Question

For the element with atomic number $27$. Write the electronic configuration.
1. Write down the value $n$ and $l$ for the electrons in the valence shell.
2. How many unpaired electrons are present in it.

Answer 1

We know that atomic number is the number of electrons present in the atom. For electronic configuration express the no. of electrons in their various valence shells. And subsequently calculate values of n and l and number of unpaired electrons too.

Complete step by step answer:
The element having atomic number $27$ is cobalt having symbol $CO$.
The electronic configuration of cobalt is $CO \to 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^7}$
Or we can also write it as:
$CO \to \left[ {As} \right]3{d^7}4{s^2}$
${}^\backprime n'$ is basically the principal quantum number. It describes the electron shell or energy level of an electron. The value of $n$ ranges from $1$ to the shell that contains the outermost electron of that atom.
So, the value of $n$ for cobalt is $3$ as the last electron enters in the $3d$ subshell.
${}^\backprime l'$ is known as azimuthal quantum number. It describes the subshell and gives the magnitude of the orbital angular momentum. The value of $l$ ranges from $0$ to $n - 1$
So, for cobalt $l$ varies from $0$ to $2$ as the valence electron is in the ${}^\backprime d'$ subshell.
For,
$0 \to s$
$1 \to p$
$2 \to d$
It follows this sequence.
The number of unpaired electrons in cobalt are $3$ as there are a total of $7$ electrons out of which two will get paired.

Note:
The electronic configuration should be written according to the rule of increasing energies of shells/ orbitals so that correct values of n, l and unpaired electrons are calculated.