- What is the electron configuration of the central metal ion in the complex [Co(NH₃)₆]³⁺ and how many unpaired electrons does it have?
Answer: [Ar] 3d⁶, 4 unpaired electrons
First, determine the oxidation state of cobalt. Since NH₃ is a neutral ligand and the overall complex charge is +3, the cobalt ion must be Co³⁺. The electron configuration of neutral cobalt (Co, Z=27) is [Ar] 4s²3d⁷. When forming Co³⁺, three electrons are removed (4s electrons are removed first, then one 3d electron), giving [Ar] 3d⁶. In the d⁶ configuration, assuming a high-spin complex (typical for IB level), the six d-electrons are distributed as: ↑↓ ↑ ↑ ↑ ↑ across the five d-orbitals, resulting in 4 unpaired electrons.
- A neutral atom in its ground state is in Period 3 and has the same number of unpaired electrons as a nitrogen atom. What is the condensed electron configuration of this element?
Answer: [Ne] 3s²3p³
First, determine the number of unpaired electrons in a nitrogen atom (N, [He] 2s²2p³). According to Hund's rule, the three electrons in the 2p sublevel will occupy separate orbitals with parallel spins, resulting in 3 unpaired electrons. Next, find a Period 3 element with 3 unpaired electrons. The valence electrons are in the n=3 level. A configuration ending in 3p³ will have 3 unpaired electrons. This corresponds to the element phosphorus (P), with the configuration [Ne] 3s²3p³. The other options for Period 3 elements are Al (3p¹, 1 unpaired e⁻), Cl (3p⁵, 1 unpaired e⁻), and Ar (3p⁶, 0 unpaired e⁻).
- Which species contains the greatest number of unpaired electrons in its ground state?
Answer: Cr
To determine the number of unpaired electrons, we examine the orbital diagrams for the valence shell.
A. N ([He] 2s²2p³): The three p-electrons occupy three separate orbitals with parallel spins (Hund's rule), so there are 3 unpaired electrons.
B. S ([Ne] 3s²3p⁴): The four p-electrons result in one filled p-orbital and two singly-occupied p-orbitals, so there are 2 unpaired electrons.
C. Cr ([Ar] 4s¹3d⁵): Due to the stability of half-filled sublevels, chromium's configuration is an exception. It has one unpaired electron in the 4s orbital and five unpaired electrons in the 3d orbitals, for a total of 6 unpaired electrons.
D. Cu ([Ar] 4s¹3d¹⁰) -> Cu⁺ ([Ar] 3d¹⁰): All d-orbitals are filled, so there are 0 unpaired electrons.
- What is the total number of electrons occupying p-orbitals in a single bromide ion, Br⁻?
Answer: 18
A bromine atom (Br, Z=35) has the configuration [Ar] 4s²3d¹⁰4p⁵. A bromide ion, Br⁻, gains one electron to achieve the configuration [Ar] 4s²3d¹⁰4p⁶. To find the total number of p-electrons, we must count them in all occupied principal energy levels. The configuration of argon (Ar) is 1s²2s²2p⁶3s²3p⁶. So, the Br⁻ ion has 6 p-electrons in the n=2 shell (2p⁶), 6 p-electrons in the n=3 shell (3p⁶), and 6 p-electrons in the n=4 shell (4p⁶). The total is 6 + 6 + 6 = 18 p-electrons.
- An element in period 4 has the same number of unpaired electrons in its ground state as it does when it forms its most stable +2 cation. Which element could this be?
Answer: Manganese (Mn), because both Mn and Mn²⁺ have 5 unpaired electrons
Mn has configuration [Ar] 3d⁵ 4s² with 5 unpaired electrons (following Hund's rule in d⁵). Mn²⁺ has configuration [Ar] 3d⁵ (loses both 4s electrons) and retains 5 unpaired electrons. Fe ([Ar] 3d⁶ 4s²) has 4 unpaired electrons, but Fe²⁺ ([Ar] 3d⁶) has only 4 unpaired electrons due to pairing. Ni ([Ar] 3d⁸ 4s²) has 2 unpaired electrons, but Ni²⁺ ([Ar] 3d⁸) also has 2 unpaired electrons. Zn ([Ar] 3d¹⁰ 4s²) and Zn²⁺ ([Ar] 3d¹⁰) both have 0 unpaired electrons. However, Mn is the clearest example where the d⁵ configuration provides maximum stability with all unpaired electrons.
- The successive ionization energies of aluminium show a large increase between the third and fourth ionization energies. What is the primary reason for this observation?
Answer: The fourth electron is removed from a principal energy level with a lower value of n.
The electron configuration of aluminium is [Ne] 3s²3p¹. The first three ionization energies correspond to the removal of the three valence electrons from the n=3 shell. The fourth electron must be removed from the n=2 shell (from the 2p⁶ sublevel). Electrons in a lower principal energy level are much closer to the nucleus and experience significantly less shielding from other electrons. This results in a much stronger electrostatic attraction to the nucleus, requiring a substantially larger amount of energy for removal. While A and B are true statements, C provides the most fundamental physical reason for the large jump in energy.
- The first ionization energy corresponds to the removal of the most loosely held electron. Which element has a first ionization energy that is lower than that of the element immediately preceding it in the same period?
Answer: B
Generally, first ionization energy (IE) increases across a period. However, there are exceptions. The element preceding boron (B) is beryllium (Be). Be has the configuration [He] 2s². Boron has the configuration [He] 2s²2p¹. The electron removed from B is a 2p electron. The electron removed from Be is a 2s electron. The 2p electron is in a higher energy sublevel and is slightly more shielded by the 2s electrons, making it easier to remove than a 2s electron from Be, despite Boron's higher nuclear charge. This causes a dip in the IE trend, so B has a lower first IE than Be.
- Element Y has the electron configuration [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁵. When element Y gains one electron, what type of orbital does the electron enter, and what is the resulting electron configuration?
Answer: The electron enters a 6p orbital; configuration becomes [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁶
Element Y is astatine (At) with one electron short of a complete 6p subshell. When gaining an electron, it will complete the 6p subshell to achieve the stable noble gas configuration [Rn]. The electron enters the available 6p orbital, giving [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁶. Choice B incorrectly assumes the next higher d orbital. Choice C incorrectly suggests the 7s orbital. Choice D wrongly suggests adding to the already filled 5d subshell.
- The first five successive ionization energies (in kJ mol⁻¹) for an element are 786, 1577, 3232, 4356, and 16091. To which group of the periodic table does this element most likely belong?
Answer: Group 14
Successive ionization energies increase as more electrons are removed. A large, disproportionate jump in ionization energy occurs when an electron is removed from a new, inner principal energy level (a core electron). The energies are approximately 786, 1577, 3232, 4356, and 16091. The largest increase is between the 4th IE (4356) and the 5th IE (16091). This indicates that the first four electrons are valence electrons and the fifth is a core electron. An element with four valence electrons belongs to Group 14.
- Vanadium (V, Z=23) is a transition element. What is the total number of electrons in the highest principal quantum number shell (n) and the total number of valence electrons, respectively, for a vanadium atom?
Answer: 2, 5
The ground state electron configuration of vanadium (V, Z=23) is [Ar] 4s²3d³. The highest principal quantum number (n) is 4. There are two electrons in this shell, in the 4s orbital. The valence electrons for a transition metal are typically considered all electrons outside the noble gas core. In this case, they are the 4s and 3d electrons. The total number of valence electrons is 2 + 3 = 5. Therefore, the respective numbers are 2 and 5.
- An element X has the ground state electron configuration [Ar] 3d⁵ 4s¹. When this element forms its most common cation in aqueous solution, what is the electron configuration of the resulting ion?
Answer: [Ar] 3d³
Element X is chromium (Cr), which has the unusual electron configuration [Ar] 3d⁵ 4s¹ due to the stability of half-filled d orbitals. Chromium's most common oxidation state is +3, forming Cr³⁺. When forming cations, electrons are removed first from the 4s orbital, then from the 3d orbitals. For Cr³⁺: remove 1 electron from 4s and 2 electrons from 3d, giving [Ar] 3d³. Choice A represents Cr¹⁺ (only 4s electron removed), choice B represents Cr²⁺, and choice D would require adding an electron.
- Which of the following electron configurations represents a possible excited state of a neutral silicon (Si) atom?
Answer: 1s²2s²2p⁶3s²3p¹4s¹
The ground state configuration of silicon (Si, Z=14) is 1s²2s²2p⁶3s²3p². An excited state involves one or more electrons being promoted to a higher energy orbital, while the total number of electrons remains the same.
A is the ground state.
B is the configuration of an Al atom, not Si.
C shows one of the 3p electrons promoted to the 4s orbital. This is a valid excited state for Si as it has the correct number of electrons (14).
D is an impossible configuration because the s-sublevel can only hold a maximum of two electrons.
- Which of the following statements correctly explains why the electron configuration of copper is [Ar] 3d¹⁰ 4s¹ rather than the expected [Ar] 3d⁹ 4s²?
Answer: A completely filled d subshell provides extra stability due to symmetrical electron distribution and exchange energy
The anomalous electron configuration of copper results from the extra stability gained by having completely filled d orbitals. The d¹⁰ configuration provides maximum exchange energy and symmetrical electron distribution. Choice A is incorrect because 4s fills before 3d in most transition metals. Choice C is wrong as 4s can hold two electrons. Choice D confuses atomic electron configuration with ionization in metallic bonding.
- Which of the following represents the correct electron configuration for the most stable form of chromium in the +6 oxidation state?
Answer: [Ar] 3d⁰ (all electrons removed from 3d and 4s orbitals)
When you encounter electron configuration questions involving transition metals in high oxidation states, focus on which electrons are removed and in what order. For chromium in the +6 oxidation state, you need to remove 6 electrons from neutral chromium.
Neutral chromium has the electron configuration [Ar] 4s¹ 3d⁵ (note the unusual configuration where chromium has only one 4s electron to achieve half-filled d orbitals for extra stability). To reach the +6 oxidation state, you must remove 6 electrons total. Since 4s electrons are removed before 3d electrons when forming ions, you remove the 1 electron from 4s and then 5 electrons from 3d, leaving [Ar] 3d⁰.
Option A introduces a nonsensical "negative electron configuration" concept that doesn't exist in chemistry. You cannot have negative electrons in an orbital.
Option B ([Ar] 3d⁴) only accounts for removing 2 electrons, which would give Cr²⁺, not Cr⁶⁺. This represents a common error of not removing enough electrons.
Option C ([Ne] 3s² 3p⁶) incorrectly suggests removing electrons from the 3s and 3p orbitals of the argon core, which would require enormous energy and doesn't occur under normal chemical conditions. Core electrons are not involved in typical oxidation states.
Option D correctly shows [Ar] 3d⁰, representing the removal of all valence electrons (1 from 4s + 5 from 3d = 6 total).
Study tip: For transition metal ions, always remove 4s electrons first, then 3d electrons. Count carefully to match the oxidation state with the total electrons removed.
- An atom has 26 electrons and exhibits paramagnetism in its ground state. If this atom loses three electrons to form a cation, how many unpaired electrons does the resulting ion contain?
Answer: 5 unpaired electrons
The atom with 26 electrons is iron (Fe) with configuration [Ar] 3d⁶ 4s². When Fe loses 3 electrons to form Fe³⁺, it loses the two 4s electrons and one 3d electron, resulting in [Ar] 3d⁵. The d⁵ configuration has 5 unpaired electrons (one in each d orbital following Hund's rule). Choice A would be d¹, choice B would be d³, and choice D would be an impossible d⁴ configuration with 4 unpaired electrons.
- The emission spectrum of an element arises from electron transitions between discrete energy levels. If a line in the spectrum has a frequency of 4.57 × 10¹⁴ Hz, what is the energy, in Joules, of one photon of this light? (Planck's constant, h = 6.63 × 10⁻³⁴ J s)
Answer: 3.03 × 10⁻¹⁹ J
The energy of a photon (E) is related to its frequency (ν) by the Planck-Einstein relation, E = hν. Using the given values: E = (6.63 × 10⁻³⁴ J s) × (4.57 × 10¹⁴ s⁻¹) = 3.0299 × 10⁻¹⁹ J. Rounding to three significant figures gives 3.03 × 10⁻¹⁹ J. Distractor A results from dividing frequency by Planck's constant. Distractor B results from dividing Planck's constant by frequency. Other distractors represent common calculation errors.
- An orbital box diagram shows the arrangement of electrons in an atom's sublevels. Which of the following arrangements for a p-sublevel is forbidden by the Pauli exclusion principle?
Answer: Two electrons with parallel spins in one orbital; the other two orbitals are empty.
The Pauli exclusion principle states that no two electrons in the same atom can have the same set of four quantum numbers. This implies that if two electrons occupy the same orbital, they must have opposite spins. Choice B describes two electrons in the same orbital with parallel (the same) spins, which directly violates this principle. Choice A violates Hund's rule but is permitted by the Pauli principle. Choices C and D represent valid ground or excited state configurations.
- The first four ionization energies of an element X are 578, 1817, 2745, and 11577 kJ mol⁻¹. Which element is most likely to be X?
Answer: Aluminium
A large jump in successive ionization energies indicates the removal of an electron from a core shell. The ionization energies are IE₁=578, IE₂=1817, IE₃=2745, IE₄=11577. The jump from IE₃ to IE₄ (2745 to 11577) is significantly larger than the preceding jumps. This implies that the element has three valence electrons. After these three are removed, the fourth electron is removed from a stable, inner electron shell. The element with three valence electrons among the choices is Aluminium (Al), which is in Group 13.
- A student writes the electron configuration of Fe³⁺ as [Ar] 3d³ 4s². What error did the student make, and what is the correct configuration?
Answer: The student removed electrons from the wrong orbitals; correct configuration is [Ar] 3d⁵
Fe has the configuration [Ar] 3d⁶ 4s². To form Fe³⁺, three electrons must be removed: the two 4s electrons first, then one 3d electron, giving [Ar] 3d⁵. The student incorrectly kept the 4s² and removed too many d electrons. The error is removing electrons from the wrong orbitals (should remove 4s before 3d). Choice A suggests the student didn't remove enough electrons, but they removed 3 as required. Choice C is wrong about the noble gas core. Choice D misidentifies the error type.
- An atom has the ground-state electron configuration [Kr] 5s²4d¹⁰5p⁴. In which period and block of the periodic table is this element located?
Answer: Period 5, p-block
The period number is determined by the highest principal energy level (n) occupied by electrons. In this configuration, the highest n value is 5 (from 5s² and 5p⁴), so the element is in Period 5. The block is determined by the sublevel of the highest energy electron(s). Here, the last electrons are added to the 5p sublevel, placing the element in the p-block.
