Periodic Table & Periodicity — Page 25 | JEE Chemistry

Q241

A s-block element (M) reacts with oxygen to form an oxide of the formula MO2. The oxide is pale yellow in colour and paramagnetic. The element (M) is :

A Mg

B Na

C Ca

D K

Correct Answer

Option D

Solution

The element (M) is potassium (K).

It reacts with O2 to form KO2, which is paramagnetic in nature.

All other elements form oxides or peroxides which are diamagnetic in nature.

Q242

Given below are two statements : Statement (I) : The

4 \mathrm{f}

and

5 \mathrm{f}

- series of elements are placed separately in the Periodic table to preserve the principle of classification. Statement (II) : S-block elements can be found in pure form in nature. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is false but Statement II is true

B Both Statement I and Statement II are true

C Statement I is true but Statement II is false

D Both Statement I and Statement II are false

Correct Answer

Option C

Solution

Statement I is correct because the 4f and 5f series elements, commonly referred to as the lanthanides and actinides, respectively, have unique properties and configurations that distinguish them from other elements.

Placing them separately in the Periodic table helps to maintain the orderly classification based on atomic numbers and avoids disrupting the pattern established by the periodic law.Statement II is incorrect because s-block elements, which include Group 1 (alkali metals) and Group 2 (alkaline earth metals), are indeed highly reactive.

They typically do not exist in their pure form in nature due to their reactivity.

Instead, they are mostly found in compound forms, combined with other elements.

For instance, sodium (Na) and potassium (K) from Group 1 react with water and are usually found as salts, not in their metallic, pure state.

Q243

Match List I with List II .tg .tg LIST I LIST II A. Melting Point

[\mathrm{K}]

I.

\mathrm{T} 1>\mathrm{In}>\mathrm{Ga}>\mathrm{A} 1>\mathrm{B}

B. Ionic Radius

[\mathrm{M}^{+3} / \mathrm{pm}]

II.

\mathrm{B}>\mathrm{T} 1>\mathrm{Al} \approx \mathrm{Ga}>\mathrm{In}

C.

\Delta_{\mathrm{i}} \mathrm{H}_1[\mathrm{~kJ} \mathrm{~mol}^{-1}]

III.

\mathrm{T} 1>\mathrm{In}>\mathrm{Al}>\mathrm{Ga}>\mathrm{B}

D. Atomic Radius [pm] IV.

\mathrm{B}>\mathrm{A} 1>\mathrm{T} 1>\mathrm{In}>\mathrm{Ga}

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option A

Solution

Melting point :

\mathrm{B}>\mathrm{A} \ell>\mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}

Ionic radius

(\mathrm{M}^{+3} / \mathrm{pm}): \mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}>\mathrm{A} \ell>\mathrm{B}

\left(\Delta_{\mathrm{IE}} \mathrm{H}\right)_1\left[\frac{\mathrm{kJ}}{\mathrm{mol}}\right]: \mathrm{B}>\mathrm{T} \ell>\mathrm{A} \ell \approx \mathrm{Ga}>\mathrm{In}

Atomic radius (in

\mathrm{pm}

) :

\mathrm{T} \ell>\mathrm{In}>\mathrm{A} \ell>\mathrm{Ga}>\mathrm{B}

Q244

Which of the following statements are correct? A. The process of adding an electron to a neutral gaseous atom is always exothermic. B. The process of removing an electron from an isolated gaseous atom is always endothermic. C. The

1^{\text{st }}

ionization energy of boron is less than that of beryllium. D. The electronegativity of C is 2.5 in

\mathrm{CH}_4

and

\mathrm{CCl}_4

E. Li is the most electropositive among elements of group I. Choose the correct answer from the options given below:

A A, C and D Only

B B and C Only

C B and D Only

D B, C and E Only

Correct Answer

Option B

Solution

(A) The process of adding an $\mathrm{e}^{-}$ to a neutral gaseous atom is not always exothermic it may be exothermic or endothermic.

$\begin{array}{cc}\text{ (C) } \mathrm{Be} & \mathrm{B} \\ 1 \mathrm{~s}^2 2 s^2 & 1 s^2 2 s^2 2 p^1\end{array}$ In Be 2s subshell in fully filled So, need high energy to remove $\mathrm{e}^{-}$ as compared to B . due to partially positive charge $z_{\text{eff }} \uparrow, \mathrm{EN} \uparrow$ So, EN of $\mathrm{C} \Rightarrow \mathrm{CCl}_4>\mathrm{CH}_4$ (E) Cs is most electropositive.

Q245

The main oxides formed on combustion of Li, Na and K in excess of air are, respectively :

A LiO2, Na2O2 and K2O

B Li2O2, Na2O2 and KO2

C Li2O, Na2O2 and KO2

D Li2O, Na2O and KO2

Correct Answer

Option C

Solution

On heating with excess of air

Li,

Na

and

K

forms following oxides

4Li + {O_2}\,\,\overset{\,}\longrightarrow \,\,\mathop {2L{i_2}O\,\,\,\,\,\,\,\,\,\,}\limits_{Lithium\,\,\,\,monoxide} \,

2Na + {O_2}\,\,\overset{{575\,\,K}}\longrightarrow \,\,\mathop {N{a_2}{O_2}\,\,\,\,\,\,\,\,\,\,\,}\limits_{Sodium\,\,\,peroxide}

K + {O_2}\,\,\overset{\,}\longrightarrow \,\,\mathop {K{O_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\limits_{Potassium\,\,\,\sup eroxise}

Q246

B has a smaller first ionization enthalpy than Be. Consider the following statements : (I) It is easier to remove 2p electron than 2s electron (II) 2p electron of B is more shielded from the nucleus by the inner core of electrons than the 2s electrons of Be. (III) 2s electron has more penetration power than 2p electron. (IV) atomic radius of B is more than Be (Atomic number B = 5, Be = 4) The correct statements are :

A (I), (III) and (IV)

B (II), (III) and (IV)

C (I), (II) and (IV)

D (I), (II) and (III)

Correct Answer

Option D

Solution

1st I.E. of Be > B In case of Be, electron is removed from 2s orbital which has more penetration power, while in case of B electron is removed from 2p orbital which has less penetration power. 2p electron of B is more shielded from nucleus by the inner electrons than 2s electrons of Be $\therefore$ It is easier to remove 2p electron than 2s electron.

Q247

The correct order of electron affinity is :

A F > Cl > O

B F > O > Cl

C Cl > F > O

D O > F > Cl

Correct Answer

Option C

Solution

Electron affinity means tendency of gaining an electron by an atom.

In a period from left ot right the electron affinity increases and in a group it decreases from top to bottom.

So according to this theory Fluorine(F) should have most electron affinity.

But when an electron is added to the F atom, electron comes to the 2p orbital and for Cl atom electron is added in 3p orbital.

As 2p orbital is closer to the nucleus than 3p orbital as 3p orbital is larger in size, so when a new electron comes to 2p orbital then it will face a strong repulsion force by the nucleus than if electron comes to 3p orbital.

So, F have lesser tendency of gaining electron than Cl.

Note : In entire periodic table Cl have highest electron affinity.

Q248

Which of the following statements are correct? (A) Both LiCl and MgCl2 are soluble in ethanol. (B) The oxides Li2O and MgO combine with excess of oxygen to give superoxide. (C) LiF is less soluble in water than other alkali metal fluorides. (D) Li2O is more soluble in water than other alkali metal oxides. Choose the most appropriate answer from the options given below :

A (A) and (C) only

B (A), (C) and (D) only

C (B) and (C) only

D (A) and (D) only

Correct Answer

Option A

Solution

(A) Both LiCl and MgCl2 are soluble in ethanol (B) Li and Mg do not form superoxide (C) LiF has high lattice energy (D) Li2O is least soluble in water than another alkali metal oxides

Q249

Given below are the oxides: Na2O, As2O3, N2O, NO and Cl2O7 Number of amphoteric oxides is :

A 0

B 1

C 2

D 3

Correct Answer

Option B

Solution

Oxides $\mathrm{Na}_2 \mathrm{O} \longrightarrow$ Basic $\mathrm{As}_2 \mathrm{O}_3 \longrightarrow$ Amphoteric $\mathrm{N}_2 \mathrm{O} \longrightarrow$ Neutral $\mathrm{NO} \longrightarrow$ Neutral $\mathrm{Cl}_2 \mathrm{O}_7 \longrightarrow$ Acidic Hence, only one amphoteric oxide is present.

Q250

Match with : $$ salt

List - I		List - II
(b)	Na	(ii)	Most abundant element in cell fluid
(c)	K	(iii)	Bicarbonate salt used in fire extinguisher
(d)	Cs	(iv)	Carbonate salt decomposes easily on heating

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

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

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

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

Correct Answer

Option A

Solution

(a) CsI salt is poor water soluble due to it's low hydration energy (b) NaHCO3 is used in fire extinguisher (c) K is most abundant element in cell fluid (d) Li2CO3 decomposes easily due to high covalent character caused by small size Li+ cation.