Question

The electronic configuration $\left[Kr\right]4d^{10}$ represents ____
($Z$ for$Sr=38$, $Sn=50$, $Te=52$, $Ag=47$)

• A. $S{r}^{2+}$
• B. $S{n}^{2+}$
• C. $T{e}^{2+}$
• D. $A{g}^{+}$

Answer 1

The correct option is D $A{g}^{+}$

(a) $S{r}^{2+}$
The atomic number of Strontium is $\left(Sr\right)=38$
Electronic configuration of $Sr:\left[Kr\right]5s^2$
The electronic configuration of $S{r}^{2+}\left(36\text{electrons}\right)\text{is}:\left[Kr\right]$


(b) $S{n}^{2+}$
The atomic number of Tin $\left(Sn\right)=50$
Electronic configuration of $Sn=\left[Kr\right]4d^{10}5s^{2}5p^2$
The electronic configuration of $S{n}^{2+}$ ( with 48 electrons) $=\left[Kr\right]4d^{10}5s^2$

(c) $T{e}^{2+}$
The atomic number of Tellurium $\left(Te\right)=52$
Electronic configuration of $Te=\left[Kr\right]4d^{10}5s^{2}5p^4$
Electronic configuration of $T{e}^{2+}$ ( 50 electrons) $=\left[Kr\right]4d^{10}5s^{2}5p^2$

(d) $A{g}^{+}$
The atomic number of silver $\left(Ag\right)=47$
Electronic configuration of $Ag=\left[Kr\right]4d^{10}5s^1$
Electronic configuration of $A{g}^{+}$ ( with 46 electrons) $=\left[Kr\right]4d^{10}$

Hence, The electronic configuration $\left[Kr\right]4d^{10}$ represents the $A{g}^{+}$ ion.