Question

Which of the following will show paramagnetism?

• A. $Ar$
• B. $Ti$
• C. $S{c}^{3+}$
• D. $V^{5+}$

Answer 1

The correct option is B $Ti$

The elements or ions which have unpaired electrons have the ability to show the paramagnetic behaviour.

The electronic configuration of $Ar$ is
$1s^{2}2s^{2}2p^{6}3s^{2}3p^6$
So number of unpaired electrin in $Ar=0$

The electronic configuration of $Ti$ is
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$
So number of unpaired electrin in $Ti=2$

The electronic configuration of $S{c}^{3+}$ is
$1s^{2}2s^{2}2p^{6}3s^{2}3p^6$
So number of unpaired electrin in $S{c}^{3+}=0$

The electronic configuration of $V^{5+}$ is
$1s^{2}2s^{2}2p^{6}3s^{2}3p^6$
So number of unpaired electrin in $V^{5+}=0$