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(B) To determine the number of unpaired $d$-electrons,we look at the electronic configuration of each ion:$Zn^{+}$ $(Z=30)$: $[Ar] \, 3d^{10} 4s^1$. N

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$Fe^{2+}$

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(B) To determine the number of unpaired $d$-electrons,we look at the electronic configuration of each ion:$Zn^{+}$ $(Z=30)$: $[Ar] \, 3d^{10} 4s^1$. Number of unpaired $d$-electrons = $0$.$Fe^{2+}$ $(Z=26)$: $[Ar] \, 3d^6 4s^0$. The $3d^6$ configuration has $4$ unpaired electrons.$N^{3+}$ $(Z=7)$: $1s^2 2s^2 2p^6$. This ion has no $d$-electrons.$Cu^{+}$ $(Z=29)$: $[Ar] \, 3d^{10} 4s^0$. Number of unpaired $d$-electrons = $0$.Comparing these,$Fe^{2+}$ has the most unpaired $d$-electrons $(4)$.

List-$I$	List-$II$
$A$. Titanium	$I$. Non-metal
$B$. Fluorine	$II$. Transition metal
$C$. Tellurium	$III$. Lanthanoid
$D$. Dysprosium	$IV$. Metalloid

List-$I$ (Element)	List-$II$ (Block)
$A$. $Cd$	$I$. $f$-block
$B$. $Eu$	$II$. $s$-block
$C$. $Se$	$III$. $d$-block
$D$. $Ba$	$IV$. $p$-block

Which of the following has the most unpaired $d$-electrons?

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Niobium and tantalum metals are used in making surgical instruments because they are

Match the following:
List-$I$ List-$II$
$A$. Titanium $I$. Non-metal
$B$. Fluorine $II$. Transition metal
$C$. Tellurium $III$. Lanthanoid
$D$. Dysprosium $IV$. Metalloid

Technetium,the element below manganese in the Periodic Table,would be expected to have high values for its:
$(I)$ melting point
$(II)$ boiling point
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Match the following:
List-$I$ (Element) List-$II$ (Block)
$A$. $Cd$ $I$. $f$-block
$B$. $Eu$ $II$. $s$-block
$C$. $Se$ $III$. $d$-block
$D$. $Ba$ $IV$. $p$-block

The correct answer is:

$Zn$ and $Hg$ do not show variable valency like other $d$-block elements because ...

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