Q. Calculate the ionization energy of hydrogen in eV if the energy level is -13.6 eV.
-
A.
13.6 eV
-
B.
1.24 eV
-
C.
3.4 eV
-
D.
0.85 eV
Solution
Ionization energy is the energy required to remove an electron from the ground state. For hydrogen, it is 13.6 eV.
Correct Answer:
A
— 13.6 eV
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Q. Calculate the ionization energy of hydrogen in eV if the energy of the electron in the ground state is -13.6 eV.
-
A.
13.6 eV
-
B.
1.24 eV
-
C.
3.4 eV
-
D.
27.2 eV
Solution
Ionization energy is the energy required to remove the electron from the ground state. For hydrogen, it is 13.6 eV.
Correct Answer:
A
— 13.6 eV
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Q. If the atomic radius of chlorine is 99 pm, what is the atomic radius of potassium in pm?
-
A.
227 pm
-
B.
196 pm
-
C.
186 pm
-
D.
210 pm
Solution
Atomic radius increases down a group. Potassium (K) has a radius of approximately 227 pm.
Correct Answer:
A
— 227 pm
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Q. What is the effective nuclear charge (Z_eff) for a 3s electron in sodium (Na)?
Solution
Z_eff = Z - S; for Na, Z = 11, S (shielding from 1s and 2s) ≈ 3. Thus, Z_eff = 11 - 3 = 8.
Correct Answer:
D
— 8
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Q. What is the effective nuclear charge (Z_eff) for a 3s electron in sodium (Z=11)?
Solution
Using Slater's rules, Z_eff = Z - S; for Na, S = 3 (from 1s and 2s electrons). Thus, Z_eff = 11 - 3 = 8.
Correct Answer:
C
— 8
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Q. What is the electron affinity of chlorine in kJ/mol?
-
A.
-349 kJ/mol
-
B.
-328 kJ/mol
-
C.
-300 kJ/mol
-
D.
-400 kJ/mol
Solution
The electron affinity of chlorine is approximately -349 kJ/mol.
Correct Answer:
A
— -349 kJ/mol
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Q. What is the electron configuration of the sulfur ion (S^2-)?
-
A.
[Ne] 3s² 3p⁴
-
B.
[Ne] 3s² 3p⁶
-
C.
[Ne] 3s² 3p³
-
D.
[He] 2s² 2p⁶
Solution
Sulfur (S) has 16 electrons. S^2- has 18 electrons, resulting in the configuration [Ne] 3s² 3p⁶.
Correct Answer:
B
— [Ne] 3s² 3p⁶
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Q. What is the electronegativity difference between sodium (Na) and chlorine (Cl)?
-
A.
0.8
-
B.
1.0
-
C.
2.1
-
D.
3.0
Solution
Electronegativity of Na ≈ 0.9 and Cl ≈ 3.0. Difference = 3.0 - 0.9 = 2.1.
Correct Answer:
C
— 2.1
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Q. What is the electronegativity of fluorine on the Pauling scale?
-
A.
3.0
-
B.
3.5
-
C.
4.0
-
D.
4.5
Solution
The electronegativity of fluorine is 4.0 on the Pauling scale.
Correct Answer:
C
— 4.0
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Q. What is the first ionization energy of lithium (Li) in kJ/mol?
-
A.
520 kJ/mol
-
B.
750 kJ/mol
-
C.
1000 kJ/mol
-
D.
1500 kJ/mol
Solution
The first ionization energy of lithium is approximately 520 kJ/mol.
Correct Answer:
A
— 520 kJ/mol
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Q. What is the first ionization energy of magnesium (Mg) in kJ/mol?
-
A.
738 kJ/mol
-
B.
1000 kJ/mol
-
C.
580 kJ/mol
-
D.
1200 kJ/mol
Solution
The first ionization energy of magnesium is approximately 738 kJ/mol.
Correct Answer:
A
— 738 kJ/mol
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Q. What is the maximum number of electrons that can occupy the 3d subshell?
Solution
The maximum number of electrons in a subshell is given by 2n^2. For d (n=3), it is 2(3^2) = 18, but only 10 can occupy 3d.
Correct Answer:
C
— 10
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Q. What is the principal quantum number (n) for the outermost electron in rubidium (Rb)?
Solution
Rubidium (Rb) has an atomic number of 37, and its outermost electron is in the 5th shell (n=5).
Correct Answer:
D
— 4
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Q. What is the principal quantum number for the outermost electron in potassium (K)?
Solution
Potassium has an electron configuration of [Ar] 4s1, so the principal quantum number for the outermost electron is 4.
Correct Answer:
C
— 4
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Q. What is the wavelength of an electron moving with a velocity of 1 x 10^6 m/s? (h = 6.626 x 10^-34 J·s)
-
A.
6.63 x 10^-28 m
-
B.
6.63 x 10^-34 m
-
C.
6.63 x 10^-22 m
-
D.
6.63 x 10^-30 m
Solution
Using de Broglie's equation, λ = h/p; p = mv = (9.11 x 10^-31 kg)(1 x 10^6 m/s). λ = 6.626 x 10^-34 / (9.11 x 10^-31)(1 x 10^6) = 6.63 x 10^-28 m.
Correct Answer:
A
— 6.63 x 10^-28 m
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