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joja [24]
2 years ago
3

Illustrate and describe the sequence in which ten electrons occupy the five orbitals related to an atoms d sublevel

Chemistry
2 answers:
krok68 [10]2 years ago
4 0

<span>Actually a d sublevel contains 5 orbitals. So in this case, the electrons would first fill each of the five orbitals so remaining electrons are 5. Now this remaining 5 electrons would pair with the filled orbital. That is, the 6th electron would pair its spin with the 1st orbital, the 7th electron would pair with the one in the 2nd orbital, and so on.</span>

Ann [662]2 years ago
4 0

Answer:

In an atom, subshell is defined as the set of states in a given shell that have the same azimuthal quantum number (ℓ).  

For a d-subshell, the value of azimuthal quantum number (ℓ) is 3. Also, a d-subshell contains <em>5 atomic orbitals, </em>namely d_{{z}^{2}},\, d_{xy},\, d_{yz},\, d_{xz},\, d_{{x}^{2}-{y}^{2}}, <em>that can be occupied by 2 electrons each.</em>

The filling of electrons in a subshell, such as the d-subshell, of a given shell is governed by the <em>Hund's Rule of maximum multiplicity.</em>

<u><em>According to this rule, firstly all the 5 atomic orbitals of the d-subshell are singly filled and then the electrons are paired.</em></u>

Since, a d-subshell can contain 10 electrons.

<u><em>Therefore, firstly the 5 atomic orbitals are occupied with one electron each</em></u>

d^{1}_{{z}^{2}},\, d^{1}_{xy},\, d^{1}_{yz},\, d^{1}_{xz},\, d^{1}_{{x}^{2}-{y}^{2}}

<u><em>Then the remaining five electrons, results in the pairing of electrons in each d-atomic orbital.</em></u>

d^{2}_{{z}^{2}},\, d^{2}_{xy},\, d^{2}_{yz},\, d^{2}_{xz},\, d^{2}_{{x}^{2}-{y}^{2}}

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2. A compound with the following composition by mass: 24.0% C, 7.0% H, 38.0% F, and 31.0% P. what is the empirical formula
Svetach [21]

Answer:

C₂H₇F₂P

Explanation:

Given parameters:

Composition by mass:

                C = 24%

                H = 7%

                 F  = 38%

                 P  = 31%

Unknown:

Empirical formula of compound;

Solution :

The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;

                                   C                          H                         F                   P

% composition

by mass                     24                          7                        38                  31

Molar mass                 12                           1                         19                  31

Number of

moles                       24/12                          7/1                    38/19           31/31

                                     2                               7                       2                   1

Dividing

by the

smallest                      2/1                             7/1                       2/1                1/1

                                     2                                7                        2                   1

           Empirical formula        C₂H₇F₂P

5 0
2 years ago
A sample of an unknown substance has a mass of .158 kg if 2,510 J of heat is required to heat the substance from 32°C to 61°C wh
ziro4ka [17]

Answer: 0.548J/g°C

Explanation:

Q = s × m × DeltaT

Q = Heat (J)

S = Specific Heat Capacity

M = mass (g)

DeltaT = Change in temperature (°C)

0.158Kg x 1000 = 158g

2.510J = s x 158g x (61°C-32°C)

2.510J/(158g x 29°C) = s

S = 0.54779.... J/g°C

S = 0.548 J/g°C

7 0
2 years ago
Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-conta
Leokris [45]

Answer:MnCO3+2H2O----->MnO2+ HCO3-+2e-+3H+

Explanation:The equation to be balanced is

MnCO3 ------> MnO2+HCO3-

The oxidation number of Mn changes from +2 in MnCO3 to +4 in MnO2

Therefore two electrons must be added to the right as shown below:

MnCO3 -------> MnO2+ HCO3-+ 2e-Now,there is one negative charge HCO3- and 1 negative charge on the two electrons making a total of -3 charges on the right. There is zero charge on the left.

To balance the equation,add3H+on the right,to cancel out the charges.

MnCO3 --------> MnO2+HCO3-+2e-+3H+

Adding H2O to balance Hydrogen and Oxygen atoms:

MnCO3+2H2O ------->MnO2+HCO3-+2e-+3H+

3 0
2 years ago
What is the name of the functional group that is attached to this hydrocarbon? The first and last of a chain of three carbons ar
disa [49]

Answer:

Ketone

Explanation:

As you are stating here, we have a carbonated chain of three carbons, and the first and last has 3 Hydrogens, then this means that we have CH₃ . The center carbon is a carbon double bonded to oxygen.

In general terms this belongs to the carbonyl group. However, this alone does not represent a functional group, but when it's in a chain with other radycals or chains, it becomes a functional group.

In this case, the molecule you are talking here is the following:

CH₃ - CO - CH₃

This molecule is known as the Acetone, and has the general form of:

R - CO - R'

Which belongs to a ketone as a functional group.

4 0
2 years ago
You have a 16.0-oz. (473-mL) glass of lemonade with a concentration of 2.66 M. The lemonade sits out on your counter for a coupl
salantis [7]

<u>Answer:</u> The new concentration of lemonade is 3.90 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

Molarity of lemonade solution = 2.66 M

Volume of solution = 473 mL

Putting values in equation 1, we get:

2.66M=\frac{\text{Moles of lemonade}\times 1000}{473}\\\\\text{Moles of lemonade}=\frac{2.66\times 473}{1000}=1.26mol

Now, calculating the new concentration of lemonade by using equation 1:

Moles of lemonade = 1.26 moles

Volume of solution = (473 - 150) mL = 323 mL

Putting values in equation 1, we get:

\text{New concentration of lemonade}=\frac{1.26\times 1000}{323}\\\\\text{New concentration of lemonade}=3.90M

Hence, the new concentration of lemonade is 3.90 M

7 0
2 years ago
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