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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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8. A 220 mL sample of helium gas is in a cylinder with a movable piston at 105 kPa and 275K. The piston
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The sample has a new pressure of 274kPa. If at 105 kPa and 275K, a 220 mL sample of helium gas is contained in a cylinder with a moving piston. The sample is pushed till it has a 95.0 mL volume and 310K .

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From ideal gas equation

P₁V₁/T₁ =P₂V₂/T₂

105×220÷275 = P₂ ×95÷310

P₂= (105×220×310)÷(275×95)

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brainly.com/question/28837405

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Keesha performed a chemical reaction and the products looked quite different from the reactants. She knew the amount of matter h
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the spotlight effect.

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The spotlight effect is a tendency to think that people get noticed more often than they really do. It is an overestimation of the situation regarding the concern of getting observed. It concerns the self-confidence of an individual. For example, an individual feels that everybody in a party would notice him for a bad pair of shoes while in reality, it does not concern them.

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For the following electron-transfer reaction:
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Answer:

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2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

Explanation:

Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)

In the oxidation half reaction, the oxidation number increases:

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The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.

Silver changes from Ag⁺ to Ag.

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

To balance the hole reaction, we need to multiply by 2, the second half reaction:

Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2

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