DAT Survey of the Natural Sciences Quiz: Atomic Structure And Periodicity
20 questions · exam conditions
0:00
Atomic Structure And PeriodicityQuestion 1 of 20

Which of the following molecules or ions exhibits resonance?

SiH₄
H₂S
SO₂
PCl₃
← Back to quizzes

DAT Survey of the Natural Sciences Quiz

DAT Survey of the Natural Sciences Quiz: Atomic Structure And Periodicity

Practice Atomic Structure And Periodicity in DAT Survey of the Natural Sciences with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Atomic Structure And Periodicity, giving you a quick way to practice the rules, question types, and explanations that matter most for DAT Survey of the Natural Sciences.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which of the following molecules or ions exhibits resonance?

  1. SiH₄
  2. H₂S
  3. SO₂ (correct answer)
  4. PCl₃
Explanation: Resonance occurs when you can draw multiple valid Lewis structures for the same molecule by moving electrons (but not atoms) to different positions. The key is identifying molecules where electrons can be delocalized across multiple bonds. For SO₂ (choice C), sulfur forms a bent molecule with oxygen atoms. You can draw the Lewis structure with single bonds and lone pairs, but you can also draw equivalent structures where double bonds form with different oxygen atoms. The electrons are delocalized between the S-O bonds, creating resonance. This delocalization actually strengthens the bonds and makes them shorter than typical single bonds. Choice A (SiH₄) is wrong because silicon tetrahydride has a simple tetrahedral structure with four equivalent Si-H single bonds. There's no way to move electrons to create alternative structures—the electrons are localized. Choice B (H₂S) is incorrect because hydrogen sulfide has a bent structure similar to water, but with only single bonds between sulfur and hydrogen. The lone pairs on sulfur don't participate in bonding, so no resonance structures are possible. Choice D (PCl₃) is wrong because phosphorus trichloride forms a trigonal pyramidal molecule with three single P-Cl bonds. Like the other incorrect choices, there's no electron delocalization possible—all bonds are discrete single bonds. When evaluating resonance, look for molecules containing atoms that can form multiple bonds (like sulfur, nitrogen, or carbon) and check if you can draw equivalent structures by redistributing electrons. Molecules with only single bonds to hydrogen or halogens rarely exhibit resonance.

Question 2

Which of the following sets of quantum numbers (n, l, m_l, m_s) is not permissible for an electron in an atom?

  1. (3, 2, -1, +1/2)
  2. (4, 0, 0, -1/2)
  3. (5, 3, -2, -1/2)
  4. (2, 2, 1, +1/2) (correct answer)
Explanation: When you encounter quantum numbers questions, you need to understand the rules that govern which combinations are allowed for electrons in atoms. Each electron is described by four quantum numbers that must follow specific constraints. The principal quantum number nn can be any positive integer (1, 2, 3...). The angular momentum quantum number ll must be less than nn, ranging from 0 to (n−1)(n-1). The magnetic quantum number mlm_l ranges from −l-l to +l+l. The spin quantum number msm_s is always either +1/2+1/2 or −1/2-1/2. Option D violates the fundamental rule for ll. Here, n=2n = 2 and l=2l = 2, but ll cannot equal nn. When n=2n = 2, the maximum allowed value for ll is 1, so l=2l = 2 is impossible. This makes D the impermissible set. Option A is valid: n=3n = 3 allows l=2l = 2, and l=2l = 2 allows ml=−1m_l = -1. Option B works because n=4n = 4 permits l=0l = 0, and when l=0l = 0, mlm_l must be 0. Option C is acceptable since n=5n = 5 allows l=3l = 3, and l=3l = 3 permits ml=−2m_l = -2. For quantum numbers questions, always check the l<nl < n rule first—it's the most commonly violated constraint. Remember that ll values correspond to subshells: l=0l = 0 (s), l=1l = 1 (p), l=2l = 2 (d), l=3l = 3 (f). You can't have a d subshell in the second shell because that would require l=2l = 2 when n=2n = 2.

Question 3

According to the principles of the photoelectric effect, increasing the intensity of monochromatic light shining on a metal surface, while keeping the frequency above the threshold, will result in which of the following outcomes?

  1. No change in electron number or maximum kinetic energy.
  2. Higher maximum kinetic energy, but same number of electrons.
  3. Both more electrons and higher maximum kinetic energy.
  4. More ejected electrons, but same maximum kinetic energy. (correct answer)
Explanation: When you encounter photoelectric effect questions, focus on Einstein's key insight: light behaves as discrete packets of energy called photons. The energy of each photon depends only on frequency (E=hfE = hf), not intensity. In the photoelectric effect, each photon can eject at most one electron. The maximum kinetic energy of ejected electrons follows the equation KEmax=hf−ϕKE_{max} = hf - \phi, where ϕ\phi is the work function (minimum energy needed to remove an electron). Notice that maximum kinetic energy depends only on frequency, not intensity. Intensity refers to the number of photons hitting the surface per unit time. More photons means more opportunities for electron ejection, but each individual photon-electron interaction remains unchanged. Since frequency stays constant in this question, each photon carries the same energy, giving ejected electrons the same maximum kinetic energy. Choice A is wrong because increasing intensity definitely increases the number of photons, so more electrons will be ejected. Choice B incorrectly suggests that intensity affects maximum kinetic energy – this contradicts Einstein's model where only frequency determines photon energy. Choice C combines both errors from A and B, incorrectly linking intensity to maximum kinetic energy. Choice D correctly recognizes that more photons (higher intensity) eject more electrons, while constant frequency means constant maximum kinetic energy per electron. Remember this distinction for DAT questions: frequency controls the energy per photon (and thus maximum kinetic energy), while intensity controls the number of photons (and thus the number of ejected electrons).

Question 4

Which of the following statements correctly compares the radii of an ion and its parent atom?

  1. Cations are larger than their parent atoms due to increased electron-electron repulsion.
  2. Anions are smaller than their parent atoms because effective nuclear charge increases.
  3. Anions have the same radius as their parent atoms since proton number is constant.
  4. Cations are smaller than their parent atoms because of higher effective nuclear charge. (correct answer)
Explanation: When atoms gain or lose electrons to form ions, their sizes change predictably due to shifts in the balance between nuclear attraction and electron-electron repulsion. The key concept here is effective nuclear charge — how strongly the nucleus attracts the remaining electrons. When an atom loses electrons to form a cation, two important changes occur: there are fewer electrons experiencing repulsion from each other, and the same number of protons now attracts fewer electrons. This means each remaining electron feels a stronger pull toward the nucleus, causing the ion to contract significantly. Additionally, cations often lose their entire outermost electron shell, making them much smaller than their parent atoms. Let's examine why the other options are incorrect: Choice A is backwards — cations are smaller, not larger, and they have decreased electron-electron repulsion since they've lost electrons. Choice B misapplies the effective nuclear charge concept to anions. When atoms gain electrons to form anions, the effective nuclear charge per electron actually decreases because you're adding more electrons without adding protons. Choice C ignores the fundamental principle that electron-electron repulsion increases when you add electrons to form anions. Even though the proton number stays constant, the additional electrons repel each other, causing the ion to expand. Choice D correctly identifies that cations are smaller due to higher effective nuclear charge per remaining electron. Study tip: Remember the pattern: "Cations contract, anions expand." Losing electrons (cations) means stronger nuclear pull and smaller size; gaining electrons (anions) means more repulsion and larger size.

Question 5

Which of the following species is predicted to have the smallest bond angle?

  1. H₂O (correct answer)
  2. NH₃
  3. CH₄
  4. BF₃
Explanation: When you encounter questions about molecular geometry and bond angles, you need to apply VSEPR (Valence Shell Electron Pair Repulsion) theory, which predicts molecular shapes based on electron pair repulsion around the central atom. To determine bond angles, analyze each molecule's electron geometry. H₂O has 2 bonding pairs and 2 lone pairs around oxygen, creating a bent molecular geometry. The lone pairs occupy more space than bonding pairs, compressing the H-O-H bond angle to approximately 104.5°. NH₃ has 3 bonding pairs and 1 lone pair around nitrogen, forming a trigonal pyramidal shape. The single lone pair compresses the H-N-H bond angles to about 107°. CH₄ has 4 bonding pairs and no lone pairs around carbon, creating a perfect tetrahedral geometry with bond angles of 109.5°. BF₃ has 3 bonding pairs and no lone pairs around boron, forming a trigonal planar geometry with 120° bond angles. Looking at each option: A) H₂O has the smallest bond angle at 104.5° due to two lone pairs creating maximum compression. B) NH₃ has bond angles of 107° - smaller than tetrahedral but larger than water's. C) CH₄ maintains the ideal tetrahedral angle of 109.5° with no lone pair compression. D) BF₃ has the largest bond angles at 120° in its planar structure. Remember this key principle: lone pairs repel more strongly than bonding pairs, so molecules with more lone pairs on the central atom will have more compressed bond angles. Water, with two lone pairs, experiences maximum compression.

Question 6

Which of the following molecules is nonpolar, despite containing polar covalent bonds?

  1. CCl₄ (correct answer)
  2. NH₃
  3. H₂O
  4. CHCl₃
Explanation: When you encounter questions about molecular polarity, remember that a molecule can contain polar bonds yet still be nonpolar overall if its geometry causes the bond dipoles to cancel out. Carbon tetrachloride (CCl₄) exemplifies this principle perfectly. Each C-Cl bond is polar because chlorine is more electronegative than carbon, creating individual dipoles pointing from carbon toward each chlorine atom. However, CCl₄ has a tetrahedral geometry with the carbon at the center and four chlorine atoms positioned symmetrically around it. These four dipoles point in opposite directions and cancel each other out completely, making the overall molecule nonpolar despite its polar bonds. Looking at the incorrect options: NH₃ (option B) has a trigonal pyramidal shape due to its lone pair of electrons, preventing the N-H dipoles from canceling, so it's polar. H₂O (option C) has a bent molecular geometry because of two lone pairs on oxygen, causing the O-H dipoles to add rather than cancel, making it highly polar. CHCl₃ (option D) has an asymmetrical tetrahedral structure where the C-H bond differs significantly from the three C-Cl bonds, resulting in a net dipole and polar molecule. The key strategy here is to visualize molecular geometry using VSEPR theory. Symmetrical molecules like CCl₄, CO₂, and BF₃ often have canceling dipoles despite polar bonds, while asymmetrical molecules retain their polarity. Always consider both bond polarity and molecular shape when determining overall polarity.

Question 7

What is the predicted bond order of the dinitrogen molecule, N₂?

  1. 1
  2. 1.5
  3. 2
  4. 3 (correct answer)
Explanation: When you encounter questions about bond order, you're dealing with molecular orbital theory, which explains how atomic orbitals combine to form molecular orbitals and determine bond strength. To find the bond order of N₂, you need to consider nitrogen's electron configuration and how the electrons fill molecular orbitals. Each nitrogen atom has 7 electrons, giving N₂ a total of 14 electrons. These fill the molecular orbitals in order of increasing energy: σ1s2\sigma_{1s}^2, σ1s∗2\sigma_{1s}^*{}^2, σ2s2\sigma_{2s}^2, σ2s∗2\sigma_{2s}^*{}^2, π2px2\pi_{2p_x}^2, π2py2\pi_{2p_y}^2, σ2pz2\sigma_{2p_z}^2. Bond order equals (bonding electrons - antibonding electrons) ÷ 2. For N₂: (10 bonding electrons - 4 antibonding electrons) ÷ 2 = 3. This triple bond makes N₂ extremely stable, which explains why nitrogen gas is so unreactive. Choice A (bond order = 1) would represent a single bond, like you'd see in F₂. Choice B (bond order = 1.5) occurs in molecules like NO, where unpaired electrons create fractional bond orders. Choice C (bond order = 2) describes double bonds found in molecules like O₂. These are all too low for N₂. The correct answer is D, reflecting N₂'s triple bond. Remember that bond order correlates with bond strength and stability. When you see dinitrogen on the DAT, immediately think "triple bond" - it's one of the strongest bonds in chemistry and a favorite example of molecular orbital theory in action.

Question 8

A photon has a wavelength of 450 nm. Which of the following statements is true regarding a photon with a wavelength of 600 nm?

  1. The 600 nm photon possesses a higher frequency and travels at a faster speed.
  2. The 600 nm photon possesses a lower frequency and higher energy.
  3. The 600 nm photon possesses a higher frequency and lower energy.
  4. The 600 nm photon possesses a lower frequency and lower energy. (correct answer)
Explanation: When analyzing photon properties, you need to understand the fundamental relationships between wavelength, frequency, energy, and speed. All electromagnetic radiation (including photons) travels at the same speed in a vacuum—the speed of light (c = 3.0 × 10⁸ m/s). The key relationships are: c=λνc = \lambda \nu (where λ is wavelength and ν is frequency) and E=hνE = h\nu (where E is energy and h is Planck's constant). Since the speed of light is constant, wavelength and frequency are inversely related—as wavelength increases, frequency must decrease. Comparing the 600 nm photon to the 450 nm photon: since 600 nm is longer than 450 nm, the 600 nm photon has a lower frequency. Because energy is directly proportional to frequency, lower frequency means lower energy as well. Answer choice A is incorrect because longer wavelength means lower frequency, and all photons travel at the same speed in vacuum. Choice B incorrectly states that the 600 nm photon has both lower frequency and higher energy—this violates the direct relationship between frequency and energy. Choice C incorrectly claims the 600 nm photon has higher frequency, when longer wavelengths actually correspond to lower frequencies. Answer D correctly identifies that the 600 nm photon has both lower frequency and lower energy compared to the 450 nm photon. Remember this inverse relationship: longer wavelength = lower frequency = lower energy. This pattern appears frequently on science exams, so always check whether wavelength increases or decreases when comparing photons.

Question 9

An orbital diagram shows two electrons in a single 2p orbital, both with an upward-pointing spin arrow. This configuration violates which fundamental principle?

  1. Pauli Exclusion Principle (correct answer)
  2. Hund's Rule
  3. Aufbau Principle
  4. Heisenberg Uncertainty Principle
Explanation: When you encounter electron configuration problems, you're dealing with the fundamental rules that govern how electrons occupy atomic orbitals. This question tests your understanding of what happens when electrons pair up in orbitals. The Pauli Exclusion Principle states that no two electrons in an atom can have identical quantum numbers. Since electrons in the same orbital share the same spatial quantum numbers (n, l, and m), they must differ in their spin quantum number. This means paired electrons must have opposite spins - one "spin up" (↑) and one "spin down" (↓). The described configuration violates this principle because both electrons have the same spin direction, making their quantum numbers identical. Looking at the wrong answers: B) Hund's Rule requires electrons to singly occupy each orbital in a subshell before pairing occurs, but this question describes electrons that are already paired in one orbital, so Hund's Rule isn't the issue here. C) The Aufbau Principle governs the order in which orbitals are filled (lowest energy first), but since we're told these are 2p electrons, we're not dealing with orbital filling sequence. D) The Heisenberg Uncertainty Principle relates to the impossibility of simultaneously knowing an electron's exact position and momentum, which is unrelated to spin configurations. Remember this key distinction: Pauli deals with electron pairing and spin, while Hund's Rule deals with how electrons initially occupy empty orbitals. When you see same-spin electrons paired together, immediately think Pauli Exclusion Principle violation.

Question 10

How many sigma (σ) and pi (π) bonds are present in a molecule of acetylene (C₂H₂)?

  1. 2 σ, 2 π
  2. 2 σ, 3 π
  3. 3 σ, 2 π (correct answer)
  4. 5 σ, 0 π
Explanation: When analyzing molecular bonding, you need to understand that sigma (σ) bonds form from direct orbital overlap along the bond axis, while pi (π) bonds form from sideways overlap of p orbitals. Every single bond contains one σ bond, double bonds contain one σ and one π bond, and triple bonds contain one σ and two π bonds. To solve this, first draw acetylene's structure: H-C≡C-H. You'll see two C-H single bonds and one C≡C triple bond. Now count systematically: each C-H bond contributes 1 σ bond (2 total), and the C≡C triple bond contributes 1 σ bond plus 2 π bonds. This gives you 3 σ bonds and 2 π bonds total. Choice A (2 σ, 2 π) incorrectly counts only the bonds in the triple bond itself, forgetting the two C-H single bonds. Choice B (2 σ, 3 π) makes the common error of thinking a triple bond contains 3 π bonds rather than 1 σ and 2 π bonds. Choice D (5 σ, 0 π) treats the triple bond as if it were three separate single bonds, completely misunderstanding multiple bond structure. The correct answer is C (3 σ, 2 π). For DAT success, remember this pattern: always count every single bond as contributing one σ bond, then add the π bonds from multiple bonds separately. When you see linear molecules like acetylene, immediately recognize the triple bond and remember it's always 1 σ + 2 π, never 3 σ bonds.

Question 11

Which statement best describes the trend in effective nuclear charge (Zeff) across a period of the periodic table?

  1. Zeff increases because the number of protons increases while the number of core shielding electrons is constant. (correct answer)
  2. Zeff remains relatively constant because the number of protons and core electrons increase at the same rate.
  3. Zeff decreases because the number of core electrons increases, which enhances electron shielding.
  4. Zeff decreases because the increasing atomic radius weakens the nucleus's pull on valence electrons.
Explanation: When you encounter questions about effective nuclear charge trends, focus on understanding what's actually changing as you move across a period and how those changes affect the nuclear attraction experienced by valence electrons. Effective nuclear charge (ZeffZ_{eff}) represents the net positive charge that valence electrons experience after accounting for shielding by inner electrons. As you move across a period from left to right, two key changes occur: the number of protons in the nucleus increases by one with each element, and electrons are added to the same valence shell (not to inner shells). The correct answer is A because it captures this fundamental relationship. Each additional proton increases the nuclear charge by +1, while the core (inner shell) electrons remain constant across the period. Since valence electrons don't shield each other very effectively, the net result is an increasing attraction between the nucleus and valence electrons. Answer B incorrectly suggests that core electrons increase at the same rate as protons, but core electrons don't change as you move across a period. Answer C makes the opposite error, claiming that core electrons increase and enhance shielding, which doesn't happen within a period. Answer D confuses cause and effect—while atomic radius does decrease across a period, this is a consequence of increasing ZeffZ_{eff}, not the cause of decreasing ZeffZ_{eff}. Remember this pattern: across a period, ZeffZ_{eff} increases because nuclear charge increases faster than electron shielding. This drives many other periodic trends, including decreasing atomic radius and increasing ionization energy.

Question 12

The valence shell electron configuration of an element is ns²np⁴. In which group and period of the periodic table would this element be found?

  1. Group 4, Period n
  2. Group 14, Period n
  3. Group 16, Period n (correct answer)
  4. Group 2, Period n
Explanation: When you encounter electron configuration problems, you need to connect the valence shell configuration to both group number and period number on the periodic table. The configuration ns²np⁴ tells you this element has 6 valence electrons total (2 from the s orbital + 4 from the p orbital). For main group elements, the group number equals the number of valence electrons. Since there are 6 valence electrons, this element belongs to Group 6 in the older numbering system, which corresponds to Group 16 in the modern IUPAC numbering system. The period number corresponds directly to the principal quantum number n in the valence shell. If the valence configuration is ns²np⁴, then the element is in period n. Looking at the answer choices: Choice A incorrectly suggests Group 4, which would correspond to 4 valence electrons (like carbon's ns²np²). Choice B suggests Group 14, which corresponds to 4 valence electrons in the modern system - again wrong for our 6-electron configuration. Choice D suggests Group 2, which represents elements with only 2 valence electrons (ns² configuration like magnesium). Choice C correctly identifies Group 16, Period n. Group 16 elements (oxygen family) all have 6 valence electrons and follow the ns²np⁴ pattern. Remember this key relationship: for main group elements, count the valence electrons to find the group number (using modern IUPAC numbering), and the highest principal quantum number n gives you the period. This systematic approach works for any valence electron configuration problem.

Question 13

What is the hybridization of the central carbon atom in a molecule of carbon dioxide, CO₂?

  1. sp (correct answer)
  2. sp²
  3. sp³
  4. sp³d
Explanation: When you encounter hybridization questions, you need to determine the electron geometry around the central atom by counting electron domains (bonding pairs and lone pairs). For carbon dioxide (CO₂), the central carbon atom forms two double bonds with the oxygen atoms. Each double bond counts as one electron domain, regardless of whether it's a single, double, or triple bond. Carbon has no lone pairs in this molecule, so you have exactly two electron domains around the central carbon. The number of electron domains determines hybridization: two domains require sp hybridization. This creates a linear molecular geometry with a bond angle of 180°, which matches CO₂'s known structure. Looking at the wrong answers: Choice B (sp²) would require three electron domains and result in trigonal planar geometry with 120° bond angles - this doesn't match CO₂'s linear structure. Choice C (sp³) needs four electron domains and produces tetrahedral geometry with 109.5° bond angles, which is completely wrong for CO₂. Choice D (sp³d) requires five electron domains and involves d orbitals, which carbon doesn't use in its ground state bonding. The correct answer is A (sp hybridization). Study tip: Remember the pattern: 2 electron domains = sp (linear), 3 domains = sp² (trigonal planar), 4 domains = sp³ (tetrahedral). Count all bonds as single domains regardless of bond order, and don't forget to include lone pairs on the central atom in your count.

Question 14

Which property is characteristic of halogens (Group 17) but not of alkali metals (Group 1)?

  1. They are highly reactive elements found naturally only in compounds.
  2. They tend to form ions with a charge of +1 in ionic compounds.
  3. They exist as diatomic molecules in their elemental state at standard temperature and pressure. (correct answer)
  4. They have low first ionization energies compared to other elements in their respective periods.
Explanation: When comparing different groups on the periodic table, focus on their unique chemical and physical properties that arise from their electron configurations and resulting bonding behavior. Halogens (Group 17) have seven valence electrons and need one more to achieve a stable octet. In their elemental form, they achieve this by sharing electrons in covalent bonds, forming diatomic molecules like F₂, Cl₂, Br₂, and I₂. This is why choice C is correct - halogens exist as diatomic molecules at standard conditions. Now let's examine why the other options don't distinguish these groups: Choice A incorrectly suggests this applies only to halogens. Both halogens and alkali metals are so reactive that they're never found as free elements in nature - they only exist in compounds like NaCl or CaF₂. Choice B describes alkali metals, not halogens. Alkali metals lose their single valence electron to form +1 cations (like Na⁺), while halogens gain an electron to form -1 anions (like Cl⁻). Choice D also describes alkali metals rather than halogens. Alkali metals have the lowest ionization energies in their periods because they easily lose their single valence electron. Halogens actually have high ionization energies since they strongly attract electrons. Remember this pattern: when distinguishing element groups, focus on their unique structural features in elemental form. Halogens forming diatomic molecules is a distinctive characteristic that sets them apart from metals, which typically form metallic lattices or exist as individual atoms in the gas phase.

Question 15

Consider the resonance structures for the thiocyanate ion, SCN⁻. Based on formal charge analysis, which structure is the most significant contributor to the resonance hybrid?

  1. A structure where the central carbon atom bears a negative formal charge
  2. The structure with single bond between S and C, and triple bond between C and N
  3. The structure with triple bond between S and C, and single bond between C and N
  4. The structure with double bonds between S and C, and between C and N (correct answer)
Explanation: When analyzing resonance structures, you need to evaluate formal charges to determine which structure contributes most significantly to the resonance hybrid. The most stable structures have formal charges closest to zero and place negative charges on the most electronegative atoms. For the thiocyanate ion SCN⁻, let's examine the formal charges. Using the formula: Formal Charge = (valence electrons) - (nonbonding electrons) - ½(bonding electrons), the structure with double bonds between S-C and C-N gives formal charges of 0, 0, and -1 for S, C, and N respectively. This places the negative charge on nitrogen, the most electronegative atom in the ion, while keeping formal charges minimal. Choice A is incorrect because having a negative formal charge on the central carbon atom is less favorable than placing it on the more electronegative nitrogen. Choice B (single S-C, triple C-N bonds) results in a positive formal charge on sulfur and negative on carbon, which is unfavorable since sulfur is more electronegative than carbon. Choice C (triple S-C, single C-N bonds) places a large positive formal charge on nitrogen, the most electronegative atom, making this structure highly unstable and essentially non-contributing. The structure in choice D minimizes formal charges while placing the negative charge on the most electronegative atom, making it the major contributor to the resonance hybrid. Study tip: Always calculate formal charges for each atom in resonance structures. The most stable contributor typically has the smallest formal charges and places negative charges on the most electronegative atoms.

Question 16

Which of the following atoms or ions is paramagnetic in its ground state?

  1. Ca
  2. Zn²⁺
  3. Ne
  4. N (correct answer)
Explanation: When you encounter questions about paramagnetism, you need to determine whether an atom or ion has unpaired electrons. Paramagnetic substances are attracted to magnetic fields because they contain unpaired electrons, while diamagnetic substances (with all electrons paired) are weakly repelled. To solve this, write out the electron configurations for each option. Nitrogen (N) has 7 electrons with the configuration 1s22s22p31s^2 2s^2 2p^3. In the 2p subshell, Hund's rule requires electrons to occupy orbitals singly before pairing up. So the three 2p electrons occupy separate orbitals, leaving nitrogen with three unpaired electrons and making it paramagnetic. Let's examine why the other options are diamagnetic: Calcium (A) has 20 electrons with configuration [Ar]4s2[Ar] 4s^2, where all electrons are paired. Zinc ion Zn²⁺ (B) loses its two 4s electrons from neutral zinc, leaving the configuration [Ar]3d10[Ar] 3d^{10} with all d electrons paired. Neon (C) is a noble gas with configuration 1s22s22p61s^2 2s^2 2p^6, where all electrons are completely paired in filled shells. The key strategy is to focus on partially filled subshells, especially p, d, and f orbitals. Remember Hund's rule: electrons will occupy orbitals singly before pairing up. This means atoms with odd numbers of electrons in a subshell (like nitrogen's 2p³) will typically have unpaired electrons. For transition metals, consider both the neutral atom and any ions formed by losing electrons, as electron removal can change the magnetic properties.

Question 17

Which of the following covalent bonds is the most polar?

  1. C-H
  2. F-H (correct answer)
  3. O-H
  4. N-H
Explanation: When you encounter questions about bond polarity, you need to consider the difference in electronegativity between the two atoms forming the bond. The greater this difference, the more polar the bond becomes. Electronegativity measures an atom's ability to attract electrons in a chemical bond. On the Pauling scale, fluorine has the highest electronegativity at 4.0, followed by oxygen (3.4), nitrogen (3.0), carbon (2.6), and hydrogen (2.2). To determine bond polarity, calculate the electronegativity difference for each bond. For F-H (choice B), the difference is 4.0 - 2.2 = 1.8, making this the most polar bond. The large difference means fluorine strongly pulls electron density away from hydrogen, creating a significant dipole moment. Choice A (C-H) has an electronegativity difference of only 0.4 (2.6 - 2.2), making it nearly nonpolar. Choice C (O-H) shows a difference of 1.2 (3.4 - 2.2), creating moderate polarity but less than F-H. Choice D (N-H) has a difference of 0.8 (3.0 - 2.2), making it less polar than both O-H and F-H bonds. The F-H bond's high polarity explains why hydrogen fluoride is such a strong acid and why fluorine-containing compounds often have unique properties in biological systems. Remember this pattern: fluorine creates the most polar bonds with other elements due to its extreme electronegativity. When comparing bond polarities, always look for fluorine first, then oxygen, then nitrogen as the most electronegative partners.

Question 18

The molecular geometry of ammonia (NH₃) is best described as which of the following?

  1. Tetrahedral
  2. Trigonal pyramidal (correct answer)
  3. Trigonal planar
  4. T-shaped
Explanation: When determining molecular geometry, you need to consider both the bonding electron pairs and lone pairs around the central atom, as both affect the three-dimensional shape of the molecule. For ammonia (NH₃), nitrogen is the central atom with 5 valence electrons. It forms three single bonds with hydrogen atoms, using 3 electrons, leaving one lone pair. This gives you 4 electron pairs total around nitrogen: 3 bonding pairs and 1 lone pair. These 4 electron pairs arrange themselves in a tetrahedral electron geometry to minimize repulsion. However, molecular geometry describes only the positions of atoms, not electron pairs. Since one of the four positions is occupied by a lone pair (which you can't "see"), the three hydrogen atoms form a trigonal pyramidal shape around nitrogen. Think of it as a three-legged stool with the nitrogen slightly above the plane of the hydrogens. Choice A (tetrahedral) describes the electron pair geometry, not the molecular geometry. This is a common trap - tetrahedral would only be the molecular geometry if all four electron pairs were bonding pairs. Choice C (trigonal planar) would occur if nitrogen had three bonding pairs and no lone pairs, making it flat. Choice D (T-shaped) typically occurs with five electron pairs around the central atom, where two positions are occupied by lone pairs. Remember: molecular geometry questions require you to count both bonding and lone pairs, then describe only the arrangement of atoms. Always distinguish between electron geometry and molecular geometry.

Question 19

Which of the following elements exhibits the most pronounced metallic character?

  1. Beryllium (Be)
  2. Boron (B)
  3. Carbon (C)
  4. Cesium (Cs) (correct answer)
Explanation: When you encounter questions about metallic character, you're being tested on periodic trends and how an element's position affects its properties. Metallic character refers to how readily an element loses electrons to form positive ions, along with properties like electrical conductivity and metallic bonding. Metallic character increases as you move down a group (column) and decreases as you move across a period (row) from left to right. This happens because atoms get larger going down a group, making outer electrons easier to remove, while atoms get smaller and hold electrons more tightly going across a period. Looking at the positions: Cesium (Cs) sits in Group 1, Period 6 - the bottom-left region of the periodic table where metallic character is maximized. Cesium is an alkali metal with the largest atomic radius among these choices, making its single valence electron extremely easy to remove. This gives it the strongest metallic properties. Option A, Beryllium (Be), is a metal but sits in Period 2, so it's much smaller than cesium and holds its electrons more tightly. Option B, Boron (B), is a metalloid that exhibits both metallic and nonmetallic properties, placing it between metals and nonmetals in character. Option C, Carbon (C), is a nonmetal that rarely loses electrons and instead tends to share them in covalent bonds. For DAT chemistry questions, remember that metallic character follows the same trend as atomic size: increases going down and left on the periodic table. The bottom-left elements are always the most metallic.

Question 20

What is the molecular geometry of the methane molecule, CH₄?

  1. Square planar
  2. Trigonal pyramidal
  3. Tetrahedral (correct answer)
  4. See-saw
Explanation: When you encounter molecular geometry questions, you need to determine the three-dimensional arrangement of atoms around a central atom using VSEPR (Valence Shell Electron Pair Repulsion) theory. For methane (CH₄), start by identifying the central atom (carbon) and counting electron pairs around it. Carbon has 4 valence electrons and forms 4 single bonds with hydrogen atoms, giving it 4 bonding electron pairs and no lone pairs. According to VSEPR theory, these 4 electron pairs arrange themselves as far apart as possible in three-dimensional space to minimize repulsion. This arrangement creates a tetrahedral geometry, where the carbon atom sits at the center and the four hydrogen atoms occupy the corners of a tetrahedron. The bond angles are approximately 109.5°, making answer C correct. Looking at the wrong choices: Answer A (square planar) occurs when you have 4 bonding pairs and 2 lone pairs, typically seen in compounds like XeF₄. Answer B (trigonal pyramidal) results from 3 bonding pairs and 1 lone pair, as in ammonia (NH₃) - the lone pair pushes the bonding pairs down, creating a pyramid shape. Answer D (see-saw) occurs with 4 bonding pairs and 1 lone pair, found in molecules like SF₄. Remember this pattern: for molecules with 4 bonding pairs and no lone pairs around the central atom, the geometry is always tetrahedral. Count bonding and lone pairs first, then apply VSEPR theory to predict the shape.