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Match the following with their condensed electronic configuration:

$\begin{array} {ll} \text{Element} & \text{Electron Configuration}\\ 1.\; Zr & a.\;[Ar]3d^2 \\ 2. \; V^{3+} & b.\; [Kr]5s^24d^2 \\ 3. \;Mo^{3+} & c.\;[Kr]4d^2 \end{array}$

$\begin {array} {1 1} (1) \rightarrow a; \quad (2) \rightarrow b; \quad (3) \rightarrow c \\ (1) \rightarrow b; \quad (2) \rightarrow a; \quad (3) \rightarrow c \\ (1) \rightarrow b; \quad (2) \rightarrow c; \quad (3) \rightarrow a \\ (1) \rightarrow c; \quad (2) \rightarrow a; \quad (3) \rightarrow b\end {array}$

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Answer: $(1) \rightarrow b; \quad (2) \rightarrow a; \quad (3) \rightarrow c$
The elements are in Period 4 and 5 so the general configuration is $ns^2(n-1)d^x$
1. Zr is the second elements in 4d series $\rightarrow [Kr]5s^24d^2$
2. V is the third element in the 3d series. To form $V^{3+}$, 3 electrons are lost (two 4s and one 3d) $\rightarrow [Ar] 3d^2$
3. Mo lies below Cr in Group 6B(6) so we expect the same exception as for Cr. Mo is hence $[Kr] 5s^14d^5$. To form $Mo^{3+}$, Mo loses one 5s and two of the 4d electroncs $\rightarrow [Kr] 4d^3$
answered Aug 2, 2014 by balaji.thirumalai
 

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