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3 years ago

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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The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of NaOH needed is 36.2 mL

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of sulfurous acid = 0.104 g

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Volume of solution = 250 mL

Putting values in above equation, we get:

$\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M$

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL$

Putting values in above equation, we get:

$2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL$

Hence, the volume of NaOH needed is 36.2 mL

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