Arrange the following ions in the increasing order of their total number of bonds (σ + π):
A. MnO42–
B. MnO4–
C. Cr2O72–
D. CrO42–
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → D, A, B, C
In the oxoanions of chromium and manganese, the metal forms a σ bond to every oxygen atom. Each terminal oxygen that does not carry a negative charge is joined by a double bond (one σ + one dπ–pπ bond), while each \( \ce{O^{-}} \) and each bridging oxygen has only single bonds. The oxidation state of the metal tells how many double bonds there are:
(A) \( \ce{MnO4^{2-}} \): x + 4(−2) = −2, so Mn is +6. Tetrahedral: 2 Mn=O and 2 \( \ce{Mn-O^{-}} \), so 4 σ + 2 π = 6 bonds.
(B) \( \ce{MnO4^{-}} \): x + 4(−2) = −1, so Mn is +7. Tetrahedral: 3 Mn=O and 1 \( \ce{Mn-O^{-}} \), so 4 σ + 3 π = 7 bonds.
(C) \( \ce{Cr2O7^{2-}} \): 2x + 7(−2) = −2, so Cr is +6. Two tetrahedra share one oxygen (Cr–O–Cr); each Cr has 2 Cr=O and 1 \( \ce{Cr-O^{-}} \), so 8 σ + 4 π = 12 bonds.
(D) \( \ce{CrO4^{2-}} \): x + 4(−2) = −2, so Cr is +6. Tetrahedral: 2 Cr=O and 2 \( \ce{Cr-O^{-}} \), so 4 σ + 2 π = 6 bonds.
Shortcut: \( \ce{Cr2O7^{2-}} \) (C) has the most bonds and \( \ce{MnO4^{-}} \) (B) the next most, so the order must end with B, C. Only Option (2) does.
Note: It seems that there is a mistake in the original NTA paper, which asks for a strictly increasing order although MnO42− (A) and CrO42− (D) have the same number of bonds (6 each).