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The incorrect set of quantum numbers $(n, l, m, s)$ for an electron in $3 p$ orbital is
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1913 Upvotes
Verified Answer
The correct answer is:
$3,1,-2,-1 / 2$
For an electron in 3p-orbital, the values of
$$
\begin{aligned}
n & =3 \\
l & =0, \quad 1, \quad 2(\because \text { value of } l \text { are form } 0 \text { to } n-1) \\
\therefore m & =0 \quad-1,0,+1,-2,-1,0,+1,+2
\end{aligned}
$$
( $\because$ For each value of $l$, the value of $n$ varies from $-l$
$$
\text { to } 0 \text { to }+l \text { ) }
$$
Also, for every value of $m$, the values of $s$ can be $+1 / 2$ and $-1 / 2$.
$\therefore$ Option (b) is incorrect, as for $n=3$, the value of $m$ cannot be -2 .
$$
\begin{aligned}
n & =3 \\
l & =0, \quad 1, \quad 2(\because \text { value of } l \text { are form } 0 \text { to } n-1) \\
\therefore m & =0 \quad-1,0,+1,-2,-1,0,+1,+2
\end{aligned}
$$
( $\because$ For each value of $l$, the value of $n$ varies from $-l$
$$
\text { to } 0 \text { to }+l \text { ) }
$$
Also, for every value of $m$, the values of $s$ can be $+1 / 2$ and $-1 / 2$.
$\therefore$ Option (b) is incorrect, as for $n=3$, the value of $m$ cannot be -2 .
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