An element 'y' present in its ground state, the value of principal quantum number and azimuthal quantum number of the last electron of element 'y' is n = 3, and l = 1 and spin multiplicity for a given element is 4. Select incorrect statements based on the above information.

Spin multiplicity =|(2s+1)|=4, indicates, three e−1s, with same spin, i.e.,212+12+12+1=4n=3,l=1⇒3p, with 3e−1 s with same spin, ⇒3p3 Element ' y ' electronic configuration =1s2,2s22p6,3s2,3p3Thus element 'y' - Phosphorous (Z = 15)(3rd period, 15 group) mmm=n(n+2)=3(3+2)=15BM