1. Find oxidation state and $d$-electron count of the metal in each complex

Since $CN^-$ is a strong-field ligand, it usually causes pairing of electrons.

2. Complex 1: $[Fe(CN)_6]^{4-}$

Let oxidation state of Fe be $x$.

$x+6(-1)=-4 \Rightarrow x=+2$

So, $Fe^{2+}$ is:

$Fe: [Ar]3d^64s^2 \Rightarrow Fe^{2+}: 3d^6$

In octahedral strong field, $d^6$ becomes low spin:

$t_{2g}^6 e_g^0$

All electrons are paired.

• Diamagnetic

3. Complex 2: $[Fe(CN)_6]^{3-}$

$x+6(-1)=-3 \Rightarrow x=+3$

So, $Fe^{3+}$ is $3d^5$.

With strong-field $CN^-$ in octahedral field, it is low spin:

$t_{2g}^5 e_g^0$

This has 1 unpaired electron.

• Paramagnetic

4. Complex 3: $[Ti(CN)_6]^{3-}$

$x+6(-1)=-3 \Rightarrow x=+3$

So, $Ti^{3+}$ is:

$Ti: [Ar]3d^24s^2 \Rightarrow Ti^{3+}: 3d^1$

Octahedral splitting gives:

$t_{2g}^1 e_g^0$

There is 1 unpaired electron.

• Paramagnetic

5. Complex 4: $[Ni(CN)_4]^{2-}$

$x+4(-1)=-2 \Rightarrow x=+2$

So, $Ni^{2+}$ is:

$Ni: [Ar]3d^84s^2 \Rightarrow Ni^{2+}: 3d^8$

Because $CN^-$ is a strong-field ligand, this $4$-coordinate complex is square planar.

Square planar $d^8$ complexes have all electrons paired.

• Diamagnetic

6. Complex 5: $[Co(CN)_6]^{3-}$

$x+6(-1)=-3 \Rightarrow x=+3$

So, $Co^{3+}$ is:

$Co: [Ar]3d^74s^2 \Rightarrow Co^{3+}: 3d^6$

With strong-field $CN^-$ in octahedral geometry, it is low spin:

$t_{2g}^6 e_g^0$

All electrons are paired.

• Diamagnetic

7. Count paramagnetic complexes

Paramagnetic complexes are:

1. $[Fe(CN)_6]^{3-}$
2. $[Ti(CN)_6]^{3-}$

So, the number of paramagnetic complexes is

$\boxed{2}$

8. Comparison with stored correct answer

Stored correct answer = $2$

Derived answer = $2$

They match.