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Mkey [24]
2 years ago
13

How many electrons are present in the nonbonding π molecular orbital of the allyl anion? a. 2 b. 1 c. 3 d. 0

Chemistry
1 answer:
Nostrana [21]2 years ago
6 0

The number of electrons, which are present in the nonbonding π molecular orbital of the allyl anion is "0".

Anions, cations, but also allylic radicals have always been frequently mentioned as reaction intermediates. Each one has three adjacent sp^{2}-hybridized carbon centers, and they all rely on resonance for stability. Two resonance structures would be used to present each species, with the charge as well as unpaired electron scattered across both the 1,3 and 0 positions.

The Aufbau principle states that these orbitals would fill up based on the order of stability, therefore a typical pi bond, will have 2 electrons in the Pi orbital as well as zero in the Pi* orbital.

Therefore, the correct answer will be option (d)

To know more about allyl anion

brainly.com/question/14286167

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2 years ago
1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?
garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l  or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

                     2HI(aq)    +    BaCO3(s)   ⇒   BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

                     197 g of BaCO3 ----------------- 1 mol

                     5.97 g                -----------------   x

                     x = (5.97 x 1) /197

                    x = 0.03 mol of BaCO3

                    2 moles of HI ----------------  1 mol of BaCO3

                    x                     ----------------  0.03 mol of BaCO3

                    x = (0.03 x 2) / 1

                   x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ?    NaoH 0.757 M

Co⁺² Volume = 167 ml   0.548 M

             CoSO4(aq) + 2NaOH(aq)   ⇒   Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x  volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

                                 1 mol of CoSO4 -------------- 2 moles of NaOH

                                0.092 moles      ---------------   x

                                x = (0.092 x 2) /1

                               x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

                             = 0.183 / 0.757

                            = 0.241 l  or 241 ml of NaOH

6 0
3 years ago
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