymbol{Ar^+}$

$Ar^+$ loses 1 electron, so electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^5$. The 3p orbital has 5 electrons (p orbital can hold 6). So unpaired electrons = 1 (since 3p has one unpaired electron: $\uparrow \downarrow, \uparrow \downarrow, \uparrow$).

Step3: Electron configuration of $\boldsymbol{Ar^{2+}}$

$Ar^{2+}$ loses 2 electrons, electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^4$. The 3p orbital has 4 electrons: $\uparrow \downarrow, \uparrow \downarrow, \uparrow, \uparrow$? Wait, no: p orbitals are three, each can hold 2. So 3p⁴: first two orbitals have 2 (paired), third has 2? No, electron filling follows Hund's rule: fill singly first. So 3p⁴: $\uparrow \downarrow, \uparrow, \uparrow$? Wait, no: 3p has three orbitals. So 4 electrons: first orbital: $\uparrow \downarrow$, second: $\uparrow$, third: $\uparrow$. Wait, no, Hund's rule: electrons fill orbitals singly with parallel spins before pairing. So for 3p⁴: three orbitals, so first three electrons: $\uparrow, \uparrow, \uparrow$ (each in separate orbitals), then the fourth pairs with one: $\uparrow \downarrow, \uparrow, \uparrow$. So unpaired electrons = 2.

Step4: Electron configuration of $\boldsymbol{Ar^{3+}}$

$Ar^{3+}$ loses 3 electrons, electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^3$. The 3p orbital has 3 electrons. By Hund's rule, each of the three p orbitals gets one electron (parallel spins), so unpaired electrons = 3.

Answer:

$Ar^{3+}$ has the most unpaired electrons (3), compared to $Ar$ (0), $Ar^+$ (1), $Ar^{2+}$ (2). So the answer is $Ar^{3+}$.