Question

In the ground state electronic configuration of $Mn(Z=25)$:

• A. the number of electrons having $n+l$ = $4$ are $5$
• B. the number of electrons having $m=0$ are $13$
• C. the electron is first filled in $3d$ then in $4s$ subshell
• D. the number of electrons having $l=1$ are $10$

Answer 1

The correct option is B the number of electrons having $m=0$ are $13$

Ground state electronic configuration for $Mn$ is:
$Mn=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^2$
Now let’s check n+l value for each subshell involved i.e
$1s\Rightarrow 1+0=1$
$2s\Rightarrow 2+0=2$
$2p\Rightarrow 2+1=3$
$3s\Rightarrow 3+0=3$
$3p\Rightarrow 3+1=4$
$3d\Rightarrow 3+2=5$
$4s\Rightarrow 4+0=4$
So there are two cases for which $n+l=4$ i.e. $3p$ and $4s$. Hence, total no of electrons are $8$. So $\left(a\right)$ option is incorrect.
$m$ will be zero for electrons in all $s$ subshell, one orbital each in each $p$ subshell and d subshell. For Mn, total number of electrons having m = 0 is $13$. Hence $\left(b\right)$ option is correct.

In ground state electron is first filled in $4s$ than in $3d$ because $4s$ is of lower energy than $3d$ as per $(n+l)$ rule. Hence $\left(c\right)$ is incorrect.

The electrons having $l=1$ ($p$ subshell) are $12$. Hence $\left(d\right)$ option is incorrect.