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Varvara68 [4.7K]
4 years ago
15

Which of the subshells below do not exist due to the constraints upon the angular momentum quantum number?A) 2dB) 2sC) 2pD) all

of the aboveE) none of the above
Chemistry
1 answer:
Gnom [1K]4 years ago
7 0

Answer:

  • Option A): <em>Due to the constraints upton the angular momentum quantum number, the subshell </em><u><em>2d</em></u><em> does not exist.</em>

Explanation:

The <em>angular momentum quantum number</em>, identified with the letter l (lowercase L),  number is the second quantum number.

This number identifies the shape of the orbital or <em>kind of subshell</em>.

The possible values of the angular momentum quantum number, l, are constrained by the value of the principal quantum number n: l can take values from 0 to n - 1.

So, you can use this guide:

Principal quantum   Angular momentum         Shape of the orbital

number, n                 quantum number, l

         1                                      0                             s

         2                                     0, 1                          s, p

         3                                     0, 1, 2                      s, p, d

Hence,

  • <u>the subshell 2d (n = 2, l = 2) is not feasible</u>.

  • 2s (option B) is possible: n = 2, l = 0

  • 2p (option C) is possible: n = 2, l = 1

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Which tool should Darla and Rob use to measure the force that pulls a car down the map
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C. A Spring Scale

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2 years ago
Electronic transitions (i.e., absorption of uv or visible light) of the conjugated molecule butadiene can be approximated using
dlinn [17]

Answer:

2.51 Angstroms

Explanation:

For a particle in a one dimensional box, the energy level, En, is given by the expression:

En = n²π² ħ² / 2ma²

where n is the energy level, ħ²  is Planck constant divided into 2π, m is the mass of the electron ( 9.1  x 10⁻³¹ Kg ), and a is the length of the one dimensional box.

We can calculate  the change in energy, ΔE, from n = 2 to n= 3  since we know the wavelength of the transition  ( ΔE = h c/λ ) and then substitute this value for the expresion of the ΔE for a particle in a box and solve for the  length a.

λ = 207 nm x 1 x 10⁻⁹ m/nm = 2.07 x 10⁻⁷ m      ( SI units )

ΔE = 6.626 x 10⁻³⁴ J·s x  3 x 10⁸ m/s  / 2.07 x 10⁻⁷ m

ΔE   = 9.60 x 10⁻¹⁹ J

ΔE(2⇒3) = ( 3 - 2 )  x  π² x ( 6.626 x 10⁻³⁴ J·s / 2π )² / ( 2 x 9.1 x 10⁻³¹ Kg x a² )

9.60 x 10⁻¹⁹ J  =  π² x( 6.626 x 10⁻³⁴ J·s / 2π )² / ( 2 x 9.1 x 10⁻³¹ Kg x a² )

⇒ a = 2.51 x 10⁻¹⁰ m

Converting to Angstroms:

a = 2.51 x 10⁻¹⁰ m x 1 x 10¹⁰ Angstrom / m = 2.51 Angstroms

6 0
4 years ago
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