Periodic Table Class 10 ICSE Important Questions with Answers

Answer: (b) Decreases.

Along a period the nuclear charge increases while electrons are added to the same shell, so the increasing pull of the nucleus draws the electron cloud inward and the atomic radius decreases.

Answer: (b) Atomic number.

Moseley's work showed atomic number ($Z$Z) is the fundamental property; the modern periodic law states that properties are a periodic function of atomic number, so elements are arranged by increasing $Z$Z.

Answer: (b) Period 3, group 17.

Three occupied shells give period 3. It has 7 valence electrons, placing it in group 17 (the halogens); the element is chlorine, $Cl$Cl.

Answer: (b) $Na < Si < Cl$Na < Si < Cl.

$Na$Na ($2,8,1$2,8,1), $Si$Si ($2,8,4$2,8,4) and $Cl$Cl ($2,8,7$2,8,7) all have three shells. Across the period the nuclear charge increases while electrons enter the same shell, so atomic size decreases and the outer electrons are held more tightly. More energy is needed to remove one, so ionisation potential rises from $Na$Na to $Cl$Cl.

Answer: (a) Both statements are true and R correctly explains A. $Na$Na ($2,8,1$2,8,1) and $K$K ($2,8,8,1$2,8,8,1) are in the same group; $K$K has one extra shell, so its atomic size is larger.

(i) Ionisation energy: the minimum amount of energy required to remove the most loosely bound (outermost) electron from an isolated neutral gaseous atom to form a positively charged gaseous ion.

(ii) Electron affinity: the amount of energy released when a neutral gaseous atom gains an electron to form a negatively charged gaseous ion.

Increasing order of metallic character: $Li < Na < K$Li < Na < K.

Reason: These are group 1 elements. Down the group the atomic size increases and the outermost electron is held less tightly, so it is lost more easily. The greater the ease of losing electrons, the more metallic (electropositive) the element, so metallic character increases from $Li$Li to $K$K.

Across a period, the nuclear charge increases by one unit for each successive element while electrons are added to the same shell, so the effective nuclear pull increases.

(i) Atomic size decreases because the greater nuclear charge pulls the same-shell electrons closer.

(ii) Electron affinity generally increases because the smaller atom with higher nuclear charge attracts an incoming electron more strongly, releasing more energy.

(iii) Non-metallic character increases because the tendency to gain electrons (electronegativity) rises, so elements become more electron-accepting towards the right.

(i) Most metallic: $Na$Na (sodium). It lies at the extreme left with only one valence electron, which is lost most easily, making it the most electropositive.

(ii) Highest ionisation energy: $Ar$Ar (argon). Being a noble gas with a stable, completely filled octet, it holds its electrons most tightly, so the most energy is needed to remove an electron. (Among reactive elements, $Cl$Cl has the highest.)

(iii) Smallest atomic size (excluding $Ar$Ar): $Cl$Cl (chlorine), because it has the highest nuclear charge among the reactive period-3 elements pulling the same shell inward.

All three elements lie in the same period, so trends across a period apply. A is in group 1 (left), B in group 16, C in group 17 (right).

(i) Largest atomic size: A, because atomic size is greatest on the left of a period.

(ii) Most non-metallic: C, since non-metallic character increases across a period and C is farthest right (a halogen).

(iii) Increasing ionisation energy: $A < B < C$A < B < C, because ionisation energy increases from left to right across the period.

(a) Modern periodic law: The physical and chemical properties of elements are a periodic function of their atomic numbers.

(b) The law is based on the work of Henry Moseley.

(c) Element $M$M, $Z = 12$Z = 12 (magnesium):

(i) Electronic configuration: $2,8,2$2,8,2.

(ii) Three occupied shells give period 3; 2 valence electrons place it in group 2.

(iii) It has only 2 loosely held valence electrons which it loses easily, so it is a metal (electropositive).

(iv) It forms $M^{2+}$M^2+ ions; with $O^{2-}$O^2- the oxide is $MgO$MgO.

(i) Noble gases have a stable, completely filled octet (or duplet for $He$He); this stable configuration holds electrons very tightly, so removing an electron needs very high energy.

(ii) $Li$Li ($2,1$2,1) and $F$F ($2,7$2,7) are in the same period; $F$F has a greater nuclear charge (9 protons vs 3) pulling the same shell inward, so $F$F is smaller than $Li$Li.

(iii) In the very small $F$F atom, the incoming electron enters a compact 2nd shell where strong inter-electronic repulsion opposes its addition; $Cl$Cl is slightly larger, has less repulsion, and so accepts the electron with a greater release of energy, giving it a higher electron affinity.

(iv) Down group 1 the atomic size increases and the valence electron is held less tightly, so it is lost more readily; the ease of losing the electron raises reactivity.

(v) When an atom loses electrons to form a cation, it often loses its outermost shell and the same nuclear charge now acts on fewer electrons, pulling them closer, so the cation is smaller than the parent atom.

All four elements have three occupied shells, so they belong to period 3.

(i) S ($2,8,8$2,8,8) is the noble gas, because its outermost shell is completely filled with 8 electrons (a stable octet), so it is chemically unreactive.

(ii) P ($2,8,1$2,8,1) has the largest atomic size; it is on the extreme left (group 1) where nuclear pull on the outer shell is least.

(iii) R ($2,8,7$2,8,7) is the most non-metallic; with 7 valence electrons it strongly tends to gain one electron. It belongs to group 17 (the halogens).

(iv) P forms $P^{+}$P^+ (loses 1 electron) and R forms $R^{-}$R^- (gains 1 electron), so the compound is $PR$PR.