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Unit #2 Atomic Structure
Multiple Concepts Questions or Multiple Choice Questions.
The quantum number ‘m’ of a free gaseous atom is associated with:
a) the effective volume of the orbital b) the shape of the orbital c) the spatial orientation of the orbital d) the energy of the orbital in the absence of a magnetic fieldWhen 3d subshell is completely filled, the next entering electron goes into:
a) 4f b) 4s c) 4p d) 4dQuantum number values for 2p orbitals are:
a) n = 2, l = 1 b) n = 1, l = 2 c) n = 1, l = 0 d) n = 2, l = 0An electron having the set of values: n = 4, l = 0, m = 0 and s = +1/2 lies in:
a) 2s b) 3s c) 4s d) 5sThe quantum number values for the fourth electron of ⁹⁴Be atom are:
a) 1, 0, 0 b) 2, 0, 0 c) 2, 1, 0 d) 1, 1, 1The correct order of first ionization energies is:
a) F > He > Mg > N > O b) He > F > N > O > Mg c) He > O > F > N > Mg d) N > F > He > O > MgA p orbital has a characteristic shape with how many lobes?
a) 1 b) 2 c) 3 d) 4The three p orbitals in a given energy level are oriented:
a) Along the same axis b) At 45° to each other c) Mutually perpendicular to each other along the x, y, and z axes d) In a complex tetrahedral arrangementHow many d orbitals are there in a given energy level?
a) 1 b) 3 c) 5 d) 7Which of the following species is predicted to have the highest bond order?
a) N₂ b) O₂ c) F₂ d) Ne₂Which of the following species has a zero bond order according to MOT?
a) H₂⁺ b) He₂⁺ c) He₂²⁺ d) He₂Which of the following molecular geometries has bond angles of approximately 120°?
a) Tetrahedral b) Trigonal planar c) Bent (V-shaped) d) Trigonal pyramidal
Short Questions
There are three orientations of p-orbital due to three values of magnetic quantum number. Justify it.
For a certain value of l there are (2l + 1) integral values of m as follows. - l………0 ……..+ l.
For p-orbitals, the azimuthal quantum number l =1, the magnetic quantum number ‘m’ can have values m= -1, 0, +1. These three values of 𝑚 represent three different orientations of the p-orbital in space. This means there are three p-orbitals, called 𝑝x, 𝑝y, and 𝑝z oriented along the x, y, and z axes respectively.
Size of Mg is bigger than Al, but ionization energy of Mg is more than that of Al. Why?
Magnesium (Mg) has a bigger atomic size than Aluminium (Al), but its ionization energy is higher. This is due to the difference in their electronic configurations:
Mg: 1s² 2s² 2p⁶ 3s² Al: 1s² 2s² 2p⁶ 3s² 3p¹
In Mg, the outermost electrons are paired in the 3s orbital. In Al, the outermost electron is unpaired in the 3p orbital. Paired electrons are more stable, so more energy is needed to remove one of them from Mg. On the other hand, the unpaired 3p¹ electron in Al is less stable and easier to remove. Therefore, Mg has higher ionization energy than Al, even though its atomic size is larger.
‘I3’ of Mg is much bigger than its ‘I2’. Justify
The first two electrons in magnesium are removed from the 3s shell. After their removal, magnesium attains a stable octet configuration (like neon), which is energetically very stable. The third electron is then removed from the inner 2p shell, which is more strongly held by the nucleus. Therefore, I₃ is much larger than I₂.
Among the elements Li, K, Ca, S and Kr which one has the lowest first ionization energy? Which has the highest first ionization.
Potassium (K) has the lowest first ionization energy because it is a Group 1 metal with a large atomic size and only one valence electron, which is easily removed.
Krypton (Kr) has the highest first ionization energy because it is a noble gas with a completely filled stable octet, and its electrons are tightly held by the nucleus.
Consider the electronic configuration of the potassium atom (atomic number 19).
(i) Write the full electronic configuration of potassium using the s, p, d, f notation.
The full electronic configuration of potassium is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹.
(ii) Explain why the 4s subshell is filled before the 3d subshell in potassium, even though the principal quantum number of the 3d subshell is lower.
According to the Aufbau principle, electrons fill orbitals in the order of increasing energy. The energy of a subshell is determined by the n + l rule. For 4s: n + l = 4 + 0 = 4 and for 3d: n + l = 3 + 2 = 5.
Since 4s has a lower n + l value, it has lower energy and is filled before the 3d subshell in potassium.
(i) An atom of element X has an atomic number of 17 and a mass number of 35. Determine the number of protons, neutrons, and electrons in this atom.
Protons = Atomic number = 17, Electrons = Protons = 17 (because the atom is neutral)
Neutrons = Mass number – Atomic number = 35 – 17 = 18.
(ii) If this element forms an ion with a charge of -1, how many protons, neutrons, and electrons will be present in the ion?
Protons = 17 (same), Neutrons = 18 (same)
Electrons = Atomic number − Charge. Electrons = 17 − (–1) = 17 + 1 = 18 (one extra electron is gained)
In the ground state of mercury 80Hg (i) How many electrons occupy atomic orbitals with n = 3?
80Hg = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10.
For n=3, the orbitals are: 3s, 3p, and 3d. Electrons in n=3 orbitals=2+6+10=18
(ii) How many electrons occupy 4d atomic orbitals? 10 electrons.
(iii) How many electrons occupy 4pz atomic orbital? 2 electrons.
(iv) How many electrons have spin “up” (s=-1/2)?
Since electrons fill orbitals in pairs (spin up + spin down), half the electrons have spin up.
So, Electrons with spin up= 80/2=40
The successive ionization energies for an unknown element are
I1 = 896 kJ/mol I2 = 1752 kJ/mol I3 = 14,807 kJ/mol I4 = 17,948 kJ/mol
To which family in the periodic table, does the unknown element most likely belong?
The large jump between the second and third ionization energies shows that the element has 2 valence electrons. Therefore, the unknown element most likely belongs to the alkaline earth metals (Group 2) family.
Consider the following ionization energies for aluminum:
(i) Account for the trend in the values of the ionization energies.
Ionization energies increase because removing each electron makes the ion more positively charged, increasing the attraction between nucleus and remaining electrons.
(ii) Explain the large increase between I3 and I4.
The large jump occurs because the first three electrons are valence electrons, while the fourth electron is from a stable inner shell, requiring much more energy to remove.
(iii) List the four aluminum ions given in order of increasing size, and explain your ordering.
Al4+< Al3+< Al2+< Al+. Ionic size increases with decreasing positive charge due to weaker attraction between the nucleus and electrons, allowing the electrons to spread out more.
State the general order of filling orbitals up to the 4p subshell.
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p
Explain why the 4s subshell is filled before the 3d subshell, according to the Aufbau principle
According to the Aufbau principle, electrons fill orbitals starting with the lowest energy. The energy of a subshell is determined by the n + l rule. For 4s: n + l = 4 + 0 = 4 and for 3d: n + l = 3 + 2 = 5. Since 4s has a lower n + l value, it has lower energy and is filled before the 3d subshell.
Draw the orbital box diagram for the valence electrons of a phosphorus atom (atomic number 15), ensuring that your diagram adheres to Hund’s rule and the Pauli Exclusion Principle.
List of Some Elements mostly used in textbook
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