Total Number of Paramagnetic Species Among Given Coordination Compounds

To determine the total number of paramagnetic compounds among the given species, we examine the electronic configuration and magnetic behavior of each species individually:

1. $[\text{Ni}(\text{CO})_4]$:

   • The oxidation state of nickel is $0$.
   • The ground state electronic configuration of $\text{Ni}^0$ is $[\text{Ar}] 3d^8 4s^2$.
   • Since $\text{CO}$ is a strong field ligand, pairing of electrons takes place, resulting in a $3d^{10} 4s^0$ configuration ($sp^3$ hybridization).
   • Number of unpaired electrons ($n$) = $0$.
   • Therefore, it is diamagnetic.

2. $[\text{NiCl}_4]^{2-}$:

   • The oxidation state of nickel is $+2$.
   • The electronic configuration of $\text{Ni}^{2+}$ is $[\text{Ar}] 3d^8$.
   • $\text{Cl}^-$ is a weak field ligand, forming a tetrahedral complex ($sp^3$ hybridization) without pairing of $3d$ electrons.
   • Number of unpaired electrons ($n$) = $2$.
   • Therefore, it is paramagnetic.

3. $[\text{Co}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}$:

   • The oxidation state of cobalt is $+3$.
   • The electronic configuration of $\text{Co}^{3+}$ is $[\text{Ar}] 3d^6$.
   • $\text{Co}^{3+}$ forms a low-spin octahedral complex with $\text{NH}_3$ as the main ligand, causing pairing of electrons ($t_{2g}^6 e_g^0$).
   • Number of unpaired electrons ($n$) = $0$.
   • Therefore, it is diamagnetic.

4. $\text{Na}_3[\text{CoF}_6]$:

   • The complex ion is $[\text{CoF}_6]^{3-}$, where the oxidation state of cobalt is $+3$.
   • The electronic configuration of $\text{Co}^{3+}$ is $[\text{Ar}] 3d^6$.
   • $\text{F}^-$ is a weak field ligand, forming a high-spin octahedral complex ($t_{2g}^4 e_g^2$).
   • Number of unpaired electrons ($n$) = $4$.
   • Therefore, it is paramagnetic.

5. $\text{Na}_2\text{O}_2$:

   • Contains the peroxide ion, $\text{O}_2^{2-}$.
   • The molecular orbital electronic configuration for $\text{O}_2^{2-}$ (18 electrons) is: \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi^*_{2p_x}^2 = \pi^*_{2p_y}^2)
   • All electrons are paired ($n = 0$).
   • Therefore, it is diamagnetic.

6. $\text{CsO}_2$:

   • Contains the superoxide ion, $\text{O}_2^-$.
   • The molecular orbital electronic configuration for $\text{O}_2^-$ (17 electrons) is: \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi^*_{2p_x}^2 = \pi^*_{2p_y}^1)
   • It contains one unpaired electron ($n = 1$) in the antibonding $\pi^*$ orbital.
   • Therefore, it is paramagnetic.

Summary:

The paramagnetic species are:

1. $[\text{NiCl}_4]^{2-}$
2. $\text{Na}_3[\text{CoF}_6]$
3. $\text{CsO}_2$

The total number of paramagnetic compounds is $3$.

Thus, the correct option is B.