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2 years ago

Which of the following is not a valid set of quantum numbers

Chemistry

2 answers:

zhuklara [117]2 years ago

7 0

Answer: C. n = 3, l = 3, m = 3

Explanation:

1. Principle Quantum Numbers: This quantum number describes the size of the orbital. It is represented by n. n = 1,2,3,4....

2. Azimuthal Quantum Number: This quantum number describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

3. Magnetic Quantum Number: This quantum number describes the orientation of the orbitals. It is represented as $m_l$ . The value of this quantum number ranges from $(-l\text{ to }+l)$ . When l = 2, the value of $m_l$ will be -2, -1, 0, +1, +2.

A. n = 2, l = 1 , m = 0 is possible.

B. n = 1, l = 0, m = 0 is possible.

C. n = 3, l = 3, m = 3 is not possible as for n = 3, l = 0, 1 and 2 and m = 0,+1, +2, -1 and -2.

ASHA 777 [7]2 years ago

3 0

I believe the answer is C, n = 3, l = 3, m = 3. The magnetic quantum number, or

ml, can only take values that range from −l to +l, as you can see in the table above.

For option C), the angular momentum quantum number of equal to ++2, which means that ml can have a maximum value of +2. Since it is given as having a value of +3**, this set of quantum numbers is not a valid one.

The other three sets are valid and can correctly describe an electron.

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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$