Super-hard periodic table quiz: f-block and relativity

Q: What does the super-hard f-block quiz test?

Twenty conceptual questions on lanthanide configurations, Eu²⁺/Yb²⁺/Ce⁴⁺, relativistic gold and mercury, and the island of stability. Six near-twin choices each.

Q: Can I print the super-hard f-block quiz?

Yes. Print or download a blank paper from this page. After you submit, a marked paper is available.

Q: Where are my answers stored?

Only in this browser, under a ptw.quiz.f-block key. Incomplete attempts wipe when you leave. Nothing is sent to a server.

Question 1

Lanthanum’s ground-state configuration is:

Accepted Answer

[Xe] 5d¹ 6s²

Answer

[Xe] 4f¹ 6s²

Answer

[Xe] 4f¹ 5d¹ 6s¹

Answer

[Xe] 4f⁷ 6s²

Answer

[Xe] 6s² 6p¹

Answer

[Kr] 5d¹ 6s²

Question 2

Lutetium’s ground-state configuration is:

Accepted Answer

[Xe] 4f¹⁴ 5d¹ 6s²

Answer

[Xe] 4f¹⁴ 6s²

Answer

[Xe] 4f¹³ 5d² 6s²

Answer

[Xe] 5d¹ 6s²

Answer

[Xe] 4f¹⁴ 5d²

Answer

[Xe] 4f⁷ 5d¹ 6s²

Question 3

Actinium’s ground-state configuration is:

Accepted Answer

[Rn] 6d¹ 7s²

Answer

[Rn] 5f¹ 7s²

Answer

[Rn] 5f¹ 6d¹ 7s¹

Answer

[Xe] 6d¹ 7s²

Answer

[Rn] 7s² 7p¹

Answer

[Rn] 5f⁷ 7s²

Question 4

Thorium’s ground-state configuration is:

Accepted Answer

[Rn] 6d² 7s²

Answer

[Rn] 5f² 7s²

Answer

[Rn] 5f¹ 6d¹ 7s²

Answer

[Rn] 5f⁰ 7s²

Answer

[Xe] 6d² 7s²

Answer

[Rn] 6d¹ 7s²

Question 5

Eu²⁺ and Yb²⁺ are unusually stable for Ln²⁺ mainly because:

Accepted Answer

they have 4f⁷ and 4f¹⁴, so half-filled and filled 4f shells

Answer

europium and ytterbium never form +3

Answer

they are 5f ions, not 4f

Answer

both are d¹⁰ s⁰ like zinc

Answer

the +2 ions are 4f⁰

Answer

Ln²⁺ is the default for every lanthanide

Question 6

Ce⁴⁺ is accessible mainly because:

Accepted Answer

Ce⁴⁺ is 4f⁰, isoelectronic with Xe

Answer

cerium is 4f¹⁴ in the atom

Answer

cerium cannot form +3

Answer

Ce⁴⁺ is 4f⁷

Answer

cerium is an actinide

Answer

all lanthanides prefer +4

Question 7

Gd³⁺ is 4f⁷ rather than 4f⁸ or 4f⁶ because:

Accepted Answer

gadolinium metal is 4f⁷ 5d¹ 6s², so Gd³⁺ is 4f⁷

Answer

Gd³⁺ must be 4f¹⁴

Answer

gadolinium is 5f, not 4f

Answer

the ion keeps 6s²

Answer

Gd³⁺ is 4f⁸ 5d⁻¹

Answer

all +3 lanthanides are 4f⁷

Question 8

The predicted island of stability for superheavy nuclei is centered near:

Accepted Answer

Z = 114, N = 184

Answer

Z = 26, N = 30 (iron-56)

Answer

Z = 82, N = 126 (lead-208), which is already the last stable closed shell

Answer

Z = 92, N = 146 (uranium)

Answer

Z = 2, N = 2 (helium-4)

Answer

there is no predicted closed shell above lead

Question 9

Oganesson is not expected to behave like a typical noble gas mainly because:

Accepted Answer

relativistic 7p splitting and a diffuse 7p₃/₂ set make Og more reactive than Xe

Answer

it has only two valence electrons

Answer

IUPAC placed Og in group 1

Answer

oganesson has been isolated in kilogram amounts and is a metal

Answer

Og is 5f¹⁴ 6d¹⁰ 7s² with no p electrons

Answer

noble-gas chemistry stops at argon

Question 10

CsAu is a salt-like auride mainly because:

Accepted Answer

relativistic 6s stabilization lets gold accept an electron and form Au⁻

Answer

gold has a closed 3d¹⁰ shell like zinc

Answer

cesium is more electronegative than gold

Answer

gold(I) is d⁴

Answer

non-relativistic gold already has the highest electron affinity in group 11

Answer

Au⁻ is 4s², so it is a noble gas

Question 11

The mercurous ion is Hg₂²⁺ with an Hg–Hg bond, not two Hg⁺, mainly because:

Accepted Answer

relativistic 6s character supports an Hg–Hg σ bond; the +1 state is dimeric

Answer

mercury has no 6s electrons

Answer

Hg⁺ would be d⁴

Answer

mercury never forms +1 in any form

Answer

Hg₂²⁺ is isoelectronic with He₂

Answer

the ion is actually two isolated Hg⁺ ions in the crystal

Question 12

The inert-pair effect in Tl, Pb, and Bi is mainly:

Accepted Answer

relativistic ns² stabilization, so +1 (Tl) and +2 (Pb) compete with the group valence

Answer

a 3d contraction, as in gallium

Answer

Pauli exclusion forbidding +3 in period 6

Answer

the 4f shell being empty in those atoms

Answer

a nuclear magic-number effect

Answer

identical to the reason chromium is d⁵ s¹

Question 13

Astatine is more metallic than iodine mainly because:

Accepted Answer

metallic character increases down group 17; At is also less electronegative

Answer

astatine has a higher ionization energy than iodine

Answer

astatine is a noble gas

Answer

iodine has no p electrons

Answer

astatine is lighter than fluorine

Answer

all halogens are metals

Question 14

Copernicium is predicted to be more inert than mercury mainly because:

Accepted Answer

relativistic 7s² contraction and stabilization weaken metallic bonding even more than in Hg

Answer

Cn has no valence electrons

Answer

copernicium is a noble-gas configuration in a non-relativistic calculation

Answer

Cn is 5d⁸ 6s²

Answer

IUPAC moved Cn to group 18

Answer

mercury is already 7s²

Question 15

High-spin versus low-spin octahedral d⁶ is decided by:

Accepted Answer

whether the ligand-field splitting Δo exceeds the spin-pairing energy

Answer

whether n+ℓ is even

Answer

the nuclear magic numbers of the metal

Answer

whether the complex is square planar

Answer

the mass number A alone

Answer

Hund’s rule forbidding pairing in every complex

Question 16

O₂ is paramagnetic because MO theory puts:

Accepted Answer

two unpaired electrons in the π* set

Answer

all twelve valence electrons in bonding σ orbitals

Answer

an empty π* set and a triple bond

Answer

O₂ in a closed-shell ¹Σ state as the ground term

Answer

the extra electrons in 3s

Answer

unpaired protons in the nuclei

Question 17

The 18-electron count in Cr(CO)₆ is:

Accepted Answer

Cr(0) d⁶ plus six CO ligands donating two each: 6 + 12 = 18

Answer

chromium’s atomic number, 24, minus six

Answer

six CO ligands only, with the metal contributing nothing

Answer

Cr⁶⁺ plus six CO

Answer

a 16-electron square-planar count

Answer

the same as Fe(CO)₅, which is 16 electrons

Question 18

The N≡N bond is so strong mainly because:

Accepted Answer

2p–2p overlap in a short triple bond is excellent, and there is no lone-pair crowding like F₂

Answer

nitrogen is larger than carbon, so overlap is poor

Answer

N₂ is ionic

Answer

the bond order is 1

Answer

nitrogen has empty d orbitals that add a fourth bond

Answer

N₂ is paramagnetic, like O₂

Question 19

Pb²⁺ is more common than Pb⁴⁺ in aqueous chemistry mainly because of:

Accepted Answer

the inert 6s² pair; Pb(IV) is strongly oxidizing

Answer

lead having only two valence electrons in the atom

Answer

lead being an alkali metal

Answer

Pb⁴⁺ violating the octet

Answer

nuclear magic number 82 forbidding +4

Answer

Pb²⁺ being 4f¹⁴ 5d¹⁰ 6s⁰

Question 20

Why is 4f contraction along the lanthanides so steady?

Accepted Answer

4f orbitals are core-like, penetrate poorly to ligands, and shield each other badly, so Z_eff on 6s/5d rises smoothly

Answer

4f electrons are more diffuse than 6s, so size jumps at every element

Answer

each lanthanide adds a full new principal shell

Answer

5f is filling, not 4f

Answer

the contraction is actually an expansion after gadolinium

Answer

relativistic 4f expansion cancels Z entirely, so radii are identical

Super-hard periodic table quiz: f-block and relativity

Common questions

What does the super-hard f-block quiz test?

Can I print the super-hard f-block quiz?

Where are my answers stored?

Super-hard periodic table quiz: f-block and relativity

Common questions

What does the super-hard f-block quiz test?

Can I print the super-hard f-block quiz?

Where are my answers stored?