An element has an atomic number of 17. The nu | vedclass

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(C) The electronic configuration of an element with atomic number $17$ $(Cl)$ is $1s^2 2s^2 2p^6 3s^2 3p^5$.The valence shell is the $n=3$ shell.The o

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$3$

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(C) The electronic configuration of an element with atomic number $17$ $(Cl)$ is $1s^2 2s^2 2p^6 3s^2 3p^5$.The valence shell is the $n=3$ shell.The orbitals in the valence shell are $3s$ and $3p$.The $3s$ orbital contains $2$ electrons (one pair).The $3p$ subshell has $3$ orbitals $(3p_x, 3p_y, 3p_z)$.With $5$ electrons in the $3p$ subshell,the distribution is $(3p_x)^2 (3p_y)^2 (3p_z)^1$.Thus,there are two pairs in the $3p$ subshell and one pair in the $3s$ orbital.Total number of orbitals containing electron pairs = $1$ $(3s)$ + $2$ $(3p_x, 3p_y)$ = $3$.

List-$I$ (Element)	List-$II$ (Electronic Configuration)
$A. N$	$I. [Ar] 3d^{10} 4s^2 4p^5$
$B. S$	$II. [Ne] 3s^2 3p^4$
$C. Br$	$III. [He] 2s^2 2p^3$
$D. Kr$	$IV. [Ar] 3d^{10} 4s^2 4p^6$

An element has an atomic number of $17$. The number of orbitals containing electron pairs in the valence shell will be .......

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Match List-$I$ with List-$II$.

List-$I$ (Element) List-$II$ (Electronic Configuration)

$A. N$ $I. [Ar] 3d^{10} 4s^2 4p^5$

$B. S$ $II. [Ne] 3s^2 3p^4$

$C. Br$ $III. [He] 2s^2 2p^3$

$D. Kr$ $IV. [Ar] 3d^{10} 4s^2 4p^6$

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