Periodic Table & Periodicity

NEET Chemistry · 94 questions · Page 7 of 10 · Click an option or "Show Solution" to reveal answer

Q61
The ionic radius of Na+ ions is 1.02 Ao\mathop A\limits^o . The ionic radii (in Ao\mathop A\limits^o ) of Mg2+ and Al3+, respectively, are
A 1.05 and 0.99
B 0.72 and 0.54
C 0.85 and 0.99
D 0.68 and 0.72
Correct Answer
Option B
Solution

The ionic radii (in

Ao\mathop A\limits^o

) of

Mg2+\text{Mg}^{2+}

and

Al3+\text{Al}^{3+}

, respectively, are: Option B:

0.72Ao0.72 \mathop A\limits^o

for

Mg2+\text{Mg}^{2+}
0.54Ao0.54 \mathop A\limits^o

for

Al3+\text{Al}^{3+}
Q62
Given below are two statements : Statement I : None of the alkaline earth metal hydroxides dissolve in alkali. Statement II : Solubility of alkaline earth metal hydroxides in water increases down the group. In the light of the above statements, choose the most appropriate answer from the options given below :
A Statement I is correct but Statement II is incorrect.
B Statement I is incorrect but Statement II is correct.
C Statement I and Statement II both are incorrect.
D Statement I and Statement II both are correct.
Correct Answer
Option B
Solution

Statement - I is incorrect Be(OH)2 dissolve in alkali due to it's amphoteric nature.

Statement - II is correct Solubility of alkaline earth metal hydroxide in water increases down the group due to rapid decreases in lattice energy as compared to hydration energy.

Q63
Inert gases have positive electron gain enthalpy. Its correct order is :
A He < Ne < Kr < Xe
B He < Xe < Kr < Ne
C Xe < Kr < Ne < He
D He < Kr < Xe < Ne
Correct Answer
Option B
Solution

.tg .tg  Electron  gain \begin{gathered}\text{ Electron } \\\text{ gain }\end{gathered} He\mathrm{He} Ne\mathrm{Ne} Ar\mathrm{Ar} Kr\mathrm{Kr} Xe\mathrm{Xe}  Enthalpyl /kJ mol1\begin{gathered}\text{ Enthalpyl } / \\\mathrm{kJ} ~\mathrm{mol}^{-1}\end{gathered} 48 116 96 96 77

Q64
Consider the following elements In,Tl,Al,Pb,Sn\mathrm{In}, \mathrm{Tl}, \mathrm{Al}, \mathrm{Pb}, \mathrm{Sn} and Ge . The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are
A +4 and +1
B +4 and +3
C +1 and +4
D +2 and +3
Correct Answer
Option B
Solution

Among Al,In,Tl,Ge,Sn,Pb\mathrm{Al}, \mathrm{In}, \mathrm{Tl}, \mathrm{Ge}, \mathrm{Sn}, \mathrm{Pb}, the metal having highest IE1\mathrm{IE}_1 is Ge and lowest IE1\mathrm{IE}_1 is In\mathrm{In}.

Most stable oxidation state of Ge is +4 and In is +3 .

Q65
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.

Q66
Given below are two statements : Statement (I) : The 4f4 \mathrm{f} and 5f5 \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.

Q67
Match List I with List II .tg .tg LIST I LIST II A. Melting Point [K][\mathrm{K}] I. T1>In>Ga>A1>B\mathrm{T} 1>\mathrm{In}>\mathrm{Ga}>\mathrm{A} 1>\mathrm{B} B. Ionic Radius [M+3/pm][\mathrm{M}^{+3} / \mathrm{pm}] II. B>T1>AlGa>In\mathrm{B}>\mathrm{T} 1>\mathrm{Al} \approx \mathrm{Ga}>\mathrm{In} C. ΔiH1[ kJ mol1]\Delta_{\mathrm{i}} \mathrm{H}_1[\mathrm{~kJ} \mathrm{~mol}^{-1}] III. T1>In>Al>Ga>B\mathrm{T} 1>\mathrm{In}>\mathrm{Al}>\mathrm{Ga}>\mathrm{B} D. Atomic Radius [pm] IV. B>A1>T1>In>Ga\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 :

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

Ionic radius

(M+3/pm):T>In>Ga>A>B(\mathrm{M}^{+3} / \mathrm{pm}): \mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}>\mathrm{A} \ell>\mathrm{B}
(ΔIEH)1[kJmol]:B>T>AGa>In\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

pm\mathrm{pm}

) :

T>In>A>Ga>B\mathrm{T} \ell>\mathrm{In}>\mathrm{A} \ell>\mathrm{Ga}>\mathrm{B}
Q68
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 1st 1^{\text{st }} ionization energy of boron is less than that of beryllium. D. The electronegativity of C is 2.5 in CH4\mathrm{CH}_4 and CCl4\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 e\mathrm{e}^{-}to a neutral gaseous atom is not always exothermic it may be exothermic or endothermic.

 (C) BeB1 s22s21s22s22p1\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 e\mathrm{e}^{-}as compared to B . due to partially positive charge zeff ,ENz_{\text{eff }} \uparrow, \mathrm{EN} \uparrow So, EN of CCCl4>CH4\mathrm{C} \Rightarrow \mathrm{CCl}_4>\mathrm{CH}_4 (E) Cs is most electropositive.

Q69
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,Li,
NaNa

and

KK

forms following oxides

4Li+O22Li2OLithiummonoxide4Li + {O_2}\,\,\overset{\,}\longrightarrow \,\,\mathop {2L{i_2}O\,\,\,\,\,\,\,\,\,\,}\limits_{Lithium\,\,\,\,monoxide} \,
2Na+O2575KNa2O2Sodiumperoxide2Na + {O_2}\,\,\overset{{575\,\,K}}\longrightarrow \,\,\mathop {N{a_2}{O_2}\,\,\,\,\,\,\,\,\,\,\,}\limits_{Sodium\,\,\,peroxide}
K+O2KO2PotassiumsuperoxiseK + {O_2}\,\,\overset{\,}\longrightarrow \,\,\mathop {K{O_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\limits_{Potassium\,\,\,\sup eroxise}
Q70
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.

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