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nekit [7.7K]
2 years ago
5

In the quantum theory, the magnetic orbital momentum quantum number is most directly associated with which property of orbitals?

a. orientation, b. shape, c. size d. energyIn the quantum theory, the angular orbital momentum quantum number is most directly associated with which property of orbitals?a. shapeb. energyc. orientationd. size
Chemistry
1 answer:
kolezko [41]2 years ago
8 0

Answer: In the quantum theory, the magnetic orbital momentum quantum number is most directly associated with orientation of orbitals

In the quantum theory, angular orbital momentum quantum number is most directly associated with shape of orbitals.

Explanation:

Principle Quantum Number : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....

Angular orbital momentum quantum number or Azimuthal Quantum Number : It 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...

Magnetic Quantum Number : It 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.

Spin Quantum number : It describes the direction of electron spin. This is represented as m_s The value of this is +\frac{1}{2} for upward spin and -\frac{1}{2} for downward spin.

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51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
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Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

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