Atomic Structure and Chemical Bonding

81 soru

Soru 21Soru

Magnesium oxide (MgOMgO) has a significantly higher melting point than sodium chloride (NaClNaCl) because the electrostatic forces of attraction between its divalent ions (Mg2+Mg^{2+} and O2O^{2-}) are substantially stronger than those between the monovalent ions (Na+Na^+ and ClCl^-).

Cevabı ve açıklamayı göster

Cevap: True

Cevap

True
The statement is true because the electrostatic force holding an electrovalent lattice together scales with the product of the ionic charges. Divalent magnesium (Mg2+Mg^{2+}) and oxide (O2O^{2-}) ions form a lattice with much higher lattice energy than monovalent sodium (Na+Na^+) and chloride (ClCl^-) ions, giving magnesium oxide a much higher melting point.

Adım Adım Çözüm

1
Determine the ionic charges of the component ions in both compounds
MgOMgO is composed of Mg2+Mg^{2+} and O2O^{2-} ions (divalent), while NaClNaCl is composed of Na+Na^+ and ClCl^- ions (monovalent).
The magnitude of ionic charges directly dictates the strength of electrostatic forces in an electrovalent crystal lattice.
2
Apply Coulomb's Law to compare lattice attraction strength
The attraction force scales with the charge product: for MgOMgO, (+2)×(2)=4|(+2) \times (-2)| = 4; for NaClNaCl, (+1)×(1)=1|(+1) \times (-1)| = 1.
A fourfold increase in charge product produces significantly stronger ionic bonds and greater lattice energy.
3
Correlate lattice energy with melting point
Greater thermal energy is required to overcome the electrostatic forces in MgOMgO than in NaClNaCl, resulting in a vastly higher melting point.
Melting point directly reflects the energy required to break down the solid giant ionic lattice structure.

Anahtar Kavram

Lattice Energy and Ion Charge Dependency in Ionic Compounds
Soru 22Soru

Which of the following chemical species contains both covalent and dative (coordinate) covalent bonds?

Cevabı ve açıklamayı göster

Cevap: Ammonium ion (NH4+NH_4^+)

Cevap

Ammonium ion (NH4+NH_4^+) contains three covalent bonds and one dative (coordinate) covalent bond.
The ammonium ion (NH4+NH_4^+) is formed when a neutral ammonia molecule (NH3NH_3), which has three single covalent bonds and one unshared lone pair on the nitrogen atom, donates its lone pair of electrons to an electron-deficient hydrogen ion (H+H^+). Therefore, NH4+NH_4^+ contains both three standard covalent bonds and one dative bond.

Adım Adım Çözüm

1
Analyze the electronic configuration and bonding of ammonia (NH3NH_3)
In NH3NH_3, the central nitrogen atom forms 3 covalent bonds with hydrogen atoms, sharing three of its valence electrons, leaving one unshared lone pair of electrons.
Understanding the starting structure is essential to see where the unshared electron pair originates.
2
Examine the reaction of ammonia with a hydrogen ion (H+H^+)
The hydrogen ion (H+H^+) has an empty 1s1s orbital and requires two electrons to achieve a stable duplet configuration.
A dative bond requires an electron donor (species with a lone pair) and an electron acceptor (species with an empty orbital).
3
Determine the bond types in NH4+NH_4^+
Nitrogen donates its lone pair to the incoming H+H^+ ion forming a coordinate bond, while retaining the three original covalent bonds.
This confirms the co-existence of 3 ordinary covalent bonds and 1 dative bond within the ammonium ion.

Anahtar Kavram

Covalent and Coordinate (Dative) Bonding in Polyatomic Ions
Tahmini Süre:1m 0s
Soru 23Soru

Match each chemical species in Column A with its correct bonding description in Column B.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Hydronium ion (H3O+H_3O^+)
Ammonium ion (NH4+NH_4^+)
Aluminum chloride dimer (Al2Cl6Al_2Cl_6)
Carbon monoxide (COCO)

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

The correct pairings are: Hydronium ion (H3O+H_3O^+) matches with oxygen donating a lone pair to a proton; Ammonium ion (NH4+NH_4^+) matches with nitrogen donating a lone pair to a proton; Aluminum chloride dimer (Al2Cl6Al_2Cl_6) matches with bridging chlorine atoms donating lone pairs to electron-deficient aluminum centers; Carbon monoxide (COCO) matches with a triple bond containing two covalent bonds and one coordinate bond from oxygen to carbon.
Each chemical species correctly corresponds to its specific electron pair donor-acceptor behavior: H3O+H_3O^+ relies on oxygen lone pair donation to H+H^+, NH4+NH_4^+ relies on nitrogen lone pair donation to H+H^+, Al2Cl6Al_2Cl_6 uses bridging chlorine lone pairs to complete aluminum octets, and COCO uses a dative bond from oxygen to carbon alongside two covalent bonds.

Adım Adım Çözüm

1
Analyze the coordinate bonding in simple protonated ions (H3O+H_3O^+ and NH4+NH_4^+).
In H3O+H_3O^+, oxygen in H2OH_2O acts as the electron pair donor to H+H^+. In NH4+NH_4^+, nitrogen in NH3NH_3 acts as the electron pair donor to H+H^+.
Distinguishing between oxygen and nitrogen donor atoms correctly pairs the protonated species.
2
Examine the structure of the aluminum chloride dimer (Al2Cl6Al_2Cl_6).
Monomeric AlCl3AlCl_3 is electron-deficient (6 valence electrons around aluminum). Dimerization occurs when a lone pair from a chlorine atom on one AlCl3AlCl_3 molecule is donated to the aluminum atom of another AlCl3AlCl_3 molecule, creating two coordinate bridges.
This bridge satisfies the octet rule for both aluminum atoms in Al2Cl6Al_2Cl_6.
3
Determine the bonding in carbon monoxide (COCO).
Carbon has 4 valence electrons and oxygen has 6. To complete octets for both atoms, oxygen shares two pairs covalently and donates one lone pair to form a coordinate bond, creating a triple bond (COC \equiv O).
This establishes that oxygen is the donor and carbon is the acceptor in the dative component of the triple bond.

Anahtar Kavram

Coordinate (Dative) Covalent Bonding and Electron Pair Donation
Soru 24Soru

Element YY exists naturally as two isotopes, 63Y^{63}Y and 65Y^{65}Y. If the relative atomic mass of element YY is 63.563.5, what is the percentage abundance of the lighter isotope, 63Y^{63}Y?

Cevabı ve açıklamayı göster

Cevap: 75%75\%

Cevap

The percentage abundance of the lighter isotope 63Y^{63}Y is 75%75\%.
The percentage abundance is calculated using the weighted average formula: RAM=(m1×x1)+(m2×x2)100\text{RAM} = \frac{(m_1 \times x_1) + (m_2 \times x_2)}{100}. Substituting the given values yields 63.5=63x+65(100x)10063.5 = \frac{63x + 65(100-x)}{100}, which simplifies to 2x=150-2x = -150, giving x=75%x = 75\% for 63Y^{63}Y.

Adım Adım Çözüm

1
Define variables for isotopic abundances.
Let the abundance of 63Y^{63}Y be x%x\%. The abundance of 65Y^{65}Y is therefore (100x)%(100 - x)\%.
The sum of all relative isotopic abundances for a natural element equals 100%100\%.
2
Set up the weighted average formula for relative atomic mass.
Relative Atomic Mass=(63×x)+65×(100x)100=63.5\text{Relative Atomic Mass} = \frac{(63 \times x) + 65 \times (100 - x)}{100} = 63.5
Relative atomic mass is the weighted average of the atomic masses of the naturally occurring isotopes.
3
Solve the equation for xx.
63x+650065x=6350    2x=150    x=7563x + 6500 - 65x = 6350 \implies -2x = -150 \implies x = 75
Simplifying algebraic terms gives the percentage abundance of the lighter isotope.

Anahtar Kavram

Isotopic Abundance and Relative Atomic Mass Calculation
Tahmini Süre:1m 30s
Soru 25Soru

A tripositive ion, X3+X^{3+}, has a mass number of 5656 and contains 2323 electrons. How many neutrons are present in the nucleus of an atom of element XX?

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

30
To find the number of neutrons, first determine the atomic number (number of protons) of element XX. The ion X3+X^{3+} carries a +3+3 charge because it lost 3 electrons. Since X3+X^{3+} has 23 electrons, the neutral atom XX has 23+3=2623 + 3 = 26 electrons, which means it has 26 protons. The mass number (A=56A = 56) is the sum of protons (ZZ) and neutrons (NN). Thus, N=5626=30N = 56 - 26 = 30.

Adım Adım Çözüm

1
Determine the atomic number (number of protons) of element XX
Protons (ZZ) = 26
The tripositive ion X3+X^{3+} has lost 3 electrons. The neutral atom has 23+3=2623 + 3 = 26 electrons, which equals its proton count.
2
Calculate the number of neutrons
Neutrons (NN) = 30
Subtract the atomic number from the mass number: N=AZ=5626=30N = A - Z = 56 - 26 = 30.

Anahtar Kavram

Calculation of subatomic particles in ions using atomic number and mass number relationships
Soru 26Soru

Complete the statement below regarding the chemical reaction between ammonia and boron trifluoride.

Aşağıdaki boşlukları doldurun

In the formation of the adduct between ammonia (NH3NH_3) and boron trifluoride (BF3BF_3), the nitrogen atom contributes both electrons to form the chemical bond with boron. This type of linkage is known as a bond, and nitrogen functions as the electron pair .
Cevabı ve açıklamayı göster

Cevap

Blank 1 should be filled with 'dative covalent' (or 'coordinate covalent'), and Blank 2 should be filled with 'donor' (or 'Lewis base').
When NH3NH_3 reacts with BF3BF_3, nitrogen donates its unshared lone pair of electrons into the empty p-orbital of the electron-deficient boron atom. A covalent bond where one atom contributes both bonding electrons is a dative (coordinate covalent) bond, and the species donating the electron pair is the donor (or Lewis base).

Adım Adım Çözüm

1
Examine the valence electron configurations of the reacting species.
Ammonia (NH3NH_3) has a non-bonding lone pair on the nitrogen atom, whereas boron trifluoride (BF3BF_3) has an incomplete octet with only six valence electrons around the central boron atom.
Determining the presence of unshared pairs and electron deficiency identifies the roles of each atom.
2
Identify the nature of the bond formed between nitrogen and boron.
Nitrogen shares its unshared lone pair to complete the octet of the boron atom, forming a bond where both shared electrons originate from nitrogen.
A covalent bond in which one atom provides both shared electrons is defined as a dative or coordinate covalent bond.
3
Assign the appropriate terminology to the electron-donating species.
The atom or molecule supplying the shared electron pair is termed the electron pair donor or Lewis base.
By definition in Lewis acid-base theory, the species donating the electron pair is the donor.

Anahtar Kavram

Dative (Coordinate) Covalent Bonding and Lewis Acid-Base Theory
Soru 27Soru

A sulfide ion is represented by the symbol 1632S2^{32}_{16}\text{S}^{2-}. How many neutrons, protons, and electrons are present in this ion?

Cevabı ve açıklamayı göster

Cevap: 16 neutrons, 16 protons, and 18 electrons

Cevap

16 neutrons, 16 protons, and 18 electrons
For the nuclide symbol 1632S2^{32}_{16}\text{S}^{2-}, the atomic number is 16, which dictates that the species has 16 protons. The mass number is 32, so the number of neutrons is 3216=1632 - 16 = 16. Because the species is a divalent anion (S2\text{S}^{2-}), it has gained 2 electrons beyond its neutral count of 16, giving a total of 18 electrons.

Adım Adım Çözüm

1
Determine the number of protons and neutrons from the nuclide symbol
Protons = 16, Neutrons = 16
The lower subscript represents the atomic number (Z=16Z = 16), which equals the number of protons. The upper superscript represents the mass number (A=32A = 32). The number of neutrons is calculated as AZ=3216=16A - Z = 32 - 16 = 16.
2
Determine the number of electrons for the charged species
Electrons = 18
The ion carries a charge of 2-2, meaning it has gained 2 extra electrons compared to its neutral atomic state (16+2=1816 + 2 = 18).

Anahtar Kavram

Subatomic Particle Calculations in Ions
Soru 28Soru

An atom of potassium is represented by the nuclide symbol 1939K^{39}_{19}\text{K}. How many neutrons are present in the nucleus of this atom?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The nucleus of the potassium atom contains 20 neutrons.
The correct answer is 20 because the number of neutrons in an atom is found by subtracting the atomic number (19) from the mass number (39), giving 3919=2039 - 19 = 20.

Adım Adım Çözüm

1
Identify the mass number (AA) and atomic number (ZZ) from the given nuclide notation 1939K^{39}_{19}\text{K}.
Mass number A=39A = 39, Atomic number Z=19Z = 19.
In standard nuclide notation ZAX^{A}_{Z}\text{X}, the superscript represents the mass number and the subscript represents the atomic number.
2
Calculate the number of neutrons (NN) using the relationship N=AZN = A - Z.
N=3919=20N = 39 - 19 = 20.
The mass number is the sum of protons and neutrons (A=Z+NA = Z + N), so subtracting the atomic number (protons) from the mass number gives the neutron count.

Anahtar Kavram

Mass Number and Subatomic Particles
Tahmini Süre:45s
Soru 29Soru

Element XX exists naturally as three isotopes with mass numbers 2424, 2525, and 2626. The relative atomic mass of element XX is 24.3224.32. If the natural abundance of the isotope 25X^{25}X is 10%10\%, what is the percentage abundance of the isotope 26X^{26}X?

Cevabı ve açıklamayı göster

Cevap: 11

Cevap

The percentage abundance of 26X^{26}X is 11%11\%.
By setting the percentage abundance of 24X^{24}X as xx and 26X^{26}X as zz, given 25X=10%^{25}X = 10\%, we have x+z=90%x + z = 90\%. Using the relative atomic mass formula 24x+25(10)+26z=243224x + 25(10) + 26z = 2432 and substituting x=90zx = 90 - z yields 2410+2z=24322410 + 2z = 2432, solving to z=11%z = 11\%.

Adım Adım Çözüm

1
Formulate the abundance relation for the three isotopes.
The sum of abundances is x+10+z=100%x + 10 + z = 100\%, which gives x=90zx = 90 - z.
The sum of percentage abundances of all naturally occurring isotopes of an element must equal 100%.
2
Set up the weighted relative atomic mass equation.
24x+25(10)+26z=243224x + 25(10) + 26z = 2432.
Relative atomic mass is the weighted average of isotopic masses based on fractional abundance.
3
Substitute x=90zx = 90 - z and solve for zz.
24(90z)+250+26z=2432    2410+2z=2432    z=11%24(90 - z) + 250 + 26z = 2432 \implies 2410 + 2z = 2432 \implies z = 11\%.
Substituting xx reduces the equation to a single variable zz, representing the percentage abundance of 26X^{26}X.

Anahtar Kavram

Calculation of Isotopic Abundances from Relative Atomic Mass
Soru 30Soru

Element MM has a relative atomic mass of 24.3224.32. It exists naturally as three isotopes: 24M^{24}M with a relative abundance of 79%79\%, 25M^{25}M with a relative abundance of 10%10\%, and an isotope xM^{x}M with a relative abundance of 11%11\%. What is the mass number (xx) of the third isotope?

Cevabı ve açıklamayı göster

Cevap: 26

Cevap

The mass number of the third isotope is 26.
The relative atomic mass is the weighted average of isotopic mass numbers: 24.32=(24×79)+(25×10)+(x×11)10024.32 = \frac{(24 \times 79) + (25 \times 10) + (x \times 11)}{100}. Simplifying gives 2432=1896+250+11x2432 = 1896 + 250 + 11x, which simplifies to 11x=28611x = 286, yielding x=26x = 26.

Adım Adım Çözüm

1
Set up the relative atomic mass equation based on percentage abundances.
RAM=(24×0.79)+(25×0.10)+(x×0.11)\text{RAM} = (24 \times 0.79) + (25 \times 0.10) + (x \times 0.11)
The relative atomic mass of an element is the weighted average of the mass numbers of its naturally occurring isotopes.
2
Calculate the mass contributions of the first two isotopes.
24×0.79=18.9624 \times 0.79 = 18.96 and 25×0.10=2.5025 \times 0.10 = 2.50, giving a combined sum of 21.4621.46
Evaluating the contribution of known isotopes isolates the variable term.
3
Subtract the combined contribution from the given relative atomic mass to find the contribution of the third isotope.
0.11x=24.3221.46=2.860.11x = 24.32 - 21.46 = 2.86
The remaining mass contribution must come entirely from the third isotope.
4
Divide by the fractional abundance of the third isotope to solve for xx.
x=2.860.11=26x = \frac{2.86}{0.11} = 26
Dividing the mass contribution by fractional abundance yields the integer mass number.

Anahtar Kavram

Relative Atomic Mass Calculation from Isotopic Abundances
Tahmini Süre:1m 15s
Soru 31Soru

Match each quantum rule or principle with its correct statement regarding electronic configuration.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Hund's Rule of Maximum Multiplicity
Pauli Exclusion Principle
Aufbau Principle

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

Hund's Rule matches single occupancy of degenerate orbitals before pairing; Pauli Exclusion Principle matches the restriction that no two electrons share four identical quantum numbers; Aufbau Principle matches filling lowest energy orbitals first.
Hund's rule describes filling degenerate orbitals singly first with parallel spins. The Pauli exclusion principle mandates that no two electrons in an atom possess identical sets of four quantum numbers. The Aufbau principle specifies filling orbitals starting from the lowest energy level.

Adım Adım Çözüm

1
Identify the definition of Hund's Rule of Maximum Multiplicity.
Hund's rule specifies that degenerate orbitals are occupied singly first to minimize electron-electron repulsion.
Electrons in different orbitals with parallel spins lower electrostatic energy.
2
Identify the definition of the Pauli Exclusion Principle.
Pauli's principle states that an orbital can hold at most two electrons of opposite spin.
This guarantees that every electron in an atom has a unique set of four quantum numbers (n,l,ml,msn, l, m_l, m_s).
3
Identify the definition of the Aufbau Principle.
The Aufbau principle dictates that subshells fill in order of increasing energy according to the (n+l)(n+l) rule.
Ground states require minimum total electronic potential energy.

Anahtar Kavram

Fundamental Quantum Rules Governing Ground-State Electronic Configuration
Soru 32Soru

In the carbon monoxide molecule (COCO), a triple covalent linkage holds the carbon and oxygen atoms together. Which statement correctly describes the nature of the bonding pairs between the two atoms and identifies the electron pair donor for the dative (coordinate) bond?

Cevabı ve açıklamayı göster

Cevap: There are three shared electron pairs in total, with oxygen donating the lone pair required for the coordinate bond.

Cevap

The carbon monoxide molecule features three shared pairs of electrons forming a triple bond, where oxygen serves as the donor atom contributing the lone pair for the coordinate covalent bond.
In carbon monoxide (COCO), carbon contributes 2 electrons and oxygen contributes 2 electrons to form two standard single covalent bonds. To allow carbon to attain a stable octet (8 valence electrons), oxygen donates both electrons from one of its lone pairs to form a third bond (dative/coordinate bond). Thus, there are 3 shared electron pairs (a triple bond), and oxygen is the donor.

Adım Adım Çözüm

1
Determine valence electron counts for Carbon and Oxygen
Carbon (Group 14) has 4 valence electrons; Oxygen (Group 16) has 6 valence electrons.
Establishing valence electrons determines how many electrons are shared to complete octets.
2
Analyze standard covalent sharing between Carbon and Oxygen
Sharing 2 electrons from Carbon and 2 electrons from Oxygen forms 2 ordinary covalent bonds. This gives Oxygen 8 valence electrons, but leaves Carbon with only 6 valence electrons.
Two standard covalent bonds are insufficient to fulfill the octet rule for Carbon.
3
Identify the dative (coordinate) bond formation
Oxygen donates one of its remaining lone pairs into Carbon's vacant orbital to form a 3rd shared bond (coordinate bond), completing Carbon's octet.
This establishes a total of 3 shared pairs (a triple bond) with Oxygen acting as the donor atom.

Anahtar Kavram

Covalent and Dative (Coordinate) Bonding in Carbon Monoxide
Tahmini Süre:1m 30s
Soru 33Soru

An element MM forms a stable electrovalent chloride with the formula MCl2MCl_2. If the dipositive cation M2+M^{2+} has the electronic configuration 1s22s22p63s23p61s^2 2s^2 2p^6 3s^2 3p^6, which of the following statements correctly explains the electrical conductivity and lattice properties of MCl2MCl_2?

Cevabı ve açıklamayı göster

Cevap: It does not conduct electricity in the solid state because its ions are held in fixed positions, but conducts in the molten state due to mobile M2+M^{2+} and ClCl^- ions.

Cevap

The compound does not conduct electricity in the solid state because its ions are held in fixed positions within the lattice, but conducts in the molten state due to mobile M2+M^{2+} and ClCl^- ions.
In giant electrovalent (ionic) lattices like CaCl2CaCl_2, ions are immobilized in fixed positions in the solid state, making the solid a non-conductor. Upon melting, the electrostatic lattice forces are overcome, producing free mobile M2+M^{2+} and ClCl^- ions that conduct electricity.

Adım Adım Çözüm

1
Identify the element and ion structure
The ion M2+M^{2+} has 18 electrons (1s22s22p63s23p61s^2 2s^2 2p^6 3s^2 3p^6), corresponding to a neutral calcium atom (CaCa, atomic number 20). The compound formed is calcium chloride (CaCl2CaCl_2).
Determining the electronic structure confirms MCl2MCl_2 is a typical giant ionic (electrovalent) lattice.
2
Analyze solid-state properties
In the solid state, M2+M^{2+} cations and ClCl^- anions are locked in a rigid three-dimensional crystal lattice by strong omnidirectional electrostatic forces of attraction. Because the ions cannot move, the solid is an electrical insulator.
Conduction of electricity requires mobile charge carriers.
3
Analyze molten-state properties
When heated to its high melting point, thermal energy overcomes the lattice energy, allowing the M2+M^{2+} and ClCl^- ions to move freely and carry electrical current.
Liquid ionic compounds conduct electricity via migration of ions toward oppositely charged electrodes.

Anahtar Kavram

Ionic Lattice Properties and Conduction Mechanism
Soru 34Soru

Consider three Period 3 elements: sodium (NaNa), magnesium (MgMg), and aluminium (AlAl). As one moves from NaNa to AlAl across the period, there is a notable increase in both melting point and electrical conductivity per mole of metal. Which of the following best accounts for this observed trend in metallic bond strength and physical properties?

Cevabı ve açıklamayı göster

Cevap: The number of delocalized valence electrons contributed per atom increases while cationic radius decreases, increasing electrostatic attraction and mobile charge density.

Cevap

The trend is best explained by the increase in the number of delocalized valence electrons contributed per atom combined with a smaller cationic radius, which increases electrostatic attraction and mobile charge carrier density.
Metallic bonding consists of electrostatic attractions between fixed positive metal cations and a surrounding delocalized sea of valence electrons. Moving from sodium to aluminium, each atom donates more valence electrons (Na=1eNa = 1e^-, Mg=2eMg = 2e^-, Al=3eAl = 3e^-) into the electron sea while the ionic radius decreases (Na+>Mg2+>Al3+Na^+ > Mg^{2+} > Al^{3+}). The combination of higher cationic charge, smaller ionic radius, and greater electron density increases the electrostatic attraction, raising both melting points and electrical conductivity.

Adım Adım Çözüm

1
Analyze the structural factors determining metallic bond strength.
Metallic bond strength depends directly on two main factors: (1) the charge on the metal cation (number of delocalized electrons donated per atom) and (2) the cationic radius (distance between cations and delocalized electrons).
Strength of electrostatic attraction follows Coulomb's law: Fq1q2r2F \propto \frac{q_1 q_2}{r^2}.
2
Compare valence electron contributions across Period 3 metals.
Sodium ([Ne]3s1[Ne]3s^1) donates 1 electron per atom (Na+Na^+), Magnesium ([Ne]3s2[Ne]3s^2) donates 2 electrons per atom (Mg2+Mg^{2+}), and Aluminium ([Ne]3s3[Ne]3s^3) donates 3 electrons per atom (Al3+Al^{3+}).
Higher delocalized electron count yields greater mobile charge density for electrical conductivity.
3
Compare cationic radii across the period.
Ionic radii decrease across the period: Na+(102 pm)>Mg2+(72 pm)>Al3+(54 pm)Na^+ (102\text{ pm}) > Mg^{2+} (72\text{ pm}) > Al^{3+} (54\text{ pm}).
Smaller cations allow delocalized electrons to approach closer to positively charged nuclei, dramatically strengthening electrostatic attraction.

Anahtar Kavram

Factors affecting metallic bond strength and properties (charge density and delocalized electron count)
Soru 35Soru

Match each microscopic structural feature of metallic bonding on the left with the macroscopic physical property of metals on the right that directly results from it.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Presence of a mobile sea of delocalized valence electrons
Non-directional electrostatic bonding between cation layers and free electrons
Strong multi-directional electrostatic attraction throughout the giant metallic lattice

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

The mobile sea of delocalized electrons accounts for high electrical and thermal conductivity. Non-directional bonding between cation layers accounts for malleability and ductility. Strong multi-directional electrostatic attraction accounts for high melting and boiling points.
Each physical property of metals stems directly from the electron-sea structural model. Electrical and thermal conductivity rely on mobile delocalized valence electrons. Malleability and ductility arise from non-directional bonding that allows cation layers to slide smoothly past each other under force. High melting and boiling points are due to the strong electrostatic forces holding the giant metallic lattice together.

Adım Adım Çözüm

1
Link electrical/thermal transport to microscopic charge carriers
Free-moving (delocalized) electrons drift when an electric field or thermal gradient is applied, explaining high conductivity.
Electrical conduction requires mobile charged particles, which in metals are delocalized electrons.
2
Analyze the mechanical behavior of metal cation layers during deformation
When hammered or drawn into wires, cation layers slide without repulsive disruption because the electron sea flexibly adjusts.
Non-directional bonding prevents catastrophic cleavage, providing malleability and ductility.
3
Connect lattice stability to thermal energy requirements for melting
Substantial thermal energy is necessary to overcome strong electrostatic attraction between positive ions and negative electrons.
High bond energy across the giant metallic structure leads directly to elevated melting and boiling points.

Anahtar Kavram

Relationship between metallic bonding structure (electron sea model) and physical properties of metals
Soru 36Soru

Complete the statement below regarding chemical bonding.

Aşağıdaki boşlukları doldurun

A chemical bond formed when a single atom donates both electrons of a shared pair is known as a coordinate covalent bond or a bond.
Cevabı ve açıklamayı göster

Cevap

dative
In standard chemistry terminology, a coordinate covalent bond and a dative bond are synonymous. Both refer to a covalent linkage where one atom contributes both electrons to the shared pair.

Adım Adım Çözüm

1
Recall the definition and alternative chemical terminology for coordinate covalent bonding.
A coordinate bond is also termed a dative covalent bond.
Both terms describe a special type of covalent bond where the shared pair of electrons is provided exclusively by a single donor atom possessing a lone pair.

Anahtar Kavram

Definition and terminology of coordinate (dative) covalent bonding
Soru 37Soru

Which of the following best explains why solid sodium chloride (NaClNaCl) does not conduct electricity, whereas molten sodium chloride conducts electricity readily?

Cevabı ve açıklamayı göster

Cevap: In the solid state, the ions are held in fixed positions within the crystal lattice and cannot move freely.

Cevap

In the solid state, the ions are held in fixed positions within the crystal lattice and cannot move freely.
Ionic (electrovalent) compounds conduct electricity only when charge carriers (ions) are free to move. In the solid state, ions are locked tightly into fixed positions within the giant ionic lattice, preventing electrical conduction. When melted, the high thermal energy breaks the rigid lattice structure, enabling the positive (Na+Na^+) and negative (ClCl^-) ions to move freely and carry electrical current.

Adım Adım Çözüm

1
Identify the nature of charge carriers in ionic compounds.
Electrical conductivity in electrovalent (ionic) compounds relies on mobile ions (Na+Na^+ and ClCl^-), not free electrons.
Ionic compounds do not contain free delocalized electrons like metals.
2
Analyze the structural state of solid sodium chloride.
In solid NaClNaCl, electrostatic forces lock ions in fixed positions in a giant 3D crystal lattice structure.
Because ions cannot move from place to place, solid NaClNaCl acts as an electrical insulator.
3
Compare the solid state with the molten state.
When melted (molten state), thermal energy overcomes the rigid lattice forces, allowing Na+Na^+ and ClCl^- ions to move freely toward electrodes.
Mobile ions enable the flow of electric current in the liquid/molten state.

Anahtar Kavram

Electrical Conductivity of Electrovalent Compounds
Soru 38Soru

Magnesium oxide (MgOMgO) has a significantly higher melting point (2,852C2,852^\circ\text{C}) than sodium fluoride (NaFNaF) (996C996^\circ\text{C}) primarily because the product of the ionic charges in MgOMgO is four times greater than in NaFNaF, resulting in stronger electrostatic forces within the crystal lattice. Is this statement true or false?

Cevabı ve açıklamayı göster

Cevap: True

Cevap

The statement is true because the strength of electrovalent bonding and lattice energy increases proportionally with the product of the ionic charges.
The statement is correct because electrovalent bond strength is governed by Coulomb's Law of electrostatic attraction. Doubly charged cations and anions (Mg2+Mg^{2+} and O2O^{2-}) experience an electrostatic force four times stronger than singly charged ions (Na+Na^+ and FF^-) of comparable size, yielding a substantially higher lattice energy and melting point.

Adım Adım Çözüm

1
Identify the constituent ions and their respective charges for both ionic compounds.
In MgOMgO, the ions are Mg2+Mg^{2+} and O2O^{2-}. In NaFNaF, the ions are Na+Na^+ and FF^-.
Electrostatic attraction depends fundamentally on the magnitude of nuclear charges transferred during ionic bond formation.
2
Calculate the product of the ionic charges for each compound.
For MgOMgO: (+2)×(2)=4|(+2) \times (-2)| = 4. For NaFNaF: (+1)×(1)=1|(+1) \times (-1)| = 1.
Coulomb's Law states that electrostatic force is proportional to the product of the charges (q1q2q_1 q_2).
3
Relate the charge product to lattice energy and physical properties such as melting point.
A higher lattice energy requires more thermal energy to break the rigid giant ionic lattice, leading to a much higher melting point for MgOMgO than NaFNaF.
Melting point directly reflects the magnitude of the electrostatic attraction holding the ions in their solid lattice positions.

Anahtar Kavram

Factors Affecting Ionic Lattice Energy and Melting Points
Soru 39Soru

Match each physical property of metallic elements listed on the left with the atomic-scale mechanism on the right that best accounts for it.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

High thermal conductivity
Malleability and ductility
High melting point and tensile strength
Metallic luster and opacity

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

High thermal conductivity matches rapid transfer of kinetic energy by mobile delocalized electrons; Malleability and ductility matches non-directional electrostatic attractions permitting cation layers to slide; High melting point matches strong multi-directional electrostatic attraction between cations and the electron sea; Metallic luster matches oscillation and re-emission of incident photons by surface delocalized electrons.
Each macro-level property corresponds directly to specific behaviors of the delocalized electron sea and cation lattice: thermal conduction relies on mobile electron kinetic transport; malleability depends on non-directional bonding allowing cation layers to slip; high melting points result from strong multi-directional electrostatic attractions; and luster is caused by surface electron excitation and photon re-emission.

Adım Adım Çözüm

1
Analyze the microscopic origin of thermal transport in metals.
Identify mobile delocalized valence electrons as the primary carriers of thermal kinetic energy.
Delocalized electrons move rapidly through the lattice when a temperature gradient is applied, transferring kinetic energy much faster than localized atomic vibrations.
2
Examine the mechanism of mechanical deformation under applied stress.
Identify non-directional electrostatic attraction enabling cation layers to slide without fracture.
Unlike ionic crystals where sliding brings like charges into repelling contact, metallic electron clouds shield shifting cations, preserving lattice cohesion.
3
Evaluate the structural requirements for melting and high mechanical strength.
Identify strong multi-directional electrostatic binding throughout the 3D lattice.
Overcoming the structural stability requires substantial energy to disrupt the strong net electrostatic pull between positive metal ions and delocalized electrons.
4
Determine the interaction of metal surfaces with electromagnetic radiation.
Connect surface delocalized electrons to photon absorption and rapid re-emission.
Unbound surface valence electrons absorb incoming light energy and immediately vibrate and re-emit the photons, producing specular reflection.

Anahtar Kavram

Connecting macroscopic physical properties of metals to the delocalized electron sea model and non-directional metallic bonding.
Soru 40Soru

Complete the following statement describing the bonding and structural change during the dimerization of aluminium chloride.

Aşağıdaki boşlukları doldurun

In the formation of the dimer Al2Cl6Al_2Cl_6 from monomeric AlCl3AlCl_3, a chlorine atom from one molecule donates a lone pair of electrons to the electron-deficient aluminium atom of another, resulting in a total of coordinate bonds in the dimer. This coordination causes the electron pair geometry around each aluminium atom to become .
Cevabı ve açıklamayı göster

Cevap

Blank 1: 2 (or two); Blank 2: tetrahedral
In the dimerization of AlCl3AlCl_3 into Al2Cl6Al_2Cl_6, two bridging chlorine atoms act as Lewis bases by donating lone pairs to the electron-deficient aluminium atoms, creating 2 coordinate bonds. Because each aluminium atom now shares 4 electron pairs, the hybridization changes from sp2sp^2 to sp3sp^3, resulting in a tetrahedral geometry around each aluminium atom.

Adım Adım Çözüm

1
Analyze the electron configuration and geometry of monomeric AlCl3AlCl_3.
Aluminium has 3 valence electrons and forms 3 single covalent bonds with three chlorine atoms, leaving aluminium with 6 valence electrons (an incomplete octet) and a trigonal planar geometry with sp2sp^2 hybridization.
Identifying the electron deficiency in monomeric AlCl3AlCl_3 explains why it undergoes dimerization.
2
Determine the number of coordinate (dative) bonds formed in the Al2Cl6Al_2Cl_6 dimer.
To complete its octet, each aluminium atom accepts one lone pair of electrons from a chlorine atom belonging to the neighboring AlCl3AlCl_3 molecule. Since there are two aluminium atoms being bridged, exactly 2 coordinate covalent bonds are formed in Al2Cl6Al_2Cl_6.
Two bridging chlorine atoms each contribute a lone pair to form two separate dative bonds.
3
Determine the resulting molecular geometry around each aluminium atom in Al2Cl6Al_2Cl_6.
Each aluminium atom becomes surrounded by 4 electron pair bonds (3 single covalent bonds + 1 coordinate bond), shifting its hybridization from sp2sp^2 to sp3sp^3 and yielding a tetrahedral geometry.
According to VSEPR theory, four bonding pairs arrange themselves to minimize repulsion in a 3D tetrahedral orientation.

Anahtar Kavram

Dative Covalent Bonding and Geometric Transition in Dimeric Aluminium Chloride (Al2Cl6Al_2Cl_6)
Tahmini Süre:1m 30s
ÖncekiSayfa 2 / 5Sonraki
Atomic Structure and Chemical Bonding Alıştırma Soruları — JAMB UTME — Sayfa 2 | Examkin