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Group 15 Flashcard Notes
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1
Question
Which elements are in Group 15 and how are they generally classified (metal, nonmetal, metalloid)?
Page 1
Answer
Group 15 elements: Nitrogen (N) — nonmetal; Phosphorus (P) — nonmetal; Arsenic (As) — metalloid; Antimony (Sb) — metalloid; Bismuth (Bi) — metal.
2
Question
What are the common ionic forms (salts) mentioned for nitrates of sodium and potassium?
Page 1
Answer
Sodium nitrate: NaNO_3 (Chile saltpetre). Potassium nitrate: KNO_3 (saltpetre/fertilizer).
3
Question
What is fluorapatite's formula given in the notes?
Page 1
Answer
Fluorapatite composition is written as 3[Ca_3(PO_4)_2 · CaF_2] (showing the calcium phosphate and calcium fluoride units).
4
Question
How does ionization energy (I.E.) change down Group 15?
Page 1
Answer
Ionization energy decreases down Group 15.
5
Question
How does electronegativity (E.N.) change down Group 15 and what exception is noted?
Page 1
Answer
Electronegativity generally decreases down Group 15. Exception: Sb and Bi may show similar electronegativity values; Bi can have poor shielding leading to some increase in effective nuclear attraction compared to Sb.
6
Question
How does atomic size (atomic radius) change down Group 15?
Page 1
Answer
Atomic size increases down Group 15.
7
Question
How does boiling point (B.Pt) trend down Group 15?
Page 1
Answer
Boiling points increase down Group 15.
8
Question
What is the general trend for density down Group 15 according to the notes?
Page 1
Answer
Density generally increases down Group 15 (no exception noted in the notes).
9
Question
What is mentioned about melting point (M.Pt) trends for Group 15?
Page 2
Answer
There are exceptions, but the notes state melting point shows a trend with network structures giving the highest M.Pt, then metallic, then van der Waals solids. The M.Pt order given: As > Sb > Bi > P_4 > N_2 (As highest, N_2 lowest).
10
Question
Why is As given a higher melting point than P_4 and N_2 in the notes?
Page 2
Answer
Because As and Sb form network (covalent) structures which have higher melting points than metallic or molecular solids; P_4 and N_2 are molecular (held by van der Waals forces) and thus melt at much lower temperatures.
11
Question
What types of bonding hold N_2 and P_4 together, according to the notes?
Page 2
Answer
N_2 and P_4 are held together by van der Waals forces (intermolecular forces) since they are molecular species.
12
Question
What is the trend of electron affinity (E.A.) or electronegativity (E.N.) down Group 15 given on the second page diagram?
Page 2
Answer
Electronegativity (E.N.) decreases down the group (N highest, Bi lowest) — the diagram shows N > P > As > Sb > Bi with E.N. decreasing downward.
13
Question
List the anomalous properties of nitrogen mentioned.
Page 3
Answer
Anomalous properties of nitrogen include: small size, high electronegativity, high ionization energy, and absence of d-orbitals. These lead to unique behavior like forming strong N≡N triple bonds and multiple π–π bonding abilities not seen in heavier group 15 elements.
14
Question
Why can nitrogen form multiple π–π bonds while heavier group 15 elements typically do not?
Page 3
Answer
Nitrogen has small size and good p-orbital overlap, enabling effective p–p π bonding forming N≡N and other multiple bonds. Heavier elements have larger, more diffuse orbitals, so p–p overlap is poor and they usually form single bonds.
15
Question
Compare bond energies of N_2 and P_4. Which is greater and why?
Page 3
Answer
Bond energy of N_2 is greater than that of P_4. N_2 contains an N≡N triple bond which is very strong; P–P bonds in P_4 are single bonds only, so overall bond energy is lower.
16
Question
Why might a single N–N bond be weaker than a single P–P bond despite N_2 having a stronger bond overall?
Page 3
Answer
A single N–N bond can be weakened by lone pair–lone pair repulsion on the small nitrogen atoms; the small size places lone pairs close together causing repulsion that lowers single-bond strength compared to P–P single bonds.
17
Question
Write the general reactions (formation) of hydrides for Group 15 elements shown in the notes.
Page 4
Answer
General formation: N_2 + 3H_2 → 2NH_3 (exothermic); P + 3H_2 → PH_3; AsH_3, SbH_3, BiH_3 are also listed (these heavier hydrides form endothermically according to the notes).
18
Question
Which hydride in Group 15 is noted as the strongest hydride and thermodynamically stable?
Page 4
Answer
Ammonia (NH_3) is noted as the strongest hydride and thermodynamically stable (formation releases energy).
19
Question
How does the N–H to Bi–H bond length and bond energy trend change down the group?
Page 4
Answer
Bond length increases down the group (N–H < P–H < As–H < Sb–H < Bi–H), and bond energy decreases down the group (N–H has the strongest bond, Bi–H the weakest).
20
Question
What does the 'Draper rule' (Dragon rule as in the notes) say regarding bonding of 2nd-row elements vs 3rd/4th-row elements when forming hydrides?
Page 4
Answer
The notes say: for 2nd-row elements (like P in period 3?), hybridization (sp^3) is not always used — instead pure orbitals may react to form bonds with hydrogen; for heavier elements (3rd/4th row), no hybridization may be seen, and pure orbitals (like p) react to form bonds with hydrogen. The point made is heavier elements tend to use their valence orbitals without forming idealized sp^3 hybridization.
21
Question
What trend in bond angle is given for hydrides NH_3, PH_3, AsH_3, SbH_3?
Page 4
Answer
Bond angles trend: NH_3 ≈ 107° > PH_3 ≈ 93° > AsH_3 ≈ 92° > SbH_3 ≈ 91° (bond angle decreases down the group).
22
Question
Why is NH_3 bond angle larger (≈107°) than PH_3 and heavier hydrides?
Page 4
Answer
Because in NH_3 the lone pair occupies an sp^3-like orbital with significant s-character causing greater repulsion and expanding the H–N–H angle. For heavier atoms, lone pairs occupy orbitals with more p-character (or are in pure p orbitals per the note), so bond angles are smaller.
23
Question
How is basicity of Group 15 hydrides ordered according to the notes?
Page 5
Answer
Basicity order (from strongest base to weakest): NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3.
24
Question
How is reducing nature of Group 15 hydrides ordered in the notes?
Page 5
Answer
Reducing ability order (increasing reducing nature): NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3. So BiH_3 is the strongest reducing agent among them.
25
Question
Why is BiH_3 considered the strongest reducing hydride among Group 15 hydrides?
Page 5
Answer
Because the Bi–H bond is weakest (longest bond length down the group), so BiH_3 can lose hydrogen most easily and thus acts as a better reducing agent.
26
Question
What is the expected (ideal) melting point trend for Group 15 hydrides based on van der Waals forces and size?
Page 6
Answer
Ideally, melting point would increase down the series as van der Waals forces increase with molecular size: NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3. The notes mark this as 'WRONG' due to additional effects (hydrogen bonding in NH_3).
27
Question
What actual melting point order for Group 15 hydrides is noted, and why does ammonia (NH_3) differ?
Page 6
Answer
Actual order in the notes: PH_3 < AsH_3 < SbH_3 < (BiH_3 data not reliable) < NH_3. NH_3 has an anomalously higher melting/boiling point due to hydrogen bonding, making it more associated and raising its M.Pt compared with heavier hydrides.
28
Question
How are oxides of Group 15 elements classified (acidic/basic) depending on the element type?
Page 6
Answer
Nonmetal oxides (N, P) are acidic. Metalloid oxides (As, Sb) are amphoteric. Metal oxide (Bi) is basic. Also, oxides in higher oxidation states are more acidic than those in lower oxidation states.
29
Question
What two types of oxides can Group 15 elements form as listed?
Page 6
Answer
They can form E_2O_3-type oxides and E_2O_5-type oxides (where E is the Group 15 element), indicating lower and higher oxidation state oxides respectively.
30
Question
In reactivity with halogens, what formulae types do Group 15 elements form?
Page 7
Answer
Group 15 elements form trihalides (BX_3) and pentahalides (BX_5) depending on oxidation state (typically +3 and +5 species).