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Chemistry · 2024

JEE Main · 30 January 2024, Shift 1 · Q66

Match List - I with List - II. Choose the correct answer from the options given below:

Match List - I with List - II.
List - IList - II
SpeciesElectronic distribution
(A)$\displaystyle \mathrm{Cr}^{+2}$(I)$\displaystyle 3 \mathrm{~d}^8$
(B)$\displaystyle \mathrm{Mn}^{+}$(II)$\displaystyle 3 \mathrm{~d}^3 4 \mathrm{~s}^1$
(C)$\displaystyle \mathrm{Ni}^{+2}$(III)$\displaystyle 3 \mathrm{~d}^4$
(D)$\displaystyle \mathrm{V}^{+}$(IV)$\displaystyle 3 \mathrm{~d}^5 4 \mathrm{~s}^1$
Choose the correct answer from the options given below :
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JEE Main 2024 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.