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pychu [463]
3 years ago
9

To show the electron configuration for an atom, when would it be better to use an orbital notation than to use a written configu

ration with numbers, letters, and superscripts?
Chemistry
2 answers:
anzhelika [568]3 years ago
7 0
I believe it would be better to use an orbital designation than the written configuration, if the number of electrons in the ground state of the atom are quite high for the given element, as above 50, for instance.

This saves space and also one can see the discrete quantized energy levels associated with the subshells of the main energy levels if written in orbital designation.
Harrizon [31]3 years ago
6 0

Answer:

It would be better to use an orbital notation when the electron configuration aims to show the spin.

Explanation:

When we have the electron configuration for an atom, we can use a written configuration with numbers, letters, and superscripts. Those will show the energetic level, the letter S, P, d, F and the number of electrons in each one of those. Nevertheless, it is not possible to study the spin or other important properties through this representation form.

Due to the above, when it is necessary to see al the quantum numbers in an electron configuration we have to use the orbital notation diagram. This diagram shows the same information as the written previously presented, energetic level, letter, amount of electrons and besides that, it shows the different orientations and the spin of every electron.  

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The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation

R = molar gas constant

K = A(e^(-Ea/RT))

Taking natural log of both sides

In K = In A - (Ea/RT)

In K = (-Ea/R)(1/T) + In A

Comparing this to the equation of a straight line; y = mx + c

y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A

a) From the question, m = (-Ea/R) = -1.10 × (10^4) K

(-Ea/R) = -1.10 × (10^4) = -11000

R = 8.314 J/K.mol

Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol

b) c = In A = 33.5

A = e^33.5 = (3.54 × (10^14))/s

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K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s

QED!

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