NEET Chemistry: Inorganic Chemistry · Chapter 1

Classification of Elements and Periodicity in Properties

What to remember

The modern periodic law says that the properties of elements are a periodic function of their atomic numbers. The long form table has 7 periods and 18 groups.
Across a period (left to right) atomic radius decreases, while ionisation enthalpy, electronegativity and non-metallic character increase. Down a group the trends reverse.
Two exceptions in ionisation enthalpy are Be > B and N > O. Two exceptions in electron gain enthalpy are Cl more negative than F, and S more negative than O.
Oxides change from basic to amphoteric to acidic across a period, and become more basic down a group. Second-period elements behave differently from their group members, and show diagonal relationships (Li–Mg, Be–Al, B–Si).

Why elements are classified

More than 100 elements are known. Grouping them by similar properties makes it easy to study their chemistry and to predict the properties of unknown elements.

Early attempts at classification

Döbereiner's triads (1829). Three elements with similar properties are arranged in order of atomic mass. The atomic mass of the middle element is about the average of the other two.

Examples: Li, Na, K; Ca, Sr, Ba; Cl, Br, I.
Check: Li (6.9) and K (39.1) give an average of 23.0, close to Na (23.0).
Only a few triads could be found, so the idea failed as a general method.

Newlands' law of octaves (1865). When elements are arranged in increasing atomic mass, the eighth element shows properties similar to the first, like the notes of a music scale. It worked only for lighter elements up to calcium and was not accepted for heavier ones.

Mendeleev's periodic law (1869). The properties of elements are a periodic function of their atomic masses.

Elements were placed in horizontal rows (periods) and vertical columns (groups).
He left gaps for undiscovered elements and predicted their properties. Eka-aluminium was later found as gallium (Ga), and eka-silicon as germanium (Ge).
He also reversed the order of some pairs when properties demanded it, for example Te and I.
Limitations: the position of hydrogen was doubtful; isotopes of the same element have different atomic masses but no separate places; some elements of higher atomic mass came before lower ones (for example Ar before K, Co before Ni); similar and dissimilar elements were sometimes placed together; the lanthanoids and actinoids had no proper place.

Modern periodic law and the long form of the table

Modern periodic law (Moseley). The physical and chemical properties of elements are periodic functions of their atomic numbers. The reason for periodicity is the repetition of similar outer electronic configurations.

The long form (or modern) table has:

7 periods (horizontal rows) and 18 groups (vertical columns).
The period number equals the principal quantum number n of the valence shell.
Period lengths: 1st = 2, 2nd = 8, 3rd = 8, 4th = 18, 5th = 18, 6th = 32, 7th = 32 elements.
Lanthanoids and actinoids are shown in two separate rows at the bottom of the table.

Period	Filling orbitals	Number of elements
1	1s	2
2	2s, 2p	8
3	3s, 3p	8
4	4s, 3d, 4p	18
5	5s, 4d, 5p	18
6	6s, 4f, 5d, 6p	32
7	7s, 5f, 6d, 7p	32

Finding the position of an element from Z. Write the electronic configuration.

Period = the highest n.
For s-block: group = number of valence electrons (1 or 2).
For p-block: group = 10 + (number of electrons in ns and np).
For d-block: group = number of electrons in (n−1)d + ns.

*Worked example 1.* An element has atomic number 33. Configuration: [Ar] 3d¹⁰ 4s² 4p³. Highest n = 4, so period 4. The ns and np electrons are 2 + 3 = 5, so group = 10 + 5 = 15. It is arsenic, a p-block element.

*Worked example 2.* An element has Z = 26. Configuration: [Ar] 3d⁶ 4s². Period 4. Group = 6 + 2 = 8. It is iron, a d-block element.

Nomenclature of elements with Z > 100

Until names are approved, the IUPAC uses a systematic name built from numerical roots, with the ending "ium".

Digit	0	1	2	3	4	5	6	7	8	9
Root	nil	un	bi	tri	quad	pent	hex	sept	oct	enn
Symbol letter	n	u	b	t	q	p	h	s	o	e

The roots are joined in the order of the digits of Z and ended with "ium".
The symbol is made from the first letters of the roots.
Examples: Z = 101: un + nil + un + ium = Unnilunium, symbol Unu. Z = 105: Unnilpentium, symbol Unp. Z = 110: Ununnilium, symbol Uun. Z = 120: Unbinilium, symbol Ubn.

*Worked example 3.* The element named ununoctium has digits un (1), un (1), oct (8): Z = 118. The symbol is Uuo.

Electronic configuration and the blocks

Elements are grouped into blocks by the subshell that receives the last electron.

Block	Groups	General outer configuration	Characteristics
s-block	1, 2	ns¹⁻²	Reactive metals (except H); low ionisation enthalpy; form ionic compounds; strong reducing agents
p-block	13 to 18	ns²np¹⁻⁶	Include non-metals, metalloids and some metals; non-metallic character increases to the right
d-block	3 to 12	(n−1)d¹⁻¹⁰ ns⁰⁻²	Transition metals; variable oxidation states; coloured ions; form complexes
f-block	Lanthanoids, actinoids	(n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹ ns²	Inner transition elements; most actinoids are radioactive

Group 1 elements are the alkali metals, group 2 the alkaline earth metals, group 16 the chalcogens, group 17 the halogens and group 18 the noble gases. Helium is placed in group 18 although it is 1s².
Metals form about three-fourths of the known elements. They are lustrous, conduct heat and electricity, and form cations.
Non-metals are placed at the top right of the table. They form anions and acidic oxides.
Metalloids (B, Si, Ge, As, Sb, Te) lie along the zig-zag line between metals and non-metals and show properties of both.
Metallic character increases down a group and decreases across a period.

Periodic trends in physical properties

Atomic radius. There are three kinds.

Covalent radius: half the distance between the nuclei of two atoms of the same element joined by a single covalent bond (for non-metals).
Metallic radius: half the distance between the nuclei of two adjacent atoms in a metallic crystal.
van der Waals radius: half the distance between the nuclei of two non-bonded atoms of neighbouring molecules. It is the largest of the three for the same element, which is why the noble gases show a large radius in a period.

Trend: across a period the radius decreases because the nuclear charge increases while the electrons enter the same shell. Down a group the radius increases because new shells are added.

Ionic radius.

A cation is smaller than its atom (fewer electrons, same nuclear charge). Example: Na⁺ is smaller than Na.
An anion is larger than its atom (more electron repulsion). Example: Cl⁻ is larger than Cl.
Isoelectronic species have the same number of electrons. The one with the larger nuclear charge has the smaller radius. Order for 10 electrons: O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺.

Ionisation enthalpy (ΔiH). The energy needed to remove the outermost electron from an isolated gaseous atom, X(g) → X⁺(g) + e⁻, in kJ mol⁻¹.

ΔiH₂ > ΔiH₁, because removing an electron from a cation is harder.
Across a period it increases in general; down a group it decreases.
Exceptions: Be > B and Mg > Al (the removed electron in B and Al is from a p orbital, which is better shielded and less penetrating than s); N > O and P > S (half-filled p³ is extra stable, and O has paired-electron repulsion).

Element (period 2)	Li	Be	B	C	N	O	F	Ne
ΔiH₁ (kJ mol⁻¹)	520	899	801	1086	1402	1314	1681	2081

Order for period 2: Li < B < Be < C < O < N < F < Ne.

Electron gain enthalpy (ΔegH). The enthalpy change when an isolated gaseous atom gains an electron, X(g) + e⁻ → X⁻(g). It is negative when energy is released.

Across a period it becomes more negative (halogens highest); down a group it generally becomes less negative.
Exceptions: Cl (−349 kJ mol⁻¹) is more negative than F (−328 kJ mol⁻¹), and S (−200) is more negative than O (−141). In the small 2p subshell of F and O, the added electron feels strong repulsion from the electrons already present.
Order for halogens: Cl > F > Br > I.
Noble gases and elements such as Be, Mg and N have positive or near-zero values. The second electron gain enthalpy of any atom (O⁻ → O²⁻) is positive.

*Worked example 4.* The ΔegH of Cl is −349 kJ mol⁻¹. Energy released when 0.5 mol of gaseous Cl atoms each gain one electron = 0.5 × 349 = 174.5 kJ.

Electronegativity. The ability of a bonded atom to attract the shared electron pair. It has no unit on the Pauling scale (F = 4.0, O = 3.5, N = 3.0, Cl = 3.2, C = 2.5, H = 2.1). It increases across a period and decreases down a group.

Valence and oxidation state. Valence is the combining capacity of an element. The oxidation state is the charge an atom would have if all bonds were ionic. Group 1 shows +1 and group 2 shows +2. A p-block element can show a positive state equal to its number of valence electrons and a negative state equal to (8 − number of valence electrons). For example, group 17 shows −1 and +7; group 16 shows −2 and +6.

Summary of trends

Property	Across a period (left to right)	Down a group
Atomic radius	Decreases	Increases
Ionisation enthalpy	Increases	Decreases
Electron gain enthalpy	More negative	Less negative (with exceptions)
Electronegativity	Increases	Decreases
Metallic character	Decreases	Increases
Basic character of oxides	Decreases	Increases

Periodic trends in chemical properties

Anomalous behaviour of second-period elements. Li, Be, B, C, N, O and F differ from the rest of their groups. The causes are small size, large charge-to-radius ratio, high electronegativity and the absence of d orbitals. Their valence shell has only four orbitals (2s and 2p), so the maximum covalence is 4. They can also form pπ–pπ multiple bonds (for example C=C, N≡N, O=O), which the heavier members rarely form.

Diagonal relationship. Elements placed diagonally in periods 2 and 3 show similar properties because they have similar ionic radii (or charge-to-radius ratio) and similar electronegativity.

Pair	Common features
Li and Mg	Both form nitrides with N₂; carbonates and hydroxides decompose on heating; chlorides are soluble in ethanol
Be and Al	Both form amphoteric oxides; chlorides are Lewis acids; both are passivated by nitric acid
B and Si	Both are metalloids; both form acidic oxides and covalent halides that are hydrolysed

Nature of oxides.

Across a period the oxides change from strongly basic to amphoteric to acidic. For period 3: Na₂O (strongly basic), MgO (basic), Al₂O₃ (amphoteric), SiO₂ (weakly acidic), P₄O₁₀, SO₃ and Cl₂O₇ (strongly acidic).
Down a group the oxides become more basic. In group 1 all oxides are basic.
Amphoteric oxides react with both acids and bases, for example Al₂O₃ and ZnO. Neutral oxides, such as CO, NO and N₂O, react with neither.
A higher oxidation state of the same element gives a more acidic oxide. For example Mn₂O₇ is acidic, whereas MnO is basic.

Reactivity. In group 1 reactivity increases down the group (Cs most reactive metal). In group 17 reactivity decreases down the group (F most reactive non-metal). Across a period the strongest reducing agent is at the left (group 1), and the strongest oxidising agent is at the right (group 17).

NEET traps

1. Be has a higher ionisation enthalpy than B, and N higher than O, even though the general trend rises across a period.
2. Cl has a more negative electron gain enthalpy than F, but F is more electronegative than Cl.
3. Noble gases have a large van der Waals radius, not a small covalent radius, when the radius is listed in a period.
4. An isoelectronic series: the ion with the highest nuclear charge is the smallest (Al³⁺, not O²⁻).
5. Mendeleev arranged elements by atomic mass; the modern table uses atomic number.
6. Eka-aluminium is gallium and eka-silicon is germanium, not the other way round.
7. Second ionisation enthalpy is always larger than the first, but for Na the jump is enormous (inner shell), while for Mg it is small.
8. Electron gain enthalpy is not the same as electronegativity: one is an energy of an isolated atom, the other is a tendency inside a bond.
9. Al₂O₃ is amphoteric, not acidic, and SiO₂ is acidic, not amphoteric.
10. Helium belongs to the s subshell by configuration (1s²) but is placed in group 18.

One-liners

1. Döbereiner's triads: Li–Na–K, Ca–Sr–Ba, Cl–Br–I.
2. Newlands' law of octaves works only up to calcium.
3. Mendeleev predicted eka-aluminium (Ga) and eka-silicon (Ge).
4. Moseley showed that atomic number is the basis of periodicity.
5. The long form table has 7 periods and 18 groups.
6. The 6th and 7th periods each have 32 elements.
7. The IUPAC root for 5 is pent and for 8 is oct.
8. Element 118 is ununoctium (Uuo) in the systematic scheme.
9. s-block elements are in groups 1 and 2; d-block in groups 3 to 12.
10. Metalloids are B, Si, Ge, As, Sb and Te.
11. Cations are smaller and anions are larger than their parent atoms.
12. Fluorine has the highest electronegativity (4.0 on the Pauling scale).
13. Second electron gain enthalpy of oxygen is positive.
14. Al₂O₃ and ZnO are amphoteric oxides.
15. Li shows a diagonal relationship with Mg, Be with Al, and B with Si.

Practice questions

Which of the following sets of elements forms a Döbereiner triad?
1. Ca, Sr, Ba
2. N, O, F
3. Na, Mg, Al
4. Be, B, C
Answer
A. Ca, Sr, Ba
Ca, Sr and Ba have similar properties and the atomic mass of Sr is about the average of Ca and Ba.
Mendeleev's eka-silicon was later discovered as
1. scandium
2. germanium
3. gallium
4. arsenic
Answer
B. germanium
Eka-silicon is germanium; eka-aluminium is gallium.
What is the IUPAC systematic name of the element with atomic number 115?
1. Ununhexium
2. Unnilpentium
3. Ununpentium
4. Ununquadium
Answer
C. Ununpentium
115 = un (1), un (1), pent (5), so Ununpentium.
Which species has the largest radius?
1. Na⁺
2. F⁻
3. Mg²⁺
4. O²⁻
Answer
D. O²⁻
All are isoelectronic with 10 electrons; O²⁻ has the lowest nuclear charge (+8), so it is the largest.
Which is the correct order of first ionisation enthalpy among Li, Be, B, C, N and O?
1. Li < B < Be < C < N < O
2. Li < B < Be < C < O < N
3. B < Li < Be < C < O < N
4. Li < Be < B < C < N < O
Answer
B. Li < B < Be < C < O < N
Be > B (2s electron removed more difficult than 2p) and N > O (half-filled 2p³), giving Li < B < Be < C < O < N.
Which element has the highest electronegativity on the Pauling scale?
1. Fluorine
2. Chlorine
3. Oxygen
4. Nitrogen
Answer
A. Fluorine
F has the value 4.0, the highest.
Which element has the most negative electron gain enthalpy?
1. Bromine
2. Iodine
3. Chlorine
4. Fluorine
Answer
C. Chlorine
Cl (−349 kJ mol⁻¹) is more negative than F (−328 kJ mol⁻¹) because the small 2p subshell of F repels the incoming electron.
Elements with general outer configuration ns²np¹ to ns²np⁶ belong to the
1. f-block
2. d-block
3. s-block
4. p-block
Answer
D. p-block
These are groups 13 to 18, the p-block.
Which of the following oxides is amphoteric?
1. Na₂O
2. Al₂O₃
3. MgO
4. SiO₂
Answer
B. Al₂O₃
Al₂O₃ reacts with both acids and bases. MgO and Na₂O are basic, SiO₂ is acidic.
Which pair of elements shows a diagonal relationship?
1. Be and Mg
2. Li and Na
3. Be and Al
4. B and Al
Answer
C. Be and Al
Li–Mg, Be–Al and B–Si are the diagonal pairs.
Which of the following is a metalloid?
1. Tin
2. Aluminium
3. Bismuth
4. Antimony
Answer
D. Antimony
B, Si, Ge, As, Sb and Te are metalloids.
How many elements are present in the sixth period of the long form table?
1. 32
2. 18
3. 24
4. 8
Answer
A. 32
The sixth period fills 6s, 4f, 5d and 6p, giving 2 + 14 + 10 + 6 = 32 elements.
Which of the following ions is NOT isoelectronic with Ne?
1. Mg²⁺
2. F⁻
3. Na⁺
4. Cl⁻
Answer
D. Cl⁻
Cl⁻ has 18 electrons; the others have 10.
Which of the following has the highest second ionisation enthalpy?
1. Si
2. Na
3. Mg
4. Al
Answer
B. Na
Removing the second electron from Na⁺ breaks a noble gas (neon) core, which needs much more energy.
The van der Waals radius of an element, compared with its covalent radius, is
1. equal
2. half of it
3. larger
4. smaller
Answer
C. larger
van der Waals radius is measured between non-bonded atoms, so it is larger than the covalent radius.
Given below are two statements: Statement I: Newlands' law of octaves worked only for elements up to calcium. Statement II: Mendeleev's periodic law was based on atomic number.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
I is correct; Mendeleev used atomic mass, so II is incorrect (atomic number is Moseley's basis).
Given below are two statements: Statement I: The first ionisation enthalpy of Be is higher than that of B. Statement II: In boron the electron is removed from a 2p orbital, which is less penetrating than the 2s orbital of Be.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are correct; the 2p electron of B is easier to remove.
Given below are two statements: Statement I: The electron gain enthalpy of F is more negative than that of Cl. Statement II: Fluorine is more electronegative than chlorine.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
I is incorrect (Cl is more negative). II is correct (F = 4.0, Cl = 3.2).
Given below are two statements: Statement I: The second electron gain enthalpy of oxygen is positive. Statement II: The O⁻ ion repels the incoming electron.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Adding an electron to O⁻ requires energy because of repulsion, so both are correct.
Which of the following statements are correct? 1. Metallic character decreases across a period. 2. Atomic radius increases down a group. 3. Ionisation enthalpy increases down a group.
1. 1 and 2 only
2. 1 and 3 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
Ionisation enthalpy decreases down a group, so 3 is wrong.
Which of the following statements are correct? 1. Al³⁺ is smaller than Mg²⁺. 2. The first ionisation enthalpy of Mg is higher than that of Al. 3. The first ionisation enthalpy of P is lower than that of S.
1. 1 and 3 only
2. 2 and 3 only
3. 1, 2 and 3
4. 1 and 2 only
Answer
D. 1 and 2 only
Al³⁺ and Mg²⁺ are isoelectronic, so higher charge gives a smaller radius. Mg > Al. P > S because of the half-filled 3p³, so 3 is wrong.
Given below are two statements: Statement I: Eka-aluminium was discovered as germanium. Statement II: Mendeleev left gaps in his table for undiscovered elements.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Eka-aluminium is gallium, so I is incorrect. II is correct.
Given below are two statements: Statement I: The seventh period has 18 elements. Statement II: Metalloids are placed in the s-block.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
The seventh period has 32 elements; metalloids lie in the p-block. Both are incorrect.
Which of the following statements are correct? 1. Li and Mg both form nitrides directly with nitrogen. 2. The oxides of Be and Al are amphoteric. 3. B and Si both show strongly metallic character.
1. 1 only
2. 2 and 3 only
3. 1 and 2 only
4. 1, 2 and 3
Answer
C. 1 and 2 only
B and Si are metalloids, so 3 is wrong. 1 and 2 are correct diagonal-relationship features.
Given below are two statements: Statement I: Basic character of oxides increases down a group. Statement II: Metallic character of elements increases down a group.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are correct and the second explains the first.
Assertion (A): Atomic radius decreases across a period. Reason (R): The nuclear charge increases while the electrons enter the same shell.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Greater nuclear charge pulls the same shell closer.
Assertion (A): Chlorine has a more negative electron gain enthalpy than fluorine. Reason (R): Chlorine is more electronegative than fluorine.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
A is true; R is false because F (4.0) is more electronegative than Cl (3.2).
Assertion (A): Li shows a diagonal relationship with Mg. Reason (R): Li⁺ and Mg²⁺ have similar charge-to-radius ratios.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Similar polarising power gives similar properties.
Assertion (A): The first ionisation enthalpy of Be is lower than that of B. Reason (R): In Be the electron removed is from the 2s orbital, which penetrates more than 2p.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
A is false (Be > B). R is a true statement but it explains why Be is higher.
Assertion (A): The Cl⁻ ion is larger than the Cl atom. Reason (R): Chlorine has the highest electron gain enthalpy in the table.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both true. The radius grows because of extra electron repulsion, not because of the ΔegH value.
Match List I with List II. List I: a. Döbereiner b. Newlands c. Mendeleev d. Moseley. List II: i. Atomic number as the basis of periodicity ii. Triads iii. Law of octaves iv. Atomic mass as the basis of periodicity.
1. a-ii, b-iii, c-i, d-iv
2. a-iii, b-ii, c-iv, d-i
3. a-ii, b-iii, c-iv, d-i
4. a-iv, b-iii, c-ii, d-i
Answer
C. a-ii, b-iii, c-iv, d-i
Döbereiner = triads, Newlands = octaves, Mendeleev = atomic mass, Moseley = atomic number.
Match List I with List II. List I: a. Unnilunium b. Unnilpentium c. Ununnilium d. Unbinilium. List II: i. Z = 120 ii. Z = 105 iii. Z = 101 iv. Z = 110.
1. a-iii, b-ii, c-i, d-iv
2. a-iii, b-ii, c-iv, d-i
3. a-iv, b-ii, c-iii, d-i
4. a-ii, b-iii, c-iv, d-i
Answer
B. a-iii, b-ii, c-iv, d-i
un-nil-un = 101; un-nil-pent = 105; un-un-nil = 110; un-bi-nil = 120.
Match List I with List II. List I: a. Na₂O b. Al₂O₃ c. P₄O₁₀ d. CO. List II: i. Neutral ii. Strongly basic iii. Amphoteric iv. Acidic.
1. a-ii, b-iii, c-i, d-iv
2. a-iii, b-ii, c-iv, d-i
3. a-ii, b-iii, c-iv, d-i
4. a-iv, b-iii, c-ii, d-i
Answer
C. a-ii, b-iii, c-iv, d-i
Na₂O is strongly basic, Al₂O₃ amphoteric, P₄O₁₀ acidic, CO neutral.
Match List I with List II. List I: a. Fe b. Na c. Cl d. Ce. List II: i. s-block ii. p-block iii. d-block iv. f-block.
1. a-iii, b-i, c-iv, d-ii
2. a-i, b-iii, c-ii, d-iv
3. a-iii, b-ii, c-i, d-iv
4. a-iii, b-i, c-ii, d-iv
Answer
D. a-iii, b-i, c-ii, d-iv
Fe is a d-block metal, Na is s-block, Cl is p-block, Ce is a lanthanoid (f-block).
Match List I with List II. List I (period 2 element): a. N b. B c. Li d. O. List II (first ionisation enthalpy, kJ mol⁻¹): i. 520 ii. 801 iii. 1314 iv. 1402.
1. a-ii, b-iv, c-i, d-iii
2. a-iv, b-ii, c-i, d-iii
3. a-iii, b-ii, c-i, d-iv
4. a-iv, b-i, c-ii, d-iii
Answer
B. a-iv, b-ii, c-i, d-iii
N = 1402, B = 801, Li = 520, O = 1314.
Match List I with List II. List I: a. Li b. Be c. B d. Al. List II (diagonal partner): i. Si ii. Al iii. Mg iv. Be.
1. a-iii, b-ii, c-iv, d-i
2. a-ii, b-iii, c-i, d-iv
3. a-i, b-ii, c-iii, d-iv
4. a-iii, b-ii, c-i, d-iv
Answer
D. a-iii, b-ii, c-i, d-iv
Li–Mg, Be–Al, B–Si, and Al–Be.
In a Döbereiner triad the atomic masses of calcium and barium are 40 and 137. The atomic mass of strontium predicted by the triad rule is
1. 68.5
2. 97.0
3. 88.5
4. 77.0
Answer
C. 88.5
Average = (40 + 137)/2 = 88.5.
An element has atomic number 35. It is placed in
1. period 4, group 17
2. period 4, group 7
3. period 5, group 17
4. period 3, group 17
Answer
A. period 4, group 17
Configuration [Ar] 3d¹⁰ 4s² 4p⁵: n = 4, group = 10 + 7 = 17.
The atomic number of the element with the systematic name ununtrium is
1. 131
2. 113
3. 103
4. 133
Answer
B. 113
un (1) un (1) tri (3) gives 113.
The electron gain enthalpy of chlorine is −349 kJ mol⁻¹. The energy released when 71 g of gaseous chlorine atoms (atomic mass 35.5) each gain one electron is
1. 349 kJ
2. 1396 kJ
3. 174.5 kJ
4. 698 kJ
Answer
D. 698 kJ
71 g = 2 mol; energy = 2 × 349 = 698 kJ.
How many of the following species are isoelectronic with argon: S²⁻, Cl⁻, K⁺, Ca²⁺, Na⁺?
1. 2
2. 3
3. 4
4. 5
Answer
C. 4
S²⁻, Cl⁻, K⁺ and Ca²⁺ each have 18 electrons. Na⁺ has 10.
An element X has successive ionisation enthalpies (kJ mol⁻¹) of 496, 4562 and 6910. X most likely belongs to
1. group 1
2. group 14
3. group 13
4. group 2
Answer
A. group 1
A very large jump after the first ionisation shows one valence electron.
The first ionisation enthalpy of sodium is 496 kJ mol⁻¹. The energy needed to convert 0.1 mol of gaseous Na atoms to Na⁺ ions is
1. 24.8 kJ
2. 49.6 kJ
3. 4.96 kJ
4. 496 kJ
Answer
B. 49.6 kJ
0.1 × 496 = 49.6 kJ.
The oxide of an element X dissolves in both dilute HCl and aqueous NaOH. X is most likely
1. P
2. Al
3. Si
4. Na
Answer
B. Al
Only Al₂O₃ is amphoteric among these.

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