All categories

1 month ago

Which of the following is a reasonable ground-state electron configuration?

Chemistry

1 answer:

lorasvet [2.6K]1 month ago

3 0

Answer:

The correct choice is: option A.

Justification:

To address this inquiry, we need to evaluate the total number of electrons each orbital can accommodate.

Orbital Number of electrons

s 2

p 6

d 10

f 14

Provided options:

A. 1s² 2s² 2p⁶ 3s² This configuration is valid as it aligns with the permitted number of electrons in each orbital and follows the correct sequence.

B. 1s² 2s² 2p⁶ 3s² 3d⁴ This configuration is not accurate because

3d⁴ should follow 3p.

C. 1s² 2s² 2d¹⁰ 2p³ This is incorrect since 2d¹⁰ is not a valid orbital.

D. 1s² 2s^s 2p³ 2d¹⁰ This option contains two errors; s as an exponent does not exist, and 2d¹⁰ is also an invalid description.

You might be interested in

The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when

castortr0y [2906]

Answer: The cell potential for the given reaction stands at 0.50 V

Explanation:

The provided cell reaction is:

$Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)$

The half-reactions are:

Anode oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Cathode reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

Initially, we need to find the cell potential for this reaction.

Utilizing the Nernst equation:

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}$

where,

F = Faraday's constant = 96500 C

R = gas constant = 8.314 J/mol·K

T = room temperature = $25^oC=273+25=298K$

n = electrons exchanged in oxidation-reduction = 2

$E^o_{cell}$ = standard electrode potential for the cell = +0.63 V

$E_{cell}$ = cell potential for the reaction =?

$[Zn^{2+}]$ = concentration of Zn²⁺ = 3.5 M

$[Pb^{2+}]$ = $2.0\times 10^{-4}M$

Now substituting all known values into the equation, we arrive at:

$E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}$

$E_{cell}=0.50V$

So, the resulting cell potential for this reaction is 0.50 V

5 0

6 days ago

Montrel takes a trip to Central America during the winter. He finds out that the temperature in Central America is about the sam

lorasvet [2668]

Answer:

Explanation

My teacher mentioned that, for instance, Florida tends to be warm, while Central America can have both hot and cold temperatures, and a jacket isn't necessary. I don't recall many details.

6 0

1 month ago

Commercial grade fuming nitric acid contains about 90.0% HNO3 by mass with a density of 1.50 g/mL, calculate the molarity of the

Alekssandra [2891]

x is greater than or equal to 56Step-by-step explanation: He requires at least 56 additional cans. Hence, x should be a minimum of 56.

8 0

14 hours ago

Read 2 more answers

If your lungs were filled with air containing this level of lead, how many lead atoms would be in your lungs? (Assume a total lu

Tems11 [2624]

A total of 1.505×10^23 lead atoms

In the lungs, the volume of lead equals the total lung volume, which is 5.60L

1 mole corresponds to 22.4L

Thus, 5.6L of lead converts to 5.6/22.4 = 0.25 moles

According to Avogadro's law

1 mole of lead contains 6.02×10^23 lead atoms

Thus, 0.25 moles of lead equates to 0.25×6.02×10^23 = 1.505×10^23 lead atoms

6 0

18 hours ago

Why is it important to have regular supervision of the weight and measurements in the market

lions [2782]

Answer:

Oversight of weights and measures ensures correct evaluations of goods and services so that everyone receives a fair exchange in the marketplace. It also acts as a deterrent, promoting honesty among traders.

Explanation:

4 0

1 month ago