Question

The number of complexes among the following having exactly four unpaired electrons is \[ [Cr(H_2O)_6]^{2+}, \; [Mn(H_2O)_6]^{2+}, \; [Fe(H_2O)_6]^{2+}, \; [Co(H_2O)_6]^{3+}, \] \[ [Cu(H_2O)_6]^{2+}, \; [CoF_6]^{3-}, \; [Cr(CN)_6]^{4-}, \; [MnCl_4]^{2-} \]

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

For coordination compounds, always remember: \[ CN^- > NH_3 > H_2O > F^- > Cl^- \] in ligand field strength. Strong field ligands cause pairing, whereas weak field ligands generally produce high-spin complexes.

Answer 1

The Correct Option is C

Step 1: Know the goal.
We must count how many of the eight complexes have exactly four unpaired electrons. For each one we find the metal's oxidation state, its $d$ count, and whether the ligand is weak or strong field.

Step 2: Recall the ligand strength.
$H_2O$, $F^-$ and $Cl^-$ are weak field, so they do not pair electrons (high spin). $CN^-$ is strong field, so it pairs electrons (low spin).

Step 3: Test the first four.
$[Cr(H_2O)_6]^{2+}$: Cr$^{2+}$ is $d^4$, high spin gives 4 unpaired.
$[Mn(H_2O)_6]^{2+}$: Mn$^{2+}$ is $d^5$, high spin gives 5.
$[Fe(H_2O)_6]^{2+}$: Fe$^{2+}$ is $d^6$, high spin gives 4 unpaired.
$[Co(H_2O)_6]^{3+}$: Co$^{3+}$ is $d^6$, with water it is high spin giving 4 unpaired.

Step 4: Test the next four.
$[Cu(H_2O)_6]^{2+}$: Cu$^{2+}$ is $d^9$, gives 1 unpaired.
$[CoF_6]^{3-}$: Co$^{3+}$ is $d^6$, F is weak so high spin gives 4 unpaired.
$[Cr(CN)_6]^{4-}$: Cr$^{2+}$ is $d^4$, CN is strong so low spin gives 2 unpaired.
$[MnCl_4]^{2-}$: Mn$^{2+}$ is $d^5$, gives 5 unpaired.

Step 5: Pick the ones with exactly four.
Four unpaired electrons appear in $[Cr(H_2O)_6]^{2+}$, $[Fe(H_2O)_6]^{2+}$, $[Co(H_2O)_6]^{3+}$ and $[CoF_6]^{3-}$.

Step 6: Count them.
That makes four complexes with exactly four unpaired electrons.

Step 7: State the final answer.
The number of such complexes is:
\[ \boxed{4} \]