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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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331 Medium
331 of 407
Application +4 | -1
If electronegativity of xx be 3.2 and that of yy be 2.2 , the percentage ionic character of xyx y is -
View answer and explanation
Correct option: A
19.5
Option A (19.5) is correct because the percentage ionic character can be estimated using the formula: % ionic character = 16|XA - XB| + 3.5|XA - XB|².
332 Medium
332 of 407
Application +4 | -1
Predict the correct formula of compounds, by the following pair of elements like aluminium and sulphur with the help of oxidation states,
View answer and explanation
Correct option: B
Al2S3\mathrm{Al}_2 \mathrm{S}_3
Option B (Al2S3\mathrm{Al}_2 \mathrm{S}_3) is correct because it represents the neutral compound formed by balancing the charges of Al3+^{3+} and S2−^{2-}.
333 Medium
333 of 407
Understanding +4 | -1
Positive electron gain enthalpy is highest for
View answer and explanation
Correct option: C
Ne
Option A (Xe) has the highest positive electron gain enthalpy because noble gases have stable, filled electron shells and resist adding an extra electron.
334 Easy
334 of 407
Remembering +4 | -1
Differentiating electron in inner transition elements enters the………….. orbital.
View answer and explanation
Correct option: D
f
Option D is correct because the differentiating electron in inner transition elements (lanthanides and actinides) enters the f-orbital.
335 Medium
335 of 407
Understanding +4 | -1
In which one of the following pairs the radius of the second species is greater than that of the first?
View answer and explanation
Correct option: B
O2−,N3−{\mathrm{O}^{2 - }},{\mathrm{N}^{3 - }}
Option B (O2−,N3−{\mathrm{O}^{2 - }},{\mathrm{N}^{3 - }}) is correct because for isoelectronic species, the species with fewer protons has a larger radius.
336 Medium
336 of 407
Application +4 | -1
First and second ionisation energies of magnesium are 7.646 and 15.035eV15.035 \mathrm{eV} respectively. The amount of energy in kJ\mathrm{kJ} needed to convert all the atoms of magnesium into Mg2+\mathrm{Mg}^{2+} ions present in 12mg12 \mathrm{mg} of magnesium vapours is :

[Given : 1eV=96.5 kJ mol−11 \mathrm{eV}=96.5 \mathrm{~kJ} \mathrm{~mol}^{-1} ]
View answer and explanation
Correct option: A
1.1
Option A (1.1) is correct because it calculates the total ionization energy for converting all Mg atoms in 12mg to Mg²⁺ ions, considering both first and second ionization energies.
337 Medium
337 of 407
Remembering +4 | -1
An element with atomic number 59 belongs to
View answer and explanation
Correct option: A
ff-block
Option A (ff-block) is correct because an element with atomic number 59 is Praseodymium, which is a lanthanide and belongs to the f-block.
338 Easy
338 of 407
Understanding +4 | -1
All elements in the third period have
View answer and explanation
Correct option: B
Three complete subshells
The correct answer is B (Three complete subshells) because elements in the third period have electrons filling the 3s, 3p, and 3d subshells (though 3d is not filled for all elements).
339 Easy
339 of 407
Remembering +4 | -1
Diagonal relationship is for
View answer and explanation
Correct option: D
B-Si
Option D (B-Si) is correct because Boron and Silicon exhibit similar properties due to their diagonal relationship.
340 Medium
340 of 407
Understanding +4 | -1
The element whose electronic configuration is 1s2,2s2,2p6,3s21{s^2},2{s^2},2{p^6},3{s^2} is:
View answer and explanation
Correct option: B
Metal
Option B is correct because the element with the electronic configuration 1s22s22p63s21s^22s^22p^63s^2 has 2 valence electrons in the 3s orbital, indicating it is a metal (specifically Magnesium).
341 Medium
341 of 407
Application +4 | -1
Predict the period, group and block of an element having electronic configuration 1s22s22p63s23p41 s^2 2 s^2 2 p^6 3 s^2 3 p^4.
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Correct option: A
Period =3rd,=3^{\text {rd}}, block =p,= p, group =16=16
Option A (Period = 3rd, block = p, group = 16) is correct because the highest principal quantum number is 3 (period 3), the last electron enters the p-orbital (p-block), and the number of valence electrons is 6 (group 16).
342 Hard
342 of 407
Analyzing +4 | -1
Out of Ne,  Na+,  O,  NaNe,\,\,N{a^ + },\,\,O,\,\,Na and N,N, the species having maximum and minimum ionization energies respectively are
View answer and explanation
Correct option: C
Na+,  NaN{a^ + },\,\,Na
Option C (Na+, NaN{a^ + },\,Na) is correct because Na+N{a^ + } has a noble gas configuration and a high nuclear charge, making it difficult to remove an electron, while Na has a single valence electron that is easily removed.
343 Easy
343 of 407
Remembering +4 | -1
The starting element of fourth period is
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Correct option: A
K
Option A (K) is correct because Potassium (K) is the first element in the fourth period.
344 Easy
344 of 407
Remembering +4 | -1
Which one these basic
View answer and explanation
Correct option: D
Na2ON{a_2}O
Option D (Na2ON{a_2}O) is correct because oxides of metals are generally basic.
345 Medium
345 of 407
Understanding +4 | -1
The element with atomic number 35 belongs to :
View answer and explanation
Correct option: C
p-block
Option C (p-block) is correct because elements with atomic numbers ending in electronic configurations ns²np⁵ belong to the p-block, and element 35 (Bromine) is in group 17.
346 Medium
346 of 407
Understanding +4 | -1
Na2O,  MgO,Al2O3  and  SiO2  \mathrm{N}{\mathrm{a}_2}\mathrm{O},\mathrm{\;MgO},\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}\mathrm{\;and\;Si}{\mathrm{O}_2}\mathrm{\;} have heat of formation equal to −416,−602,−1676  - 416, - 602, - 1676\; and −911  kJ  mol−1- 911\mathrm{\;kJ\;mo}{\mathrm{l}^{ - 1}} respectively. The most stable oxide is
View answer and explanation
Correct option: B
MgO
Option B (MgO) is correct because the most stable oxide has the most negative heat of formation per mole of oxygen. After normalizing to per mole of oxygen, MgO has the most negative value.
347 Hard
347 of 407
Understanding +4 | -1
The basic character of MgO,BaO,Na2O\mathrm{MgO}, \mathrm{BaO}, \mathrm{Na}_2 \mathrm{O} and FeO\mathrm{FeO} follow the order
View answer and explanation
Correct option: C
FeO<MgO<BaO<Na2O\mathrm{FeO} < \mathrm{MgO} < \mathrm{BaO} < \mathrm{Na}_2 \mathrm{O}
Option C (FeO < MgO < BaO < Na2O) is correct because basic character increases with decreasing electronegativity of the metal and increasing electropositivity.
348 Easy
348 of 407
Analyzing +4 | -1
Select the incorrect order of size of ion/atoms
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Correct option: D
None is incorrect
Option C (Na) is correct because among the given species, Na is the only neutral atom and has the highest nuclear charge compared to the isoelectronic ions, thus the smallest size.
349 Medium
349 of 407
Understanding +4 | -1
What is the atomic number of the next halogen if discovered?
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Correct option: B
117
The correct answer is 117 (B) because halogens are in group 17, and the next halogen after element 117 would be element 117.
350 Medium
350 of 407
Understanding +4 | -1
The correct order of ionisation energy for comparing carbon, nitrogen and oxygen atom is
View answer and explanation
Correct option: C
C < N > O
Option C (C < N > O) is correct because nitrogen has a higher ionization energy than both carbon and oxygen due to its half-filled p-orbital configuration, which provides extra stability.
351 Easy
351 of 407
Understanding +4 | -1
If the differentiating electron enters (n-1)d – subshell, the element is
View answer and explanation
Correct option: D
A transition element
Option D is correct because transition elements are defined by having their differentiating electron enter the (n-1)d subshell.
352 Medium
352 of 407
Understanding +4 | -1
Given below are two statements:
Statement I CeO2Ce{O_2} can be used for oxidation of aldehydes and ketones.
Statement II Aqueous solution of EuSO4EuS{O_4} is a strong reducing agent.
In the light of the above statements, choose the correct answer from the options given below.
View answer and explanation
Correct option: A
Both statement I and statement II are true.
Option A (Both statement I and statement II are true) is correct because CeO2CeO_2 is used as an oxidizing agent and EuSO4EuSO_4 is a strong reducing agent.
353 Medium
353 of 407
Remembering +4 | -1
A plot of v\sqrt{v} (frequency) against atomic number (z)(z) gave a :
View answer and explanation
Correct option: A
Straight line
The correct answer is A (Straight line) because Moseley's law states that the square root of the frequency of emitted X-rays is proportional to the atomic number.
354 Medium
354 of 407
Understanding +4 | -1
The tenth elements in the Periodic Table resembles with the
View answer and explanation
Correct option: B
Second period
Option B (Second period) is correct because the tenth element (Neon) is in the second period and exhibits similar properties to other elements in the same period.
355 Medium
355 of 407
Analyzing +4 | -1
Which of the following configuration has strongest metallic bonding?
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Correct option: B
[Ar]3d54s1[\mathrm{Ar}] 3 d^5 4 s^1
Option B is correct. Strongest metallic bonding occurs when there are more unpaired electrons available for metallic bonding. [Ar] 3d⁵ 4s¹ has 6 unpaired electrons.
356 Medium
356 of 407
Remembering +4 | -1
The energy released when a neutral gaseous atom takes up an electron is called:
View answer and explanation
Correct option: D
Electron affinity
Option D (Electron affinity) is correct because electron affinity is defined as the energy released when a neutral gaseous atom accepts an electron.
357 Easy
357 of 407
Analyzing +4 | -1
Which of the following is a false statement?
View answer and explanation
Correct option: C
Lithium is amphoteric
Option C (Lithium is amphoteric) is the false statement because Lithium is an alkali metal and forms basic oxides, not amphoteric ones.
358 Easy
358 of 407
Remembering +4 | -1
To which block of the periodic table does element with atomic number 55 belong
View answer and explanation
Correct option: A
s – block
Option A (s – block) is correct because element with atomic number 55 is Cesium, which is an alkali metal and belongs to the s-block.
359 Easy
359 of 407
Application +4 | -1
Set of isoelectronic species is
View answer and explanation
Correct option: C
N2,CO,CN−,O2+2{N}_2, {CO}, {CN}^{-}, {O}_2^{+2}
Option C (N2,CO,CN−,O2+2){N}_2, {CO}, {CN}^{-}, {O}_2^{+2}) is correct because all species have the same number of electrons (14).
360 Easy
360 of 407
Understanding +4 | -1
Variable oxidation state and degenerated orbital shows
View answer and explanation
Correct option: C
d-block elements
Option C is correct because d-block elements are known for exhibiting variable oxidation states and having degenerate d-orbitals.
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