NEET 2018 · ChemistryPrevious Year Question

Which one is the wrong statement?

Answer: (D) The energy of the $2 s$ orbital is less than the energy of the $2 p$ orbital in case of hydrogen-like atoms. \textbf{Answer:} (D) The energy of the $2 s$ orbital is less than the energy of the $2 p$ orbital in case of hydrogen-like atoms \textbf{Solution:} In single-electron (hydrogen-like) species the orbital energy depends only on $n$ , so the $2 s$ and $2 p$ orbitals are degenerate (equal energy).

A.de Broglie's wavelength is given by $λ = \frac{h}{m v}$ , where $m$ = mass of the particle and $v$ = group velocity of the particle
B.The uncertainty principle is $Δ E \cdot Δ t \geq \frac{h}{4 π}$
C.Half-filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement
D.The energy of the $2 s$ orbital is less than the energy of the $2 p$ orbital in case of hydrogen-like atoms✓

Correct Answer

(D) The energy of the $2 s$ orbital is less than the energy of the $2 p$ orbital in case of hydrogen-like atoms

Solution & Explanation

\textbf{Answer:} (D) The energy of the $2 s$ orbital is less than the energy of the $2 p$ orbital in case of hydrogen-like atoms \textbf{Solution:} In single-electron (hydrogen-like) species the orbital energy depends only on $n$ , so the $2 s$ and $2 p$ orbitals are degenerate (equal energy). Hence statement (D) is wrong. The other statements are correct: de Broglie $λ = \frac{h}{m v}$ , Heisenberg $Δ E \cdot Δ t \geq \frac{h}{4 π}$ , and the extra stability of half/fully-filled sub-shells from exchange energy and symmetry.

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