Question

which of the following has the electron configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3d^3$?
answer:
a $v^{3+}$
b $mn^{3+}$
c $cr^{2+}$
d $sc^{2+}$
e $v^{2+}$
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Explanation:

Step1: Calculate total electrons

Count electrons from the configuration:
$2+2+6+2+6+3=21$

Step2: Find neutral atom electron count

For each ion, add lost electrons to get neutral atom's electrons (atomic number):

• A. $\text{V}^{3+}$: $21 + 3 = 24$ (V has atomic number 23, no)
• B. $\text{Mn}^{3+}$: $21 + 3 = 24$ (Mn has atomic number 25, no)
• C. $\text{Cr}^{2+}$: $21 + 2 = 23$ (Cr has atomic number 24, no)
• D. $\text{Sc}^{2+}$: $21 + 2 = 23$ (Sc has atomic number 21, no)
• E. $\text{V}^{2+}$: $21 + 2 = 23$ (V has atomic number 23, yes)

Step3: Verify V neutral configuration

Vanadium (Z=23) neutral config: $1s^22s^22p^63s^23p^63d^34s^2$
Lose 2 4s electrons to form $\text{V}^{2+}$: $1s^22s^22p^63s^23p^63d^3$

Answer:

E. $\text{V}^{2+}$