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3 years ago

Which elements have the quantum numbers (4,3,-2,+1/2) and (5,2,-1,-1/2)

Chemistry

1 answer:

nikklg [1K]3 years ago

7 0

The elements are samarium (Sm) and silver (Ag).

Quantum numbers (4,3,-2,+½)

n = 4: Principal quantum number = 4.

l = 3: Element has 4f electrons

It is conventional to list quantum numbers in decreasing order.

Hund’s rule states that all the orbitals must be half-filled with electrons having the same spin before any can receive a second electron.

We get the following table:

Element n l mₗ mₛ

La 4 3 3 +½

Ce 4 3 2 +½

Pr 4 3 1 +½

Nd 4 3 0 +½

Pm 4 3 -1 +½

Sm 4 3 -2 +½

The element is samarium, Sm.

Quantum numbers (5,2,-1,-½)

n = 5: Principal quantum number = 5.

l = 2: Element has 5d electrons

We get the following table:

Element n l mₗ mₛ

Y 5 2 2 +½

Zr 5 2 1 +½

Nb 5 2 0 +½

Mo 5 2 -1 +½

Tc 5 2 -2 +½

Ru 5 2 2 -½

Rh 5 2 1 -½

Pd 5 2 0 -½

Ag 5 2 -1 -½

The element is silver, Ag.

Note: These assignments assume that there are no exceptions in the Periodic Table.

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3 years ago

Read 2 more answers

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Umnica [9.8K]

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Learn more: brainly.com/question/1527403

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3 years ago

The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react

Kazeer [188]

Answer:

The percent yield of chloro-ethane in the reaction is 82.98%.

Explanation:

$C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl$

Moles of ethane = $\frac{300.0 g}{30 g/mol}=10 mol$

Moles of chlorine gases = $\frac{650.0 g}{71 .0 g/mol}=9.1549 mol$

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

$\frac{1}{1}\times 9.1549 mol=9.1549 mol$ of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

$\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%$

The percent yield of chloro-ethane in the reaction is 82.98%.

6 0

3 years ago