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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 211-240 of 407 questions

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211 Easy
211 of 407
Remembering +4 | -1
The elements in which electrons are progressively filled in 4f – orbital are called
View answer and explanation
Correct option: C
Lanthanoids
Option C is correct. Lanthanoids are the elements in which electrons are progressively filled in the 4f-orbital.
212 Easy
212 of 407
Understanding +4 | -1
Which group of the Periodic Table does not contain only metals?
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Correct option: D
IIIA
Option D (IIIA) is correct because Group IIIA contains metalloids and non-metals in addition to metals.
213 Easy
213 of 407
Understanding +4 | -1
Which of the following properties show gradual decrease with increase in atomic number across a period in the Periodic Table?
View answer and explanation
Correct option: D
Size of atom
Option D (Size of atom) is correct because atomic size generally decreases across a period due to increasing nuclear charge.
214 Easy
214 of 407
Understanding +4 | -1
Which of the following is the smallest in size?
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Correct option: A
Na+N{a^ + }
Option A (Na+N{a^ + }) is the smallest because, among the given isoelectronic species, it has the highest nuclear charge.
215 Hard
215 of 407
Understanding +4 | -1
The ionisation potential values of an element are in the following order I1<I2<<<<I3<I4<I5,{I_1} < {I_2} < < < < {I_3} < {I_4} < {I_5}, The element is
View answer and explanation
Correct option: D
Alkaline earth metal
Option D (Alkaline earth metal) is correct because the large jump between I3 and I4 indicates that the element has two valence electrons, characteristic of alkaline earth metals.
216 Medium
216 of 407
Analyzing +4 | -1
The incorrect order of the radii of the species is
View answer and explanation
Correct option: D
Cl−<K+<Ca2+C{l^ - } < {K^ + } < C{a^{2 + }}
Option D (Cl−<K+<Ca2+C{l^ - } < {K^ + } < C{a^{2 + }}) is the incorrect order because isoelectronic species with greater nuclear charge have smaller radii.
217 Hard
217 of 407
Understanding +4 | -1
The charge/size ratio of a cation determines its polarising power. Which one of the following sequences represents the increasing order of the polarising power of the cationic species, K+,Ca2+,Mg2+,Be2+{\mathrm{K}^ + },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{M}{\mathrm{g}^{2 + }},\mathrm{B}{\mathrm{e}^{2 + }}?
View answer and explanation
Correct option: C
K+<Ca2+<Mg2+<Be2+{\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}
Option C (K+<Ca2+<Mg2+<Be2+{\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}) is correct because polarizing power increases with increasing charge and decreasing size. Be²⁺ has the highest charge/size ratio and K⁺ has the lowest.
218 Easy
218 of 407
Understanding +4 | -1
The IUPAC symbol for the element with atomic number 119 would be
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Correct option: B
uue
Option B (uue) is correct because it follows the IUPAC naming conventions for elements with atomic numbers greater than 100.
219 Easy
219 of 407
Remembering +4 | -1
What is the name of element with atomic number 105?
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Correct option: B
Dubnium
Option B (Dubnium) is correct because the element with atomic number 105 is named Dubnium (Db).
220 Medium
220 of 407
Remembering +4 | -1
The enthalpy change accompanying in the given reaction, X(g)+e−⟶X−(g)X(g)+e^{-} \longrightarrow X^{-}(g) is defined as
View answer and explanation
Correct option: A
Electron gain enthalpy
Option A (Electron gain enthalpy) is correct because the reaction describes the addition of an electron to a gaseous atom, which is the definition of electron gain enthalpy.
221 Medium
221 of 407
Application +4 | -1
Predict the position of an element has the electronic configuration 1s22s22p63s23p63d54s11 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^1.
View answer and explanation
Correct option: A
Period 4, group 6
Option A (Period 4, group 6) is correct. The highest principal quantum number is 4 (period 4), and the number of d electrons + s electrons is 5+1 = 6 (group 6).
222 Easy
222 of 407
Remembering +4 | -1
The elements in which 4 f- orbitals are progressively filled up are called as
View answer and explanation
Correct option: C
Lanthanoids
Option C is correct because lanthanoids are the elements in which the 4f-orbitals are progressively filled.
223 Medium
223 of 407
Understanding +4 | -1
The elements having the electronic configuration [Kr]  4d10f14,  5s2p6d2,  6s2\left[ {\mathrm{Kr}} \right]\;4{d^{10}}{f^{14}},\;5{s^2}{p^6}{d^2},\;6{s^2} belongs to
View answer and explanation
Correct option: C
d-block
Option C (d-block) is correct because the last electron enters the d-orbital.
224 Easy
224 of 407
Understanding +4 | -1
Elements of a vertical group have same&nbsp;
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Correct option: A
Same valence shell electronic configuration
Option A (Same valence shell electronic configuration) is correct because elements in the same group have the same number of valence electrons and thus similar chemical properties.
225 Easy
225 of 407
Understanding +4 | -1
Which of the following pairs has both members of the same period of the Periodic Table?
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Correct option: A
Na−Cl\mathrm{Na} - \mathrm{Cl}
Option A (Na−Cl\mathrm{Na} - \mathrm{Cl}) is correct because both Sodium (Na) and Chlorine (Cl) are in the third period of the periodic table.

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226 Easy
226 of 407
Understanding +4 | -1
The elements present in each group form a family because:
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Correct option: B
They possess common pattern of electronic configuration.
Option B (They possess common pattern of electronic configuration) is correct because elements in the same group have the same number of valence electrons, leading to similar chemical properties.
227 Medium
227 of 407
Remembering +4 | -1
The first ionisation potential in electron volts of nitrogen and oxygen atoms are respectively given by
View answer and explanation
Correct option: A
14.6,13.614.6,13.6
Option A (14.6,13.614.6,13.6) is correct because nitrogen has a higher ionization potential than oxygen due to its half-filled p-orbital stability.
228 Easy
228 of 407
Understanding +4 | -1
Difference between S and S2−as  S2−{\mathrm{S}^{2 - }}\mathrm{as\;}{\mathrm{S}^{2 - }} has
View answer and explanation
Correct option: A
Larger radii and larger size
Option A (Larger radii and larger size) is correct because adding electrons to an atom increases electron-electron repulsion, leading to an expansion of the electron cloud.
229 Easy
229 of 407
Remembering +4 | -1
According to Mendeleev’s periodic law, the properties of elements were the periodic functions of
View answer and explanation
Correct option: B
Atomic mass
Option B (Atomic mass) is correct because Mendeleev's periodic law stated that the properties of elements are periodic functions of their atomic masses.
230 Medium
230 of 407
Analyzing +4 | -1
Which one of the following elements has the highest ionisation energy?
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Correct option: B
[Ne]3s23p3\left[ {\mathrm{Ne}} \right]3{s^2}3{p^3}
Option B ([Ne]3s²3p³) is correct because half-filled p orbitals have extra stability, leading to a higher ionization energy.
231 Medium
231 of 407
Remembering +4 | -1
Which elements in the periodic table should be in liquid state at 35∘C35^{\circ} {C} ?
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Correct option: D
Hg,Br2,Cs,Ga{Hg}, {Br_2}, {Cs}, {Ga}
Option D is correct. At 35°C, Mercury (Hg) and Bromine (Br₂) are liquids. Cesium (Cs) and Gallium (Ga) also have melting points below 35°C, so they are also liquids at this temperature.
232 Medium
232 of 407
Analyzing +4 | -1
In which of the following arrangements the order is not according to the property indicated against it?
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Correct option: C
B<C<N<O\mathrm{B} < C < N < O increasing first ionisation enthalphy
Option C is correct because the first ionization enthalpy of Nitrogen is higher than Oxygen due to the stable half-filled electronic configuration of Nitrogen.
233 Easy
233 of 407
Remembering +4 | -1
Eka-silicon is now known as
View answer and explanation
Correct option: C
Germanium
Option C (Germanium) is correct because Mendeleev predicted the existence of Eka-silicon, which was later discovered and named Germanium.
234 Medium
234 of 407
Remembering +4 | -1
14 elements of 6th period and 14 elements of 7th period in the periodic table are termed as respectively
View answer and explanation
Correct option: A
Lanthanoids, actinoids
Option A is correct. The 14 elements of the 6th period are called Lanthanoids, and the 14 elements of the 7th period are called Actinoids.
235 Easy
235 of 407
Application +4 | -1
First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family:
View answer and explanation
Correct option: A
1012 kJ mol−11012 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option A (1012 kJ mol−11012 \mathrm{~kJ} \mathrm{~mol}^{-1}) is correct because it corresponds to the ionization enthalpy of Phosphorus, a key component of the apatite family.
236 Hard
236 of 407
Application +4 | -1
If the radius of the 3rd {3^{{\text{rd }}}} Bohr's orbit of hydrogen atom is r3{r_3} and the radius of 4th{4^{{\text{th}}}} Bohr's orbit is r4.{r_4}. Then:
View answer and explanation
Correct option: B
r4=169r3{r_4} = \frac{{16}}{9}{r_3}
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.
237 Easy
237 of 407
Analyzing +4 | -1
The element having maximum value for its 3rd 3^{\text {rd }} ionisation enthalpy is
View answer and explanation
Correct option: B
Be\mathrm{Be}
Option B (Be) is correct because after losing two electrons, Be attains a stable noble gas configuration, making the removal of the third electron require significantly more energy.
238 Medium
238 of 407
Remembering +4 | -1
Pauling&rsquo;s equation for determining the electronegativity of an element, is XA,  XB={X_A},\;{X_B} = electronegativity values of elements and Δ=\Delta = represents polarity of A-B bond
View answer and explanation
Correct option: A
XA−XB=0.208Δ{X_A} - {X_B} = 0.208\sqrt \Delta
Option A (XA−XB=0.208Δ{X_A} - {X_B} = 0.208\sqrt \Delta) is correct because it represents Pauling's equation for determining electronegativity difference.
239 Hard
239 of 407
Analyzing +4 | -1
Correct order of second ionisation potential of C, N, O, F is
View answer and explanation
Correct option: C
O>F>N>C\mathrm{O > F > N > C}
Option C (O>F>N>C\mathrm{O > F > N > C}) is correct because after the first ionization, nitrogen achieves a stable half-filled configuration, making the second ionization easier than removing from F+.
240 Medium
240 of 407
Remembering +4 | -1
Element having atomic no. of 56 belongs to which of the following block of periodic table?
View answer and explanation
Correct option: D
ss-block
Option D (ss-block) is correct because elements with atomic number 56 (Barium) belongs to the s-block.
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