Match List I with List II.
| List I Species |
List II Electronic distribution |
||
|---|---|---|---|
| (A) | $\mathrm{Cr^{+2}}$ | (I) | $\mathrm{3d^8}$ |
| (B) | $\mathrm{Mn^+}$ | (II) | $\mathrm{3d^34s^1}$ |
| (C) | $\mathrm{Ni^{+2}}$ | (III) | $\mathrm{3d^4}$ |
| (D) | $\mathrm{V^+}$ | (IV) | $\mathrm{3d^54s^1}$ |
Choose the correct answer from the options given below :
Solution
<p>$$\begin{gathered}
{ }_{24} \mathrm{Cr} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^1 ; \mathrm{Cr}^{2+} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^4 \\
{ }_{25} \mathrm{Mn} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{s}^2 ; \mathrm{Mn}^{+} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{s}^1 \\
{ }_{28} \mathrm{Ni} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{s}^2 ; \mathrm{Ni}^{2+} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^8 \\
{ }_{23} \mathrm{V} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^3 4 \mathrm{s}^2 ; \mathrm{V}^{+} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^3 4 \mathrm{s}^1
\end{gathered}$$</p>
About this question
Subject: Chemistry · Chapter: Periodic Table and Periodicity · Topic: Periodic Trends
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