Periodic Table & Periodicity — Page 6 | NEET Chemistry

Q51

Bond dissociation energy of "E-H" bond of the "

\mathrm{H}_{2} \mathrm{E}

" hydrides of group 16 elements (given below), follows order. A.

\mathrm{O}

B.

\mathrm{S}

C. Se D.

\mathrm{Te}

Choose the correct from the options given below:

A

\mathrm{A}>\mathrm{B}>\mathrm{C}>\mathrm{D}

B

\mathrm{D}>\mathrm{C}>\mathrm{B}>\mathrm{A}

C

\mathrm{B}>\mathrm{A}>\mathrm{C}>\mathrm{D}

D

\mathrm{A}>\mathrm{B}>\mathrm{D}>\mathrm{C}

Correct Answer

Option A

Solution

The correct order of bond strength is

\mathrm{H_2O > H_2S > H_2Se > H_2Te}

Q52

The statement(s) that are correct about the species

\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}

and

\mathrm{Mg}^{2+}

. (A) All are isoelectronic (B) All have the same nuclear charge (C)

\mathrm{O}^{2-}

has the largest ionic radii (D)

\mathrm{Mg}^{2+}

has the smallest ionic radii Choose the most appropriate answer from the options given below :

A (C) and (D) only

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

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

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

Correct Answer

Option B

Solution

To answer this question, let's address each statement individually.

(A) All are isoelectronic An isoelectronic species is a group of ions or atoms which have the same number of electrons.

The electron configuration for each species is as follows:

\mathrm{O}^{2-}

: Oxygen has 8 protons and normally 8 electrons. Gaining 2 electrons gives it a total of 10 electrons.

\mathrm{F}^{-}

: Fluorine has 9 protons and normally 9 electrons. Gaining 1 electron gives it a total of 10 electrons.

\mathrm{Na}^{+}

: Sodium has 11 protons and normally 11 electrons. Losing 1 electron leaves it with 10 electrons.

\mathrm{Mg}^{2+}

: Magnesium has 12 protons and normally 12 electrons.

Losing 2 electrons leaves it with 10 electrons.

All of these ions have the same number of electrons (10), making them isoelectronic.

Therefore, statement (A) is correct.

(B) All have the same nuclear charge The nuclear charge refers to the total charge within the nucleus, which is determined by the number of protons.

The nuclear charges are:

\mathrm{O}^{2-}

: 8 protons

\mathrm{F}^{-}

: 9 protons

\mathrm{Na}^{+}

: 11 protons

\mathrm{Mg}^{2+}

: 12 protons Since the number of protons varies among these species, they do not have the same nuclear charge.

Therefore, statement (B) is incorrect.

(C)

\mathrm{O}^{2-}

has the largest ionic radii In a series of isoelectronic ions, the ionic radius decreases with increasing nuclear charge because the greater the nuclear charge, the more strongly the electrons are pulled towards the nucleus, reducing the size of the ion.

As

\mathrm{O}^{2-}

has the lowest nuclear charge among the given ions, it will have the largest ionic radius.

Therefore, statement (C) is correct.

(D)

\mathrm{Mg}^{2+}

has the smallest ionic radii Following the same logic as above,

\mathrm{Mg}^{2+}

, having the highest nuclear charge among the given isoelectronic species, will have the smallest ionic radius because its electrons are held most tightly by the nucleus.

Thus, statement (D) is correct.

Given the evaluations above, the most appropriate answer is: Option B (A), (C) and (D) only.

Q53

Which of the following electronegativity order is incorrect?

A

\mathrm{S}<\mathrm{Cl}<\mathrm{O}<\mathrm{F}

B

\mathrm{Al}<\mathrm{Si}<\mathrm{C}<\mathrm{N}

C

\mathrm{Al}<\mathrm{Mg}<\mathrm{B}<\mathrm{N}

D

\mathrm{Mg}<\mathrm{Be}<\mathrm{B}<\mathrm{N}

Correct Answer

Option C

Solution

Let’s compare the known electronegativities (Pauling scale) of the elements in each option: F ~ 3.98 O ~ 3.44 N ~ 3.04 Cl ~ 3.16 C ~ 2.55 B ~ 2.04 S ~ 2.58 (sometimes listed as 2.5–2.58) Si ~ 1.90 Al ~ 1.61 Be ~ 1.57 Mg ~ 1.31 Now, check each statement: Option A: $\mathrm{S} S (2.58) < Cl (3.16) < O (3.44) < F (3.98) This order is correct. Option B:$ \mathrm{Al} Al (1.61) < Si (1.90) < C (2.55) < N (3.04) This order is correct.

Option C: $\mathrm{Al} Actual electronegativities are Al (1.61) and Mg (1.31). The statement says “Al < Mg,” which would mean 1.61 < 1.31, which is wrong. Therefore, Option C is incorrect. Option D:$ \mathrm{Mg} Mg (1.31) < Be (1.57) < B (2.04) < N (3.04) This order is correct.

Answer: Option C is the incorrect order.

Q54

s-block element which cannot be qualitatively confirmed by the flame test is :

A Li

B Na

C Rb

D Be

Correct Answer

Option D

Solution

Beryllium does not give flame test because of its small size and high ionization energy the energy of flame is not sufficient to excite the electrons to higher energy level.

Q55

Which one of the following alkaline earth metal ions has the highest ionic mobility in its aqueous solution?

A Be2+

B Mg2+

C Ca2+

D Sr2+

Correct Answer

Option D

Solution

In aqueous solution, the ionic mobility is inversely proportional to the charge density on the ion.

Hence Sr2+ has highest ionic mobility in water.

Q56

Match the following compounds (Column-I) with their uses (Column-II) : .tg .tg S.No. Column-I S.No. Column-II (I) Ca(OH)2 (A) casts of statues (II) NaCl (B) white wash (III) CaSO4,

{1 \over 2}

H2O (C) antacid (IV) CaCO3 (D) washing soda preparation

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

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

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

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

Correct Answer

Option B

Solution

(I) Ca(OH)2 is used in white wash due to its disinfectant nature.

(II) NaCl is used in preparation of washing soda(Na2CO3).

(1) 2NH3 + H2O + CO2 $\to$ (NH4 )2CO3 (2) (NH4 )2CO3 + H2O + CO2 $\to$ 2NH4HCO3 (3) NH4HCO3 + NaCl $\to$ NH4Cl + NaHCO3 (4) 2NaHCO3 $\to$ Na2CO3 + CO2 + H2O (III) CaSO4.

{1 \over 2}

H2O (Plaster of Paris) Used for making casts of statues. (IV) CaCO3 is used as an Antacid.

Q57

Identify the correct order of standard enthalpy of formation of sodium halides :

A

\mathrm{NaI}<\mathrm{NaBr}<\mathrm{NaCl}<\mathrm{NaF}

B

\mathrm{NaI}<\mathrm{NaBr}<\mathrm{NaF}<\mathrm{NaCl}

C

\mathrm{NaF}<\mathrm{NaCl}<\mathrm{NaBr}<\mathrm{NaI}

D

\mathrm{NaCl}<\mathrm{NaF}<\mathrm{NaBr}<\mathrm{NaI}

Correct Answer

Option A

Solution

Lattice energy is directly proportional to the charges of the ions and inversely proportional to the sum of their radii: Lattice energy ∝ (Q₁Q₂) / (r₁ + r₂) As we move down the group, the size of the halide ions increases, resulting in a decrease in lattice energy.

This leads to a decrease in the exothermicity of the lattice energy release during the formation of sodium halides.

Thus, the correct order of standard enthalpy of formation of sodium halides, considering both lattice energy and electron affinity, is:

\mathrm{NaF}>\mathrm{NaCl}>\mathrm{NaBr}>\mathrm{NaI}

Q58

The five successive ionization enthalpies of an element are 800, 2427, 3658, 25024 and 32824 kJ mol–1. The number of valence electrons in the element is :

A 2

B 3

C 4

D 5

Correct Answer

Option B

Solution

There is a sudden jump after 3rd I.E. due to attainment of noble gas configuration.

So, the number of valence electrons in this element are 3.

Q59

The correct orders among the following are Atomic radius :

\mathrm{B} Electronegativity :

\mathrm{Al} Density :

\mathrm{Tl} 1st Ionisation Energy :

\mathrm{In} Choose the correct answer from the options given below:

A B and D Only

B A and B Only

C C and D Only

D A and C Only

Correct Answer

Option A

Solution

.tg .tg B Al Ga In Tl Atomic radius (pm) 88 143 135 167 170 Electronegativity 2 1.5 1.6 1.7 1.8 Density (g/cm $^3$ ) 2.35 2.7 5.9 7.31 11.85 Ionisation Energy (kJ/mol) 801 577 579 558 589

\begin{array}{lc} \text{ Radius Order } & \mathrm{T} \ell>\mathrm{In}>\mathrm{A} \ell>\mathrm{Ga}>\mathrm{B} \\ \text{ EN Order } & \mathrm{B}>\mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}>\mathrm{Al} \\ \text{ Density Order } & \mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}>\mathrm{A} \ell>\mathrm{B} \\ \mathrm{IE}_1 \text{ Order } & \mathrm{B}>\mathrm{T} \ell>\mathrm{Ga}>\mathrm{A} \ell>\mathrm{In} \end{array}

Q60

The substance not likely to contain CaCO3 is :

A calcined gypsum

B sea shells

C dolomite

D a marble statue

Correct Answer

Option A

Solution

Gypsum is

\,\,CaS{O_4}.2{H_2}O