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Periodic Table & Periodicity

JEE Chemistry · 270 questions · Page 11 of 27 · Click an option or "Show Solution" to reveal answer

Q101
Given below are two statements : Statement (I) : Both metals and non-metals exist in p and d-block elements. Statement (II) : Non-metals have higher ionisation enthalpy and higher electronegativity than the metals. In the light of the above statements, choose the most appropriate answer from the options given below :
A Both Statement I and Statement II are false
B Both Statement I and Statement II are true
C Statement I is false but Statement II is true
D Statement I is true but Statement II is false
Correct Answer
Option C
Solution

To answer this question, let's investigate both statements in detail.

Statement (I): Both metals and non-metals exist in p and d-block elements.

The periodic table is divided into blocks based on the electron configuration of the elements.

These blocks are labeled s, p, d, and f.

The p-block contains a mix of metals, non-metals, and metalloids.

For example, the p-block contains nonmetals such as oxygen (O) and nitrogen (N), as well as metals like aluminum (Al) and lead (Pb).

The d-block, also known as the transition metals block, primarily consists of metals.

However, the d-block does not typically contain elements that are traditionally classified as non-metals.

The elements in the d-block have a wide range of properties but are generally characterized by their metallic traits such as conductivity and malleability.

Therefore, Statement I is not entirely true because while both metals and nonmetals exist in the p-block of the periodic table, the d-block is principally composed of metals.

Statement (II): Non-metals have higher ionisation enthalpy and higher electronegativity than the metals.

Ionization enthalpy (or ionization energy) is the energy required to remove an electron from a gaseous atom or ion.

Non-metals generally have higher ionization energies compared to metals because non-metals have more tightly bound electrons to their nucleus.

This is partly due to the fact that non-metals tend to have higher electronegativities and smaller atomic radii, meaning that their outer electrons are closer to the nucleus and more strongly attracted to it.

Electronegativity is a measure of an atom's ability to attract and bond with electrons.

Non-metals have higher electronegativities typically because they are more eager to gain electrons to achieve a full valence shell, reflecting their position on the right side of the periodic table.

Metals, on the other hand, are more inclined to lose electrons and form positive ions, indicating their lower electronegativities.

Hence, Statement II is true as non-metals indeed have higher ionisation enthalpy and higher electronegativity than metals.

Given the analysis: Statement I is not true.

Statement II is true.

Thus, the most appropriate answer is: Option C - Statement I is false but Statement II is true.

Q102
The correct sequence of electron gain enthalpy of the elements listed below is A. Ar B. Br C. F D. S Choose the most appropriate from the options given below:
A A>D>C>B\mathrm{A>D>C>B}
B A>D>B>C\mathrm{A}>\mathrm{D}>\mathrm{B}>\mathrm{C}
C D>C>B>A\mathrm{D}>\mathrm{C}>\mathrm{B}>\mathrm{A}
D C>B>D>A\mathrm{C}>\mathrm{B}>\mathrm{D}>\mathrm{A}
Correct Answer
Option B
Solution
 Element ΔegH(kJ/mol)F−333 S−200Br−325Ar+96\begin{array}{ll} \text{ Element } & \Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{kJ} / \mathrm{mol}) \\ \mathrm{F} & -333 \\ \mathrm{~S} & -200 \\ \mathrm{Br} & -325 \\ \mathrm{Ar} & +96 \end{array}
Q103
Given below are two statements : Statement I : The electronegativity of group 14 elements from Si\mathrm{Si} to Pb\mathrm{Pb}, gradually decreases. Statement II : Group 14 contains non-metallic, metallic, as well as metalloid elements. In the light of the above statements, choose the most appropriate 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 Both Statement I and Statement II are false
D Statement I is true but Statement II is false
Correct Answer
Option A
Solution
 Gr-14  EN C2.5Si1.8Ge1.8Sn1.8 Pb1.9\begin{array}{ll} \text{ Gr-14 } & \text{ EN } \\ \mathrm{C} & 2.5 \\ \mathrm{Si} & 1.8 \\ \mathrm{Ge} & 1.8 \\ \mathrm{Sn} & 1.8 \\ \mathrm{~Pb} & 1.9 \end{array}

The electronegativity values for elements from

Si\mathrm{Si}

to

Pb\mathrm{Pb}

are almost same. So Statement

I\mathrm{I}

is false.

Q104
Given below are two statements: Statement I : Fluorine has most negative electron gain enthalpy in its group. Statement II : Oxygen has least negative electron gain enthalpy in its group. In the light of the above statements, choose the most appropriate from the options given below
A Both Statement I and Statement II are true
B Both Statement I and Statement II are false
C Statement I is false but Statement II is true
D Statement I is true but Statement II is false
Correct Answer
Option C
Solution

Statement- 1 is false because chlorine has most negative electron gain enthalpy in its group.

Q105
The element having the highest first ionization enthalpy is
A C\mathrm{C}
B Al\mathrm{Al}
C Si\mathrm{Si}
D N\mathrm{N}
Correct Answer
Option D
Solution

$$\mathrm{Al}

Q106
Anomalous behavior of oxygen is due to its
A large size and low electronegativity
B small size and high electronegativity
C small size and low electronegativity
D large size and high electronegativity
Correct Answer
Option B
Solution

Fact Based.

Q107
Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point :
A LiCl
B NaCl
C KCl
D RbCl
Correct Answer
Option B
Solution
LiClLiCl

has partly covalent character.

Other halides are ionic in nature.

Lattice energy decreases with increases of ionic radius of cation, anion being the same.

Larger is the lattice energy, the higher will be

m.m.
pt.pt.

hence

NaClNaCl

will have highest lattice energy.

Q108
Given below are two statements : Statement (I) : The oxidation state of an element in a particular compound is the charge acquired by its atom on the basis of electron gain enthalpy consideration from other atoms in the molecule. Statement (II) : pπ−pπ\mathrm{p} \pi-\mathrm{p} \pi bond formation is more prevalent in second period elements over other periods. 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 Both Statement I and Statement II are correct
C Statement I is incorrect but Statement II is correct
D Both Statement I and Statement II are incorrect
Correct Answer
Option C
Solution

To evaluate the given statements, we need to consider the concepts of oxidation states and bonding tendencies across different periods in the periodic table.

Statement (I) describes the oxidation state of an element in a compound as the hypothetical charge an atom would have if all bonds to atoms of different elements were 100% ionic.

This consideration does involve looking at how electrons are shared or transferred between atoms, but it's not solely based on electron gain enthalpy.

Rather, the oxidation state is determined by a set of rules which include, but are not limited to, electronegativity differences and known oxidation states of elements.

Therefore, the statement is somewhat misleading in its specificity about electron gain enthalpy being the sole consideration.

The electron gain enthalpy is a measure of the energy change when an electron is added to a neutral atom in the gaseous state to form a negative ion, which indirectly influences the discussion of oxidation states but is not the direct basis for their determination.

Statement (II) points out that

pπ−pπ\mathrm{p}\pi-\mathrm{p}\pi

bond formation is more prevalent among second period elements compared to elements in other periods.

This is correct and can be attributed to the smaller size of the second period elements (such as carbon, nitrogen, and oxygen), which allows their p orbitals to overlap more effectively for

pπ−pπ\mathrm{p}\pi-\mathrm{p}\pi

bonding.

As we move down the periods, the atomic size increases, which leads to less effective overlapping of p orbitals for

pπ−pπ\mathrm{p}\pi-\mathrm{p}\pi

bonding due to increased distance between the valence electrons and nucleus. This makes

pπ−pπ\mathrm{p}\pi-\mathrm{p}\pi

bonds less common and less strong in elements of higher periods compared to those in the second period.

Therefore, the most accurate assessment of the statements given the explanations above is: Statement I is not entirely accurate because it overemphasizes electron gain enthalpy in determining oxidation states.

Statement II is correct as it accurately describes the prevalence of

pπ−pπ\mathrm{p}\pi-\mathrm{p}\pi

bonding among second period elements.

Hence, the correct option is: Option C: Statement I is incorrect but Statement II is correct.

Q109
The correct order of first ionization enthalpy values of the following elements is : (A) O (B) N (C) Be (D) F (E) B Choose the correct answer from the options given below :
A A<B<D<C<E\mathrm{A}<\mathrm{B}<\mathrm{D}<\mathrm{C}<\mathrm{E}
B C<E<A<B<D\mathrm{C}<\mathrm{E}<\mathrm{A}<\mathrm{B}<\mathrm{D}
C E<C<A<B<D\mathrm{E}<\mathrm{C}<\mathrm{A}<\mathrm{B}<\mathrm{D}
D B<D<C<E<A\mathrm{B}<\mathrm{D}<\mathrm{C}<\mathrm{E}<\mathrm{A}
Correct Answer
Option C
Solution

Correct ionization enthalpy order :

\quad

B or E

Q110
The statement(s) that are correct about the species O2−,F−,Na+\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+} and Mg2+\mathrm{Mg}^{2+}. (A) All are isoelectronic (B) All have the same nuclear charge (C) O2−\mathrm{O}^{2-} has the largest ionic radii (D) Mg2+\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:

O2−\mathrm{O}^{2-}

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

F−\mathrm{F}^{-}

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

Na+\mathrm{Na}^{+}

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

Mg2+\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:

O2−\mathrm{O}^{2-}

: 8 protons

F−\mathrm{F}^{-}

: 9 protons

Na+\mathrm{Na}^{+}

: 11 protons

Mg2+\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)

O2−\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

O2−\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)

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

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

Mg2+\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.

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