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(B) The electronic configuration of $Zn$ $(Z=30)$ is $[Ar] 3d^{10} 4s^2$.In transition elements,variable oxidation states arise due to the participati

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completion of $(n-1)d$ orbitals

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(B) The electronic configuration of $Zn$ $(Z=30)$ is $[Ar] 3d^{10} 4s^2$.In transition elements,variable oxidation states arise due to the participation of both $(n-1)d$ and $ns$ electrons in bonding.Since $Zn$ has a completely filled $3d$ subshell $(3d^{10})$,the $d$-electrons are stable and do not participate in chemical bonding.Therefore,$Zn$ only exhibits a $+2$ oxidation state by losing its two $4s$ electrons,failing to show variable oxidation states.

List-$I$ (Element)	List-$II$ (Block)
$(A)$ $Ra$	$(I)$ $p$-block
$(B)$ $Uuq$	$(II)$ $s$-block
$(C)$ $Ds$	$(III)$ $f$-block
$(D)$ $Fm$	$(IV)$ $d$-block

Transition elements are usually characterised by variable oxidation states,but $Zn$ does not show this property because of:

Similar Questions

Match the following:
List-$I$ (Element) List-$II$ (Block)
$(A)$ $Ra$ $(I)$ $p$-block
$(B)$ $Uuq$ $(II)$ $s$-block
$(C)$ $Ds$ $(III)$ $f$-block
$(D)$ $Fm$ $(IV)$ $d$-block

The correct answer is

The number of $d$-electrons in $Cr^{3+}$ ion $(Z=24)$ is

Maximum oxidation state is shown by

The element with the electronic configuration $[Kr] \ 4d^{10} \ 5s^0$ is

The atomic radii of the elements are almost the same for which series?

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