Question

For the given coordination compounds:
1) \([Co(ox)_3]^{3-}\) 2) \([Fe(CN)_6]^{3-}\) 3) \([Ni(CN)_4]^{2-}\) 4) \([NiCl_4]^{2-}\)
5) \([Ni(CO)_4]\) 6) \([MnBr_4]^{2-}\) 7) \([CoF_6]^{3-}\) 8) \([Cr(H_2O)_3F_3]\)
The number of paramagnetic species are:

Answer 1

Correct Answer: 3

Step 1: Understanding the Question:
Count the number of coordination complexes that have at least one unpaired electron (paramagnetic).
Step 2: Detailed Explanation:
1) $[Co(ox)_3]^{3-}$: $Co^{3+} (d^6)$. Oxalate is a chelating strong-field ligand with $Co^{3+}$, causing pairing. $\rightarrow$ $t_{2g}^6 e_g^0$ (Diamagnetic).
2) $[Fe(CN)_6]^{3-}$: $Fe^{3+} (d^5)$. $CN^-$ is strong field. $\rightarrow$ $t_{2g}^5 e_g^0$ (1 unpaired e$^-$, Paramagnetic).
3) $[Ni(CN)_4]^{2-}$: $Ni^{2+} (d^8)$. $CN^-$ is strong field, square planar geometry. $\rightarrow$ $dsp^2$ (Diamagnetic).
4) $[NiCl_4]^{2-}$: $Ni^{2+} (d^8)$. $Cl^-$ is weak field, tetrahedral. $\rightarrow$ $sp^3$ (2 unpaired e$^-$, Paramagnetic).
5) $[Ni(CO)_4]$: $Ni^0 (3d^8 4s^2 \rightarrow 3d^{10})$. $CO$ is strong field. $\rightarrow$ $sp^3$ (Diamagnetic).
6) $[MnBr_4]^{2-}$: $Mn^{2+} (d^5)$. $Br^-$ is weak field. $\rightarrow$ $sp^3$ (5 unpaired e$^-$, Paramagnetic).
7) $[CoF_6]^{3-}$: $Co^{3+} (d^6)$. $F^-$ is weak field. $\rightarrow$ $sp^3d^2$ (4 unpaired e$^-$, Paramagnetic).
8) $[Cr(H_2O)_3F_3]$: $Cr^{3+} (d^3)$. Both $H_2O$ and $F^-$ are weak field/neutral. $\rightarrow$ $t_{2g}^3 e_g^0$ (3 unpaired e$^-$, Paramagnetic).

The paramagnetic species are 2, 4, 6, 7, and 8. Total = 5.
Step 3: Final Answer:
There are 5 paramagnetic species.