NEET PYQs

Question 1 of 54

1. Question
The element \(Z=114\) has been discovered recently. It will belong to which of the following family/group and electronic configuration? [NEET 2017]
- Carbon family, [Rn] \(5 f^{14} 6 d^{10} 7 s^{2} 7 p^{4}\)
- Oxygen family, [Rn] \(5 f^{14} 6 d^{10} 7 s^{2} 7 p^{4}\)
- Nitrogen family, [Rn] \(5 f^{14} 6 d^{10} 7 s^{2} 7 p^{6}\)
- Halogen family, [Rn] \(5 f^{14} 6 d^{10} 7 s^{2} 7 p^{5}\)
Correct

Incorrect

Hint

(a) The electronic configuration of the element with \(Z=114\) (flerovium) is \([\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^{2} 7 p^{2}\).
Hence, it belongs to carbon family which has the same outer electronic configuration.
Question 2 of 54

2. Question
In which of the following options the order of arrangement does not agree with the variation of property indicated against it? [NEET-I 2016]
- \(\mathrm{I} \lt \mathrm{Br} \lt \mathrm{Cl} \lt \mathrm{F}\) (increasing electron gain enthalpy)
- \(\mathrm{Li}\lt \mathrm{Na}\lt \mathrm{K}\lt \mathrm{Rb}\) (increasing metallic radius)
- \(\mathrm{Al}^{3+}\lt \mathrm{Mg}^{2+}\lt \mathrm{Na}^{+}\lt \mathrm{F}^{-}\)(increasing ionic size)
- \(\mathrm{B}\lt \mathrm{C}\lt \mathrm{N}\lt \mathrm{O}\) (increasing first ionisation enthalpy)
Correct

Incorrect

Hint

(a, d) The correct order of increasing negative electron gain enthalpy is : \(\mathrm{I}<\mathrm{Br}<\mathrm{F}<\mathrm{Cl}\) and the correct order of increasing first ionisation enthalpy is \(\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}\).

Explanation:
Generally ionisation energy increases across a period. But here first I.E. of O is less than the first I.E. of N. This is due to the half-filled \(2 p\) orbital in \(\mathrm{N}\left(1 s^2, 2 s^2, 2 p^3\right)\) which is more stable than the \(2 p\) orbital in \(\mathrm{O}\left(1 s^2, 2 s^2, 2 p^4\right)\).
Halogens have high electron affinities which decreases on moving down the group. However, fluorine has lower value than chlorine which is due to its small size and repulsion between the electron added and electrons already present.
Question 3 of 54

3. Question
The species Ar, \(\mathrm{K}^{+}\)and \(\mathrm{Ca}^{2+}\) contain the same number of electrons. In which order do their radii increase? [NEET 2015]
- \(\mathrm{Ca}^{2+}\lt \mathrm{K}^{+}\lt \mathrm{Ar}\)
- \(\mathrm{K}^{+}\lt \mathrm{Ar}\lt \mathrm{Ca}^{2+}\)
- \(\mathrm{Ar}\lt \mathrm{K}^{+}\lt \mathrm{Ca}^{2+}\)
- \(\mathrm{Ca}^{2+}\lt \mathrm{Ar}\lt \mathrm{K}^{+}\)
Correct

Incorrect

Hint

(a) In isoelectronic species the radius decrease with increase in nuclear charge hence increasing order of radius is \(\mathrm{Ca}^{+2}<\mathrm{K}^{+}<\mathrm{Ar}\)
Question 4 of 54

4. Question
Which of the following orders of ionic radii is correctly represented? [NEET 2014]
- \(\mathrm{H}^{-}>\mathrm{H}^{+}>\mathrm{H}\)
- \(\mathrm{Na}^{+}>\mathrm{F}^{-}>\mathrm{O}^{2-}\)
- \(\mathrm{F}^{-}>\mathrm{O}^{2-}>\mathrm{Na}^{+}\)
- \(\mathrm{Al}^{3+}>\mathrm{Mg}^{2+}>\mathrm{N}^{3-}\)
- None
Correct

Incorrect

Hint

(e) (None) Cations lose electrons and are smaller in size than the parent atom, whereas anions add electrons and are larger in size than the parent atom. Hence, the order is \(\mathrm{H}^{-}>\mathrm{H}>\mathrm{H}^{+}\).

For isoelectronic species, the ionic radii decreases with increase in atomic number i.e. nuclear charge. Hence, the correct orders are \(\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}\)and \(\mathrm{N}^{3-}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}\).
Question 5 of 54

5. Question
Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for \(\mathrm{C}, \mathrm{Ca}, \mathrm{Al}, \mathrm{F}\) and \(\mathrm{O}\) ? [NEET Karnataka 2013]
- \(\mathrm{Al}\lt \mathrm{Ca}\lt \mathrm{O}\lt \mathrm{C}\lt \mathrm{F}\)
- \(\mathrm{Al}\lt \mathrm{O}\lt \mathrm{C}\lt \mathrm{Ca}\lt \mathrm{F}\)
- \(\mathrm{C}\lt \mathrm{F}\lt \mathrm{O}\lt \mathrm{Al}\lt \mathrm{Ca}\)
- \(\mathrm{Ca}\lt \mathrm{Al}\lt \mathrm{C}\lt \mathrm{O}\lt \mathrm{F}\)
Correct

Incorrect

Hint

(d) Electron gain enthalpy becomes less negative from top to bottom in a group while it becomes more negative from left to right within a period.

Explanation:

As the nuclear charge increases, the force of attraction between the nucleus and the incoming electron increases and hence the elecron gain enthalpy becomes more negative, hence the correct order is
\(
\mathrm{Ca}<\mathrm{Al}<\mathrm{C}<\mathrm{O}<\mathrm{F}
\)
Question 6 of 54

6. Question
Identify the wrong statement in the following [NEET 2012]
- Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius.
- Amongst isoelectronic species, greater the negative charge on the anion, larger is the ionic radius.
- Atomic radius of the elements increases as one moves down the first group of the periodic table.
- Atomic radius of the elements decreases as one moves across from left to right in the \(2^{\text {nd }}\) period of the periodic table.
Correct

Incorrect

Hint

(a) As positive charge on the cation increases, effective nuclear charge increases. Thus atomic size decreases.
Question 7 of 54

7. Question
What is the value of electron gain enthalpy of \(\mathrm{Na}^{+}\) if \(I E_{1}\) of \(\mathrm{Na}=5.1 \mathrm{~eV}\) ? [NEET 2011]
- \(-5.1 \mathrm{~eV}\)
- \(-10.2 \mathrm{~eV}\)
- \(+2.55 \mathrm{~eV}\)
- \(+10.2 \mathrm{~eV}\)
Correct

Incorrect

Hint

(a) The electron gain enthalpy of sodium cation is equal in magnitude and opposite in sign to the first ionization energy of sodium atom. It is equal to \(-5.1 \mathrm{~eV}\).

\(\mathrm{Na} \rightarrow \mathrm{Na}^{+}+e^{-} ; \Delta H=5.1 \mathrm{~eV}\)
\(
\mathrm{Na}^{+}+e^{-} \rightarrow \mathrm{Na} ; \Delta H=-5.1 \mathrm{~eV}
\)
Question 8 of 54

8. Question
The correct order of the decreasing ionic radii among the following isoelectronic species is [NEET 2010]
- \(\mathrm{Ca}^{2+}>\mathrm{K}^{+}>\mathrm{S}^{2-}>\mathrm{Cl}^{-}\)
- \(\mathrm{Cl}^{-}>\mathrm{S}^{2-}>\mathrm{Ca}^{2+}>\mathrm{K}^{+}\)
- \(\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}\)
- \(\mathrm{K}^{+}>\mathrm{Ca}^{2+}>\mathrm{Cl}^{-}>\mathrm{S}^{2-}\)
Correct

Incorrect

Hint

(c) \(\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}\)
Among isoelectronic species, ionic radii increases with increase in negative charge. This happens because effective nuclear charge \(\left(Z_{e f f}\right)\) decreases. Similarly, ionic radii decreases with increase in positive charge as \(Z_{\text {eff }}\) increases.
Question 9 of 54

9. Question
Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements \(\mathrm{O}, \mathrm{S}, \mathrm{F}\) and \(\mathrm{Cl}\) ? [NEET 2010]
- \(\mathrm{Cl}\lt \mathrm{F}\lt \mathrm{O}\lt \mathrm{S}\)
- \(\mathrm{O}\lt \mathrm{S}\lt \mathrm{F}\lt \mathrm{Cl}\)
- \(\mathrm{F}\lt \mathrm{S}\lt \mathrm{O}\lt \mathrm{Cl}\)
- \(\mathrm{S}\lt \mathrm{O}\lt \mathrm{Cl}\lt \mathrm{F}\)
Correct

Incorrect

Hint

(b) \( \mathrm{Cl}\) atom has the highest electron affinity in the periodic table. \( F\) being a member of group 17 has higher electron gain enthalpy than \( \mathrm{S}\) which belong: to group 16. This in turn is higher than the electror affinity of \( \mathrm{O}\) atom. Thus,
\(
\mathrm{Cl}>\mathrm{F}>\mathrm{S}>\mathrm{O}
\)

It is worth noting that the electron gain enthalpy of oxygen and fluorine, the members of the second period, have less negative values than the elements sulphur and chlorine of the third period.
This is due to small size of the atoms of oxygen and fluorine. As a result, there is a strong inter-electronic repulsion when extra electron is added to these atoms, i.e., electron density is high and the addition of an extra electron is not easy.
Question 10 of 54

10. Question
Among the elements \(\mathrm{Ca}, \mathrm{Mg}, \mathrm{P}\) and \(\mathrm{Cl}\), the order of increasing atomic radii is [NEET 2010]
- \(\mathrm{Mg}\lt \mathrm{Ca}\lt \mathrm{Cl}\lt \mathrm{P}\)
- \(\mathrm{Cl}\lt \mathrm{P}\lt \mathrm{Mg}\lt \mathrm{Ca}\)
- \(\mathrm{P}\lt \mathrm{Cl}\lt \mathrm{Ca} \lt \mathrm{Mg}\)
- \(\mathrm{Ca}\lt \mathrm{Mg}\lt \mathrm{P}\lt \mathrm{Cl}\)
Correct

Incorrect

Hint

(b) The atomic radii decrease on moving from left to right in a period, thus order of sizes for \(\mathrm{Cl}, \mathrm{P}\) and \(\mathrm{Mg}\) is \(\mathrm{Cl}<\mathrm{P}<\mathrm{Mg}\). Down the group size increases. Thus overall order is : \(\mathrm{Cl}<\mathrm{P}<\mathrm{Mg}<\mathrm{Ca}\)
Question 11 of 54

11. Question
Among the following which one has the highest cation to anion size ratio? [NEET 2010]
- \(\mathrm{CsI}\)
- \(\mathrm{CsF}\)
- \(\mathrm{LiF}\)
- \(\mathrm{NaF}\)
Correct

Incorrect

Hint

(b) The cation to anion size ratio will be maximum when the cation is of largest size and the anion is of smallest size. Among the given species, \(\mathrm{Cs}^{+}\)has maximum size among given cations and \(\mathrm{F}^{-}\) has smallest size among given anions, thus CsF has highest \(r_{c} / r_{a}\) ratio.
Question 12 of 54

12. Question
Amongst the elements with following electronic configurations, which one of them may have the highest ionisation energy? [NEET 2009]
- \(\mathrm{Ne}\left[3 s^{2} 3 p^{2}\right]\)
- \(\operatorname{Ar}\left[3 d^{10} 4 s^{2} 4 p^{3}\right]\)
- \(\mathrm{Ne}\left[3 s^{2} 3 p^{1}\right]\)
- \(\mathrm{Ne}\left[3 s^{2} 3 p^{3}\right]\)
Correct

Incorrect

Hint

(d) Among options (a), (c) and (d), option (d) has the highest ionisation energy because of extra stability associated with half-filled \(3 p\)-orbital. In option (b), the presence of \(3 d^{10}\) electrons offers shielding effect, as a result the \(4 p^{3}\) electrons do not experience much nuclear charge and hence the electrons can be removed easily.
Question 13 of 54

13. Question
Which one of the following arrangements does not give the correct picture of the trends indicated against it? [NEET 2008]
- \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\) : Bond dissociation energy
- \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\) : Electronegativity
- \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\) : Oxidizing power
- \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\) : Electron gain enthalpy
Correct

Incorrect

Hint

(a,d) In case of diatomic molecules \(\left(X_{2}\right)\) of halogens the bond dissociation energy decreases in the order:

\(
\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{F}_2>\mathrm{I}_2
\)
The oxidising power, electronegativity and reactivity decrease in the order:
\(
\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2
\)
Electron gain enthalpy of halogens follows the given order :
\(
\mathrm{Cl}_2>\mathrm{F}_2>\mathrm{Br}_2>\mathrm{I}_2
\)
The low value of electron gain enthalpy (electron enthalpy) of fluorine is probably due to small size of fluorine atom.
Question 14 of 54

14. Question
Identify the correct order of the size of the following: [NEET 2007]
- \(\mathrm{Ca}^{2+}\lt \mathrm{K}^{+}\lt \mathrm{Ar}\lt \mathrm{Cl}^{-}\lt \mathrm{S}^{2-}\)
- \(\mathrm{Ar}\lt \mathrm{Ca}^{2+}\lt \mathrm{K}^{+}\lt \mathrm{Cl}^{-}\lt \mathrm{S}^{2-}\)
- \(\mathrm{Ca}^{2+}\lt \mathrm{Ar}\lt \mathrm{K}^{+}\lt \mathrm{Cl}^{-}\lt \mathrm{S}^{2-}\)
- \(\mathrm{Ca}^{2+}\lt \mathrm{K}^{+}\lt \mathrm{Ar}\lt \mathrm{S}^{2-}\lt \mathrm{Cl}^{-}\)
Correct

Incorrect

Hint

(a) Among isoelectronic ions, ionic radii of anions is more than that of cations. Further size of the anion increases with increase in negative charge and size of the cation decreases with increase in positive charge.
Question 15 of 54

15. Question
With which of the following electronic configuration an atom has the lowest ionisation enthalpy? [NEET 2007]
- \(1 s^{2} 2 s^{2} 2 p^{3}\)
- \(1 s^{2} 2 s^{2} 2 p^{5} 3 s^{1}\)
- \(1 s^{2} 2 s^{2} 2 p^{6}\)
- \(1 s^{2} 2 s^{2} 2 p^{5}\)
Correct

Incorrect

Hint

(b) The larger the atomic size, smaller is the value of the ionisation enthalpy. Again higher the screening effect, lesser is the value of ionisation potential. Hence option (b) has lowest ionisation enthalpy.
Question 16 of 54

16. Question
Which one of the following ionic species has the greatest proton affinity to form stable compound? [NEET 2007]
- \(\mathrm{NH}_2^{-}\)
- \(\mathrm{F}^{-}\)
- \(\mathrm{I}^{-}\)
- \(\mathrm{HS}^{-}\)
Correct

Incorrect

Hint

(a) In going from left to right across a period in the periodic table, the basicity (i.e. proton affinity) decreases as the electronegativity of the atom possessing the lone pair of electrons increases. Hence basicity of \(\mathrm{NH}_2^{-}\) is higher than \(\mathrm{F}^{-}\). In moving down a group, as the atomic mass increases, basicity decreases. Hence \(\mathrm{F}^{-}\)is more basic than \(\mathrm{I}^{-}\)and \(\mathrm{HO}^{-}\) is more basic than \(\mathrm{HS}^{-}\). Hence among the given ionic species, \(\mathrm{NH}_{2}^{-}\)has maximum proton affinity.
Question 17 of 54

17. Question
Which one of the following orders is not in accordance with the property stated against it? [NEET 2006]
- \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\) : Bond dissociation energy
- \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\) : Oxidising power
- \(\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}>\mathrm{HF}:\) Acidic property in water
- \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\) : Electronegativity
Correct

Incorrect

Hint

(a) \(X-X\) bond Bond dissociation
\(
\begin{array}{cccc}
& \mathrm{F}-\mathrm{F} & \mathrm{Cl}-\mathrm{Cl} & \mathrm{Br}-\mathrm{Br} & \mathrm{I}-\mathrm{I} \\
\text {Bond dissociation energy} (\mathrm{kcal} / \mathrm{mol}) & 38 & 57 & 45.5 & 35.6
\end{array}
\)

The lower value of bond dissociation energy of fluorine is due to the high inter-electronic repulsion between non-bonding electrons in the \(2 p\)-orbitals of fluorine. As a result \(\mathrm{F}-\mathrm{F}\) bond is weaker in comparison to \(\mathrm{Cl}-\mathrm{Cl}\) and \(\mathrm{Br}-\mathrm{Br}\) bonds.
Question 18 of 54

18. Question
Which one of the following arrangements represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species? [NEET 2005]
- \(\mathrm{S}\lt\mathrm{O}\lt \mathrm{Cl}\lt \mathrm{F}\)
- \(\mathrm{Cl}\lt \mathrm{F}\lt \mathrm{S}\lt \mathrm{O}\)
- \(\mathrm{F}\lt \mathrm{Cl}\lt \mathrm{O}\lt \mathrm{S}\)
- \(\mathrm{O}\lt \mathrm{S}\lt \mathrm{F}\lt \mathrm{Cl}\)
Correct

Incorrect

Hint

(d) The molar enthalpy change accompanying the addition of an electron to an atom (or ion) is known as electron gain enthalpy.

Generally it increases on moving from left to right in a period and in a group it decreases as the size increases.
Exception: Because of the small size of F, electron-electron repulsion present in its relatively compact \(2 p\)-subshell, do not easily allow the addition of an extra electron. On the other hand, \(\mathrm{Cl}\) because of its comparatively bigger size than \(\mathrm{F}\), allows the addition of an extra electron more easily.

\(
\mathrm{O}< \quad \mathrm{S}< \quad \mathrm{F}< \quad \mathrm{Cl}
\)
\(
\begin{array}{llll}
-1.48 & -2.0 & -3.6 & -3.8
\end{array}
\)
Question 19 of 54

19. Question
Ionic radii are [NEET 2004]
- inversely proportional to effective nuclear charge
- inversely proportional to square of effective nuclear charge
- directly proportional to effective nuclear charge
- directly proportional to square of effective nuclear charge.
Correct

Incorrect

Hint

(a) Ionic radius in the \(n^{\text {th }}\) orbit is given by
\(
r_{n}=\frac{n^{2} a_{0}}{Z *} \quad \text { or }, \quad r_{n} \propto \frac{1}{Z *}
\)

Where \(n\) is principal quantum number, \(a_0\) the Bohr’s radius of \(\mathrm{H}\)-atom and \(\mathrm{Z}^*\), the effective nuclear charge.
Question 20 of 54

20. Question
The ions \(\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}\) and \(\mathrm{Al}^{3+}\) are isoelectronic. Their ionic radii show [NEET 2003]
- a significant increase from \(\mathrm{O}^{2-}\) to \(\mathrm{Al}^{3+}\)
- a significant decrease from \(\mathrm{O}^{2-}\) to \(\mathrm{Al}^{3+}\)
- an increase from \(\mathrm{O}^{2-}\) to \(\mathrm{F}^{-}\)and then decrease from \(\mathrm{Na}^{+}\)to \(\mathrm{Al}^{3+}\)
- a decrease from \(\mathrm{O}^{2-}\) to \(\mathrm{F}^{-}\)and then increase from \(\mathrm{Na}^{+}\)to \(\mathrm{Al}^{3+}\).
Correct

Incorrect

Hint

(b) Amongst isoelectronic ions, ionic radii of anions is more than that of cations. Further size of the anion increases with increase in -ve charge and size of cation decreases with increase in +ve charge. Hence, correct order is
\(
\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+} .
\)
Question 21 of 54

21. Question
Which statement is wrong? [NEET 2000]
- Bond energy of \(\mathrm{F}_{2}\lt \mathrm{Cl}_{2}\)
- Electronegativity of \(\mathrm{F}\lt \mathrm{Cl}\)
- \(\mathrm{F}\) is more oxidising than \(\mathrm{Cl}\)
- Electron affinity of \(\mathrm{Cl}\lt \mathrm{F}\)
Correct

Incorrect

Hint

(a) Due to more repulsion in between nonbonding electron pair \((2 p)\) of two fluorines (due to small size of \(\mathrm{F}\)-atom) in comparison to non-bonding electron pair ( \(3 p\) ) in chlorine, the bond energy of \(F_{2}\) is less than \(\mathrm{Cl}_{2}\).

\(B E\left(\mathrm{~F}_2\right)=158.5 \mathrm{~kJ} / \mathrm{mole}\) and
\(B E\left(\mathrm{Cl}_2\right)=242.6 \mathrm{~kJ} / \mathrm{mole}\).
Question 22 of 54

22. Question
Which of the following elements has the maximum electron affinity? [NEET 1999]
- I
- Br
- Cl
- F
Correct

Incorrect

Hint

(c) Among the halogens the electron affinity value of ‘ \(F\) ‘ should be maximum. But due to small size the 7-electrons in its valence shell are much more crowded, so that it feels difficulty in entry of new electrons. Thus, the E.A. value is slightly lower than chlorine and the order is
\(
\mathrm{I}<\mathrm{Br}<\mathrm{F}<\mathrm{Cl}
\)
Question 23 of 54

23. Question
The first ionization potentials \((\mathrm{eV})\) of \(\mathrm{Be}\) and \(B\) respectively are [NEET 1998]
- \(8.29,8.29\)
- \(9.32,9.32\)
- \(8.29,9.32\)
- \(9.32,8.29\)
Correct

Incorrect

Hint

\(
\text { (d) }{ }_4 \mathrm{Be} \rightarrow 1 s^2 2 s^2,{ }_5 \mathrm{B} \rightarrow 1 s^2 2 s^2 2 p^1
\)

Due to stable fully-filled ‘ \(s\) ‘-orbital arrangement of electrons in ‘Be’ atom, more energy is required to remove an electron from the valence shell than ‘B’atom. Therefore ‘ \(\mathrm{Be}\) ‘ has higher ionisaton potential than ‘ \(\mathrm{B}\) ‘.
Question 24 of 54

24. Question
Which one of the following is correct order of the size of iodine species? [NEET 1997]
- \(\mathrm{I}^{+}>\mathrm{I}^{-}>\mathrm{I}\)
- \(\mathrm{I}^{-}>\mathrm{I}>\mathrm{I}^{+}\)
- \(\mathrm{I}>\mathrm{I}^{-}>\mathrm{I}^{+}\)
- \(\mathrm{I}>\mathrm{I}^{+}>\mathrm{I}^{-}\)
Correct

Incorrect

Hint

(b) Positive ion is always smaller and negative ion is always larger than the parent atom.
Question 25 of 54

25. Question
Which of the following ion is the largest in size? [NEET 1996]
- \(\mathrm{K}^{+}\)
- \(\mathrm{Ca}^{2+}\)
- \(\mathrm{Cl}^{-}\)
- \(\mathrm{S}^{2-}\)
Correct

Incorrect

Hint

(d) Since all of these ions contain 18 electrons each, so these are isoelectronic. For isoelectronic ions, smaller the positive nuclear charge, greater is the size of the ion
Question 26 of 54

26. Question
Which of the following has the smallest size? [NEET 1996]
- \(\mathrm{Al}^{3+}\)
- \(\mathrm{F}^{-}\)
- \(\mathrm{Na}^{+}\)
- \(\mathrm{Mg}^{2+}\)
Correct

Incorrect

Hint

(a) These are isoelectronic ions (ions with same number of electrons) and for isoelectronic ions, greater the positive nuclear charge, greater is the force of attraction on the electrons by the nucleus and the smaller is the size of the ion. Thus \(\mathrm{Al}^{3+}\) has the smallest size.
Question 27 of 54

27. Question
The electronics configuration of an element is \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}\). What is the atomic number of the element, which is just below the above element in the periodic table? [NEET 1995]
- 33
- 34
- 36
- 49
Correct

Incorrect

Hint

(a) Atomic no. of given element \(=15\), thus it belongs to \(5^{\text {th }}\) group.

Now, atomic no. of the element below the above element \(=15+18=33\)
Question 28 of 54

28. Question
One would expect proton to have very large [NEET 1993]
- charge
- Ionisation potential
- Hydration enthalpy
- radius
Correct

Incorrect

Hint

(c) Proton \(\left(\mathrm{H}^{+}\right)\)being very small in size would have very large hydration energy.
Question 29 of 54

29. Question
\(\mathrm{Na}^{+}, \mathrm{Mg}^{2+}, \mathrm{Al}^{3+}\) and \(\mathrm{Si}^{4+}\) are isoelectronic. the order of their ionic size is [NEET 1993]
- \(Na ^{+}> Mg ^{2+} \lt Al ^{3+} \lt Si ^{4+}\)
- \(Na ^{+} \lt Mg ^{2+}> Al ^{3+}> Si ^{4+}\)
- \(Na ^{+}> Mg ^{2+}> Al ^{3+}> Si ^{4+}\)
- \(Na ^{+}> Mg ^{2+}> Al ^{3+} \lt Si ^{4+}\)
Correct

Incorrect

Hint

(c) In isoelectronic ions, the size of the cation decreases as the magnitude of the positive charge increases.
Question 30 of 54

30. Question
If the atomic number of an element is 33 , it will be placed in the periodic table in the [NEET 1993]
- first group
- third group
- fifth group
- seventh group
Correct

Incorrect

Hint

(c) Electronic configuration of an element is
\(
1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{3}
\)
Hence it lies in fifth or \(15^{\text {th }}\) group.
Question 31 of 54

31. Question
In the periodic table from left to right in a period, the atomic volume [NEET 1993]
- decreases
- increases
- remains same
- first decreases then increases
Correct

Incorrect

Hint

(d) Atomic volume is the volume occupied by one gram of an element. Within a period from left to right, atomic volume first decreases and then increases.
Question 32 of 54

32. Question
Which electronic configuration of an element has abnormally high difference between second and third ionization energy? [NEET 1993]
- \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{1}\)
- \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{1} 3 p^{1}\)
- \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2} 3 p^{2}\)
- \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}\)
Correct

Incorrect

Hint

(d) Abnormally high difference between \(2^{\text {nd }}\) and \(3^{\mathrm{rd}}\) ionisation energy means that the element has two valence electrons, which is a case in configuration (d).
Question 33 of 54

33. Question
One of the characteristic properties of nonmetals is that they [NEET 1993]
- are reducing agents
- form basic oxides
- form cations by electron gain
- are electronegative
Correct

Incorrect

Hint

(d) Non-metals easily gain electrons and hence, they form negative ions, so they are electronegative in nature.
Question 34 of 54

34. Question
Pauling’s electronegativity values for elements are useful in predicting [NEET 1989]
- polarity of the molecules
- position in the E.M.F. series
- coordination numbers
- dipole moments.
Correct

Incorrect

Hint

(a) Pauling introduced the electronegativity concept. He introduced the idea that the ionic character of a bond varies with the difference in electronegativity. A large difference in electronegativity leads to a bond with high degree of polar character, i.e. the bond is predominantly ionic or vice versa.
Question 35 of 54

35. Question
The electronic configuration of four elements are given below. Which elements does not belong to the same family as others? [NEET 1989]
- \([\mathrm{Xe}] 4 f^{14} 5 d^{10} 1 s^{2}\)
- \([\mathrm{Kr}] 4 d^{10} 5 s^{2}\)
- \([\mathrm{Ne}] 3 s^{2} 3 p^{5}\)
- \([\mathrm{Ar}] 3 d^{10} 4 s^{2}\)
Correct

Incorrect

Hint

(c) Elements (a), (b) and (d) belong to the same group since each one of them has two electrons in valence shell. In contrast, element (c) has seven electrons in the valence shell, and hence it lies in other group.
Question 36 of 54

36. Question
In the periodic table, with the increase in atomic number, the metallic character of an element [NEET 1989]
- decreases in a period and increases in a group
- increases in a period and decreases in a group
- increases both in a period and the group
- decreases in a period and the group
Correct

Incorrect

Hint

(a) Metallic character decreases in a period and increases in a group.
Question 37 of 54

37. Question
Match the oxide given in column \(\mathrm{A}\) with its property given in column \(\mathrm{B}\) [NEET Odisha, 2019]
\(
\begin{array}{|l|l|}
\hline \text { Column I } & \text { Column II } \\
\hline \text { (i) } \mathrm{Na}_2 \mathrm{O} & \text { (a) Neutral } \\
\hline \text { (ii) } \mathrm{Al}_2 \mathrm{O}_3 & \text { (b) Basic } \\
\hline \text { (iii) } \mathrm{N}_2 \mathrm{O} & \text { (c) Acidic } \\
\hline \text { (iv) } \mathrm{Cl}_2 \mathrm{O}_7 & \text { (d) Amphoteric } \\
\hline
\end{array}
\)
Which of the following options has all correct pairs?
- (i)-(B), (ii)-(D), (iii)-(A), (iv)-(C)
- (i)-(B), (ii)-(A), (iii)-(D), (iv)-(C)
- (i)-(C), (ii)-(B), (iii)-(A), (iv)-(D)
- (i)-(A), (ii)-(D), (iii)-(B), (iv)-(C)
Correct

Incorrect

Hint

(a) \(\mathrm{Na}_{2} \mathrm{O}\) basic oxide
\(\mathrm{Al}_{2} \mathrm{O}_{3}\) amphoteric oxide
\(\mathrm{N}_{2} \mathrm{O}\) neutral
\(\mathrm{Cl}_{2} \mathrm{O}_{7}\) acidic oxide

Elements of group. 1 and 2 form basic oxides while group. 13 and 14 elements form amphoteric oxides. Elements of group. 16 and 17 form acidic oxides. Acidic character of oxides increases from left to right in the periodic table.
Question 38 of 54

38. Question
For the second period elements the correct increasing order of first ionisation enthalpy is: [NEET 2019]
- \(\mathrm{Li} \lt \mathrm{Be} \lt \mathrm{B} \lt \mathrm{C} \lt \mathrm{N} \lt \mathrm{O} \lt \mathrm{F} \lt \mathrm{Ne}\)
- \(\mathrm{Li} \lt \mathrm{B} \lt \mathrm{Be} \lt \mathrm{C} \lt \mathrm{O} \lt \mathrm{N} \lt \mathrm{F} \lt \mathrm{Ne}\)
- \(\mathrm{Li} \lt \mathrm{B} \lt \mathrm{Be} \lt \mathrm{C} \lt \mathrm{N} \lt \mathrm{O} \lt \mathrm{F} \lt \mathrm{Ne}\)
- \(\mathrm{Li} \lt \mathrm{Be} \lt \mathrm{B} \lt \mathrm{C} \lt \mathrm{O} \lt \mathrm{N} \lt \mathrm{F} \lt \mathrm{Ne}\)
Correct

Incorrect

Hint

(b) Consider the stability of electronic configuration after loss of one electron.

As we go from left to right in a period ionisation energy increases. ‘ \(\mathrm{Be}\) ‘ and ‘ \(\mathrm{N}\) ‘ comparatively more stable valence subshell than ‘ \(\mathrm{B}\) ‘ and ‘ \(\mathrm{O}\) ‘.

Hence, the correct order of first ionisation enthalpy is:
\(
L i<B<B e<C<O<N<F<N e
\)
Question 39 of 54

39. Question
The correct order of atomic radii in group 13 elements is [NEET 2018]
- \(\mathrm{B} \lt \mathrm{Al} \lt \operatorname{In} \lt \mathrm{Ga} \lt \mathrm{Tl}\)
- \(\mathrm{B} \lt \mathrm{Al} \lt \mathrm{Ga} \lt \operatorname{In} \lt \mathrm{Tl}\)
- \(\mathrm{B} \lt \mathrm{Ga} \lt \mathrm{Al} \lt \text { In } \lt \mathrm{Tl}\)
- \(\mathrm{B} \lt \mathrm{Ga} \lt \mathrm{Al} \lt \mathrm{Tl} \lt \text { In }\)
Correct

Incorrect

Hint

(c) Due to poor shielding effect of \(3 \mathrm{~d}\) electrons in \(\mathrm{Ga}\), the atomic radii of \(\mathrm{Ga}<\mathrm{Al}\). Thus, the correct order of atomic radii is

\(\mathrm{B}<\) \(\mathrm{Ga}<\mathrm{Al} <\mathrm{In} <\mathrm{Tl}\).
Question 40 of 54

40. Question
Identify the incorrect match. [NEET 2020]
\(
\begin{array}{|l|c|c|c|}
\hline & \text { Name } & & \text { IUPAC Official Name } \\
\hline \text { a) } & \text { Unnilunium } & \text { i) } & \text { Mendelevium } \\
\hline \text { b) } & \text { Unniltrium } & \text { ii) } & \text { Lawrencium } \\
\hline \text { c) } & \text { Unnilhexium } & \text { iii) } & \text { Seaborgium } \\
\hline \text { d) } & \text { Unununnium } & \text { iv) } & \text { Darmstadtium } \\
\hline
\end{array}
\)
- (b), (ii)
- (c), (iii)
- (d), (iv)
- (a), (i)
Correct

Incorrect

Hint

(c) Unununium \((Z=111)\), it is Roentgenium (Rg) not Darmstadtium.

The IUPAC name for Unununnium ( \(Z=111\) ) is Roentgenium not Darmstadtium
Question 41 of 54

41. Question
Match the following: [NEET 2020]
\(
\begin{array}{|l|c|c|c|}
\hline & \text { Oxide } & & \text { Nature } \\
\hline \text { (a) } & \mathrm{CO} & \text { (i) } & \text { Basic } \\
\hline \text { (b) } & \mathrm{BaO} & \text { (ii) } & \text { Neutral } \\
\hline \text { (c) } & \mathrm{Al}_2 \mathrm{O}_3 & \text { (iii) } & \text { Acidic } \\
\hline \text { (d) } & \mathrm{Cl}_2 \mathrm{O}_7 & \text { (iv) } & \text { Amphoteric } \\
\hline
\end{array}
\)
Which of the following is correct option?
- \((a)-(i i),(b)-(i),(c)-(i v),(d)-(i i i)\)
- \( (a)-(i i i),(b)-(i v),(c)-(i),(d)-(i i) \)
- \( (a)-(i v),(b)-(i i i),(c)-(i i),(d)-(i) \)
- \( (a)-(i),(b)-(i i),(c)-(i i i),(d)-(i v) \)
Correct

Incorrect

Hint

(a) \(\mathrm{CO}:\) Neutral oxide
\(\mathrm{BaO}\) : Basic oxide
\(\mathrm{Al}_2 \mathrm{O}_3\) : Amphoteric oxide
\(\mathrm{Cl}_2 \mathrm{O}_7\) : Acidic oxide
Question 42 of 54

42. Question
From the following pairs of ion which one is not an iso-electronic pair? [NEET 2021]
- \(\mathrm{O}^{2-}, \mathrm{F}^{-}\)
- \(\mathrm{Na}^{+}, \mathrm{Mg}^{2+}\)
- \(\mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}\)
- \(\mathrm{Fe}^{2+}, \mathrm{Mn}^{2+}\)
Correct

Incorrect

Hint

(d)

\(\mathrm{O}^{2-}\) and \(\mathrm{F}^{-}\)are iso-electronic pair.
\(\mathrm{Na}^{+}\)and \(\mathrm{Mg}^{2+}\) are iso-electronic pair.
\(\mathrm{Mn}^{2+}\) and \(\mathrm{Fe}^{3+}\) are iso-electronic pair.
\(\mathrm{Fe}^{2+}\) and \(\mathrm{Mn}^{2+}\) are not iso-electronic pair.
Question 43 of 54

43. Question
Gadolinium has a low value of third ionisation enthalpy because of [NEET 2022]
- high exchange enthalpy
- high electronegativity
- high basic character
- small size
Correct

Incorrect

Hint

(a)

\(
\begin{aligned}
{ }_{64} \mathrm{Gd} & =[\mathrm{Xe}] 6 \mathrm{s}^2 4 \mathrm{f}^7 5 \mathrm{d}^1 \\
\mathrm{Gd}^{+2} & =[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{d}^1
\end{aligned}
\)

After losing \(5 \mathrm{~d}\) electron \(4 \mathrm{f}\) has maximum exchange energy so \(\mathrm{Gd}\) has low value of Third Ionisation energy.

Note:

Electronic configuration of Gadolinium
\(
\text { Gd :- }[X e] 4 f^7 5 d^1 6 s^2
\)

In case of \(3^{\text {rd }}\) ionisation enthalpy electron will be removed from \(5 \mathrm{~d}\) and resultant configuration will be \([\mathrm{Xe}] 4 \mathrm{f}^7\) that is stable electronic configuration as it will have high exchange energy, hence less energy will be required to remove \(3^{\text {rd }}\) electron.
Question 44 of 54

44. Question
Which of the following is correctly matched? [NEET 2023 Manipur]
- Basic oxides \(\Rightarrow \mathrm{In}_2 \mathrm{O}_3, \mathrm{~K}_2 \mathrm{O}, \mathrm{SnO}_2\)
- Neutral oxides \(\Rightarrow \mathrm{CO}, \mathrm{NO}_2, \mathrm{~N}_2 \mathrm{O}\)
- Acidic oxides \(\Rightarrow \mathrm{Mn}_2 \mathrm{O}_7, \mathrm{SO}_2, \mathrm{TeO}_3\)
- Amphoteric oxides \(\Rightarrow \mathrm{BeO}, \mathrm{Ga}_2 \mathrm{O}_3, \mathrm{GeO}\)
Correct

Incorrect

Hint

(c)

\(
\mathrm{Mn}_2 \mathrm{O}_7, \mathrm{SO}_2, \mathrm{TeO}_3 \text { are acidic oxides. }
\)
Question 45 of 54

45. Question
The correct sequence given below containing neutral, acidic, basic and amphoteric oxide each, respectively, is [NEET 2023 Manipur]
- \(\mathrm{NO}, \mathrm{ZnO}, \mathrm{CO}_2, \mathrm{CaO}\)
- \(\mathrm{ZnO}, \mathrm{NO}, \mathrm{CaO}, \mathrm{CO}_2\)
- \(\mathrm{NO}, \mathrm{CO}_2, \mathrm{ZnO}, \mathrm{CaO}\)
- \(\mathrm{NO}, \mathrm{CO}_2, \mathrm{CaO}, \mathrm{ZnO}\)
Correct

Incorrect

Hint

(d)

\(\mathrm{NO} \rightarrow\) neutral
\(\mathrm{CaO} \rightarrow\) Basic
\(\mathrm{CO}_2 \rightarrow\) Acidic
\(\mathrm{ZnO} \rightarrow\) Amphoteric
Question 46 of 54

46. Question
he correct order of first ionization enthalpy for the given four elements is : [NEET 2022 Phase 2]
- \(\mathrm{C} \lt \mathrm{F} \lt \mathrm{N} \lt \mathrm{O}\)
- \(\mathrm{C} \lt \mathrm{N} \lt \mathrm{F} \lt \mathrm{O}\)
- \(\mathrm{C} \lt \mathrm{N} \lt \mathrm{O} \lt \mathrm{F}\)
- \(\mathrm{C} \lt \mathrm{O} \lt \mathrm{N} \lt \mathrm{F}\)
Correct

Incorrect

Hint

(d) Generally, on moving left to right in a period. First ionization enthalpy of elements increases due to increase in effective nuclear charge.
Due to more stable half-filled outer electronic configuration \(\left(2 s^2 2 p^3\right)\) of \(\mathrm{N}\), its first ionization enthalpy is more than 0.

So, correct order of IE is : \(\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}\)
Question 47 of 54

47. Question
Decrease in size from left to right in actinoid series is greater and gradual than that in lanthanoid series due to [NEET 2022 Phase 2]
- \(5 f\) orbitals have greater shielding effect
- \(4 f\) orbitals are penultimate
- \(4 f\) orbitals have greater shielding effect
- \(5 f\) orbitals have poor shielding effect
Correct

Incorrect

Hint

(d) Due to more diffused nature of \(5 \mathrm{f}\) orbitals as compared to \(4 \mathrm{f}\) orbitals the shielding effect of \(5 \mathrm{f}\) is poor, resulting in the decrease in size from left to right in actinoid series which is greater and gradual than that in lanthanoid series.
Question 48 of 54

48. Question
Fluorine is a stronger oxidising agent than chlorine because : [NEET 2022 Phase 2]
(a) F-F bond has a low enthalpy of dissociation.
(b) Fluoride ion \(\left(\mathrm{F}^{-}\right)\)has high hydration enthalpy.
(c) Electron gain enthalpy of fluorine is less negative than chlorine.
(d) Fluorine has a very small size.
Choose the most appropriate answer from the options given :
- (b) and (c) only
- (a) and (b) only
- (a) and (c) only
- (a) and (d) only
Correct

Incorrect

Hint

(b) Fluorine is a stronger oxidising agent than chlorine due to
(i) Low dissociation enthalpy of F-F bond
(ii) High hydration enthalpy of \(\mathrm{F}^{-}\)ion
Question 49 of 54

49. Question
If first ionization enthalpies of elements \(X\) and \(Y\) are \(419 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(590 \mathrm{~kJ} \mathrm{~mol}^{-1}\), respectively and second ionization enthalpies of \(X\) and \(Y\) are \(3069 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(1145 \mathrm{~kJ}\) \(\mathrm{~mol}^{-1}\), respectively. Then correct statement is : [NEET 2022 Phase 2]
- Both \(X\) and \(Y\) are alkaline earth metals
- \(X\) is an alkali metal and \(Y\) is an alkaline earth metal
- \(X\) is an alkaline earth metal and \(Y\) is an alkali metal
- Both \(X\) and \(Y\) are alkali metals
Correct

Incorrect

Hint

(b) As it can be observed from given data of question, in case of element ‘ \(X\) ‘ there is huge difference between \(\mathrm{IP}_1\) and \(\mathrm{IP}_2\) hence it will have one electron in outermost shell and will be alkali metal.

While for ‘ \(Y\) ‘ difference is not that high hence it will be alkaline earth metal.
Question 50 of 54

50. Question
The IUPAC name of an element with atomic number 119 is [NEET 2022 Phase 1]
- ununennium
- unnilennium
- unununnium
- ununoctium
Correct

Incorrect

Hint

(a) IUPAC name of element : 119 : ununennium
Question 51 of 54

51. Question
The correct decreasing order of atomic radii (pm) of \(Li , Be , B\) and C is [NEET 2024 (Re-Examination)]
- \(Be > Li > B > C\)
- \(Li > Be > B > C\)
- \(C > B > Be > Li\)
- \(Li > C > Be > B\)
Correct

Incorrect

Hint

(b) As the atomic number in a period increases, the effective nuclear charge also increases hence, atomic radii along the period decreases.
Correct order of atomic radii
\(
Li > Be > B > C
\)
Question 52 of 54

52. Question
Match List I with List II : [NEET 2024 (Re-Examination)]
\(
\begin{array}{|l|l|l|l|}
\hline & \text { List I (Atom/Molecule) } & & \text { List II (Property)} \\
\hline \text { A. } & \text { Nitrogen atom } & \text { I. } & \text { Paramagnetic } \\
\hline \text { B. } & \text { Fluorine molecule } & \text { II. } & \text { Most relative element in group 18 } \\
\hline \text { C. } & \text { Oxygen molecule } & \text { III. } & \text { Element with highest ionisation enthalpy in group 15 } \\
\hline \text { D. } & \text { Xenon atom } & \text { IV. } & \text { Strongest oxidising agent } \\
\hline
\end{array}
\)
Identify the correct answer from the options given below :
- A-III, B-I, C-IV, D-II
- A-I, B-IV, C-III, D-II
- A-II, B-IV, C-I, D-III
- A-III, B-IV, C-I, D-II
Correct

Incorrect

Hint

(d)

\(
\text { The correct match is A-III, B-IV, C-I, D-II }
\)
Question 53 of 54

53. Question
Arrange the following elements in increasing order of first ionization enthalpy: \(Li , Be , B , C , N\)
Choose the correct answer from the options given below: [NEET 2024]
- \(Li \lt Be \lt B \lt C \lt N\)
- \(Li \lt B \lt Be \lt C \lt N\)
- \(Li \lt Be \lt C \lt B \lt N\)
- \(Li \lt Be \lt N \lt B \lt C\)
Correct

Incorrect

Hint

(b) The first ionization enthalpy, also known as ionization energy, is the energy required to remove the most loosely bound electron from a neutral atom in the gaseous phase to form a cation. The trend of first ionization energies generally increases across a period from left to right in the periodic table. This is due to the increasing nuclear charge and the decreasing atomic radius, which cause the valence electrons to be attracted more strongly to the nucleus.

However, there are notable exceptions based on the electron configuration stability and electron pairing in orbitals. Let’s analyze the given elements:
Lithium (Li): Being the first element in the period, it has the smallest nuclear charge and only one electron in its outer shell, which makes it easy to remove an electron.
Beryllium (Be): This element has two electrons in the \(2 s\) orbital. The removal of one electron slightly disturbs the fully filled \(2 s\) sub-shell, creating more stability than having an unpaired electron. Therefore, Be has a higher ionization energy than Li.
Boron (B): This element has a half-filled 2 p orbital configuration (one electron in the 2 p orbital), which is relatively less stable compared to a full or empty p orbital, leading to a slightly lower ionization energy than Be .
Carbon (C): With two electrons in separate 2 p orbitals (following Hund’s rule), \(C\) experiences more effective nuclear shielding and electron-electron repulsion compared to a single electron in Boron’s \(2 p\) orbital. This makes it relatively easier to remove an electron from \(B\) than from C , but harder than removing one from Be .
Nitrogen ( \(N\) ): It has exactly half-filled 2p orbitals, which provides extra stability and hence has a higher ionization energy than Carbon. Contrarily, the configuration of three \(p\) electrons is stable owing to the exchange energy and symmetric distribution in space.
Given these points, we can order the elements by increasing first ionization enthalpy as follows:
\(
Li < B < Be < C < N \text { Answer is Option b }
\)
Question 54 of 54

54. Question
Arrange the following elements in increasing order of electronegativity:
\(
N , O , F , C , Si
\)
Choose the correct answer from the options given below : [NEET 2024]
- \(Si \lt C \lt N \lt O \lt F\)
- \(Si \lt C \lt O \lt N \lt F\)
- \(O \lt F \lt N \lt C \lt Si\)
- \(F \lt O \lt N \lt C \lt Si\)
Correct

Incorrect

Hint

(a) Electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons (or electron density) towards itself in a chemical bond. The Pauling scale is the most commonly used scale to measure electronegativity. According to this scale:

The electronegativity of Fluorine (F) is the highest among the elements at around 3.98.
Oxygen (0) follows next with an electronegativity of about 3.44 .
Nitrogen \(( N )\) has an electronegativity of approximately 3.04 .
Carbon (C) has an electronegativity value close to 2.55 .
Silicon (Si), being further down the group in the periodic table than Carbon, has a lower electronegativity of about 1.90 .
Based on these values, we can arrange the elements in order of increasing electronegativity as follows:
\(
Si < C < N < O < F
\)
Thus, considering the options provided:
Option A: \(Si < C < N < O < F\) is the correct answer since it correctly ranks the elements from the lowest to the highest electronegativity.

CLASS-XI NCERT CHEMISTRY

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