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Match the following with their condensed electronic configuration:

$\begin{array} {ll} \text{Element} & \text{Electron Configuration}\\ (1)\; Cr^{2+} & a.\;[Ar]3d^9 \\ (2) \; Zn^{2+} & b.\;[Ar]3d^7 \\ (3) \;Ni^{3+} & c.\; [Ar]3d^{10}\\ (4) \; Cu^{2+} & d.\;[Ar]3d^4 \end{array}$

$\begin {array} {1 1} (1) \rightarrow b\; \quad (2) \rightarrow c \; \quad (3) \rightarrow d\; \quad (4) \rightarrow c \\ (1) \rightarrow d\; \quad (2) \rightarrow c \; \quad (3) \rightarrow b\; \quad (4) \rightarrow a \\ (1) \rightarrow d\; \quad (2) \rightarrow b \; \quad (3) \rightarrow a\; \quad (4) \rightarrow c \\ (1) \rightarrow c\; \quad (2) \rightarrow a \; \quad (3) \rightarrow d\; \quad (4) \rightarrow b\end {array}$

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Answer: $(1) \rightarrow d\; \quad (2) \rightarrow c \; \quad (3) \rightarrow b\; \quad (4) \rightarrow a$
The elements are in Period 4 so the general configuration is $ns^2(n-1)d^x$
1. Cr is the 4th element in the series but there is an exception made for Cr because the listed electronic configurations are energetically favoured $\rightarrow 3d^54s^1$. To form $Cr^{2+}$, it must lose one 4s and one 3d $\rightarrow [Ar] 3d^4$
2. Zn is the last element in the series and to form $Zn^{2+}$ it must lose two 4s electrons $\rightarrow [Ar] 3d^{10}$
2. Ni is the 8th element in the series and to form $Ni^{3+}$ it must lose two 4s electrons and one 3d $\rightarrow [Ar] 3d^{7}$
4. Cu is the 9th element in the series but there is an exception made for Cu because the listed electronic configurations are energetically favoured $\rightarrow 3d^{10}4s^1$. To form $Cu^{2+}$, it must lose one 4s and one 3d $\rightarrow [Ar] 3d^9$
answered Aug 2, 2014 by balaji.thirumalai
edited Aug 2, 2014 by balaji.thirumalai
 

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