The complex(es) having metal-metal bond order \( \ge 3.5 \) is/are [Given: The atomic numbers of Mo, Cr, Mn, and Re are 42, 24, 25, and 75, respectively.]
The complex(es) having metal-metal bond order \( \ge 3.5 \) is/are [Given: The atomic numbers of Mo, Cr, Mn, and Re are 42, 24, 25, and 75, respectively.]
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[Mo\( _2 \)(\(\mu\)-SO\( _4 )\)\(_4\)(H\( _2 \)O)\(_2\)]\(^{3-}\)
- [Mn\( _2 \)(CO)\(_{10}\)]
- [Cr\( _2 \)(\(\mu\)-O\( _2 \)CCH\( _3 \))\(_4\)]
- [Mo\( _2 \)(\(\mu\)-HPO\( _4 \))\(_4\)(H\( _2 \)O)\(_2\)]\(^{2-}\)
The Correct Option is A, C
Solution and Explanation
To determine the metal–metal bond order, we need to consider the electronic configurations of the metal ions and the number of electrons available for metal–metal bonding.
(A) [Mo2(μ-SO4)4(H2O)2]3−
The overall charge is 3−. Each sulfate (SO42−) has a charge of 2−, and two water molecules are neutral.
So, the charge on the Mo2 core is: −3 − 4(−2) = +5 → Each Mo has an oxidation state of +2.5.
Mo (atomic) has configuration [Kr] 4d5 5s1; Mo2.5+ effectively has d3.5 configuration. So Mo2 has 7 d-electrons.
Known quadruple bonded Mo2 complexes have σ2 π4 δ2 configuration. Here: σ2 π4 δ1 → bond order = 3.5.
✅ Bond order ≈ 3.5
(B) [Mn2(CO)10]
Mn(0): [Ar] 3d5 4s2 → 7 valence electrons per Mn.
Each Mn is bonded to 5 CO ligands: 10 electrons total from COs + 1 from Mn–Mn bond = 18 electrons.
This implies a Mn–Mn single bond.
✅ Bond order = 1
(C) [Cr2(μ-O2CCH3)4]
Cr(II): [Ar] 3d5 4s1 → d4
Cr2 = 8 d-electrons → Known to form a quadruple bond with σ2 π4 δ2
✅ Bond order = 4
(D) [Mo2(μ-HPO4)4(H2O)2]2−
Overall charge = 2−; each HPO42− contributes 2− → core charge = −2 − 4(−2) = +6
Each Mo is in +3 oxidation state → d3 configuration → Mo2: 6 d-electrons
Likely σ2 π4 configuration → bond order = 3
✅ Bond order ≈ 3
Conclusion:
The complexes with metal–metal bond orders ≥ 3.5 are:
(A) [Mo2(μ-SO4)4(H2O)2]3− → BO = 3.5
(C) [Cr2(μ-O2CCH3)4] → BO = 4
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