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NEET • CHEMISTRY • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Classification of elements and periodicity in properties practice · Mixed difficulty

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Showing 361-390 of 407 questions

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361 Easy
361 of 407
Remembering +4 | -1
Which of the following is a transition element?
View answer and explanation
Correct option: C
NiNi
Option C (NiNi) is correct because Nickel is a d-block element and exhibits properties characteristic of transition elements.
362 Easy
362 of 407
Analyzing +4 | -1
The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine is:
View answer and explanation
Correct option: C
O>F>N>C
Option C (O>F>N>C) is correct because after the first ionization, oxygen attains a more stable half-filled configuration, requiring more energy for the second ionization.
363 Easy
363 of 407
Analyzing +4 | -1
Be and Al exhibit diagonal relationship. Which of the following statement about them is/are not true?
I. Both react with HCl to liberate H2{H_2}
II.They are made passive by HNO3HN{O_3}
III.Their carbides given acetylene on treatment with water
IV.Their oxides are amphoteric
View answer and explanation
Correct option: D
(iii) only
Option D ((iii) only) is correct because Be and Al carbides produce methane, not acetylene, upon hydrolysis.
364 Easy
364 of 407
Application +4 | -1
The compound of vanadium has magnetic moment of 1.73 BM. The vanadium chloride has the formula:
View answer and explanation
Correct option: C
VCl4VC{l_4}
The correct answer is C (VCl4VC{l_4}) because a magnetic moment of 1.73 BM corresponds to one unpaired electron, indicating V has a +4 oxidation state in VCl4VC{l_4}.
365 Easy
365 of 407
Analyzing +4 | -1
Assertion(A) Cu2+{Cu}^{2+} is diamagnetic.

Reason(R) All the orbitals present in Cu2+{Cu}^{2+} ion are doubly occupied.
View answer and explanation
Correct option: D
If both (A) and (R) are incorrect.
Option D is correct because both the assertion and the reason are incorrect. Cu2+{Cu}^{2+} is paramagnetic due to unpaired electrons, and not all orbitals are doubly occupied.
366 Medium
366 of 407
Understanding +4 | -1
A pp -block element in which last electron enters into ss-orbitals of valence shell instead of pp -orbital is:
View answer and explanation
Correct option: D
He\mathrm{He}
The correct answer is D (He) because Helium is a p-block element where the last electron enters the s-orbital instead of the p-orbital.
367 Easy
367 of 407
Remembering +4 | -1
Which of the following is amphoteric oxide
View answer and explanation
Correct option: B
Al2O3A{l_2}{O_3}
Option B (Al2O3A{l_2}{O_3}) is correct because amphoteric oxides can react with both acids and bases.
368 Hard
368 of 407
Understanding +4 | -1
The correct order of electron affinity is
View answer and explanation
Correct option: C
Al<N<O<ClAl < N < O < Cl
Option C (Al<N<O<ClAl < N < O < Cl) is correct because electron affinity generally increases across a period and decreases down a group, with exceptions due to electronic configurations.
369 Easy
369 of 407
Remembering +4 | -1
The charges on the colloidal CdSCdS sol and TiO2Ti{O_2} sol are, respectively
View answer and explanation
Correct option: D
negative and positive
Option D (negative and positive) is correct because metal sulfides like CdS typically form negatively charged sols, while metal oxides like TiO2Ti{O_2} usually form positively charged sols.
370 Medium
370 of 407
Understanding +4 | -1
The correct order of ionic radii is
View answer and explanation
Correct option: B
Al+>Al2+>Al3+A{l^ + } > A{l^{2 + }} > A{l^{3 + }}
Option B (Al+>Al2+>Al3+A{l^ + } > A{l^{2 + }} > A{l^{3 + }}) is correct because as the positive charge on the aluminum ion increases, the ionic radius decreases due to an increase in effective nuclear charge.
371 Easy
371 of 407
Remembering +4 | -1
Element with atomic number 35 belongs to
View answer and explanation
Correct option: C
17th group
Option C (17th group) is correct because an element with atomic number 35 has 7 valence electrons, placing it in the 17th group (halogens).
372 Medium
372 of 407
Understanding +4 | -1
Consider the following statements:

(i) The discovery of inert gases later on did not disturb Mendeleev's arrangement.

(ii) In the present periodic table, periodicity in the properties of elements is related to the periodicity in their electronic configurations.

Which of these statement(s) is/are correct?
View answer and explanation
Correct option: C
Both (i) and (ii)
Option C (Both (i) and (ii)) is correct because the discovery of inert gases fit into Mendeleev's table without disrupting it, and the modern periodic table's properties are linked to electronic configurations.
373 Easy
373 of 407
Remembering +4 | -1
Atomic radii of fluorine and neon in angstrom unit are respectively given by:
View answer and explanation
Correct option: A
0.72,1.600.72,\mathrm{ }1.60
Option A (0.72, 1.600.72,\, 1.60) is correct because fluorine is a smaller atom with a covalent radius, while neon is larger with a van der Waals radius.
374 Easy
374 of 407
Understanding +4 | -1
An element whose IUPAC name is ununtrium (Uut), belongs to
View answer and explanation
Correct option: B
P – block
Option B (P – block) is correct because Ununtrium (Uut), now known as Moscovium (Mc), belongs to group 15, which is in the p-block.
375 Easy
375 of 407
Understanding +4 | -1
An element has the configuration 1s2,2s2,2p6,3s2,3p2. 1{s^2},2{s^2},2{p^6},3{s^2},3{p^2}.\, To which block in the long form of the periodic table, does this belong?
View answer and explanation
Correct option: B
p-block
Option B (p-block) is correct because the last electron enters the p-orbital in the given electronic configuration.
376 Medium
376 of 407
Application +4 | -1
Aqueous solutions of two compounds M−O−HM-{O}-{H} and M′−O−HM^{\prime}-{O}-{H} have been prepared in two different beakers. If the electronegativity of M=3.5,M′=1.72M=3.5, M^{\prime}=1.72, O=3.0{O}=3.0 and H=2.1{H}=2.1, then the solutions respectively are
View answer and explanation
Correct option: B
acidic, basic
Option B (acidic, basic) is correct because the electronegativity difference between M and O is higher, leading to acidic properties, while a lower electronegativity difference leads to basic properties.
377 Hard
377 of 407
Understanding +4 | -1
In the following reactions, ZnOZnO is respectively acting as a/an
i) ZnO+Na2O→Na2ZnO2ZnO + N{a_2}O \to N{a_2}Zn{O_2} ii) ZnO+CO2→ZnCO3ZnO + C{O_2} \to ZnC{O_3}
View answer and explanation
Correct option: D
Acid and base
Option B (Na+ > Li+ > Mg2+ > Be2+) is correct because ionic radii decrease with increasing nuclear charge for isoelectronic species and increase down a group.
378 Medium
378 of 407
Application +4 | -1
The amount of energy released when 10610^6 atoms of iodine in vapour state is converted into I−{I}^{-} ion is 4.8×10−13 J4.8 \times 10^{-13} \mathrm{~J}. What is the electron affinity of iodine in kJ/mole?\mathrm{kJ} / \mathrm{mole}?
View answer and explanation
Correct option: B
289 kJ289 \mathrm{~kJ}
Option B (289 kJ/mole) is correct because it accurately converts the energy released per atom to energy released per mole, which represents the electron affinity.
379 Medium
379 of 407
Understanding +4 | -1
The electronic configuration of the element with maximum electron affinity is
View answer and explanation
Correct option: C
1s2,2s2,2p6,3s2,3p51{s^2},2{s^2},2{p^6},3{s^2},3{p^5}
Option C (1s2,2s2,2p6,3s2,3p51{s^2},2{s^2},2{p^6},3{s^2},3{p^5}) is correct because elements with electronic configurations close to noble gases (like halogens) have high electron affinity.
380 Medium
380 of 407
Application +4 | -1
The values of electronegativity of atom A and B are 1.20 and 4.0 respectively. The percentage of ionic character of A-B bond is
View answer and explanation
Correct option: D
0.736
Option D (0.736) is correct because the percentage of ionic character can be estimated using the Hannay-Smith equation or a similar empirical relationship based on the electronegativity difference.
381 Easy
381 of 407
Application +4 | -1
If each orbital can take maximum of three electrons, the number of elements in the third period of the periodic table will be :
View answer and explanation
Correct option: A
12
Option A (12) is correct because if each orbital can take a maximum of three electrons, the third period will have 12 elements.
382 Medium
382 of 407
Remembering +4 | -1
In Mendeleev's periodic table, iodine with lower atomic weight than that of tellurium was placed alongwith
View answer and explanation
Correct option: B
Fluorine, chlorine and bromine
Option B (Fluorine, chlorine and bromine) is correct because Iodine was placed with the halogen family (F, Cl, Br) due to its similar chemical properties, despite its atomic weight being less than Tellurium.
383 Easy
383 of 407
Understanding +4 | -1
In the Periodic Table, going down in the fluorine group
View answer and explanation
Correct option: B
Ionic radii will increases
Option B (Ionic radii will increases) is correct because as you go down a group, the number of electron shells increases, leading to larger ionic radii.
384 Easy
384 of 407
Understanding +4 | -1
The smallest among the following ions is
View answer and explanation
Correct option: D
Al3+\mathrm{A}{\mathrm{l}^{3 + }}
Option D (Al3+\mathrm{A}{\mathrm{l}^{3 + }}) is correct because, among the given isoelectronic species, it has the highest nuclear charge, leading to the smallest ionic radius.
385 Easy
385 of 407
Analyzing +4 | -1
The least stable ion among the following is
View answer and explanation
Correct option: B
Be−B{e^ - }
Option B (Be−Be^−) is correct because adding an electron to Be, which already has a stable, filled s-orbital configuration (2s22s^2), requires significant energy.
386 Easy
386 of 407
Application +4 | -1
The formula of the binary compound formed by the element with atomic number 120 and sulphur is:
View answer and explanation
Correct option: A
XSXS
Option A (XSXS) is correct because element 120 belongs to the alkaline earth metals (Group 2), forming a +2 ion, while sulfur forms a -2 ion, resulting in a 1:1 ratio.
387 Hard
387 of 407
Understanding +4 | -1
The ionisation potential of 'X+{X^ + }' ion is equal to
View answer and explanation
Correct option: D
The EA of X2+{X^{2 + }} ion
The correct answer is the EA of X2+{X^{2 + }} ion because the ionization potential of X+{X^ + } is the energy required to remove an electron from X+{X^ + }, which is the same as the electron affinity of X2+{X^{2 + }}.
388 Medium
388 of 407
Understanding +4 | -1
Zinc is not considered as a transition metal because
View answer and explanation
Correct option: C
It has no unpaired electrons
The correct answer is C (It has no unpaired electrons) because Zinc is not considered a transition metal since it has a completely filled d-orbital (d10d^{10}) and thus no unpaired electrons in its common oxidation states.
389 Medium
389 of 407
Understanding +4 | -1
Chlorine atom tends to acquire the structure of:
View answer and explanation
Correct option: C
Ar\mathrm{Ar}
Option C (Ar\mathrm{Ar}) is correct because chlorine has 17 electrons and needs to gain one electron to achieve the stable electronic configuration of Argon (18 electrons).
390 Medium
390 of 407
Analyzing +4 | -1
Which of the following relation is correct?
View answer and explanation
Correct option: A
Ist  IE  of  C>Ist  IE  of  B{\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;C} > {\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;B}
Option A (Ist  IE  of  C>Ist  IE  of  B{\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;C} > {\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;B}) is correct because carbon has a higher effective nuclear charge and a more stable electronic configuration compared to boron, resulting in a higher first ionization energy.
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