Question

A symmetrical bridged complex cation made of Co(III), NH$_3$ molecules and oxygen (in the proper ligand form) is found to have the following composition: Co = 36.875%; NH$_3$ = 53.125%; O = 10%. The complex cation exists in three ionic forms with cationic charges $(A): (n^+); (B) : (n-1)^+$ and $(C): (n-2)^+$ such that $O - O$ bond length in all of them is found to be more than that in O$_2[PtF_6]$. Calculate the value of 'n' (Nearest Integer). (Atomic weight: Co = 59).

Answer 1

6

Answer 2

The complex cation is symmetrical and bridged, containing Co(III), NH$_3$, and oxygen. Let the empirical formula be $Co_a(NH_3)_bO_c$.

Using the given percentage composition and atomic weights:

Moles of Co $\propto \frac{36.875}{59} \approx 0.625$

Moles of NH$_3 \propto \frac{53.125}{17} \approx 3.125$

Moles of O $\propto \frac{10}{16} = 0.625$

Dividing by the smallest value (0.625), we get the ratio Co:NH$_3$:O = 1:5:1.

The empirical formula is $Co(NH_3)_5O$.

Since it is a symmetrical bridged complex cation with Co(III), it likely contains two Co atoms. Scaling the empirical formula by 2, we get $Co_2(NH_3)_{10}O_2$.

Let's verify the percentage composition for $[Co_2(NH_3)_{10}O_2]$:

Molar mass = $2 \times 59 + 10 \times 17 + 2 \times 16 = 118 + 170 + 32 = 320$ g/mol.

$\%$ Co = $(118/320) \times 100 = 36.875\%$

$\%$ NH$_3$ = $(170/320) \times 100 = 53.125\%$

$\%$ O = $(32/320) \times 100 = 10\%$

The composition matches. Thus, the complex cation has the formula $[Co_2(NH_3)_{10}O_2]^{n+}$.

The complex exists in three ionic forms with charges $(A): n^+$, $(B): (n-1)^+$, and $(C): (n-2)^+$. This suggests a redox series involving the bridging ligand. The oxygen atoms are likely part of a bridging diatomic oxygen species ($O_2$).

Let the bridging ligand be $(O_2)^q$. The charge of the complex is given by the sum of the charges of the metal ions and ligands. Co is in the +3 oxidation state, and NH$_3$ is neutral (charge 0).

Charge of complex = $2 \times (+3) + 10 \times 0 + q = 6 + q$.

So, $n = 6 + q$.

The three ionic forms correspond to the bridging ligand having charges $q$, $q-1$, and $q-2$. The charges of the complexes are $6+q$, $6+q-1$, and $6+q-2$.

The problem states that the O-O bond length in all three forms is more than that in O$_2[PtF_6]$. O$_2[PtF_6]$ contains the dioxygenyl cation $O_2^+$, which has a bond order of 2.5 and a bond length of approximately 112 pm.

Bond lengths of common oxygen species with O-O bonds:

$O_2^+$ (dioxygenyl): BO = 2.5, BL $\approx 112$ pm

$O_2$ (dioxygen): BO = 2, BL $\approx 121$ pm

$O_2^-$ (superoxide): BO = 1.5, BL $\approx 128$ pm

$O_2^{2-}$ (peroxide): BO = 1, BL $\approx 149$ pm

Since the O-O bond length in all three forms is greater than 112 pm ($O_2^+$), the bond order in the bridging ligand in all three forms must be less than 2.5. Possible bridging ligands are $O_2$, $O_2^-$, and $O_2^{2-}$, with charges 0, -1, and -2 respectively. Their bond orders are 2, 1.5, and 1, and their bond lengths are 121 pm, 128 pm, and 149 pm, all greater than 112 pm.

Let the three states of the bridging ligand be $L_A, L_B, L_C$ with charges $q_A, q_B, q_C$ such that $\{q_A, q_B, q_C\} = \{q, q-1, q-2\}$ and $\{q_A, q_B, q_C\} = \{0, -1, -2\}$.

Matching these two sets of charges, we find that the only possibility is $\{q, q-1, q-2\} = \{0, -1, -2\}$.

This implies $q=0$, $q-1=-1$, $q-2=-2$.

So, the three forms of the complex contain the bridging ligands with charges 0, -1, and -2.

The complex with charge $n^+$ contains the ligand with charge $q$. If $q=0$, then $n = 6+q = 6+0 = 6$.

The complex with charge $(n-1)^+ = 5^+$ contains the ligand with charge $q-1 = -1$.

The complex with charge $(n-2)^+ = 4^+$ contains the ligand with charge $q-2 = -2$.

Let's check the consistency:

Form (A): Charge $n^+=6^+$. Bridging ligand charge $q=0$ ($O_2$). Complex: $[Co_2(NH_3)_{10}(O_2)]^6$. Co oxidation state: $2x + 0 = 6 \implies x=+3$. Consistent with Co(III). O-O bond length is for $O_2$ (BO=2, ~121 pm), which is > 112 pm.

Form (B): Charge $(n-1)^+=5^+$. Bridging ligand charge $q-1=-1$ ($O_2^-$). Complex: $[Co_2(NH_3)_{10}(O_2)]^5$. Co oxidation state: $2x - 1 = 5 \implies 2x=6 \implies x=+3$. Consistent with Co(III). O-O bond length is for $O_2^-$ (BO=1.5, ~128 pm), which is > 112 pm.

Form (C): Charge $(n-2)^+=4^+$. Bridging ligand charge $q-2=-2$ ($O_2^{2-}$). Complex: $[Co_2(NH_3)_{10}(O_2)]^4$. Co oxidation state: $2x - 2 = 4 \implies 2x=6 \implies x=+3$. Consistent with Co(III). O-O bond length is for $O_2^{2-}$ (BO=1, ~149 pm), which is > 112 pm.

All conditions are satisfied with $n=6$.