Is the reaction 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3AlCl3(s) a redox reaction?

Q: What is the correct answer for CBSE11?

The correct answer is B. Let's determine oxidation states: In BCl3, Cl is -1, so B is +3. In LiAlH4, Li is +1, H is -1 (metal hydride), so Al + 4(-1) = -1 (for AlH4-), Al = +3. In B2H6, H is +1 (non-metal hydride), so 2B + 6(+1) = 0, B = -3. In LiCl, Li is +1, Cl is -1. In AlCl3, Cl is -1, Al is +3. Boron changes from +3 to -3 (reduction). Aluminum changes from +3 to +3 (no change). Lithium changes from +1 to +1 (no chang

Q: Is the reaction 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3AlCl3(s) a redox reaction?

The correct answer is B. Let's determine oxidation states: In BCl3, Cl is -1, so B is +3. In LiAlH4, Li is +1, H is -1 (metal hydride), so Al + 4(-1) = -1 (for AlH4-), Al = +3. In B2H6, H is +1 (non-metal hydride), so 2B + 6(+1) = 0, B = -3. In LiCl, Li is +1, Cl is -1. In AlCl3, Cl is -1, Al is +3. Boron changes from +3 to -3 (reduction). Aluminum changes from +3 to +3 (no change). Lithium changes from +1 to +1 (no chang

Q: What are the options for this CBSE11 question?

A. No, all elements maintain their oxidation states. · B. Yes, boron is reduced and aluminum is oxidized. · C. No, it's a double displacement reaction. · D. Yes, lithium is oxidized and chlorine is reduced.

Question

Is the reaction 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3AlCl3(s) a redox reaction?

FirstInTest · Accepted Answer

Let's determine oxidation states: In BCl3, Cl is -1, so B is +3. In LiAlH4, Li is +1, H is -1 (metal hydride), so Al + 4(-1) = -1 (for AlH4-), Al = +3. In B2H6, H is +1 (non-metal hydride), so 2B + 6(+1) = 0, B = -3. In LiCl, Li is +1, Cl is -1. In AlCl3, Cl is -1, Al is +3. Boron changes from +3 to -3 (reduction). Aluminum changes from +3 to +3 (no change). Lithium changes from +1 to +1 (no change). Chlorine changes from -1 to -1 (no change). So, boron is reduced. To have a redox reaction, something must be oxidized. Let's recheck LiAlH4. Li is +1. Al is +3. H is -1. This is correct. Let's recheck B2H6. It's diborane. H is usually +1 with non-metals. So B is -3. This is correct. So B is reduced. What is oxidized? Let's check the source. The document lists this reaction as an exercise to justify if it's redox. If B is reduced, something else must be oxidized. In LiAlH4, Al is +3, H is -1. In B2H6, B is -3, H is +1. This is tricky. Let's re-evaluate the oxidation states in LiAlH4. Li is +1. Al is +3. H is -1. This is correct. In B2H6, H is +1. So B is -3. So B is reduced from +3 to -3. What is oxidized? In LiAlH4, H is -1. In B2H6, H is +1. So hydrogen is oxidized from -1 to +1. Therefore, it is a redox reaction where boron is reduced and hydrogen is oxidized.

Answer

No, all elements maintain their oxidation states.

Answer

Yes, boron is reduced and aluminum is oxidized.

Answer

No, it's a double displacement reaction.

Answer

Yes, lithium is oxidized and chlorine is reduced.

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