Metal-Metal Bond Order Analysis

This problem requires identifying coordination complexes with a metal-metal (M-M) bond order of at least 3.5 ($\ge 3.5$). We will determine the bond order for each complex based on its d-electron count and known molecular orbital diagrams.

Calculating Bond Order

The M-M bond order calculation relies on the number of d-electrons involved in bonding between the two metal centers. For dinuclear complexes, particularly those with $D_{4h}$ symmetry or paddlewheel structures, a common molecular orbital (MO) energy sequence includes $\sigma$, $\pi$, and $\delta$ bonding orbitals. The bond order is derived from the electron configuration within these MOs.

Specific electron counts and resulting bond orders:

• 7 d-electrons: Filling $\sigma^2 \pi^4 \delta^1$. Bond Order = $(2+4+1)/2 = 3.5$.
• 8 d-electrons: Filling $\sigma^2 \pi^4 \delta^2$. Bond Order = $(2+4+2)/2 = 4$.
• 6 d-electrons: Filling $\sigma^2 \pi^4$. Bond Order = $(2+4)/2 = 3$.
• 10 d-electrons: Filling $\sigma^2 \pi^4 \delta^2 \delta^{*2}$. Bond Order = $(2+4+2 - 2)/2 = 3$.

Analyzing Each Complex

Option A: $[Mo_2(\mu-SO_4)_4(H_2O)_2]^{3-}$

Molybdenum (Mo) is a Group 6 element ($4d^5 5s^1$). We determine the oxidation state (OS) of Mo. Let the OS be $x$. The equation is $2x + 4(\text{charge of } SO_4) + 2(0) = -3$. Assuming the bridging sulfate ($SO_4$) has a charge of -2, we get $2x + 4(-2) = -3$. Solving for $x$, we find the average Mo OS is $+2.5$.

Each Mo atom has $6 - 2.5 = 3.5$ d-electrons. The total number of d-electrons in the dimer is $2 \times 3.5 = 7$. With 7 d-electrons, the M-M bond order is 3.5. This satisfies the condition ($\ge 3.5$).

Option B: $[Mn_2(CO)_{10}]$

Manganese (Mn) is a Group 7 element ($4d^5 5s^2$). The oxidation state is 0. Each Mn atom has 5 d-electrons. The total number of d-electrons is $2 \times 5 = 10$.

With 10 d-electrons, the standard MO configuration leads to a bond order of 3. This does not satisfy the condition ($\ge 3.5$).

Option C: $[Cr_2(\mu-O_2CCH_3)_4]$

Chromium (Cr) is a Group 6 element ($4d^5 5s^1$). The complex is neutral. Acetate ($CH_3COO^-$) has a charge of -1. Let the OS of Cr be $x$. The equation is $2x + 4(-1) = 0$. Solving for $x$, we find the Cr OS is $+2$.

Each Cr atom has $6 - 2 = 4$ d-electrons. The total number of d-electrons is $2 \times 4 = 8$. With 8 d-electrons, the M-M bond order is 4. This satisfies the condition ($\ge 3.5$).

Option D: $[Mo_2(\mu-HPO_4)_4(H_2O)_2]^{2-}$

Molybdenum (Mo) is a Group 6 element. If the bridging hydrogen phosphate ($HPO_4$) ligand carries a -1 charge: $2(\text{Mo OS}) + 4(-1) + 2(0) = -2$. This gives Mo OS = +1. Each Mo has $6 - 1 = 5$ d-electrons (Total = 10). The bond order is 3. If $HPO_4$ carries a -2 charge: $2(\text{Mo OS}) + 4(-2) + 2(0) = -2$. This gives Mo OS = +3. Each Mo has $6 - 3 = 3$ d-electrons (Total = 6). The bond order is 3.

In either likely scenario, the M-M bond order is 3. This does not satisfy the condition ($\ge 3.5$).

Conclusion

Based on the calculations, complexes A ($[Mo_2(\mu-SO_4)_4(H_2O)_2]^{3-}$, BO=3.5) and C ($[Cr_2(\mu-O_2CCH_3)_4]$, BO=4) exhibit a metal-metal bond order greater than or equal to 3.5.