Why do ions of the first transition series exhibit paramagnetism? State the calculated and experimental observed magnetic moments of these ions.

(N/A) Paramagnetism arises due to the presence of unpaired electrons.
Each unpaired electron undergoes two types of motion: $(i)$ spin angular momentum and $(ii)$ orbital angular momentum. For the first series of transition metal compounds,the contribution of orbital angular momentum is effectively quenched,and thus its contribution to the magnetic moment value is not significant. Therefore,the magnetic moment value is determined by the number of unpaired electrons present.
The magnetic moment is calculated using the 'spin-only' formula,which is as follows:
$\mu = \sqrt{n(n+2)} \ BM$,where $n =$ number of unpaired electrons.
$\mu =$ magnetic moment (unit is Bohr Magneton $(BM)$).
Thus,as the number of unpaired electrons increases,the value of the magnetic moment increases.
The table below provides the calculated and experimental observed magnetic moment values:
| Ion | Electronic Configuration | Unpaired Electrons | Calculated Magnetic Moment $(BM)$ | Observed Magnetic Moment $(BM)$ |
| :--- | :--- | :--- | :--- | :--- |
| $Sc^{3+}$ | $3d^0$ | $0$ | $0$ | $0$ |
| $Ti^{3+}$ | $3d^1$ | $1$ | $1.73$ | $1.75$ |
| $Ti^{2+}$ | $3d^2$ | $2$ | $2.84$ | $2.76$ |
| $V^{2+}$ | $3d^3$ | $3$ | $3.87$ | $3.86$ |
| $Cr^{2+}$ | $3d^4$ | $4$ | $4.90$ | $4.80$ |
| $Mn^{2+}$ | $3d^5$ | $5$ | $5.92$ | $5.96$ |
| $Fe^{2+}$ | $3d^6$ | $4$ | $4.90$ | $5.3 - 5.5$ |
| $Co^{2+}$ | $3d^7$ | $3$ | $3.87$ | $4.4 - 5.2$ |
| $Ni^{2+}$ | $3d^8$ | $2$ | $2.84$ | $2.9 - 3.4$ |
| $Cu^{2+}$ | $3d^9$ | $1$ | $1.73$ | $1.8 - 2.2$ |
| $Zn^{2+}$ | $3d^{10}$ | $0$ | $0$ | $0$ |