Question

How many of the following ions are paramagnetic
 $ S{c^{ + 3}} $ , $ T{i^{ + 2}} $ , $ Z{n^{ + 2}} $ , $ N{i^{ + 2}} $ , $ C{r^{ + 3}} $ , $ C{u^{ + 2}} $ , $ C{u^ + } $ , $ F{e^{ + 2}} $

Answer 1

Hint: Paramagnetic salts usually contain transition metal ions which have one or more unpaired electrons in their electronic configuration. The more the number of unpaired electrons, the more likely the atom or molecule is to show paramagnetism

Complete answer:
To find if the given species are paramagnetic or diamagnetic, write the electronic configurations of the species and check for unpaired electrons.
The electronic configuration of $ Sc $ is $ [Ar]3{d^1}4{s^2} $ . If three electrons are removed, $ S{c^{ + 3}} $ is formed and its electronic configuration is $ [Ar] $ . There are no unpaired electrons, so $ S{c^{ + 3}} $ is diamagnetic.
The electronic configuration of $ Ti $ is $ [Ar]3{d^2}4{s^2} $ . If two electrons are removed, $ T{i^{ + 2}} $ is formed and its electronic configuration is $ [Ar]3{d^2} $ . There are two unpaired electrons, so $ T{i^{ + 2}} $ is paramagnetic.
The electronic configuration of $ Zn $ is $ [Ar]3{d^{10}}4{s^2} $ . If two electrons are removed, $ Z{n^{ + 2}} $ is formed and its electronic configuration is $ [Ar]3{d^{10}} $ . There are no unpaired electrons, so $ Z{n^{ + 2}} $ is diamagnetic.
The electronic configuration of $ N{i^{ + 2}} $ is $ [Ar]3{d^8}4{s^2} $ . If two electrons are removed, $ N{i^{ + 2}} $ is formed and its electronic configuration is $ [Ar]3{d^8} $ . There are two unpaired electrons, so $ N{i^{ + 2}} $ is paramagnetic.
 The electronic configuration of $ C{r^{ + 3}} $ is $ [Ar]3{d^3} $ . There are three unpaired electrons, so $ C{r^{ + 3}} $ is paramagnetic.
 The electronic configuration of $ C{u^{ + 2}} $ is $ [Ar]3{d^9} $ .There is an unpaired electrons, so $ C{u^{ + 2}} $ is paramagnetic.
The electronic configuration of $ C{u^ + } $ is $ [Ar]3{d^{10}} $ . There are no unpaired electrons, so $ C{u^ + } $ is diamagnetic.
The electronic configuration of $ F{e^{ + 2}} $ is $ [Ar]3{d^6} $ . There are four unpaired electrons, so $ F{e^{ + 2}} $ is paramagnetic.
Therefore, five ions are paramagnetic.

Note:
More number of unpaired electrons makes the atom or ion or molecule more paramagnetic. Because these unpaired electrons align themselves in a fixed way with the orientation of the applied magnetic field and create magnetic dipole moments around each atom or molecule.