Question

In $Cr$ atom the number of electrons having $n+m=3$ are:
(A) $5$
(B) $6$
(C) $7$
(D) $8$

Answer 1

Hint : We know that the quantum number is a set of numbers with the help of which we can locate the position of an electron in an atom. There are four quantum numbers in total and using them we can calculate the number of electrons in a shell and subshell of an element.

Complete Step By Step Answer:
As we know that we must recall the electronic configuration of chromium atoms. Also, you must be familiar with the values of the angular quantum number of various orbitals and there are four quantum numbers including: Principal quantum number tells us about the size of the electron cloud and energy level of the electron and is represented by The shells $K,L,M,N,....$ have $n=1,2,3,4,.....$ respectively. The azimuthal quantum number tells us about the shape of electron cloud and subshell to which the electron belongs, represented as $m$ and subshells $s,p,d,f$ . Thus, the number of nodes increases when the principal quantum number increases. It is essential to note that chromium has an exceptional electronic configuration. Considering the general trend, the electronic configuration of chromium atom is expected to be $Cr\Rightarrow 1{{s}^{2}}2{{s}^{2}}\underset{~~\uparrow ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\uparrow }{\mathop{2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{5}}4{{s}^{1}}}}\,$ But, one electron from the orbital shifts to the orbital. This completes five electrons in the orbital making it half filled. Half-filled orbitals have greater stability due to symmetry and greater exchange energy.
since we have subshells which are $m+n=3$ ;
$\Rightarrow 6+1=7$
Therefore, the correct answer is option C.

Note :
Remember that the probability of finding an electron is said to be zero in case of radial nodes. Angular nodes are equal to the value of azimuthal quantum number, and since the value for azimuthal quantum number for d orbital is two, the value for angular nodes also has to be $2.$