Given below are two statements:
Statement I: The metallic radius of Na is 1.86 Å and the ionic radius of Na$^+$ is lesser than 1.86 Å.
Statement II: Ions are always smaller in size than the corresponding elements.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: The metallic radius of Na is 1.86 Å and the ionic radius of Na$^+$ is lesser than 1.86 Å.
Statement II: Ions are always smaller in size than the corresponding elements.
In the light of the above statements, choose the correct answer from the options given below:
- Statement I is correct but Statement II is false
- Both Statement I and Statement II are true
- Both Statement I and Statement II are false
- Statement I is incorrect but Statement II is true
The Correct Option is A
Approach Solution - 1
Step 1: Analysis of Statement I
- The metallic radius of N a (neutral sodium atom) is 1.86 ̊A.
- The ionic radius of Na+ is smaller than the neutral atom because Na+ has one less electron, resulting in reduced electron-electron repulsion and greater effective nuclear charge on the remaining electrons. - Thus, Statement I is correct.
Step 2: Analysis of Statement II - While cations (Na+) are always smaller than their corresponding neutral atoms, anions (e.g., Cl-) are larger than their corresponding neutral atoms due to increased electron-electron repulsion in the outer shell.
- Hence, ions are not always smaller than the corresponding elements.
- Thus, Statement II is false.
Step 3: Conclusion - Statement I is correct, but Statement II is false.
Final Answer: (1)
Approach Solution -2
Statement I: The metallic radius of Na is 1.86 Å and the ionic radius of Na⁺ is lesser than 1.86 Å.
Explanation: When sodium loses one electron to form Na⁺, the outermost shell (3s¹) is removed completely, resulting in a smaller electron cloud. The number of protons remains the same, but the number of electrons decreases, increasing the effective nuclear charge. Hence, Na⁺ is much smaller than Na.
Therefore, Statement I is correct.
Step 2: Analyze Statement II.
Statement II: Ions are always smaller in size than the corresponding elements.
Explanation: This is not always true. While cations (like Na⁺) are smaller than their parent atoms, anions (like Cl⁻) are larger than their parent atoms due to increased electron-electron repulsion and decreased effective nuclear charge per electron.
Therefore, Statement II is false.
Step 3: Conclusion.
Statement I is correct but Statement II is false.
Final Answer: Statement I is correct but Statement II is false.
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Trends in periodic table Questions
- Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Both rhombic and monoclinic sulphur exist as S$_8$ while oxygen exists as O$_2$.
{Reason (R): Oxygen forms $p\pi$-$p\pi$ multiple bonds with itself and other elements having small size and high electronegativity like C, N, which is not possible for sulphur.
In the light of the above statements, choose the most appropriate answer from the options given below: - The correct statements from the following are:
(A) The decreasing order of atomic radii of group 13 elements is Tl $>$ In $>$ Ga $>$ Al $>$ B.
(B) Down the group 13 electronegativity decreases from top to bottom.
(C) Al dissolves in dil. HCl and liberate H$_2$, but conc. HNO$_3$ renders Al passive by forming a protective oxide layer on the surface.
(D) All elements of group 13 exhibits highly stable +1 oxidation state.
(E) Hybridisation of Al in [Al(H$_2$O)$_6$]$^{3+}$ ion is sp$^3$d$^2$.
Choose the correct answer from the options given below. - Given below are two statements:
Statement I: The metallic radius of Al is less than that of Ga.
Statement II: The ionic radius of Al\(^{3+}\) is less than that of Ga\(^{3+}\).
In the light of the above statements, choose the most appropriate answer from the options given below: Match List -I with List-II
LIST-I (Atomic number) LIST-II (Block of periodic table) (A) 37 (K) I. p-block (B) 78 (Pt) II. d-block (C) 52 (Te) III. f-block (D) 65 (Tb) IV. s-block Choose the correct answer from the options given below:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.