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5. Determine the wavelength of the line in the hydrogen atom spectrum corresponding to the ni = 3 to nf = 4 transition. a. 1,094 nm d. 335 nm b. 1,875 nm e. 109 nm c. 656 nm 6. The change in energy of a one-electron atom or ion for an electronic transition from the initial energy level ni to the final energy level nf is given by ?E = -(2.18 × 10^-18 J) Z^2 (1/nf^2 - 1/ni^2) Which of the following species will have the longest wavelength emission line for the transition between the ni = 2 and nf = 1 levels? a. H d. Be3+ b. He+ e. B4+ c. Li2+ 7. How many orbitals with unique shapes and orientations are possible for the n = 2 shell? a. 1 d. 4 b. 2 e. 5 c. 3 8. Which of the following is a possible set of quantum numbers for a 3d orbital? a. n = 3, l = 1, ml = -1 d. n = 3, l = 2, ml = 0 b. n = 3, l = 0, ml = 0 e. n = 3, l = 1, ml = 2 c. n = 3, l = 2, ml = 3 9. Which listing has the orbitals in order of increasing energy in a multielectron atom? a. 3s < 3p < 3d < 4s d. 2px < 2py < 2pz b. 5s < 4d < 5p e. 3s < 2p < 4s c. 5s < 3d < 5p 10. Which of the following elements would you expect to have the greatest first ionization energy? a. C d. Be b. O e. B c. Li

          5. Determine the wavelength of the line in the hydrogen atom spectrum corresponding to the ni = 3 to nf = 4 transition.
a. 1,094 nm
d. 335 nm
b. 1,875 nm
e. 109 nm
c. 656 nm
6. The change in energy of a one-electron atom or ion for an electronic transition from the initial energy level ni to the final energy level nf is given by
?E = -(2.18 &times; 10^-18 J) Z^2 (1/nf^2 - 1/ni^2)
Which of the following species will have the longest wavelength emission line for the transition between the ni = 2 and nf = 1 levels?
a. H
d. Be3+
b. He+
e. B4+
c. Li2+
7. How many orbitals with unique shapes and orientations are possible for the n = 2 shell?
a. 1
d. 4
b. 2
e. 5
c. 3
8. Which of the following is a possible set of quantum numbers for a 3d orbital?
a. n = 3, l = 1, ml = -1
d. n = 3, l = 2, ml = 0
b. n = 3, l = 0, ml = 0
e. n = 3, l = 1, ml = 2
c. n = 3, l = 2, ml = 3
9. Which listing has the orbitals in order of increasing energy in a multielectron atom?
a. 3s < 3p < 3d < 4s
d. 2px < 2py < 2pz
b. 5s < 4d < 5p
e. 3s < 2p < 4s
c. 5s < 3d < 5p
10. Which of the following elements would you expect to have the greatest first ionization energy?
a. C
d. Be
b. O
e. B
c. Li

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5. Determine the wavelength of the line in the hydrogen atom spectrum corresponding to the ni = 3 to nf = 4 transition.
a. 1,094 nm
d. 335 nm
b. 1,875 nm
e. 109 nm
c. 656 nm
6. The change in energy of a one-electron atom or ion for an electronic transition from the initial energy level ni to the final energy level nf is given by
?E = -(2.18 times; 10^-18 J) Z^2 (1/nf^2 - 1/ni^2)
Which of the following species will have the longest wavelength emission line for the transition between the ni = 2 and nf = 1 levels?
a. H
d. Be3+
b. He+
e. B4+
c. Li2+
7. How many orbitals with unique shapes and orientations are possible for the n = 2 shell?
a. 1
d. 4
b. 2
e. 5
c. 3
8. Which of the following is a possible set of quantum numbers for a 3d orbital?
a. n = 3, l = 1, ml = -1
d. n = 3, l = 2, ml = 0
b. n = 3, l = 0, ml = 0
e. n = 3, l = 1, ml = 2
c. n = 3, l = 2, ml = 3
9. Which listing has the orbitals in order of increasing energy in a multielectron atom?
a. 3s < 3p < 3d < 4s
d. 2px < 2py < 2pz
b. 5s < 4d < 5p
e. 3s < 2p < 4s
c. 5s < 3d < 5p
10. Which of the following elements would you expect to have the greatest first ionization energy?
a. C
d. Be
b. O
e. B
c. Li

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Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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03/20/2022

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5. Determine the wavelength of the line in the hydrogen atom spectrum corresponding to the ni = 3 to nf = 4 transition: 1,094 nm 335 nm 1,875 nm 109 nm 656 nm 6. The change in energy of a one-electron atom or ion for an electronic transition from the initial energy level ni to the final energy level nf is given by ΔE = -(2.18 x 10^-18 J) Z^2 [1/nf^2 - 1/ni^2]. Which of the following species will have the longest wavelength emission line for the transition between the ni = 2 and nf = 1 levels? H Be3+ He+ B4+ Li2+ 7. How many orbitals with unique shapes and orientations are possible for the n = 2 shell? 8. Which of the following is a possible set of quantum numbers for a 3d orbital? n = 3, l = 2, ml = 0 b. n = 3, l = 0, ml = 0 c. n = 3, l = 2, ml = 3 d. n = 3, l = 1, ml = -1 e. n = 3, l = 1, ml = 2 9. Which listing has the orbitals in order of increasing energy in a multielectron atom? 3s < 3p < 3d < 4s 2px < 2py < 2pz b. 5s < 4d < 5p c. 5s < 3d < 5p e. 3s < 2p < 4s 10. Which of the following elements would you expect to have the greatest first ionization energy? C O Li Be B

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00:01 So according to the question, let's jump to a question one by one.

00:05 So the first one or the fifth one in this question, the wavelength is given by 1 by lambda is equals to rz square 1 by an initial square minus 1 by n final square, where r is r is ritb's constant.

00:49 And z is atomic number.

00:58 Atomic number of hydrogen is 1 in our case.

01:02 Now we have an final as fourth and an initial as 3.

01:14 By putting all the values in our formula we'll be getting 1 by lambda is equals to 109677 which is value of its constant into 1 which is atomic number of hydrogen.

01:29 And the square of 1 will be 1 and 1 by 3 square minus 1 upon 4 square square.

01:49 Now 1 by lambda is equal to 109677 multiplied by 7 x7 x7 x7 x7 x7 x10 by 1024.

02:02 So for lambda or for wavelength will be read.

02:06 Reciprocal this value and reciprocal will be lambda is equal to 1 .87 to 10x2bore minus 4 centimeters or lambda will be 1875 nanometers therefore option b is correct now let's jump to our next question that is 6 1 .1 so in this in this question, for single electron species, species, wavelength is given by 1 by lambda is equals to r z square, 1 by an initial square minus 1 by n, final square.

03:29 Now, where r is with works constant, and z is anomic number.

03:45 Now, from this relation, we can help that 1 by lambda is proportional to spire of atomic number then whichever atom has the lowest atomic number will have longest wavelength because when atomic number increases wavelength decreases so from here we can prove that option a that is hydrogen is sorry sorry correct.

04:39 Because the atomic number is lowest and wavelength is highest.

04:50 Now let's jump to next question which is seventh one.

04:56 For n equals to 2, that is a principal quantum number, there are 4 orbitals possible.

05:14 One orbital is 2s and 3 orbitals of 2p.

05:30 Now for s, we know that it is spherical in shape.

05:35 So for s, for 2s, it will be spherical in shape.

05:49 So this is a structure of 2s and for 2p, p contains three orbitals, that is p, h, p, y and b z.

05:57 So for p, we will be drawing this is y, coordinate and these are x coordinate...

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