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worty [1.4K]
3 years ago
9

Describe the rules used for writing the outer configuration for the s-block, p-block, and d-block elements?

Chemistry
2 answers:
Lorico [155]3 years ago
4 0

<u>Rules to write the electronic configuration :</u>

Electrons complete orbitals in a way to reduce the energy of the atom. Therefore, the electrons in an atom complete the principal energy levels in order of rising energy (the electrons are getting distant from the nucleus). The order of levels filled appearances like the following

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p

One way to recognise this pattern, apparently the simplest, is to refer to the periodic table and remember where each orbital block drops to rationally understand this pattern. Different way is to make a table like the one below and use vertical lines to determine which subshells resemble with each other.

  • S block: The S obstruct in the periodic table of components known as gatherings 1 and 2. There is a limit of two electrons that can possess the s orbital.
  • P Block: The P square contains group of 13, 14, 15, 16, 17, and 18, except for Helium.
  • D Block: The D block elements are found in groups 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 of the periodic table.
Wittaler [7]3 years ago
3 0

Answer: The periodic table is divided into four group of elements s-block, p-block, d-block and f-block

Elements                                               Outer Configuration

  • s-block that is alkali metal           ns¹⁻², n = 2 - 7

 

  • p-block that is metals and          ns²np¹⁻⁶, n = 2 - 6

      non-metals

  • d-block that is transition             (n - 1) d¹ ⁻ ¹⁰ns⁰ ⁻², n = 4 - 7

      metals

Explanation:

      Elements                                          Outer Configuration

  • s-block that is alkali metal                    ns¹⁻², n = 2 - 7

 

  • p-block that is metals and                 ns²np¹⁻⁶, n = 2 - 6

      non-metals

  • d-block that is transition              (n - 1) d¹ ⁻ ¹⁰ns⁰ ⁻², n = 4 - 7

      metals

       

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Molar gas constant, R = 0.0821 atm•L/mol•K

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The volume of the container that the gas occupied.

Step-by-step solution:

The volume of the container that the gas occupied can be calculated using the ideal gas equation.

\begin{gathered} PV=nRT \\  \\ \Rightarrow V=\frac{nRT}{P}=\frac{3.2mol\times0.0821atm•L/mol•K\times316.15K}{0.095atm} \\  \\ V=\frac{83.058928\text{ }atm•L}{0.095\text{ }atm}=874.30\text{ }L \end{gathered}

Therefore, the volume of the container that the gas occupied is 874.30 L

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