Question

How many electrons does ${K^{1 + }}$ have?

Answer 1

Hint:To solve this question we must know which element is denoted by the symbol $K$. It is potassium which belongs to group one, fourth period and belongs to s-block of the periodic table. It is a silvery-white metal which is soft enough to be cut with a knife with even little force. It has an atomic number of nineteen and is chemically very similar to sodium.

Complete step-by-step answer:First we must know the electronic configuration of potassium, where the first two electrons will go to the $1s$ orbital. Since $1s$ can only hold only two electrons the next two electrons will go to the $2s$ orbital. The next six electrons will go in the $2p$ orbital because the p orbital can hold up to six electrons and then put the next two electrons in the $3s$orbital. Since the $3s$ is full we will now move to the $3p$ where we shall place the next six electrons. We now shift to the 4s orbital where we put the remaining electron. Therefore the electron configuration of potassium will be
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}$
From the above electronic configuration we can find that there are $19$ electrons present in potassium which is denoted as $K$ .
Now the electronic configuration of ${K^{1 + }}$ will be,
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$
Hence from the electronic configuration of ${K^{1 + }}$ we know that the number of electrons present in it will be 18.

Note:If we know that potassium has an atomic number of nineteen we can find out the number of electrons present in ${K^{1 + }}$ . This is because the atomic number of an element will be equal to the number of protons and in a neutral atom the number of protons will be equal to the number of electrons. Hence in ${K^{1 + }}$ one electron is lost therefore it will be nineteen minus one which will be eighteen electrons.