Manganese has the oxidation number of +5 in ______ . .
(A) [MnF6]3– (C) [MnO4]2–
(B) Mn2O7 (D) [Mn(CN)6]–
Take a.
F is -1. 6 of them makes -6. We want to leave a -3 charge on the ion; therefore, Mn must be +3. +3 and -6 = +3.
b. Mn2O7. O is -2. Seven of them makes -14.
14/2 = +7 each for Mn. Check it. +14 for 2 Mn + (-14) for 7 O atoms = 0.
c. MnO4^-. O is -2. Four of them make -8. We want to leave a -1 charge on the ion; therefore, Mn must be +7. Check it. +7 + (-8) = -1 charge on the ion.
d. Mn(CN)6^-1.
CN is -1, six of them makes -6. We want to leave a -1 charge on the ion; therefore, Mn must be +5.
Check it. Mn = +5 + (-6 for CN) = -1 charge on the ion.
Well, looks like Manganese is just trying to keep things interesting with its oxidation number. Let's see which option tickles its fancy.
(A) [MnF6]3–: Nope, not this one. Manganese prefers another oxidation number here.
(B) Mn2O7: Ah, there we go! Manganese definitely feels like showing off its +5 oxidation number in this compound.
(C) [MnO4]2–: Close, but no cigar. Manganese decided to switch things up here and showcase its +7 oxidation number.
(D) [Mn(CN)6]–: Sorry, Manganese has opted for a different oxidation number in this compound. It's not giving us a +5 this time.
So, the answer is (B) Mn2O7. Manganese likes to put on a little fireworks display with that +5 oxidation number.
To determine the oxidation number of manganese (Mn) in the given compounds, we need to consider the known oxidation numbers of other elements and apply the rule that the sum of oxidation numbers in a compound must equal the charge on that compound.
Let's analyze each option:
(A) [MnF6]3–:
Since fluorine (F) generally has an oxidation number of -1, we can set up the equation:
x + 6(-1) = -3
x - 6 = -3
x = +3
(B) Mn2O7:
Since oxygen (O) generally has an oxidation number of -2, we can set up the equation:
2x + 7(-2) = 0
2x - 14 = 0
2x = 14
x = +7
(C) [MnO4]2–:
Since oxygen (O) generally has an oxidation number of -2, we can set up the equation:
x + 4(-2) = -2
x - 8 = -2
x = +6
(D) [Mn(CN)6]–:
We can consider the charges on cyanide (CN-) as a whole and set up the equation:
x + 6(-1) = -1
x - 6 = -1
x = +5
Therefore, manganese (Mn) has the oxidation number of +5 in option (D) [Mn(CN)6]–.
To determine the oxidation number of manganese (Mn) in each compound, let's analyze the known oxidation numbers of the other elements.
In compound (A) [MnF6]3–:
- Fluorine (F) has an oxidation number of -1.
- Since there are six fluoride ions, the overall oxidation number for the fluorine ions is -6.
- The compound has a charge of -3, so the oxidation number of manganese needs to balance the negative charge. Therefore, the oxidation number of manganese is +3.
In compound (B) Mn2O7:
- Oxygen (O) usually has an oxidation number of -2 in compounds. So, the oxygen ions have a combined oxidation number of -14.
- The compound has no overall charge, meaning the sum of the oxidation numbers for all elements must be equal to zero.
- Since there are two manganese ions, the combined oxidation number of the manganese ions should be +14 to offset the -14 from the oxygen ions.
- Therefore, the oxidation number of manganese in Mn2O7 is +7.
In compound (C) [MnO4]2–:
- Oxygen (O) usually has an oxidation number of -2.
- Since there are four oxygen ions, the overall oxidation number for the oxygen ions is -8.
- The compound has a charge of -2, so the oxidation number of manganese needs to balance the negative charge. Therefore, the oxidation number of manganese is +6.
In compound (D) [Mn(CN)6]–:
- Cyanide (CN) ion has a charge of -1.
- Since there are six cyanide ions, the overall charge for the cyanide ions is -6.
- The compound has a charge of -1, so the oxidation number of manganese must be +1 to balance the negative charge.
Based on these analyses, manganese has the oxidation number of +5 in compound (B) Mn2O7.
Therefore, the correct answer is (B) Mn2O7.