Periodic Table & Periodicity — Page 8 | JEE Chemistry

Q71

Match with : (Metal Ion) (Group of Qualitative analysis)

List - I		List - II
(a)	$M{n^{2 + }}$	(i)	Group - III
(b)	$A{s^{3 + }}$	(ii)	Group - IIA
(c)	$C{u^{2 + }}$	(iii)	Group - IV
(d)	$A{l^{3 + }}$	(iv)	Group - IIB Choose the most appropriate answer from the options given below :

A (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

B (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)

C (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)

D (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)

Correct Answer

Option B

Solution

Mn2+ $\to$ IV group As3+ $\to$ II B group Cu2+ $\to$ II A group Al3+ $\to$ III group Note : IVth group cations – Zn2+ ,Mn2+ ,Ni2+ , Co2+ II B group cations – As3+ , Sb3+ , Sn2+ II A group cations – Hg2+ , Pb2+ , Bi3+ , Cu2+ , Cd2+ IIIrd group cations – Al3+ , Cr3+ , Fe3+

Q72

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : The first ionization enthalpy for oxygen is lower than that of nitrogen. Reason R : The four electrons in 2p orbitals of oxygen experience more electron-electron repulsion. In the light of the above statements, choose the correct answer from the options given below.

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

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

C A is correct but R is not correct.

D A is not correct but R is correct.

Correct Answer

Option B

Solution

Nitrogen has half-filled p-orbitals which is stable. Due to this, its 1st ionization energy is more than oxygen.

Q73

Among the following, basic oxide is :

A SO3

B SiO2

C CaO

D Al2O3

Correct Answer

Option C

Solution

Since, oxides of metals are basic in nature.

Hence CaO is a basic oxide.

SO3 and SiO2 are acidic oxides and Al2O3 is a amphoteric oxide.

Q74

The correct order of increasing ionic radii is

A Mg2+ +

-

2

-

3

-

B N3

-

2

-

-

+ 2+

C F

-

+ 2

-

2+ 3

-

D Na+

-

2+ 2

-

3

-

Correct Answer

Option A

Solution

For isoelectronic species Ionic radii

\propto \frac{(-) \text{ ve charge }}{(+) \text{ ve charge }}

Hence, correct order of ionic radii is Mg2+ + – 2– 3–

Q75

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The ionic radii of O2

-

and Mg2+ are same. Reason (R) : Both O2

-

and Mg2+ are isoelectronic species. In the light of the above statements, choose the correct answer from the options given below.

A Both (A) and (R) are true and (R) is the correct explanation of (A).

B Both (A) and (R) are true but (R) is not the correct explanation of (A).

C (A) is true but (R) is false.

D (A) is false but (R) is true.

Correct Answer

Option D

Solution

Correct order of ionic radii: O–2 > Mg+2 This is because among isoelectronic species, the size of anions are greater than the size of cations.

Statement (II) is correct as both O–2 and Mg+2 are isoelectronic.

Q76

The correct order of electron gain enthalpies of Cl, F, Te and Po is

A F < Cl < Te < Po

B Po < Te < F < Cl

C Te < Po < Cl < F

D Cl < F < Te < Po

Correct Answer

Option B

Solution

Te → –190 kJ mol–1 Po → –174 kJ mol–1 F → –333 kJ mol–1 Cl → –349 kJ mol–1 Hence, correct order is Cl > F > Te > Po

Q77

The correct order of electron gain enthalpy (

-

ve value) is :

A O > S > Se > Te

B O < S < Se < Te

C O Se > Te

D O Se < Te

Correct Answer

Option C

Solution

Down the group, the size of atom increases, so electron gain enthalpy decreases.

In case of Oxygen, due to its small size, which leads to electronelectron repulsion, results in less electron gain enthalpy than sulphur.

Q78

Magnisium powder burns in air to give :

A MgO and Mg3N2

B MgO only

C Mg(NO3)2 and Mg3N2

D MgO and Mg(NO3)2

Correct Answer

Option A

Solution

Mg burn in air and produces a mixture of nitride and oxide. So, Mg3N2 and MgO are formed.

Q79

The IUPAC nomenclature of an element with electronic configuration [Rn]

5 \mathrm{f}^{14} 6 \mathrm{d}^{1} 7 \mathrm{s}^{2}

is :

A Unnilbium

B Unnilunium

C Unnilquadium

D Unniltrium

Correct Answer

Option D

Solution

The element with electronic configuration [Rn]

5 \mathrm{f}^{14} 6 \mathrm{~d}^{1} 7 \mathrm{~s}^{2}

has atomic number $\rightarrow 103$ $\therefore$ Its IUPAC name is: Unniltrium

Q80

The first ionization enthalpies of Be, B, N and O follow the order

A O < N < B < Be

B Be < B < N < O

C B < Be < N < O

D B < Be < O < N

Correct Answer

Option D

Solution

The first ionization energy increase from left to right along $2^{\text{nd }}$ period with the following exceptions

\mathrm{IE}_{1}: \mathrm{Be}>\mathrm{B}

and

\mathrm{N}>\mathrm{O}

This is due to stable configuration of

\mathrm{Be}

in comparison to

\mathrm{B}

and that of

\mathrm{N}

in comparison to

\mathrm{O}

. Hence the correct order is

\mathrm{N}>\mathrm{O}>\mathrm{Be}>\mathrm{B}