Explanation:

B is incorrect. An atomic orbital describes the spatial probability distribution of an electron, not a definite trajectory around the nucleus. Bohr’s hydrogen model uses defined electron paths, whereas the quantum mechanical orbital model does not.

Explanation:

The Lyman series of the Ultraviolet region emit energy when excited electrons fall down to n=1 shell. The Balmer series of the Visible region emit energy when excited electrons fall down to n=2 shell. The Paschen series of the Infrared region emit energy when excited electrons fall down to the n=3 shell.

Explanation:

The difference of the energy level is K(n=1) < L(n=2) < M(n=3) < N(n=4). To be absorbed the strong energy and become an excited electron, the electron must move from the lowest shell K (n=1), near the nucleus, to higher shells. To emit strong energy, excited electrons must fall from any of the high shells back to the lowest shell K (n=1).

Explanation:

Dumbbell-shaped p orbitals exist at all electron shells except n=1. Spherical shaped s orbitals are spread near nucleus with no direction. Three orientations of coordinate axes, P_x, P_y, or P_z, are features of p orbitals. Cloverleaf-shaped d orbitals appear in the transition metal region of the periodic table. Each orbital holds a maximum of 2 electrons; therefore 7 different f orbitals contain 14 electrons.

Explanation:

D is incorrect. The Aufbau principle places electrons in lower-energy available orbitals first. The usual filling sequence is 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p. This is a filling guide, with exceptions. The relative energies of 4s and 3d depend on the atom and its electron occupancy; transition-metal cations lose 4s electrons before 3d electrons.

Explanation:

A is correct. For a one-electron hydrogen-like species, orbital energy depends on the principal quantum number n (and nuclear charge). Orbitals with the same n are degenerate in this model. In many-electron atoms, electron repulsion and shielding remove this degeneracy, so both n and l affect orbital energy.

Explanation:

Solution:

The principal quantum number of n=4 contains a total of 16 (1+3+5+7) orbitals. Each orbital can hold a maximum of 2 electrons. Therefore, 16 × 2 =32 maximum electrons can be held in n=4. (Maximum number of electrons in shell: 2n² → 2n² = 32,  n=4)

Explanation:

Explanation:

The principal quantum number of n=1 has only s orbital and the principal quantum number of n=2 has s and p orbital. From the principal quantum number of n=3, d orbital begins to be involved.

Solution:

Solution:

A is correct. In the usual Aufbau filling sequence, these subshells occur as 2p → 3s → 3p → 4s → 3d. This filling sequence does not establish a universal 3d-versus-4s energy ranking for every atom or ion.

Explanation:

B is correct. Al has the ground-state electron configuration [Ne]3s²3p¹, so its differentiating electron enters 3p. Fe also has occupied 3p core electrons, but its differentiating electron enters 3d.

explanation:

Natural atoms	Ionic atoms

explanation:

Solution:

The electron configuration of Sc is 1s²2s²2p⁶3s²3p⁶4s²3d¹. The orbitals, which are l=0 of angular momentum quantum number, are 1s, 2s, 3s, and 4s. Therefore, the total electrons in 1s, 2s, 3s, and 4s orbitals are 4 ×2 = 8. There are 2 outermost electrons contained in 4s orbital.

Solution:

Si has the ground-state electron configuration 1s²2s²2p⁶3s²3p². The occupied-orbital counts are 1s²: 1, 2s²: 1, 2p⁶: 3, 3s²: 1, and 3p²: 2 (Hund’s rule). Total = 1 + 1 + 3 + 1 + 2 = 8, so B is correct.

Explanation:

I is the ground state atom and II is the excited state of P atom. When the ground state atom becomes the excited state atom, energy is absorbed. III is Cl atom. Therefore, atomic radius is P > Cl. I and II have 3 unpaired electrons and III has 1 unpaired electron.

Explanation:

D is correct. Neutral Se is [Ar]3d¹⁰4s²4p⁴; 4d is unoccupied. Its 4p subshell can accommodate two additional electrons, giving Se²⁻ = [Kr]. The figure’s phrase “vacancy electron” means unoccupied electron capacity, not missing electrons in the ground-state atom. For any p subshell, l = 1.

Explanation:

Explanation:

E = hν  

The higher frequency (ν), the higher energy (E). The order of energy from lower to higher:

Radio waves < Microwaves < Infrared light < Visible light < Ultraviolet light< X rays < Gamma rays

Explanation:

D is incorrect: visible light has a shorter wavelength than radio waves. For electromagnetic radiation in a vacuum, c = λν and photon energy E = hν. Light speed in a medium can depend on wavelength.

Explanation:

Metal: easy oxidation (reducing agent), malleability, ductility, shiny luster, and good conductivity for heat and electricity.

Nonmetal: easy reduction (oxidizing agents), brittleness, and poor conductivity for heat and electricity.

Explanation:

In the same periods, atoms have the same number of shells. Halogen group has a higher electro affinity. In the same group, atoms have similar chemical properties because of the same outermost electrons (valence electrons). Alkali metal group is the strong reducing agents and halogen group is the strong oxidizing agents.

Explanation:

A is incorrect. Hydrogen is a nonmetal, not an alkali metal. Fe is a transition element: its atom has an incomplete d subshell and it forms cations with incomplete d subshells. Zn has a filled 3d¹⁰ subshell in its atom and usual Zn²⁺ ion, so it is not classified as a transition element under the strict IUPAC definition.

Explanation:

Atoms positioned in the same groups have similar chemical properties. Fluorine and chlorine belong to the same 7A group.

Explanation:

The elements that have intermediate chemical properties between metals and nonmetals, are called semimetal or metalloids. B, Si, Ge, As, Sb, Te, and At belong to this semimetal group. Ga, Al, and Sn are metals, and Se is a nonmetal.

Explanation:

E is correct: I, II, and III apply. Neutral ground-state halogen atoms have an unpaired p electron and are paramagnetic. Their diatomic molecules (F₂, Cl₂, Br₂, and I₂) are diamagnetic. Halide anions have lone pairs that can be donated to a Lewis acid. These elements generally form −1 anions, rather than simple cations, in ordinary ionic compounds. Paramagnetism depends on unpaired electrons and is not restricted to metals.

Explanation:

E is correct: I, II, and IV apply. I states the IUPAC definition. Paramagnetism requires unpaired electrons; transition-metal species are not all paramagnetic (for example, a d¹⁰ ion can be diamagnetic). Calcium is an alkaline earth metal, whereas cobalt is a transition element. Multiple oxidation states are common among transition elements, but are not an exception-free property of every element.

Explanation:

In the periodic table, going in the direction left and down, the radius of atoms becomes bigger.

Phosphorus is positioned at the far left and down in the period table among the atoms.

Explanation:

D is correct. A cation is generally smaller than its parent neutral atom because electron removal reduces electron repulsion and increases the attraction experienced by the remaining electrons. Loss of an occupied outer shell can further reduce size, as for Rb → Rb⁺, but cation formation does not always remove an entire shell. Adding an electron to Br increases electron repulsion, so Br⁻ is larger than Br.

Explanation:

B is correct. All five species have 10 electrons. As nuclear charge increases, the same electron cloud is attracted more strongly and contracts. Their nuclear charges are Mg (12), Na (11), Ne (10), F (9), and O (8), giving Mg²⁺ < Na⁺ < Ne < F⁻ < O²⁻. This is an isoelectronic electron-cloud comparison; experimentally tabulated atomic and ionic radii use different definitions and should not be mixed.

Explanation:

The atom with fully-filled electrons in orbitals has strong first ionization energy ([Ar]). The atom with 1 outermost electron has strong second ionization energy ([Ar]4s¹). The atom with 2 outer most electrons has strong third ionization energy ([Ar]4s²).

Explanation:

In the same period with the increasing atomic number, the E_i is increased. In the same group with the increasing atomic number, the E_i is decreased. The half-filled or fully-filled electron(s) in each orbitals have a strong E_i, such as N > O or Ne > F.

Explanation:

E1–E5 are the first through fifth successive ionization energies, rounded to the nearest kJ/mol. The large jump from E2 to E3 shows that two valence electrons are removed before a core electron is removed, so X is a group 2 (alkaline earth) metal.

These values are for magnesium: 737.750, 1450.683, 7732.692, 10542.519, and 13630.480 kJ/mol. Source: Royal Society of Chemistry.

Explanation:

D represents N, with a half-filled 2p³ subshell. Adding an electron requires pairing in a 2p orbital and increases electron repulsion; the isolated ground-state N⁻ anion is not stable. The other listed atoms release energy on first attachment: Li ≈ 59.6, B ≈ 27.0, C ≈ 121.8, and O ≈ 141.0 kJ/mol. N therefore has the lowest affinity in this set. Electron affinity describes electron addition, not electron emission. With energy released taken as positive, favorable affinity corresponds to negative electron-gain enthalpy.

Explanation:

C is correct. F has a first electron affinity of about 328.2 kJ/mol, compared with about 141.0 kJ/mol for O; Ne does not form a stable isolated ground-state Ne⁻ anion. O²⁻ and F⁻ are ions, not candidates for a neutral atom’s first electron affinity. Here electron affinity is positive for energy released; the corresponding electron-gain enthalpy is negative. F has seven valence electrons in its outer shell, and addition gives the filled 2p⁶ configuration.

Explanation:

B is correct: P < Si < S < Cl. P has a half-filled 3p³ subshell, so electron pairing makes attachment less favorable than for Si. Electron affinity involves adding an electron, not emitting one. Broad periodic trends have exceptions and do not justify a universal monotonic increase across every period.

Explanation:

A is correct: F, He, He. F releases about 328.2 kJ/mol on first electron attachment; He and Ne do not form stable isolated ground-state anions. In the same non-bonded-radius dataset, He = 1.400 Å, F = 1.47 Å, and Ne = 1.54 Å. First ionization energies are He ≈ 2372.3, Ne ≈ 2080.7, and F ≈ 1681.0 kJ/mol. He is 1s² with a filled duplet, whereas Ne is 1s²2s²2p⁶ with a filled octet. He’s compact 1s shell holds its electrons particularly strongly.

Explanation:

Electronegativity is decreased going in the direction left and down in the period table.

Explanation:

A is correct. London dispersion forces arise from instantaneous and induced dipoles. Greater polarizability generally strengthens these forces for otherwise comparable species. Size, electron count, molecular shape, and contact area matter; there is no universal left-and-down rule for the dispersion forces of all atoms and molecules.

Explanation:

C is correct. Using a consistent non-bonded-radius dataset, Sc (2.15 Å) < Ca (2.31 Å) < Sr (2.49 Å) < Ba (2.68 Å). Electron affinities, expressed as energy released in kJ/mol, are Te (190.161) < Se (194.965) < S (200.410) < Cl (348.575). These specific comparisons are supported by values; periodic trends have exceptions and should not be treated as universal rules.

Explanation:

D is incorrect. Density is mass per unit volume and does not increase monotonically across a period. It depends on atomic mass, phase, and structure. Atomic radius is a different physical quantity; its general decrease across a period does not imply a universal increase in density.