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Nezavi [6.7K]
3 years ago
7

Consider a 4 cation of Manganese. There are________ electrons accounted for in the noble gas of the shorthand electron configura

tion. For the outermost electrons, there are_______ electrons in the______ s orbital and_______ electrons in the_______ d orbitals.
Chemistry
1 answer:
timurjin [86]3 years ago
6 0

Answer: There are 18 electrons accounted for in the noble gas of the shorthand electron configuration. For the outermost electrons, there are 0 electrons in the 4s orbital and 3 electrons in the 3d orbitals.

Explanation:

Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.

The electrons are filled according to Afbau's rule in order of increasing energies and thus the electronic configuration for Manganese with 25 electrons is :

Mn:25:1s^22s^22p^63s^23p^64s^23d^5

Mn^{4+}:21:1s^22s^22p^63s^23p^63d^3

Mn^{4+}:21:[Ar]3d^3

The nearest noble gas is Argon with 18 electrons.

There are 18 electrons accounted for in the noble gas of the shorthand electron configuration. For the outermost electrons, there are 0 electrons in the 4s orbital and 3 electrons in the 3d orbitals.

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For the following reaction KClO4 → KCl + 2O2 Assign oxidation states to each element on each side of the equation.
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Oxidation states number (OSN) show how many electrons gained, lost or shared during chemical reactions. 

Here are some rules of oxidation state numbers.

1. OSN of 1A elements in compounds is +1
2. OSN of Oxygen in compounds is -2
3. Sum of OSN of elements equals zero in neutral compounds.
4. OSN of free elements is zero. ex. F2, N2, O2.. all have 0 OSN.

If an element loses electrons during reaction, it is oxidized. That element is called as reducing agent.

If an element gains electrons during reaction, it is reduced. That element is called as oxidizing agent.

In the given reaction, potassium perchlorate decompose into potassium chlorate and oxygen.

a) KClO4.
K (potassium) is 1A element, OSN +1,
Oxygen in the compound has OSN -2.
Let OSN of Cl be x 
1+ x+(4*-2)=0 (since compound is neutral)
Solve the equation. X is equal to +7. 

b)Potassium perchlorate is the reactant, Potassium chloride and Oxygen are the products.
 Reactants -----> Products

c) Check out change of OSN of elements.
in KCl, OSN number of K is +1, Cl is -1. So Cl+7---->Cl-1. It takes 8 electrons. It was reduced.
in KClO4, OSN number of O is -2, in O2 it becomes O. So O-2---->O. It gives electrons. It was oxidized.
Shortly,
LOSS OF ELECTRONS=OXIDIZED 
GAIN OF ELECTRONS=REDUCED

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Answer:

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Read 2 more answers
Consider the reaction Ca(OH)2(s)→CaO(s)+H2O(l) with enthalpy of reaction ΔHrxn∘=65.2kJ/mol What is the enthalpy of formation of
Ganezh [65]

Answer:

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Explanation:

ΔH °(reaction)a=ΔH °f(product)-ΔH °f(reactant)

where,

ΔH °f(product)=is the standard heat of formation of products

-ΔH °f(reactant)=is the standard heat of formation of reactants

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The standard heat of formation of  Ca(OH )2  is  − 986.09 k J / m o l .

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

ΔH °(reaction)a=ΔH °f(CaO+H2O)-ΔH °f(Ca(OH)₂

65.2                =  ΔH °fCaO+( -265.8) -(-986.09)

                       = ΔH °fCaO-265.8+986.09

65.2                        =  ΔH °fCaO+ 720.29

65.2-720.29      =  ΔH °fCaO

-655.09 KJ/mol ==  ΔH °fCaO

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