An ion which has 18 electrons in the outermos | vedclass

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(D) The electronic configurations of the given ions are as follows:$Cu (Z=29) = [Ar] 3d^{10} 4s^1$. Therefore,$Cu^{+} = [Ar] 3d^{10}$,which has $18$ e

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$Cu^{+}$

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(D) The electronic configurations of the given ions are as follows:$Cu (Z=29) = [Ar] 3d^{10} 4s^1$. Therefore,$Cu^{+} = [Ar] 3d^{10}$,which has $18$ electrons in the outermost shell $(3s^2 3p^6 3d^{10})$.$K (Z=19) = [Ar] 4s^1$. Therefore,$K^{+} = [Ar]$,which has $8$ electrons in the outermost shell.$Cs (Z=55) = [Xe] 6s^1$. Therefore,$Cs^{+} = [Xe]$,which has $8$ electrons in the outermost shell.$Th (Z=90) = [Rn] 6d^2 7s^2$. Therefore,$Th^{4+} = [Rn]$,which has $8$ electrons in the outermost shell.Thus,$Cu^{+}$ is the ion that has $18$ electrons in its outermost shell.

Aspects	Metal
$a$. The metal which reveals a maximum number of oxidation states	$i$. Scandium
$b$. The metal although placed in $3d$ block is considered not as a transition element	$ii$. Copper
$c$. The metal which does not exhibit variable oxidation states	$iii$. Manganese
$d$. The metal which in $+1$ oxidation state in aqueous solution undergoes disproportionation	$iv$. Zinc

An ion which has $18$ electrons in the outermost shell is

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