1. Identify the compounds and electron count

Binary metal carbonyls given are:

• $Fe(CO)_5$
• $Ni(CO)_4$

In carbonyls, $CO$ is a neutral ligand donating $2$ electrons. Also, these are neutral complexes, so oxidation state of metal is $0$ in both.

• For $Fe$: atomic number $26$, valence electrons in $Fe^0 = 3d^6 4s^2 = 8$ electrons
• For $Ni$: atomic number $28$, valence electrons in $Ni^0 = 3d^8 4s^2 = 10$ electrons

Now count total electrons at metal center:

$Fe(CO)_5: 8 + 5\times 2 = 18$ $Ni(CO)_4: 10 + 4\times 2 = 18$

So option A is false, because the total is 18, not $16$.

2. Check spin nature

$CO$ is a strong field ligand. Strong field ligands cause pairing of electrons and generally form low-spin complexes. Therefore binary metal carbonyls are predominantly low spin.

So option B is correct.

3. Effect of oxidation state on M–C bond

In metal carbonyls, bonding has two components:

• $\sigma$ donation from $CO$ to metal
• $\pi$ back-bonding from filled metal $d$ orbitals to empty $\pi^*$ orbitals of $CO$

If oxidation state of the metal is lower, the metal is more electron-rich, so it can do more $\pi$ back-bonding. More back-bonding strengthens the metal–carbon bond.

Hence option C is correct.

4. Effect on C–O bond

More back-bonding puts electron density into antibonding $\pi^*$ orbital of $CO$, which weakens the $C-O$ bond.

Now if oxidation state of metal is increased, the metal becomes less electron-rich, so back-bonding decreases. Then the $C-O$ bond should become stronger, not weaker.

Therefore option D is false.

5. Final correct options

• A: False
• B: True
• C: True
• D: False

So the correct statements are:

$\boxed{B,\ C}$

6. Comparison with stored correct answer

Stored correct answer: $B, C$

My derived answer matches the stored correct answer.