Periodic Table & Periodicity — Page 27 | JEE Chemistry

Q261

Match List I with List II .tg .tg List I (Atomic number) List II (Block of periodic table) A. 37 I. p-block B. 78 II. d-block C. 52 III. f-block D. 65 IV. s-block Choose the correct answer from the options given below :

A A - II, B - IV, C - I, D - III

B A - IV, B - III, C - II, D - I

C A - IV, B - II, C - I, D - III

D A - I, B - III, C - IV, D - II

Correct Answer

Option C

Solution

.tg .tg List I (Atomic number) List II (Block of periodic table) A.

37 (Rb - Rubidium) IV. s-block B.

78 (Pt - Platinum) II. d-block C.

52 (Te - Tellurium) I. p-block D.

65 (Tb - Terbium) III. f-block

Q262

Match with . ^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}$

List - I		List - II
(B)	$\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}$	(I)	Ionisation Enthalpy
(C)	$\mathrm{B}<\mathrm{Al}<\mathrm{Mg}<\mathrm{K}$	(II)	Metallic character
(D)	$\mathrm{Si}<\mathrm{P}<\mathrm{S}<\mathrm{Cl}$	(III)	Electronegativity
()		(IV)	Ionic radii

A (A)-(IV), (B)-(I), (C)- (II), (D)-(III)

B (A)-(III), (B)-(IV), (C)- (II), (D)-(I)

C (A)-(II), (B)-(III), (C)- (IV), (D)-(I)

D (A)-(IV), (B)-(I), (C)- (III), (D)-(II)

Correct Answer

Option A

Solution

(A) $\mathrm{Al}^{3+} All four ions ($ \mathrm{Al}^{3+},\mathrm{Mg}^{2+},\mathrm{Na}^{+},\mathrm{F}^{-} $) are isoelectronic (each has 10 electrons). For isoelectronic species, ionic radii decrease as the positive nuclear charge increases, which is why the smallest ion here is$ \mathrm{Al}^{3+} $(Z=13) and the largest is$ \mathrm{F}^{-} $(Z=9). Thus, this ordering is one of increasing ionic radius.$ \boxed{ (A) \;\longrightarrow\; \text{(IV) Ionic radii} } $(B)$ \mathrm{B} Check first ionization enthalpies ( $\mathrm{IE}_1$ ): B: $\approx 801$ \,kJ/mol C: $\approx 1086$ \,kJ/mol O: $\approx 1314$ \,kJ/mol N: $\approx 1402$ \,kJ/mol Hence, the order of increasing $\mathrm{IE}_1$ is $B This matches the given sequence exactly.$ \boxed{ (B) \;\longrightarrow\; \text{(I) Ionisation enthalpy} } $(C)$ \mathrm{B} Consider metallic character (the tendency to lose electrons easily, show metallic properties).

Across a period (left to right), metallic character decreases; down a group, it increases.

B (metalloid) has the least metallic character here.

Al (group 13 metal) is more metallic than B.

Mg (group 2 metal) is typically more metallic than Al.

K (group 1 metal) is the most metallic among these.

Thus, $\mathrm{B} increasing metallic character.$ \boxed{ (C) \;\longrightarrow\; \text{(II) Metallic character} } $(D)$ \mathrm{Si} Check electronegativities: Si: $\approx 1.90$ P: $\approx 2.19$ S: $\approx 2.58$ Cl: $\approx 3.16$ They increase in the order $\mathrm{Si} which matches the given sequence for increasing electronegativity.$ \boxed{ (D) \;\longrightarrow\; \text{(III) Electronegativity} } $Final Matching$ (A) \to (IV),\quad (B) \to (I),\quad (C) \to (II),\quad (D) \to (III).

$Looking at the choices given: Option A:$ (A)-(IV), (B)-(I), (C)-(II), (D)-(III)$ This is exactly what we found.

Answer: Option A

Q263

The increasing order of the ionic radii of the given isoelectronic species is :

A Cl–, Ca2+ , K+ , S2–

B S2–, Cl–, Ca2+ , K+

C Ca2+ , K+ , Cl– , S2–

D K+, S2–, Ca2+, Cl–

Correct Answer

Option C

Solution

Iso-electronic ions have same number of electrons.

So, for iso-electronic ions, number of electrons = constant.

$\therefore$ $\sigma$ (Slaten's Constant) = Constant.

As $\sigma$ depends on number of electrons.

If a element's number of electrons increases then that element's $\sigma$ increases.

Also we know, Zeffective = Z - $\sigma$ As for iso-electronic ions, $\sigma$ (Slaten's Constant) = Constant.

So Zeffective depend on only value of Z.

If Z of an ion increases then Zeffective also increases and if Z of an ion decreases then Zeffective also decreases.

And when Zeffective increases then nuclear attraction towards outermost electrons increase and size of ion decreases.

Similarly when Zeffective decrease then nuclear attraction towards outermost electrons decreases and size of ion increases.

$\therefore$ Size $\propto$

{1 \over {{Z_{eff}}\,or\,Z}}

.tg .tg Ca2+ K+ Cl- S2- Z 20 19 17 16 $\therefore$ Ionic radius order Ca2+ + – 2–

Q264

The incorrect decreasing order of atomic radii is

A

\mathrm{Si}>\mathrm{P}>\mathrm{Cl}>\mathrm{F}

B

\mathrm{Mg}>\mathrm{Al}>\mathrm{C}>\mathrm{O}

C

\mathrm{Al}>\mathrm{B}>\mathrm{N}>\mathrm{F}

D

\mathrm{Be}>\mathrm{Mg}>\mathrm{Al}>\mathrm{Si}

Correct Answer

Option D

Solution

Let's analyze the options based on the periodic trends in atomic radii.

Recall two key trends: Atomic radii decrease as you move from left to right in a period due to increasing effective nuclear charge.

Atomic radii increase as you move down a group because additional electron shells are added.

We'll review each option: Option A:

\mathrm{Si} > \mathrm{P} > \mathrm{Cl} > \mathrm{F}

Silicon (Si), phosphorus (P), and chlorine (Cl) are in the same period, with a gradual decrease in atomic radius from left to right.

Fluorine (F) is in an earlier period (it has only 2 electron shells) and is much smaller.

Hence, the order follows the expected trend.

Option B:

\mathrm{Mg} > \mathrm{Al} > \mathrm{C} > \mathrm{O}

Magnesium (Mg) and aluminum (Al) are period 3 elements, whereas carbon (C) and oxygen (O) belong to period 2.

Period 3 elements generally have larger radii than period 2 counterparts.

The order is consistent: Mg and Al are larger than C and O, and within each period, radii decrease from left to right.

Option C:

\mathrm{Al} > \mathrm{B} > \mathrm{N} > \mathrm{F}

Aluminum (Al) is in period 3 and is expected to be larger than boron (B), nitrogen (N), and fluorine (F) from period 2.

Within period 2 (B, N, F), the atomic radii decrease from left to right.

Thus, this order is correct as well.

Option D:

\mathrm{Be} > \mathrm{Mg} > \mathrm{Al} > \mathrm{Si}

Beryllium (Be) is in period 2, while magnesium (Mg), aluminum (Al), and silicon (Si) are in period 3.

Since atomic radii increase with the number of electron shells, Be (with only 2 shells) should be smaller than Mg.

This means that stating

\mathrm{Be} > \mathrm{Mg}

is incorrect. Thus, the incorrect decreasing order of atomic radii is given in Option D.

Q265

The incorrect statement is

A The first ionization enthalpy of K is less than that of Na and Li.

B Xe does not have the lowest first ionization enthalpy in its group.

C The first ionization enthalpy of element with atomic number 37 is lower than that of the element with atomic number 38.

D The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.

Correct Answer

Option D

Solution

On moving down in a group ionisation energy decrease $\therefore 1^{\text{st }}$ ionisation enthalpy order is $\mathrm{Li}>\mathrm{Na}>\mathrm{K}$ $\mathrm{Zn}$ has more ionisation energy as compared to Ga because of their pseudo inert gas configuration.

Q266

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R Assertion A : The alkali metals and their salts impart characteristic colour to reducing flame. Reason R : Alkali metals can be detected using flame tests. In the light of the above statements, choose the most appropriate answer from the options given below :

A A is not correct but R is correct

B A is correct but R is not correct

C Both A and R are correct and R is the correct explanation of A

D Both A and R are correct but R is NOT the correct explanation of A

Correct Answer

Option A

Solution

Assertion is not correct because alkali metals and their salts impart characteristic colour to oxidising part of flame and not reducing part of flame.

Reason is correct because all alkali metals can be detected by their flame tests.

Q267

The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+ will be:

A – 5.1 eV

B –10.2 eV

C + 2.55 eV

D – 2.55 eV

Correct Answer

Option A

Solution

Ionization potential of Na means energy required to convert of Na to Na+ ion.

Na $\to$ Na+ + e- IE = 5.1 ev Electron gain enthalpy of Na+ means energy required to convert Na+ ion to Na.

Na $\to$ Na+ + e-

\Delta {H_{eq}}

= - 5.1 ev According to Lavoisier and Laplace law of thermochemistry, when we change the direction of reaction then the sign of energy also changes.

Q268

Which one of the following elements will remain as liquid inside pure boiling water?

A Li

B Br

C Cs

D Ga

Correct Answer

Option D

Solution

The melting point of gallium is very low (about 29.76°C or 85.57°F).

Therefore, gallium can melt just from the heat of one's hand.

The boiling point of gallium, on the other hand, is quite high (2400°C or 4352°F).

On the contrary, water boils at 100°C (212°F) under normal atmospheric pressure.

Therefore, inside boiling water, gallium would melt into a liquid but would not boil or evaporate, unlike the other elements listed.

Lithium (Li), Bromine (Br), and Cesium (Cs) have boiling points below 100°C, so they would turn into a gas in boiling water.

Q269

Element "E" belongs to the period 4 and group 16 of the periodic table. The valence shell electron configuration of the element, which is just above "E" in the group is

A 3s2, 3p4

B 3d10, 4s2, 4p4

C 4d10, 5s2, 5p4

D 2s2, 2p4

Correct Answer

Option A

Solution

Element E is Selenium (Se) Electronic configuration of E is [Ar] 3d10 4s2 4p4 The element which is just above ‘E’ in periodic table is sulphur, its electronic configuration is [Ne] 3s2 3p4

Q270

Which of the following statements are not correct? A. The electron gain enthalpy of

\mathrm{F}

is more negative than that of

\mathrm{Cl}

. B. Ionization enthalpy decreases in a group of periodic table. C. The electronegativity of an atom depends upon the atoms bonded to it. D.

\mathrm{Al}_{2} \mathrm{O}_{3}

and

\mathrm{NO}

are examples of amphoteric oxides. Choose the most appropriate answer from the options given below :

A A, B, C and D

B A, C and D Only

C A, B and D Only

D B and D Only

Correct Answer

Option B

Solution

Let's analyze the statements: A.

The electron gain enthalpy of $\mathrm{F}$ is more negative than that of $\mathrm{Cl}$ .

This statement is incorrect.

The electron gain enthalpy of $\mathrm{F}$ is less negative than that of $\mathrm{Cl}$ due to its small size, which leads to increased electron-electron repulsion when an electron is added.

B.

Ionization enthalpy decreases in a group of the periodic table.

This statement is correct.

Ionization enthalpy generally decreases down a group of the periodic table due to the increase in atomic size, which results in a weaker attraction between the nucleus and the outermost electrons.

C.

The electronegativity of an atom depends upon the atoms bonded to it.

This statement is incorrect.

Electronegativity is a property of an atom that depends on the effective nuclear charge and the distance of the outermost electrons from the nucleus.

While the difference in electronegativity between atoms in a bond can affect the bond's polarity, the electronegativity itself does not depend on the atoms bonded to it.

D. $\mathrm{Al}_{2} \mathrm{O}_{3}$ and $\mathrm{NO}$ are examples of amphoteric oxides.

This statement is incorrect. $\mathrm{Al}_{2} \mathrm{O}_{3}$ is an example of an amphoteric oxide, meaning it can react with both acids and bases.

However, $\mathrm{NO}$ is not an amphoteric oxide; it is a neutral oxide.

Hence, the incorrect statements are A, C, and D.