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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 181-210 of 407 questions

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181 Easy
181 of 407
Remembering +4 | -1
Atomic radii of inert gases are expressed in terms of
View answer and explanation
Correct option: B
Vanderwaal’s radius
Option B (Vanderwaal’s radius) is correct because inert gases do not form covalent or metallic bonds, so their atomic radii are measured using van der Waals radii.
182 Easy
182 of 407
Remembering +4 | -1
Maximum oxidation state (+8) is exhibited by
View answer and explanation
Correct option: B
Ru and OS
Option B (Ru and OS) is correct because Ruthenium (Ru) and Osmium (Os) can exhibit a +8 oxidation state.
183 Easy
183 of 407
Application +4 | -1
If Aufbau rule is not followed, K-19 will be placed in
View answer and explanation
Correct option: C
d-block
Option D is correct because if the Aufbau principle is not followed, the electronic configuration of K-19 would be different, potentially placing it in the d-block due to filling of the (n-1)d orbitals before the ns orbital.
184 Easy
184 of 407
Understanding +4 | -1
Which of the following sequence regarding ionisation potential of coinage metal is correct:
View answer and explanation
Correct option: C
Cu>Ag<Au\mathrm{Cu} > Ag < Au
Option C (Cu>Ag<Au\mathrm{Cu} > Ag < Au) is correct because ionization potential generally decreases down a group, but gold's relativistic effects cause it to have a higher ionization potential than silver.
185 Easy
185 of 407
Understanding +4 | -1
Given below are two statement : One is labelled as Assertion A and the other is labelled as Reason R:
Assertion A: Size of Bk3+B{k^{3 + }} ion is less than Np3+N{p^{3 + }} ion.
Reason R: The above is a consequence of the lanthanoid contraction.
In the light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: D
Both A and R are true and R is the correct explanation of A
Option D (Both A and R are true and R is the correct explanation of A) is correct because the size of Bk3+B{k^{3 + }} is indeed smaller than Np3+N{p^{3 + }} due to the actinide contraction, analogous to lanthanide contraction.
186 Medium
186 of 407
Remembering +4 | -1
The atomic orbitals used for filling in sixth period of periodic table are
View answer and explanation
Correct option: C
6s,4f,5d,6p6 s, 4 f, 5 d, 6 p
Option C (6s,4f,5d,6p6 s, 4 f, 5 d, 6 p) is correct because the filling of orbitals in the sixth period follows the order 6s, then 4f, then 5d, and finally 6p.
187 Medium
187 of 407
Remembering +4 | -1
Among the following, basic oxide is:
View answer and explanation
Correct option: C
CaOCaO
Option C (CaOCaO) is correct because oxides of alkaline earth metals are basic.
188 Easy
188 of 407
Application +4 | -1
E. N of the element A is E1{E_1} and I.E is E2{E_2}. Hence, E. A will be
View answer and explanation
Correct option: A
2E1−E22{E_1} - {E_2}
Option A (2E1−E22{E_1} - {E_2}) is correct. Electron affinity (E.A) can be related to electronegativity (E.N) and ionization energy (I.E) using Mulliken's formula: E.N = (I.E + E.A)/2. Rearranging, E.A = 2*E.N - I.E.
189 Medium
189 of 407
Understanding +4 | -1
Which pair of elements belongs to the same period in periodic table
View answer and explanation
Correct option: A
CaCa, TiTi
Option A (CaCa, TiTi) is correct because both Calcium (Ca) and Titanium (Ti) belong to the 4th period of the periodic table.
190 Easy
190 of 407
Understanding +4 | -1
Which of the following cannot be isoelectronic
View answer and explanation
Correct option: D
Two different atoms
Option D (Two different atoms) is correct because isoelectronic species have the same number of electrons, and two different atoms will inherently have different numbers of protons and therefore different numbers of electrons in their neutral state.
191 Easy
191 of 407
Understanding +4 | -1
Which is the property of non-metal?
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Correct option: A
Electronegative
Option A (Electronegative) is correct because non-metals tend to gain electrons and are therefore electronegative.
192 Medium
192 of 407
Application +4 | -1
The formation of O2−(g){O}^{2-}(\mathrm{g}) starting from O(g){O}(\mathrm{g}) is endothermic by 603 kJ mol−1603 \mathrm{~kJ} \mathrm{~mol}^{-1}. If electron affinity of O(g){O}(\mathrm{g}) is −141 kJ mol−1-141 \mathrm{~kJ} \mathrm{~mol}^{-1}, the second electron affinity of oxygen would be:
View answer and explanation
Correct option: A
+744 kJ mol−1+744 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option A (+744 kJ/mol) is correct because it correctly calculates the second electron affinity by considering the overall endothermic process and the first electron affinity.
193 Easy
193 of 407
Remembering +4 | -1
Which of the following belong to the category of transition metal
View answer and explanation
Correct option: C
Fe
Option C is correct. Iron (Fe) is a transition metal because it has a partially filled d-orbital.
194 Hard
194 of 407
Application +4 | -1
The energy required to convert all atoms present in 1.2 g1.2 \mathrm{~g} magnesium to Mg2+{Mg}^{2+} ions if IE\mathrm{IE}, and IE2\mathrm{IE}_2 of magnesium are 120 kJ mol−1120 \mathrm{~kJ} \mathrm{~mol}^{-1} and 240 kJ mol−1240 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively
View answer and explanation
Correct option: A
18 kJ18 \mathrm{~kJ}
Option A (18 kJ) is correct because it calculates the total energy required to convert 1.2g of magnesium to Mg2+ ions using the given ionization energies and the number of moles of magnesium.
195 Easy
195 of 407
Remembering +4 | -1
Of the following elements, which one has the highest electronegativity?
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Correct option: A
F
Option A (F) is correct because Fluorine has the highest electronegativity among the halogens.
196 Medium
196 of 407
Understanding +4 | -1
The formation of the oxide ion O2−(g){\mathrm{O}^{2 - }}\left( g \right) requires first an exothermic and then an endothermic step as shown below:
O(g)+e−=O−(g);  ΔH∘=−142  kJmo1−1\mathrm{O}\left( \mathrm{g} \right) + {e^ - } = {\mathrm{O}^ - }\left( \mathrm{g} \right);\mathrm{\;}\Delta H^\circ = - 142\mathrm{\;kJmo}{1^{ - 1}} ,
O(g)−+e−=O2−(g);ΔH∘=844  kJmo1−1\mathrm{O}{\left( \mathrm{g} \right)^ - } + {e^ - } = {\mathrm{O}^{2 - }}\left( \mathrm{g} \right);\Delta H^\circ = 844\mathrm{\;kJmo}{1^{ - 1}}
This is because
View answer and explanation
Correct option: C
O−{\text{O}^ - } ion will tend to resist the addition of another electron
Option C (O−{\text{O}^ - } ion will tend to resist the addition of another electron) is correct because adding an electron to a negatively charged ion (O−{\text{O}^ - }) requires energy to overcome the electrostatic repulsion, making the process endothermic.
197 Medium
197 of 407
Remembering +4 | -1
3d3 d-transition series of elements starts with Sc\mathrm{Sc} and ends with Zn\mathrm{Zn}. The electronic configuration of Sc\mathrm{Sc} and Zn\mathrm{Zn} are respectively.
View answer and explanation
Correct option: B
[Ar]3d14s2[\mathrm{Ar}] 3 d^1 4 s^2 and [Ar]3d104s2[\mathrm{Ar}] 3 d^{10} 4 s^2
Option B is correct because Scandium (Sc) has the electronic configuration [Ar]3d¹4s², and Zinc (Zn) has the electronic configuration [Ar]3d¹⁰4s².
198 Hard
198 of 407
Understanding +4 | -1
Considering the elements B,C,N,F\mathrm{B}, \mathrm{C}, \mathrm{N}, \mathrm{F} and Si\mathrm{Si}, the correct order of their non-metallic character is
View answer and explanation
Correct option: C
F >> N >> C >> B >> Si
Option C (F > N > C > B > Si) is correct because non-metallic character increases across a period and decreases down a group.
199 Medium
199 of 407
Analyzing +4 | -1
Which is mismatched regarding the position of the element as given below?
View answer and explanation
Correct option: D
Both (1) & (3)
Option D (Both (1) & (3)) is correct because element X (Z=89) belongs to d-block and 7th period and element Z(Z=115) belongs to p-block and 7th period.
200 Medium
200 of 407
Understanding +4 | -1
Which of the following acts sometimes as a metal and sometimes as a non-metal?
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Correct option: D
At\mathrm{At}
The correct answer is D (At) because Astatine (At) is a metalloid, exhibiting properties of both metals and non-metals.
201 Easy
201 of 407
Understanding +4 | -1
The element with electronic configuration 1s22s22p63s21{s^2}2{s^2}2{p^6}3{s^2} is
View answer and explanation
Correct option: C
an alkaline earth metal
Option C (an alkaline earth metal) is correct because the electronic configuration 1s22s22p63s21{s^2}2{s^2}2{p^6}3{s^2} indicates that the element has 2 valence electrons in the s-orbital of the outermost shell, which is characteristic of alkaline earth metals.
202 Hard
202 of 407
Understanding +4 | -1
The set representing the correct order of first ionisation energy is
View answer and explanation
Correct option: B
Be>Mg>Ca\mathrm{Be} > Mg > Ca
Option B (Be>Mg>Ca\mathrm{Be} > Mg > Ca) is correct because ionization energy decreases down a group, and Be, Mg, and Ca are in Group 2.
203 Easy
203 of 407
Remembering +4 | -1
The property of attracting electrons by the halogen atoms in a molecule is called
View answer and explanation
Correct option: C
Electronegtivity
Option C (Electronegtivity) is correct because electronegativity is the measure of the ability of an atom to attract electrons in a chemical bond.
204 Easy
204 of 407
Remembering +4 | -1
Which is not a scale of measuring electronegativity?
View answer and explanation
Correct option: A
Stevenson&rsquo;s scale
Option A (Stevenson’s scale) is correct because Stevenson's scale is not a recognized scale for measuring electronegativity.
205 Easy
205 of 407
Understanding +4 | -1
Correct increasing order of first ionisation potential is
View answer and explanation
Correct option: A
Na < Mg > Al < Si
Option A (Na < Mg > Al < Si) is correct because there's an exception between Mg and Al due to the electronic configuration.
206 Medium
206 of 407
Analyzing +4 | -1
The IP1,IP2,IP3,IP4,and IP5\mathrm{I}{\mathrm{P}_1},\mathrm{I}{\mathrm{P}_2},\mathrm{I}{\mathrm{P}_3},\mathrm{I}{\mathrm{P}_4},\mathrm{ and } \, {\mathrm{IP}_5} of an element are 7.1,14.3,34.57.1,\mathrm{ }14.3,\mathrm{ }34.5, 46.846.8,162.2  eV162.2 \, \, \text{eV} respectively. The element is likely to be:
View answer and explanation
Correct option: B
Si\mathrm{Si}
The element is most likely Silicon (Si) because there is a large jump in ionization energy between IP4 and IP5, indicating 4 valence electrons.
207 Medium
207 of 407
Application +4 | -1
The species with radius less than that of neon is
View answer and explanation
Correct option: A
Mg2+M{g^{2 + }}
Option B (r4=169r3{r_4} = \frac{{16}}{9}{r_3}) is correct because the radius of the nth Bohr orbit is proportional to n2n^2.
208 Medium
208 of 407
Analyzing +4 | -1
Which is correct regarding the position of the element as given below?
View answer and explanation
Correct option: A
X(Z=60)−fX(Z=60) - f-block, 6th6^{\text {th}} period
Option A (X(Z=60)−fX(Z=60) - f-block, 6th6^{\text {th}} period) is correct because element X with Z=60 (Neodymium) belongs to f-block and 6th period.
209 Easy
209 of 407
Understanding +4 | -1
Highest amount of energy will be required for the removal of the electron from
View answer and explanation
Correct option: A
s−s -subshell
Option A (s−s -subshell) is correct because s-electrons are closest to the nucleus and thus experience the strongest effective nuclear charge, requiring the highest energy for removal.
210 Easy
210 of 407
Analyzing +4 | -1
S-I: &#39;BeBe&#39; and &#39;AlAl&#39; have similar properties.
S-II: Cations of &#39;BeBe&#39; and &#39;AlAl&#39; have same polarizing power.
View answer and explanation
Correct option: A
Both I and II are correct
Option A (Both I and II are correct) is correct because Beryllium (Be) and Aluminum (Al) exhibit diagonal relationship due to similar polarizing power of their cations.
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