How many of the following compounds are diamagnetic?
$$\mathrm{Ni(CO)_4,\ [Cu(CN)_4]^{3-},\ [Zn(NH_3)_4]^{2+},\ [Co(H_2O)_6]^{3+},\ K_2MnO_4,\ [Pt(en)_2Cl_2]^{+2},\ [Fe(EDTA)]^-}$$

A compound is diamagnetic when all the electrons are paired.

1. $$Ni(CO)_4$$

CO is a neutral ligand, so Ni is in the $$0$$ oxidation state.

$$Ni^0: 3d^{10}$$

All electrons are paired. Hence, it is diamagnetic.

2. $$[Cu(CN)_4]^{3-}$$

Let the oxidation state of Cu be $$x$$.

$$x+4(-1)=-3$$

$$x=+1$$

Thus,

$$Cu^+:3d^{10}$$

All electrons are paired. Hence, it is diamagnetic.

3. $$[Zn(NH_3)_4]^{2+}$$

NH3 is a neutral ligand.

$$Zn^{2+}:3d^{10}$$

All electrons are paired. Hence, it is diamagnetic.

4. $$[Co(H_2O)_6]^{3+}$$

$$Co^{3+}:3d^6$$

Co(III) has a large octahedral splitting, and the hexaaqua complex is treated as low spin:

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

There are no unpaired electrons. Hence, it is diamagnetic.

5. $$K_2MnO_4$$

The oxidation state of Mn is:

$$2(+1)+x+4(-2)=0$$

$$x=+6$$

Thus,

$$Mn^{6+}:3d^1$$

There is one unpaired electron. Hence, it is paramagnetic.

6. $$[Pt(en)_2Cl_2]^{2+}$$

For the complex ion:

$$x+2(0)+2(-1)=+2$$

$$x=+4$$

Thus,

$$Pt^{4+}:5d^6$$

Pt(IV) is a 5d metal ion and forms a low-spin octahedral complex:

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

All electrons are paired. Hence, it is diamagnetic.

7. $$[Fe(EDTA)]^-$$

EDTA has a charge of $$-4$$.

$$x+(-4)=-1$$

$$x=+3$$

Thus,

$$Fe^{3+}:3d^5$$

The complex is high spin with five unpaired electrons. Hence, it is paramagnetic.

Therefore, the diamagnetic compounds are:

$$Ni(CO)_4,\ [Cu(CN)_4]^{3-},\ [Zn(NH_3)_4]^{2+},\ [Co(H_2O)_6]^{3+},\ [Pt(en)_2Cl_2]^{2+}$$

Hence, the number of diamagnetic compounds is:

$$\boxed{5}$$

Therefore, the answer will be 5.