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nadezda [96]
3 years ago
7

Describe the shapes and relative energies of the SPD & f atomic orbitals

Chemistry
1 answer:
Deffense [45]3 years ago
6 0

Explanation:

The shapes and relative energies of the orbitals s,p,d and f orbitals are given by the principal quantum number and the azimuthal quantum number.

The principal quantum number gives the main energy level and the azimuthal quantum number denotes the shape of the orbitals.

  • For the principal quantum number, they represent the energy levels in which the orbital is located or the average distance of the orbital from the nucleus. It takes the number n = 1,2,3,4,5,6,7......
  • The azimuthal quantum number(L) shows the shape of the orbitals in subshells accommodating electrons. The number of possible shapes is limited by the the principal quantum number.

L             Name of orbital                     shape of orbital

0                     s                                         spherical

1                      p                                         dumb-bell

2                     d                                        double dumb-bell

3                      f                                          complex

Principal                        Azimuthal                   Orbital

Quantum                       Quantum                    Designation of

Number (N)                  Number(l)                     Sublevel

       1                                   0                                   1s

       2                                  0                                   2s

                                            1                                   2p

       3                                  0                                   3s

                                            1                                    3p

                                            2                                   3d

      4                                   0                                    4s

                                           1                                      4p

                                           2                                      4d

                                           3                                      4f

Learn more:

Atomic orbitals brainly.com/question/9514863

#learnwithBrainly

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Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
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Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

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