The total number of electrons present in all | vedclass

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(B) The atomic number of Cesium $(Cs)$ is $55$. The electronic configuration of $Cs$ is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 4d^{10},

Vedclass · Accepted Answer

$10, 24, 20$

Vedclass · Answer

(B) The atomic number of Cesium $(Cs)$ is $55$. The electronic configuration of $Cs$ is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 4d^{10}, 5s^2, 5p^6, 6s^1$. For the cesium ion $(Cs^+)$,one electron is removed from the $6s$ orbital: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 4d^{10}, 5s^2, 5p^6$. Total number of electrons in $s$-orbitals $(1s, 2s, 3s, 4s, 5s)$: $2 + 2 + 2 + 2 + 2 = 10$. Total number of electrons in $p$-orbitals $(2p, 3p, 4p, 5p)$: $6 + 6 + 6 + 6 = 24$. Total number of electrons in $d$-orbitals $(3d, 4d)$: $10 + 10 = 20$. Thus,the counts are $10, 24, 20$.

Column-$I$	Column-$II$
$A$. The orbital which has two angular nodes	$P$. $4d_{x^2-y^2}$
$B$. The $d$ orbital with zero nodal plane	$Q$. $3d_{z^2}$
$C$. The orbital with two radial nodes	$R$. $4f$
$D$. The orbital with three angular nodes	$S$. $3s$

The total number of electrons present in all the $s$-orbitals,all the $p$-orbitals,and all the $d$-orbitals of the cesium ion $(Cs^+)$ are respectively:

Similar Questions

Match the column:

Column-$I$ Column-$II$

$A$. The orbital which has two angular nodes $P$. $4d_{x^2-y^2}$

$B$. The $d$ orbital with zero nodal plane $Q$. $3d_{z^2}$

$C$. The orbital with two radial nodes $R$. $4f$

$D$. The orbital with three angular nodes $S$. $3s$

Correct match will be $:-$

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The atomic number of an element is $35$ and its mass number is $81$. The number of electrons in the outermost shell is:

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