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KiRa [710]
3 years ago
8

Refer to the periodic table tool and write the electron configurations of the following elements in both long and short terms

Chemistry
1 answer:
ahrayia [7]3 years ago
5 0

Answer:-

Carbon

[He] 2s2 2p2

1s2 2s2 2p2.

potassium

[Ar] 4s1.

1s2 2s2 2p6 3s2 3p6 4s1

Explanation:-

For writing the short form of the electronic configuration we look for the nearest noble gas with atomic number less than the element in question. We subtract the atomic number of that noble gas from the atomic number of the element in question.

The extra electrons we then assign normally starting with using the row after the noble gas ends. We write the name of that noble gas in [brackets] and then write the electronic configuration.

For carbon with Z = 6 the nearest noble gas is Helium. It has the atomic number 2. Subtracting 6 – 2 we get 4 electrons. Helium lies in 1st row. Starting with 2, we get 2s2 2p2.

So the short term electronic configuration is [He] 2s2 2p2

Similarly, for potassium with Z = 19 the nearest noble gas is Argon. It has the atomic number 18. Subtracting 19-18 we get 1 electron. Argon lies in 3rd row. Starting with 4, we get 4s1.

So the short electronic configuration is [Ar] 4s1.

For long term electronic configuration we must write the electronic configuration of the noble gas as well.

So for Carbon it is 1s2 2s2 2p2.

For potassium it is 1s2 2s2 2p6 3s2 3p6 4s1

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Answer:

\rm ^{103}_{\phantom{1}40}Zr, zirconium-103.

Explanation:

In a nuclear reaction, both the mass number and atomic number will conserve.

Let ^{A}_{Z}\mathrm{X} represent the unknown particle.

The mass number of a particle is the number on the upper-left corner. The atomic number of a particle is the number on its lower-left corner under the mass number. For example, for the particle ^{A}_{Z}\mathrm{X}, A is the mass number while Z while Z is the atomic number.

Sum of mass numbers on the left-hand side of the equation:

\underbrace{239}_{^{239}_{\phantom{2}94}\mathrm{Pu}} + \underbrace{1}_{^{1}_{0}\mathrm{n}} = 240.

Note that there are three neutrons on the right-hand side of the equation. Sum of mass numbers on the right-hand side:

\underbrace{A}_{^{A}_{Z}\mathrm{X}} + \underbrace{134}_{^{134}_{\phantom{2}54}\mathrm{Xe}} + \underbrace{3\times 1}_{3\;^{1}_{0}\mathrm{n}} = A + 137.

Mass number conserves. As a result,

A + 137 = 240.

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