Identify the oxidation states of titanium (Z= | vedclass

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(C) The electronic configuration of titanium $(Z=22)$ is $[Ar] 3d^2 4s^2$. In its $Ti^{4+}$ state,it loses all $4$ valence electrons,resulting in a $3

Vedclass · Accepted Answer

$Ti^{4+}, Cu^{1+}$

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(C) The electronic configuration of titanium $(Z=22)$ is $[Ar] 3d^2 4s^2$. In its $Ti^{4+}$ state,it loses all $4$ valence electrons,resulting in a $3d^0$ configuration,which makes it colourless due to the absence of $d-d$ transitions.The electronic configuration of copper $(Z=29)$ is $[Ar] 3d^{10} 4s^1$. In its $Cu^+$ state,it loses the $4s$ electron,resulting in a $3d^{10}$ configuration. Since the $d$-subshell is completely filled,no $d-d$ transition is possible,making $Cu^+$ compounds colourless.

List-$I$	List-$II$
$(1) \ \text{Cr}$	$(A) \ [\text{Ar}] \ 3d^8 4s^0$
$(2) \ \text{Fe}^{2+}$	$(B) \ [\text{Ar}] \ 3d^{10} 4s^1$
$(3) \ \text{Ni}^{2+}$	$(C) \ [\text{Ar}] \ 3d^6 4s^0$
$(4) \ \text{Cu}$	$(D) \ [\text{Ar}] \ 3d^5 4s^1$
	$(E) \ [\text{Ar}] \ 3d^6 4s^2$

Identify the oxidation states of titanium $(Z=22)$ and copper $(Z=29)$ in their colourless compounds.

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