Periodic Table & Periodicity — Page 9 | NEET Chemistry

Q81

Correct statement about alkali metal oxides is :

A peroxides are colored.

B superoxides are paramagnetic.

C oxides are paramagnetic.

D peroxides are both colored and paramagnetic.

Correct Answer

Option B

Solution

The peroxide and oxides of alkali metals are colourless when pure.

Superoxides are paramagnetic while peroxides are diamagnetic.

Electronic configuration of $\mathrm{O}_{2}^{2-}$ (peroxide) $\left(\sigma 1 s^{2}\right)\left(\sigma^{*} 1 s^{2}\right)\left(\sigma 2 s^{2}\right)\left(\sigma^{*} 2 s^{2}\right)\left(\sigma 2 p_{z}^{2}\right)\left(\pi 2 p_{x}^{2} \pi 2 p_{y}^{2}\right)$ $\left(\pi^{*} 2 p_{x}^{2} \pi^{*} 2 p_{y}^{2}\right)$ Electronic configuration of $\mathrm{O}_{2}^{-}$ $\left(\sigma 1 s^{2}\right)\left(\sigma^{*} 1 s^{2}\right)\left(\sigma 2 s^{2}\right)\left(\sigma^{*} 2 s^{2}\right)\left(\sigma 2 p_{z}^{2}\right)\left(\pi 2 p_{x}^{2} \quad \pi 2 p_{y}^{2}\right)$ $\left(\pi^{*} 2 p_{x}^{2} \pi^{*} 2 p_{y}^{1}\right)$ superoxide is paramagnetic due 1 unpaired electron.

Q82

The correct order of electron gain enthalpy is :

A Te > Se > S > O

B S > Se > Te > O

C S > O > Se > Te

D O > S > Se > Te

Correct Answer

Option B

Solution

Oxygen is the second most electronegative element in comparison to fluorine.

In group - 16 family $(\mathrm{O}, \mathrm{S}, \mathrm{Se}, \mathrm{Te})$ , O-atom is smallest in size.

So, electron density on O-atom is very high in group -16 During addition of a free electron to gaseous $\mathrm{O}$ -atom,

\mathrm{O}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{-}(\mathrm{g})

We have to supply a significant amount of energy (endothermic) to overcome the electrostatic repulsion between the approaching electron and O-atom of very high electron density.

So, the net value of electron affinity (EA) or (negative) electron gain enthalpy $\left[\Delta_{\mathrm{eg}} H\right.$ or $\left.\left|\Delta_{\mathrm{eg}} H\right|\right]$ of oxygen decreases to a higher extent in comparison to other elements of group -16 who have larger size and lower electronegativity.

So, the correct order of EA or $\left|\Delta_{\mathrm{eg}} \mathrm{H}\right|$ of group $-16$ elements will be $\mathrm{S}$ $>\mathrm{Se}>\mathrm{Te}>\mathrm{O}$

Q83

In the industrial production of which of the following, molecular hydrogen is obtained as a byproduct?

A NaOH

B NaCl

C Na metal

D Na2CO3

Correct Answer

Option A

Solution

Molecular hydrogen is produced as a byproduct in the industrial production of

\mathrm{NaOH}

by electrolysis of aq

\mathrm{NaCl}

solution

\mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}

\begin{aligned} & \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H^{+}}+\mathrm{OH}^{-} \end{aligned}

Cathode :

2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e} \rightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}

Anode :

2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2}+2 \mathrm{e}

\mathrm{NaOH}

is crystallised from the remaining part of electrolyte.

Q84

Given below are two statements : one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

: Assertion A : The first ionisation enthalpy decreases across a period. Reason

\mathbf{R}

: The increasing nuclear charge outweighs the shielding across the period. In the light of the above statements, choose the most appropriate from the options given below :

A

\mathbf{A}

is false but

\mathbf{R}

is true

B Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

C Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

D

\mathbf{A}

is true but

\mathbf{R}

is false

Correct Answer

Option A

Solution

First ionisation energy increases along the period. Along the period

\mathrm{Z}

increases which outweighs the shielding effect.

Q85

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

Q86

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

Q87

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–

Q88

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.

Q89

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.

Q90

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.