NEET 2018 · ChemistryPrevious Year Question

Which one is a wrong statement?

Answer: (A) The electronic configuration of the $N$ atom is $1 s^{2} 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}$ . \textbf{Answer:} (A) The electronic configuration of the $N$ atom is $1 s^{2} 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}$ \textbf{Solution:} Per the official key, statement (A) is treated as wrong.

A.The electronic configuration of the $N$ atom is $1 s^{2} 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}$ ✓
B.An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers
C.Total orbital angular momentum of an electron in an $s$ orbital is equal to zero
D.The value of $m$ for $d_{z^{2}}$ is zero

Correct Answer

(A) The electronic configuration of the $N$ atom is $1 s^{2} 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}$

Solution & Explanation

\textbf{Answer:} (A) The electronic configuration of the $N$ atom is $1 s^{2} 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}$ \textbf{Solution:} Per the official key, statement (A) is treated as wrong. Statements (B), (C) and (D) are correct: an orbital needs three quantum numbers ( $n, l, m_{l}$ ) and an electron four; for an $s$ orbital $l = 0$ so the orbital angular momentum $l (l + 1) \frac{h}{2 π} = 0$ ; and $m = 0$ for $d_{z^{2}}$ .

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