Number of paramagnetic ions among the following d- and f-block...

The magnetic nature of an ion depends on the presence of unpaired electrons. Ions with at least one unpaired electron are paramagnetic, while those with all paired electrons are diamagnetic.

Let us write the electronic configurations of the given ions:

$Mn^{2+} (Z=25) : [Ar] 3d^5$ $\rightarrow$ $5$ unpaired electrons (Paramagnetic)
$Cu^{2+} (Z=29) : [Ar] 3d^9$ $\rightarrow$ $1$ unpaired electron (Paramagnetic)
$Zn^{2+} (Z=30) : [Ar] 3d^{10}$ $\rightarrow$ $0$ unpaired electrons (Diamagnetic)
$Yb^{2+} (Z=70) : [Xe] 4f^{14}$ $\rightarrow$ $0$ unpaired electrons (Diamagnetic)
$Sc^{3+} (Z=21) : [Ar] 3d^0$ $\rightarrow$ $0$ unpaired electrons (Diamagnetic)
$La^{3+} (Z=57) : [Xe] 4f^0$ $\rightarrow$ $0$ unpaired electrons (Diamagnetic)
$Gd^{3+} (Z=64) : [Xe] 4f^7$ $\rightarrow$ $7$ unpaired electrons (Paramagnetic)
$Lu^{3+} (Z=71) : [Xe] 4f^{14}$ $\rightarrow$ $0$ unpaired electrons (Diamagnetic)
$Ti^{4+} (Z=22) : [Ar] 3d^0$ $\rightarrow$ $0$ unpaired electrons (Diamagnetic)
$Ce^{4+} (Z=58) : [Xe] 4f^0$ $\rightarrow$ $0$ unpaired electrons (Diamagnetic)

The paramagnetic ions are $Mn^{2+}$, $Cu^{2+}$, and $Gd^{3+}$.

Total number of paramagnetic ions = $3$.

Answer: $3$