Chemistry, 18.10.2020 16:01 mandi795

Which electron configuration represents the electrons in an atom of sulfur in an excited state? 2 – 8 – 6 2 – 7 – 7
2 – 8 – 7
2 – 7 – 8

Answers: 3

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The relations of the quantum numbers allow to find that the correct answer for the configuration of the excited state is:

2 - 7 - 7

The electronic configuration of the elements is the distribution of electrons and different levels and sublevels fulfilling the relationships between quantum numbers.

The principal quantum number (n) goes from 0 to infinity The orbital quantum number (l) goes from 0 to n-1, in general it is represented by lera s, p, d, f The magnetic quantum number ( m_l

) ranges from -l to l The spin quantum number ( m_s

) can have two values + ½ and - ½

In the base configuration Sulfur of the periodic table is 2 - 8 - 6

This is at the lowest energy configuration, when the atom acquires energy, an electron from its shell must be promoted to the next level.

That is, an electron from level n = 2 that is full is promoted to level n = 3 since in this shell there is still room for two electrons, the configuration of the excited state is:

2- 7 - 7

Let's examine the different alternatives:

1) 2 - 8 -6

False, this is the configuration of the base state

2) 2 - 7 - 7

True. An electron is promoted to the next level due to the excitation of the atom

3) 2 - 8 -7

False, in this configuration the atom becomes an ion

4) 2 - 7 - 8

False, there is an electron transfer between two levels, but there is an extra electron that turns that atom into an ion

In conclusion using the relationships of quantum numbers we can find that the correct answer for the excited state configuration is:

2 - 7 - 7

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Answer from: carterdaw

answer.

i think it's 2-7-7

Which electron configuration represents the electrons in an atom of sulfur in an excited state? 2 –

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

Answer from: Quest

4d^55s^1 ( third choice)

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Which electron configuration represents the electrons in an atom of sulfur in an excited state? 2 –...

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