The calculated magnetic moments (spin only value) for species [FeCl4]2-, [Co(C2O4)3]3- and MnO42- respectively are:

🧠 Three Independent Spin Counts

Read the three species, find each one's unpaired count, then plug into $\mu = \sqrt{n(n+2)}$:

| Species | Metal | d-count | Geometry/Field | Unpaired | μ (BM) | |---|---|---|---|---|---| | $[\mathrm{FeCl_4}]^{2-}$ | $\mathrm{Fe^{2+}}$ | $\mathrm{d^6}$ | tetrahedral, weak field $\mathrm{Cl^-}$ → high-spin | 4 | $\sqrt{24} \approx 4.90$ | | $[\mathrm{Co(C_2O_4)_3}]^{3-}$ | $\mathrm{Co^{3+}}$ | $\mathrm{d^6}$ | octahedral, oxalate → low-spin | 0 | 0 | | $\mathrm{MnO_4^{2-}}$ | $\mathrm{Mn^{6+}}$ | $\mathrm{d^1}$ | tetrahedral oxide → 1 unpaired | 1 | $\sqrt{3} \approx 1.73$ |

Triplet: 4.90, 0, 1.73 — option (1).

🗺️ Verify Each

$[\mathrm{FeCl_4}]^{2-}$. $\mathrm{Fe^{2+}}\,(\mathrm{d^6})$ in a tetrahedral $\mathrm{Cl^-}$ field. Tetrahedral $\Delta_t$ is much smaller than octahedral $\Delta_o$ — all tetrahedral complexes (with rare exceptions) are high-spin. d⁶ tetrahedral high-spin → 4 unpaired → 4.90 BM.

$[\mathrm{Co(C_2O_4)_3}]^{3-}$. $\mathrm{Co^{3+}}\,(\mathrm{d^6})$. Oxalate is moderate-strong; with the high-charge $\mathrm{Co^{3+}}$ this is enough for low-spin → $\mathrm{t_{2g}^6\,e_g^0}$ → 0 unpaired → μ = 0.

$\mathrm{MnO_4^{2-}}$. Manganate(VI). Mn has lost 6 electrons → $\mathrm{d^1}$. One unpaired electron → μ = $\sqrt{3} \approx 1.73$ BM.

⚡ The "Tetrahedral Always High-Spin" Shortcut

In tetrahedral fields, $\Delta_t \approx \frac{4}{9}\Delta_o$, which is almost always smaller than the pairing energy. So tetrahedral = high-spin for routine first-row complexes. This collapses many calculations to just "$d^n$ atom → $n$ unpaired (for $n \le 5$) or $10-n$ unpaired".

⚠️ Permanganate vs Manganate

$\mathrm{MnO_4^-}$ (permanganate) is $\mathrm{Mn(VII)}\,\mathrm{d^0}$ → diamagnetic, μ = 0. $\mathrm{MnO_4^{2-}}$ (manganate) is $\mathrm{Mn(VI)}\,\mathrm{d^1}$ → 1 unpaired, μ = 1.73 BM.

A single charge difference flips d-count by 1 — the reason option (3) (which writes 0 for MnO₄²⁻) is wrong.

$\boxed{\text{Answer: (1) 4.90, 0 and 1.73 BM}}$