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Q. No.The element having atomic number 55 is a member of:
1. Group 1
2. Group 3
3. Group 15
4. Group 17
Subtopic: Modern Periodic Table & Periodicity |
83%
Level 1: 80%+
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Given below are two statements:
| Statement I: | The electronegativity of any given element is not constant. |
| Statement II: | Electronegativity varies depending on the element to which it is bound. |
| 1. | Statement l is incorrect but Statement ll is correct. |
| 2. | Both Statements l and ll are correct. |
| 3. | Both Statements l and ll are incorrect. |
| 4. | Statement l is correct but Statement ll is incorrect. |
Subtopic: Electronegativity |
64%
Level 2: 60%+
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Which of the given pairs shows the first element having higher electronegativity than the second?
| 1. | Al and B | 2. | Be and Li |
| 3. | Cl and O | 4. | Si and N |
Subtopic: Electronegativity |
67%
Level 2: 60%+
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What is the correct order of electron gain enthalpy from least negative to most negative for the elements C, Ca, Al, F, and O?
1. Al < Ca < O < C < F
2. Al < O < C < Ca < F
3. C < F < O < Al < Ca
4. Ca < Al < C < O < F
1. Al < Ca < O < C < F
2. Al < O < C < Ca < F
3. C < F < O < Al < Ca
4. Ca < Al < C < O < F
Subtopic: Electron Affinity (EA) |
82%
Level 1: 80%+
NEET - 2013
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The correct order of first ionization enthalpy for the given four element is:
| 1. | C < N < F < O | 2. | C < N < O < F |
| 3. | C < O < N < F | 4. | C < F < N < O |
Subtopic: Ionization Energy (IE) |
79%
Level 2: 60%+
NEET - 2022
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The first ionization enthalpies of elements X and Y are 419 kJ mol–1 and 590 kJ mol–1, respectively and the second ionization enthalpies of X and Y are 3069 kJ mol–1 and 1145 kJ mol–1, respectively. The correct statement is:
| 1. | X is an alkali metal and Y is an alkaline earth metal. |
| 2. | X is an alkaline earth metal and Y is an alkali metal. |
| 3. | Both X and Y are alkali metals. |
| 4. | Both X and Y are alkaline earth metals. |
Subtopic: Ionization Energy (IE) |
72%
Level 2: 60%+
NEET - 2022
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| Assertion (A): | The shielding effect increase as we go down the group. |
| Reason (R): | More is the number of electrons in the penultimate shell, and more is shielding. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | Both (A) and (R) are False. |
Subtopic: Screening Effect |
65%
Level 2: 60%+
Hints
Match List-I (Elements) with List-II (Atomic radii (pm)):
| List-I Elements |
List-II
Atomic radii (pm)
|
||
| (a) | Oxygen (\(O\)) | (i) | 88 |
| (b) | Carbon (\(C\)) | (ii) | 74 |
| (c) | Boron (\(B\)) | (iii) | 66 |
| (d) | Nitrogen (\(N\)) | (iv) | 77 |
| (a) | (b) | (c) | (d) | |
| 1. | (ii) | (i) | (iv) | (iii) |
| 2. | (iv) | (iii) | (ii) | (i) |
| 3. | (i) | (iv) | (iii) | (ii) |
| 4. | (iii) | (iv) | (i) | (ii) |
Subtopic: Atomic Size |
74%
Level 2: 60%+
NEET - 2022
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The IUPAC name of an element with atomic number 119 is:
1. Ununoctium
2. Ununennium
3. Unnilennium
4. Unununnium
1. Ununoctium
2. Ununennium
3. Unnilennium
4. Unununnium
Subtopic: Modern Periodic Table & Periodicity |
83%
Level 1: 80%+
NEET - 2022
Hints
The element expected to form the largest ion to achieve the nearest noble gas configuration is:
| 1. | Sodium(Na) | 2. | Oxygen(O) |
| 3. | Fluorine(F) | 4. | Nitrogen(N) |
Subtopic: Atomic Size |
57%
Level 3: 35%-60%
NEET - 2023
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