1. Check each compound for unpaired electrons

A compound is paramagnetic if it has one or more unpaired electrons.

2. $[Ni(CO)_4]$

• CO is a strong field ligand.
• Oxidation state of Ni is $0$.
• Electronic configuration of $Ni$: $Ni: [Ar]3d^8 4s^2$ So, $Ni^0 = 3d^{10}$.
• $d^{10}$ has all electrons paired.

Hence, $[Ni(CO)_4]$ is diamagnetic.

3. $[NiCl_4]^{2-}$

• Let oxidation state of Ni be $x$: $x+4(-1)=-2 \Rightarrow x=+2$
• So, $Ni^{2+}$ is $3d^8$.
• $Cl^-$ is a weak field ligand, and tetrahedral complexes are usually high spin.
• Tetrahedral $d^8$ has 2 unpaired electrons.

Hence, $[NiCl_4]^{2-}$ is paramagnetic.

4. $[Co(NH_3)_4Cl_2]Cl$

The complex ion is $[Co(NH_3)_4Cl_2]^+$.

• Let oxidation state of Co be $x$: $x+4(0)+2(-1)=+1 \Rightarrow x=+3$
• So, $Co^{3+}$ is $3d^6$.
• Coordination number $=6$, so octahedral.
• For $Co^{3+}$, even with $NH_3$ and $Cl^-$ mixed ligands, octahedral $d^6$ is generally low spin due to high pairing tendency of $Co^{3+}$.
• Low spin $d^6$: $t_{2g}^6 e_g^0$ all electrons paired.

Hence, $[Co(NH_3)_4Cl_2]Cl$ is diamagnetic.

5. $Na_3[CoF_6]$

The complex ion is $[CoF_6]^{3-}$.

• Let oxidation state of Co be $x$: $x+6(-1)=-3 \Rightarrow x=+3$
• So, $Co^{3+}$ is $3d^6$.
• $F^-$ is a weak field ligand.
• Octahedral weak field $d^6$ is high spin: $t_{2g}^4 e_g^2$
• This has 4 unpaired electrons.

Hence, $Na_3[CoF_6]$ is paramagnetic.

6. $Na_2O_2$

• This is sodium peroxide containing peroxide ion $O_2^{2-}$.
• In $O_2^{2-}$, all electrons are paired.

Hence, $Na_2O_2$ is diamagnetic.

7. $CsO_2$

• This contains superoxide ion $O_2^-$.
• $O_2^-$ has one unpaired electron in molecular orbital theory.

Hence, $CsO_2$ is paramagnetic.

8. Count paramagnetic compounds

Paramagnetic compounds are:

• $[NiCl_4]^{2-}$
• $Na_3[CoF_6]$
• $CsO_2$

Total number $=3$.

9. Compare with stored answer

Stored correct answer: B = 3

Our derived answer is also 3. So the stored answer is correct.